nihar

2010-09-14T20:43:00.001-07:00

php interview quesitions

2010-09-07T00:11:00.000-07:00

1. What does a special set of tags do in PHP?

ans- The output is displayed directly to the browser.

2. What’s the difference between include and require?

ans - It’s how they handle failures. If the file is not found by require(), it will cause a fatal error and halt the execution of the script. If the file is not found by include(), a warning will be issued, but execution will continue.

3. I am trying to assign a variable the value of 0123, but it keeps coming up with a different number, what’s the problem?

ans - PHP Interpreter treats numbers beginning with 0 as octal. Look at the similar PHP interview questions for more numeric problems.

4. Would I use print "$a dollars" or "{$a} dollars" to print out the amount of dollars in this example?

ans- In this example it wouldn’t matter, since the variable is all by itself, but if you were to print something like "{$a},000,000 mln dollars", then you definitely need to use the braces.

5. How do you define a constant?

ans- Via define() directive, like define ("MYCONSTANT", 100);

6. How do you pass a variable by value?

ans- Just like in C++, put an ampersand in front of it, like $a = &$b

7. Will comparison of string "10" and integer 11 work in PHP?

ans- Yes, internally PHP will cast everything to the integer type, so numbers 10 and 11 will be compared.

8. When are you supposed to use endif to end the conditional statement?

ans- When the original if was followed by : and then the code block without braces.

9. Explain the ternary conditional operator in PHP?

ans- Expression preceding the ? is evaluated, if it’s true, then the expression preceding the : is executed, otherwise, the expression following : is executed.

10. How do I find out the number of parameters passed into function?

ans- func_num_args() function returns the number of parameters passed in.

11. If the variable $a is equal to 5 and variable $b is equal to character a, what’s the value of $$b?

ans - 100, it’s a reference to existing variable.

12. What’s the difference between accessing a class method via -> and via ::?

ans- :: is allowed to access methods that can perform static operations, i.e. those, which do not require object initialization.

13. Are objects passed by value or by reference?

ans- Everything is passed by value.

14. How do you call a constructor for a parent class?

ans- parent::constructor($value)

15. What’s the special meaning of __sleep and __wakeup?

ans - __sleep returns the array of all the variables than need to be saved, while __wakeup retrieves them.

16. Why doesn’t the following code print the newline properly?

$str = ‘Hello, there.nHow are you?nThanks for visiting TechInterviews’;

print $str;

?>

Because inside the single quotes the n character is not interpreted as newline, just as a sequence of two characters - and n.

17. Would you initialize your strings with single quotes or double quotes?

ans- Since the data inside the single-quoted string is not parsed for variable substitution, it’s always a better idea speed-wise to initialize a string with single quotes, unless you specifically need variable substitution.

18. How come the code works, but doesn’t for two-dimensional array of mine?

ans- Any time you have an array with more than one dimension, complex parsing syntax is required. print "Contents: {$arr[1][2]}" would’ve worked.

19. What is the difference between characters 23 and x23?

ans- The first one is octal 23, the second is hex 23.

20. With a heredoc syntax, do I get variable substitution inside the heredoc contents?

ans- Yes.

21. I want to combine two variables together:

22. $var1 = 'Welcome to ';

23. $var2 = 'TechInterviews.com';

What will work faster? Code sample 1:

$var 3 = $var1.$var2;

Or code sample 2:

$var3 = "$var1$var2";

Both examples would provide the same result - $var3 equal to "Welcome to TechInterviews.com". However, Code Sample 1 will work significantly faster. Try it out with large sets of data (or via concatenating small sets a million times or so), and you will see that concatenation works significantly faster than variable substitution.

24. For printing out strings, there are echo, print and printf. Explain the differences.

ans- echo is the most primitive of them, and just outputs the contents following the construct to the screen. print is also a construct (so parentheses are optional when calling it), but it returns TRUE on successful output and FALSE if it was unable to print out the string. However, you can pass multiple parameters to echo, like:

and it will output the string "Welcome to TechInterviews!" print does not take multiple parameters. It is also generally argued that echo is faster, but usually the speed advantage is negligible, and might not be there for future versions of PHP. printf is a function, not a construct, and allows such advantages as formatted output, but it’s the slowest way to print out data out of echo, print and printf.

25. I am writing an application in PHP that outputs a printable version of driving directions. It contains some long sentences, and I am a neat freak, and would like to make sure that no line exceeds 50 characters. How do I accomplish that with PHP?

ans - On large strings that need to be formatted according to some length specifications, use wordwrap() or chunk_split().

26. What’s the output of the ucwords function in this example?

27. $formatted = ucwords("TECHINTERVIEWS IS COLLECTION OF INTERVIEW QUESTIONS");

print $formatted;

What will be printed is TECHINTERVIEWS IS COLLECTION OF INTERVIEW QUESTIONS.

ucwords() makes every first letter of every word capital, but it does not lower

ans-case anything else. To avoid this, and get a properly formatted string, it’s worth using strtolower() first.

28. What’s the difference between htmlentities() and htmlspecialchars()?

ans- htmlspecialchars only takes care of <, >, single quote ‘, double quote " and ampersand. htmlentities translates all occurrences of character sequences that have different meaning in HTML.

29. What’s the difference between md5(), crc32() and sha1() crypto on PHP?

ans- The major difference is the length of the hash generated. CRC32 is, evidently, 32 bits, while sha1() returns a 128 bit value, and md5() returns a 160 bit value. This is important when avoiding collisions.

30. So if md5() generates the most secure hash, why would you ever use the less secure crc32() and sha1()?

ans- Crypto usage in PHP is simple, but that doesn’t mean it’s free. First off, depending on the data that you’re encrypting, you might have reasons to store a 32-bit value in the database instead of the 160-bit value to save on space. Second, the more secure the crypto is, the longer is the computation time to deliver the hash value. A high volume site might be significantly slowed down, if frequent md5() generation is required.

habbul telescope photos

2010-08-29T20:54:00.000-07:00

2010-08-23T04:55:00.001-07:00

2010-08-23T04:30:00.000-07:00

jntu(k)2008 5th sem quesition papers

2010-08-21T03:12:00.000-07:00

MULTIMEDIA APPLICATION DEVELOPMENT

1. a) Explain the various multimedia software tools?

b) List and explain the image data types?

2. Enumerate the differences between NTSC video, PAL video and SECAM video?

3. a) Explain the various features of Action script?

b) List out and describe the data types and type checking in Action Script?

4. Describe OOP Application framework?

5. Explain the importance of the concept of inheritance in action script with an example?

6. a) Explain the steps of the Shannon – Fano algorithm with an example?

b) What is wavelet based coding?

7. Describe the MPEG standard?

8. Write short notes on

a) Multimedia over ATM networks

b) Media – On – Demand (MOD)

OBJECT ORIENTED ANALYSIS & DESIGN (USING UML)

1. a) What is the need for modeling?

b) Explain software development lifecycle.

2. a) List and explain types of relationships used in object-oriented modeling along with their

notation.

b) What is the importance of package? How it is modeled in UML?

3. a) What are the contents of class diagram?

b) List and explain uses of class diagram.

4. a) What is an Interaction? What kinds of messages can be modeled in UML?

b) Distinguish between sequence Diagram and Interaction Diagram.

5. Write short notes on the following:

a) Swimlanes

b) Usecases.

6. Compare the following

a) Event Vs Signal

b) Sequential substates Vs Concurrent substates.

7. a) Distinguish between nodes and components.

b) Explain how collaboration is used to model the Realization of a usecase.

8. a) Draw and explain class Diagram for unified Library system.

b) Draw and explain sequence Diagram for “Issue of Book”

SOFTWARE PROJECT MANAGEMENT

1. a) What are the drawbacks of waterfall model?

b) Based on what parameters software cost can be estimated? Explain.

2. a) List advantages and drawbacks of custom software in comparison with commercial

components.

b) Explain principles of software management.

3. a) Give an overview of lifecycle software process Artifacts.

b) Explain about elaboration phase in lifecycle process.

4. a) From technical perspective discuss about software Architecture.

b) Describe iteration’s workflow sequence.

5. a) What is the need and contents of periodic status Assessment Reviews?

b) Discuss about project planning using top-down and bottom-up approach.

6. a) List software Management team activities.

b) Write about change Management team activities in interactive process.

7. a) What are the goals of software Metrics? Discuss in detail about core Metrics.

b) With some example applications explain process variability in technical and management

dimensions.

8. a) Explain features and benefits of a modern process from the point of software project

manager.

b) Discuss about changes in modern software economics.

SOFTWARE TESTING METHODOLOGIES

1. a) Discuss in detail various phases in a Tester’s Mental life.

b) Explain the importance of bugs.

2. a) Compare and contrast between control flow graphs and flow charts.

b) Give a detailed note on path testing.

3. a) What are the steps involved in a transaction for an online information retrieval systems?

Explain with an example.

b) Explain about structured test strategies in detail.

4. a) Give a note on testing of two-dimensional domains.

b) Discuss about domain ambiguities and contradictions.

5. Explain in detail the reduction procedure for converting a flow graph into a path

expression with example.

6. a) Draw the KU chart for the specification below:

b) Write about change Management team activities in interactive process.

7. a) “Specifications are one of the most common source of ambiguities and contradictions.”

Explain with an example.

b) Explain the impact of bugs in state testing.

8. a) Discuss about node-reduction algorithm with a suitable example.

b) What are the reasons why two-dimensional array representation is not convenient for

larger graphs?.

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jntu(k)2009 5th sem quesition papers

2010-08-21T03:06:00.000-07:00

JNTU Kakinada MCA 5th Sem Regular & Supplementary Exams, November 2009

Subject code: MC512

E-COMMERCE

Answer any FIVE questions. All questions carry Equal marks

1. a) Define E-Commerce and Explain the benefits of Electronic Commerce. 6M

b) Discuss briefly about the E-commerce consumer applications. 6M

2. a) Explain briefly about Home Shopping. 4M

b) What is meant by Home entertainment? How it is related to E-Commerce and explain the size

the size of the Home Entertainment market? 8M

3. a) List of the types of Electronic payment systems. 4M

b) How the payment transaction sequence is happens in the Electronic check system and

explain its advantages 8M

4 . a) Define EDI. Explain about EDI layered Architecture. 6M

b) How EDI works? Explain. 6M

5. Explain clearly about Customization and internal Commerce. 12M

6. a) Expain how advertising is carrid out on Internet. 6M

b)Briefly discuss about Digital documents. 6M

7. What is the purpose of Search Engine? Explain about Wide Area Information Services (WAIS)

Engine. 12M

8. a) Explain the steps in desktop video processing. 6M

b) Write notes on multimedia storage Technology. 6M

Multimedia Application Development

JNTU Kakinada MCA 5th Sem Regular & Supplementary Exams, November 2009

Subject code: MC513

Answer any 5 questions

1. a) Explain about multimedia software tools.

b) Explain about www

2. a) Explain about MIDI

b) Explain about transmission of audio

3. a) What are the features of action script?

b) Write ashort note on databases with examples

4. a) What is meant by Inheritance? Explain.

b) What are the different types of exceptions? Explain.

5. What is meant by object- orient programming? explain about the applications of OOP.

6. a) Explain about variable- length coding with examples

b)Explain about arthematic coding with examples

7. a) Explain about audio compression techniques.

b) Write a short note on MPEG.

8. Explain about the following

a) MOD

b) Multimedia over IP

OBJECT ORIENTED ANALYSIS AND DESIGN (USING UML)

JNTU Kakinada MCA 5th Sem Regular & Supplementary Exams, November 2009

Subject code: MC514

ANSWER ANY FIVE QUESTIONS

1. a) What are the essential features of object oriented paradigm? Explain them briefly. (6M)

b) Explain the conceptual model of diagram of UML. Briefly explain various types of things of UML. (6M)

2. Explain about Types and Roles, Interfaces and Packages with suitable diagrams. (12 M)

3. a) Explain terms and concepts of Object diagram with example. (6 M)

b) Explain terms and concepts of Class diagram with example. (6M)

4. Briefly explain the terms and concepts of Collaboration diagram. Draw the Collaboration diagram for ATM bank transactions.

5. a) Define Activity Diagram. Briefly explain all its terms and concepts in activity Diagram. (6M)

b) How are forking and joining used in activity diagrams? Give examples . (6 M)

6. a) Define the terms Event, Signal, State machine, Process and Thread with examples. (6M)

b)Explain briefly about state chart diagrams with examples. (6M)

7. a) Explain terms and Concepts of Component Diagram with neat example. (6 M)

b) What are the common modeling techniques in Component diagram? (6M)

8. Give detail case study of United Library Application. (12 M)

SOFTWARE PROJECT MANAGEMENT

JNTU Kakinada MCA 5th Sem Regular & Supplementary Exams, November 2009

Subject code: MC515

ANSWER ANY FIVE QUESTIONS

1.a) Define progress. Show the development progress of a conventional project diagramatically. (6M)

b) what is meant by late late risk resolution? (6M)

2.a) Write the characteristics of an object oriented project. (6M)

b) In what way software process is aoverload term? Explain. (6M)

3.a) Write briefly about work break down structure. (6M)

b) Describe the life cycle process (6M)

4.a) Discuss the architecture of software with the help of a diagram. (6M)

b) show the relative levels of effort expected across the phases of life cycle. (6M)

5.a) what are milestones? What minor milestones needed to be considered when aproject is being planned? (6M)

b) what is the necessirty of periodic statu8s assesment? (6M)

6.a) Explain the roles and responsibilities in a software line of bussines organisation. (6M)

b) Illustrate the activities of software assessment team. (6M)

7.a) What are quality indicators? Briefly Explain. (6M)

b) Write the purpose and priorities of seven core metrics. (6M)

8.a) Give the risk profile of a modern project across its life cycle. (6M)

b) On what factors should the next generation software cost models focus at? (6M)

Software Testing Methodologies

JNTU Kakinada MCA 5th Sem Regular & Supplementary Exams, November 2009

Subject code: MC518

ANSWER ANY FIVE QUESTIONS

1.a) What are the goals for testing? (3M)

b) What are the differences between Testing and Debugging? (3 M)

c) Explain the Taxonomy of Bugs. (6 M)

2. a) What is a control flow and what are its components? (4 M)

b) What is path Instrumentation? Explain various Instrumentation Methods. (8 M)

3. a) Briefly Explain Data flow techniques.(6 M)

b) Explain the Data flow Model. (6 M)

4. a) Explain Nice Domains. (6 M)

b) What are 2D Domain bugs? Explain the procedure for detecting these bugs. (6 M)

5. a) How path expressions are converted into Regular expressions. Give some examples. (6 M)

b) Explain the following (6 M)

i) Path products ii) Path Sums

6. a) Explain the role of Decision tables in Test case design and how predicates and rule evaluations are taken care off. (6 M)

b) Write the procedure for specification validation (6 M)

7. a) Explain State Bugs. (6 M)

b) Describe the design guidelines to build Finite State machines into your code. (6 M)

8. a) Describe how transtive closure is used to convert an arbitrary graph into partly ordered graph. (6 M)

b) Explain how to use Apache J Meter to test the performance of the database server, when multiple users try to access the database simultaneously. (6 M)

2010-08-19T03:54:00.001-07:00

images

2010-08-18T05:10:00.001-07:00

java

2010-08-17T00:30:00.000-07:00

What is Collection API ?
The Collection API is a set of classes and interfaces that support operation on collections of objects. These classes and interfaces are more flexible, more powerful, and more regular than the vectors, arrays, and hashtables if effectively replaces.
Example of classes: HashSet, HashMap, ArrayList, LinkedList, TreeSet and TreeMap.
Example of interfaces: Collection, Set, List and Map.
Is Iterator a Class or Interface? What is its use?
Answer: Iterator is an interface which is used to step through the elements of a Collection.
What is similarities/difference between an Abstract class and Interface?
Differences are as follows:
Interfaces provide a form of multiple inheritance. A class can extend only one other class. Interfaces are limited to public methods and constants with no implementation. Abstract classes can have a partial implementation, protected parts, static methods, etc.
A Class may implement several interfaces. But in case of abstract class, a class may extend only one abstract class. Interfaces are slow as it requires extra indirection to to find corresponding method in in the actual class. Abstract classes are fast.
Similarities:

Neither Abstract classes or Interface can be instantiated.
Java Interview Questions - How to define an Abstract class?
A class containing abstract method is called Abstract class. An Abstract class can't be instantiated.
Example of Abstract class:
abstract class testAbstractClass {
protected String myString;
public String getMyString() {
return myString;
}
public abstract string anyAbstractFunction();
}
How to define an Interface in Java ?
In Java Interface defines the methods but does not implement them. Interface can include constants. A class that implements the interfaces is bound to implement all the methods defined in Interface.
Emaple of Interface:

public interface sampleInterface {
public void functionOne();

public long CONSTANT_ONE = 1000;
}
If a class is located in a package, what do you need to change in the OS environment to be able to use it?

You need to add a directory or a jar file that contains the package directories to the CLASSPATH environment variable. Let's say a class Employee belongs to a package com.xyz.hr; and is located in the file c:\dev\com\xyz\hr\Employee.java. In this case, you'd need to add c:\dev to the variable CLASSPATH. If this class contains the method main(), you could test it from a command prompt window as follows:
c:\>java com.xyz.hr.Employee
How many methods in the Serializable interface?
There is no method in the Serializable interface. The Serializable interface acts as a marker, telling the object serialization tools that your class is serializable.
How many methods in the Externalizable interface?
There are two methods in the Externalizable interface. You have to implement these two methods in order to make your class externalizable. These two methods are readExternal() and writeExternal().
What is the difference between Serializalble and Externalizable interface?
When you use Serializable interface, your class is serialized automatically by default. But you can override writeObject() and readObject() two methods to control more complex object serailization process. When you use Externalizable interface, you have a complete control over your class's serialization process.
What is a transient variable in Java?
A transient variable is a variable that may not be serialized. If you don't want some field to be serialized, you can mark that field transient or static.
Which containers use a border layout as their default layout?
The Window, Frame and Dialog classes use a border layout as their default layout.
How are Observer and Observable used?
Objects that subclass the Observable class maintain a list of observers. When an Observable object is updated, it invokes the update() method of each of its observers to notify the observers that it has changed state. The Observer interface is implemented by objects that observe Observable objects.
What is Java?
Java is an object-oriented programming language developed initially by James Gosling and colleagues at Sun Microsystems. The language, initially called Oak (named after the oak trees outside Gosling's office), was intended to replace C++, although the feature set better resembles that of Objective C. Java should not be confused with JavaScript, which shares only the name and a similar C-like syntax. Sun Microsystems currently maintains and updates Java regularly.
What does a well-written OO program look like?
A well-written OO program exhibits recurring structures that promote abstraction, flexibility, modularity and elegance.
Can you have virtual functions in Java?
Yes, all functions in Java are virtual by default. This is actually a pseudo trick question because the word "virtual" is not part of the naming convention in Java (as it is in C++, C-sharp and VB.NET), so this would be a foreign concept for someone who has only coded in Java. Virtual functions or virtual methods are functions or methods that will be redefined in derived classes.
Jack developed a program by using a Map container to hold key/value pairs. He wanted to make a change to the map. He decided to make a clone of the map in order to save the original data on side. What do you think of it? ?
If Jack made a clone of the map, any changes to the clone or the original map would be seen on both maps, because the clone of Map is a shallow copy. So Jack made a wrong decision.
What is more advisable to create a thread, by implementing a Runnable interface or by extending Thread class?
Strategically speaking, threads created by implementing Runnable interface are more advisable. If you create a thread by extending a thread class, you cannot extend any other class. If you create a thread by implementing Runnable interface, you save a space for your class to extend another class now or in future.
What is NullPointerException and how to handle it?

When an object is not initialized, the default value is null. When the following things happen, the NullPointerException is thrown:
--Calling the instance method of a null object.
--Accessing or modifying the field of a null object.
--Taking the length of a null as if it were an array.
--Accessing or modifying the slots of null as if it were an array.
--Throwing null as if it were a Throwable value.
The NullPointerException is a runtime exception. The best practice is to catch such exception even if it is not required by language design.
An application needs to load a library before it starts to run, how to code?
One option is to use a static block to load a library before anything is called. For example,
class Test {
static {
System.loadLibrary("path-to-library-file");
}
....
}
When you call new Test(), the static block will be called first before any initialization happens. Note that the static block position may matter.
How could Java classes direct program messages to the system console, but error messages, say to a file?
The class System has a variable out that represents the standard output, and the variable err that represents the standard error device. By default, they both point at the system console. This how the standard output could be re-directed:
Stream st = new Stream(new FileOutputStream("output.txt")); System.setErr(st); System.setOut(st);
What's the difference between an interface and an abstract class?
An abstract class may contain code in method bodies, which is not allowed in an interface. With abstract classes, you have to inherit your class from it and Java does not allow multiple inheritance. On the other hand, you can implement multiple interfaces in your class.
Name the containers which uses Border Layout as their default layout?
Containers which uses Border Layout as their default are: window, Frame and Dialog classes.

Java Interview Questions and Answers

What is synchronization and why is it important?
With respect to multithreading, synchronization is the capability to control the access of multiple threads to shared resources. Without synchronization, it is possible for one thread to modify a shared object while another thread is in the process of using or updating that object's value. This often causes dirty data and leads to significant errors.
What are synchronized methods and synchronized statements?
Synchronized methods are methods that are used to control access to a method or an object. A thread only executes a synchronized method after it has acquired the lock for the method's object or class. Synchronized statements are similar to synchronized methods. A synchronized statement can only be executed after a thread has acquired the lock for the object or class referenced in the synchronized statement.
What are three ways in which a thread can enter the waiting state?
A thread can enter the waiting state by invoking its sleep() method, by blocking on IO, by unsuccessfully attempting to acquire an object's lock, or by invoking an object's wait() method. It can also enter the waiting state by invoking its (deprecated) suspend() method.
Can a lock be acquired on a class?
Yes, a lock can be acquired on a class. This lock is acquired on the class's Class object.
What's new with the stop(), suspend() and resume() methods in JDK 1.2?
The stop(), suspend() and resume() methods have been deprecated in JDK 1.2.
What is the preferred size of a component?
The preferred size of a component is the minimum component size that will allow the component to display normally.
What's the difference between J2SDK 1.5 and J2SDK 5.0?
There's no difference, Sun Microsystems just re-branded this version.
What would you use to compare two String variables - the operator == or the method equals()?
I'd use the method equals() to compare the values of the Strings and the == to check if two variables point at the same instance of a String object.
What is thread?
A thread is an independent path of execution in a system.
What is multi-threading?
Multi-threading means various threads that run in a system.
How does multi-threading take place on a computer with a single CPU?
The operating system's task scheduler allocates execution time to multiple tasks. By quickly switching between executing tasks, it creates the impression that tasks execute sequentially.
How to create a thread in a program?
You have two ways to do so. First, making your class "extends" Thread class. Second, making your class "implements" Runnable interface. Put jobs in a run() method and call start() method to start the thread.
Can Java object be locked down for exclusive use by a given thread?
Yes. You can lock an object by putting it in a "synchronized" block. The locked object is inaccessible to any thread other than the one that explicitly claimed it.
Can each Java object keep track of all the threads that want to exclusively access to it?
Yes. Use Thread.currentThread() method to track the accessing thread.
Does it matter in what order catch statements for FileNotFoundException and IOExceptipon are written?
Yes, it does. The FileNoFoundException is inherited from the IOException. Exception's subclasses have to be caught first.
What invokes a thread's run() method?
After a thread is started, via its start() method of the Thread class, the JVM invokes the thread's run() method when the thread is initially executed
What is the purpose of the wait(), notify(), and notifyAll() methods?
The wait(),notify(), and notifyAll() methods are used to provide an efficient way for threads to communicate each other.
What are the high-level thread states?
The high-level thread states are ready, running, waiting, and dead.
What is the difference between yielding and sleeping?
When a task invokes its yield() method, it returns to the ready state. When a task invokes its sleep() method, it returns to the waiting state.
What happens when a thread cannot acquire a lock on an object?
If a thread attempts to execute a synchronized method or synchronized statement and is unable to acquire an object's lock, it enters the waiting state until the lock becomes available.
What is the difference between Process and Thread?
A process can contain multiple threads. In most multithreading operating systems, a process gets its own memory address space; a thread doesn't. Threads typically share the heap belonging to their parent process. For instance, a JVM runs in a single process in the host O/S. Threads in the JVM share the heap belonging to that process; that's why several threads may access the same object. Typically, even though they share a common heap, threads have their own stack space. This is how one thread's invocation of a method is kept separate from another's. This is all a gross oversimplification, but it's accurate enough at a high level. Lots of details differ between operating systems. Process vs. Thread A program vs. similar to a sequential program an run on its own vs. Cannot run on its own Unit of allocation vs. Unit of execution Have its own memory space vs. Share with others Each process has one or more threads vs. Each thread belongs to one process Expensive, need to context switch vs. Cheap, can use process memory and may not need to context switch More secure. One process cannot corrupt another process vs. Less secure. A thread can write the memory used by another thread
Can an inner class declared inside of a method access local variables of this method?
It's possible if these variables are final.
What can go wrong if you replace &emp;&emp; with &emp; in the following code: String a=null; if (a!=null && a.length()>10) {...}
A single ampersand here would lead to a NullPointerException.
What is the Vector class?
The Vector class provides the capability to implement a growable array of objects
What modifiers may be used with an inner class that is a member of an outer class?
A (non-local) inner class may be declared as public, protected, private, static, final, or abstract.
If a method is declared as protected, where may the method be accessed?
A protected method may only be accessed by classes or interfaces of the same package or by subclasses of the class in which it is declared.
What is an Iterator interface?
The Iterator interface is used to step through the elements of a Collection.
How many bits are used to represent Unicode, ASCII, UTF-16, and UTF-8 characters?
Unicode requires 16 bits and ASCII require 7 bits. Although the ASCII character set uses only 7 bits, it is usually represented as 8 bits. UTF-8 represents characters using 8, 16, and 18 bit patterns. UTF-16 uses 16-bit and larger bit patterns.
What's the main difference between a Vector and an ArrayList?
Java Vector class is internally synchronized and ArrayList is not.
What are wrapped classes?
Wrapped classes are classes that allow primitive types to be accessed as objects.
Does garbage collection guarantee that a program will not run out of memory?
No, it doesn't. It is possible for programs to use up memory resources faster than they are garbage collected. It is also possible for programs to create objects that are not subject to garbage collection.
What is the difference between preemptive scheduling and time slicing?
Under preemptive scheduling, the highest priority task executes until it enters the waiting or dead states or a higher priority task comes into existence. Under time slicing, a task executes for a predefined slice of time and then reenters the pool of ready tasks. The scheduler then determines which task should execute next, based on priority and other factors.
Name Component subclasses that support painting ?
The Canvas, Frame, Panel, and Applet classes support painting.
What is a native method?
A native method is a method that is implemented in a language other than Java.
How can you write a loop indefinitely?
for(;;)--for loop; while(true)--always true, etc.
Can an anonymous class be declared as implementing an interface and extending a class?
An anonymous class may implement an interface or extend a superclass, but may not be declared to do both.
What is the purpose of finalization?
The purpose of finalization is to give an unreachable object the opportunity to perform any cleanup processing before the object is garbage collected.
When should the method invokeLater()be used?
This method is used to ensure that Swing components are updated through the event-dispatching thread.
How many methods in Object class?
This question is not asked to test your memory. It tests you how well you know Java. Ten in total.
clone()
equals() & hashcode()
getClass()
finalize()
wait() & notify()
toString()
How does Java handle integer overflows and underflows?
It uses low order bytes of the result that can fit into the size of the type allowed by the operation.
What is the numeric promotion?
Numeric promotion is used with both unary and binary bitwise operators. This means that byte, char, and short values are converted to int values before a bitwise operator is applied.
If a binary bitwise operator has one long operand, the other operand is converted to a long value.
The type of the result of a bitwise operation is the type to which the operands have been promoted. For example:
short a = 5;
byte b = 10;
long c = 15;
The type of the result of (a+b) is int, not short or byte. The type of the result of (a+c) or (b+c) is long.
Is the numeric promotion available in other platform?
Yes. Because Java is implemented using a platform-independent virtual machine, bitwise operations always yield the same result, even when run on machines that use radically different CPUs.
What is the difference between the Boolean & operator and the && operator?
If an expression involving the Boolean & operator is evaluated, both operands are evaluated. Then the & operator is applied to the operand. When an expression involving the && operator is evaluated, the first operand is evaluated. If the first operand returns a value of true then the second operand is evaluated. The && operator is then applied to the first and second operands. If the first operand evaluates to false, the evaluation of the second operand is skipped.
Operator & has no chance to skip both sides evaluation and && operator does. If asked why, give details as above.
When is the ArithmeticException throwQuestion: What is the GregorianCalendar class?
The GregorianCalendar provides support for traditional Western calendars.
What is the SimpleTimeZone class?
The SimpleTimeZone class provides support for a Gregorian calendar.
How can a subclass call a method or a constructor defined in a superclass?
Use the following syntax: super.myMethod(); To call a constructor of the superclass, just write super(); in the first line of the subclass's constructor.
What is the Properties class?
The properties class is a subclass of Hashtable that can be read from or written to a stream. It also provides the capability to specify a set of default values to be used.
What is the purpose of the Runtime class?
The purpose of the Runtime class is to provide access to the Java runtime system.

unix command

2010-08-16T23:51:00.001-07:00

Sun Solaris Command Tips
All commands are specific to Sun Solaris operating system
UNIX backups/Restore
Commands to take backup of /usr /var /lib directories to tape and disk using "tar"
ANS: a) tar -cvf /dev/rmt/0 /usr /var /lib [Taking into tape ]
b) tar -cvf backup_file.tar /usr /var /lib [ Taking disk image]
Taking backup to a tape device attached to a remote system
ANS: a) tar -cv /var /usr /lib | rsh remote_hostname dd of=/dev/rm/0 obs=128
b) ufsdump 0ucf remote_hostname:/dev/rmt/0 /file_system
Extracting / Restore the contents of a tape /tar file
ANS: a) tar -xvf /dev/rmt/0 [ restore contents of a tar tape ]
b) tar -xvf filename.tar [ Restore contents of a tar file ]
c) ufsrestore f /dev/rmt/0 filename [Restore from ufsdump tape backup]
d) ufsrestore rf remote_hostname:/dev/rmt/0 filename [ Restore from remote host tape device ]
List contents of backup
ANS: a) tar -tvf /dev/rmt/0 [ List contents of a tar tape ]
b) tar -tvf filename.tar [ List contents of a tar file ]
c) ufsrestore tf /dev/rmt/0 [ list contents of a tape of ufsdump]
Processes and Processor
What are the types of priority classes supported by Solaris ?
ANS: timesharing, system and reat-time
How will you place a running process in real time class?
ANS: priocntl -s –c RT –I uid process_id_number
How will you start a process in real time class?
ANS : priocntl –c RT –e process_name
How will you bind a process to processor ?
Ans: pbind –b processor_no process_id
Network
What is the pocket size used by SQL*NET Version 2 on Solaris
Ans : Default is 2K
What is the procket size used by network layer TCP/IP in SQL*NET v2?
Ans : Default 1K
Inter process communication (IPC) tuning
What is post wait driver ?
The post wait driver reduces the overhead incurred by the more expensive use of semaphore operation for interprocess communication.
Memory
What is the kernel parameter which controls the UNIX buffer cache on solaris ?
Ans : bufhwm ( bufhwm is the maximum amount of physical memory, in kilobytes that can be used by
I/O buffers)
While starting an oracle process , the unix system displays an erro message like "Cannot allocate more shared memory segment for the processes" If oracle person approaches you, how will you fix this problem?
ANS: increase the value for kernel memory parameter "set semsys:seminfo_semmns" in /etc/system file. [ this is a blind way]
System configurations
How do you find out total RAM installed on your machine
Ans: $prtconf |grep size
How do you find no. of processors installed on your sun box and processing speed?
Ans: $psrinfo –v
How do you find out how many disks are available on your system?
Ans: $ format (for sysadmins)
$ iostat –E (for users – count only the disks which has disk size correctly)
System Boot Options
How do you boot sun box in single user mode?
ANS: At ok prompt type "boot -s" [ ok boot -s OR ok boot -1 OR ok boot -S ]
What are the commands to shutdown the system?
ANS: shutdown, init , halt, reboot
Devices
What is the meaning of logical name of a disk drive format like c0t0d0s0?
ANS: c0 - Controller number
t0 - SCSI bus target number
d0 - Disk number
s0 - Partition or slice number
List some of the main device types?
ANS: /dev/dsk - Disk devices
/dev/rdsk - Raw or character devices
/dev/rmt - Tape devices
/dev/term - Serial line devices
/dev/pts - Pseudo terminals
List commands to display physical devices ?
ANS: prtconf
Sysdef
Dmesg
sysinfo
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Unix Books :-
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linux commands

2010-08-16T23:33:00.001-07:00

•
An A-Z Index of the Bash command line for Linux.
adduser Add a user to the system
addgroup Add a group to the system
alias Create an alias •
apropos Search Help manual pages (man -k)
apt-get Search for and install software packages (Debian/Ubuntu)
aptitude Search for and install software packages (Debian/Ubuntu)
aspell Spell Checker
awk Find and Replace text, database sort/validate/index
b
basename Strip directory and suffix from filenames
bash GNU Bourne-Again SHell
bc Arbitrary precision calculator language
bg Send to background
break Exit from a loop •
builtin Run a shell builtin
bzip2 Compress or decompress named file(s)
c
cal Display a calendar
case Conditionally perform a command
cat Display the contents of a file
cd Change Directory
cfdisk Partition table manipulator for Linux
chgrp Change group ownership
chmod Change access permissions
chown Change file owner and group
chroot Run a command with a different root directory
chkconfig System services (runlevel)
cksum Print CRC checksum and byte counts
clear Clear terminal screen
cmp Compare two files
comm Compare two sorted files line by line
command Run a command - ignoring shell functions •
continue Resume the next iteration of a loop •
cp Copy one or more files to another location
cron Daemon to execute scheduled commands
crontab Schedule a command to run at a later time
csplit Split a file into context-determined pieces
cut Divide a file into several parts
d
date Display or change the date & time
dc Desk Calculator
dd Convert and copy a file, write disk headers, boot records
ddrescue Data recovery tool
declare Declare variables and give them attributes •
df Display free disk space
diff Display the differences between two files
diff3 Show differences among three files
dig DNS lookup
dir Briefly list directory contents
dircolors Colour setup for `ls'
dirname Convert a full pathname to just a path
dirs Display list of remembered directories
dmesg Print kernel & driver messages
du Estimate file space usage
e
echo Display message on screen •
egrep Search file(s) for lines that match an extended expression
eject Eject removable media
enable Enable and disable builtin shell commands •
env Environment variables
ethtool Ethernet card settings
eval Evaluate several commands/arguments
exec Execute a command
exit Exit the shell
expect Automate arbitrary applications accessed over a terminal
expand Convert tabs to spaces
export Set an environment variable
expr Evaluate expressions
f
false Do nothing, unsuccessfully
fdformat Low-level format a floppy disk
fdisk Partition table manipulator for Linux
fg Send job to foreground
fgrep Search file(s) for lines that match a fixed string
file Determine file type
find Search for files that meet a desired criteria
fmt Reformat paragraph text
fold Wrap text to fit a specified width.
for Expand words, and execute commands
format Format disks or tapes
free Display memory usage
fsck File system consistency check and repair
ftp File Transfer Protocol
function Define Function Macros
fuser Identify/kill the process that is accessing a file
g
gawk Find and Replace text within file(s)
getopts Parse positional parameters
grep Search file(s) for lines that match a given pattern
groups Print group names a user is in
gzip Compress or decompress named file(s)
h
hash Remember the full pathname of a name argument
head Output the first part of file(s)
help Display help for a built-in command •
history Command History
hostname Print or set system name
i
id Print user and group id's
if Conditionally perform a command
ifconfig Configure a network interface
ifdown Stop a network interface
ifup Start a network interface up
import Capture an X server screen and save the image to file
install Copy files and set attributes
j
jobs List active jobs •
join Join lines on a common field
k
kill Stop a process from running
killall Kill processes by name
l
less Display output one screen at a time
let Perform arithmetic on shell variables •
ln Make links between files
local Create variables •
locate Find files
logname Print current login name
logout Exit a login shell •
look Display lines beginning with a given string
lpc Line printer control program
lpr Off line print
lprint Print a file
lprintd Abort a print job
lprintq List the print queue
lprm Remove jobs from the print queue
ls List information about file(s)
lsof List open files
m
make Recompile a group of programs
man Help manual
mkdir Create new folder(s)
mkfifo Make FIFOs (named pipes)
mkisofs Create an hybrid ISO9660/JOLIET/HFS filesystem
mknod Make block or character special files
more Display output one screen at a time
mount Mount a file system
mtools Manipulate MS-DOS files
mtr Network diagnostics (traceroute/ping)
mv Move or rename files or directories
mmv Mass Move and rename (files)
n
netstat Networking information
nice Set the priority of a command or job
nl Number lines and write files
nohup Run a command immune to hangups
notify-send Send desktop notifications
nslookup Query Internet name servers interactively
o
open Open a file in its default application
op Operator access
p
passwd Modify a user password
paste Merge lines of files
pathchk Check file name portability
ping Test a network connection
pkill Stop processes from running
popd Restore the previous value of the current directory
pr Prepare files for printing
printcap Printer capability database
printenv Print environment variables
printf Format and print data •
ps Process status
pushd Save and then change the current directory
pwd Print Working Directory
q
quota Display disk usage and limits
quotacheck Scan a file system for disk usage
quotactl Set disk quotas
r
ram ram disk device
rcp Copy files between two machines
read Read a line from standard input •
readarray Read from stdin into an array variable •
readonly Mark variables/functions as readonly
reboot Reboot the system
rename Rename files
renice Alter priority of running processes
remsync Synchronize remote files via email
return Exit a shell function
rev Reverse lines of a file
rm Remove files
rmdir Remove folder(s)
rsync Remote file copy (Synchronize file trees)
s
screen Multiplex terminal, run remote shells via ssh
scp Secure copy (remote file copy)
sdiff Merge two files interactively
sed Stream Editor
select Accept keyboard input
seq Print numeric sequences
set Manipulate shell variables and functions
sftp Secure File Transfer Program
shift Shift positional parameters
shopt Shell Options
shutdown Shutdown or restart linux
sleep Delay for a specified time
slocate Find files
sort Sort text files
source Run commands from a file `.'
split Split a file into fixed-size pieces
ssh Secure Shell client (remote login program)
strace Trace system calls and signals
su Substitute user identity
sudo Execute a command as another user
sum Print a checksum for a file
suspend Suspend execution of this shell •
symlink Make a new name for a file
sync Synchronize data on disk with memory
t
tail Output the last part of files
tar Tape ARchiver
tee Redirect output to multiple files
test Evaluate a conditional expression
time Measure Program running time
times User and system times
touch Change file timestamps
top List processes running on the system
traceroute Trace Route to Host
trap Run a command when a signal is set(bourne)
tr Translate, squeeze, and/or delete characters
true Do nothing, successfully
tsort Topological sort
tty Print filename of terminal on stdin
type Describe a command •
u
ulimit Limit user resources •
umask Users file creation mask
umount Unmount a device
unalias Remove an alias •
uname Print system information
unexpand Convert spaces to tabs
uniq Uniquify files
units Convert units from one scale to another
unset Remove variable or function names
unshar Unpack shell archive scripts
until Execute commands (until error)
useradd Create new user account
usermod Modify user account
users List users currently logged in
uuencode Encode a binary file
uudecode Decode a file created by uuencode
v
v Verbosely list directory contents (`ls -l -b')
vdir Verbosely list directory contents (`ls -l -b')
vi Text Editor
vmstat Report virtual memory statistics
w
watch Execute/display a program periodically
wc Print byte, word, and line counts
whereis Search the user's $path, man pages and source files for a program
which Search the user's $path for a program file
while Execute commands
who Print all usernames currently logged in
whoami Print the current user id and name (`id -un')
Wget Retrieve web pages or files via HTTP, HTTPS or FTP
write Send a message to another user
x
xargs Execute utility, passing constructed argument list(s)
xdg-open Open a file or URL in the user's preferred application.
yes Print a string until interrupted
. Run a command script in the current shell
### Comment / Remark
Commands marked • are bash built-ins, these are available under all shells.

apptitude quesitions

2010-08-16T23:23:00.000-07:00

1. NUMBERS

IMPORTANT FACTS AND FORMULAE
I..Numeral : In Hindu Arabic system, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 called digits to represent any number.
A group of digits, denoting a number is called a numeral.
We represent a number, say 689745132 as shown below :

Ten Crores (108) Crores(107) Ten Lacs (Millions) (106) Lacs(105) Ten Thousands (104) Thousands (103) Hundreds (102) Tens(101) Units(100)
6 8 9 7 4 5 1 3 2
We read it as : 'Sixty-eight crores, ninety-seven lacs, forty-five thousand, one hundred and thirty-two'.
II Place Value or Local Value of a Digit in a Numeral :
In the above numeral :
Place value of 2 is (2 x 1) = 2; Place value of 3 is (3 x 10) = 30;
Place value of 1 is (1 x 100) = 100 and so on.
Place value of 6 is 6 x 108 = 600000000
III.Face Value : The face value of a digit in a numeral is the value of the digit itself at whatever place it may be. In the above numeral, the face value of 2 is 2; the face value of 3 is 3 and so on.
IV.TYPES OF NUMBERS
1.Natural Numbers : Counting numbers 1, 2, 3, 4, 5,..... are called natural
numbers.
2.Whole Numbers : All counting numbers together with zero form the set of whole
numbers. Thus,
(i) 0 is the only whole number which is not a natural number.
(ii) Every natural number is a whole number.
3.Integers : All natural numbers, 0 and negatives of counting numbers i.e.,
{…, -3,-2,-1, 0, 1, 2, 3,…..} together form the set of integers.
(i) Positive Integers : {1, 2, 3, 4, …..} is the set of all positive integers.
(ii) Negative Integers : {- 1, - 2, - 3,…..} is the set of all negative integers.
(iii) Non-Positive and Non-Negative Integers : 0 is neither positive nor
negative. So, {0, 1, 2, 3,….} represents the set of non-negative integers, while
{0, -1,-2,-3,…..} represents the set of non-positive integers.
4. Even Numbers : A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8, 10, etc.
5. Odd Numbers : A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7, 9, 11, etc.
6. Prime Numbers : A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself.
Prime numbers upto 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,
47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Prime numbers Greater than 100 : Letp be a given number greater than 100. To find out whether it is prime or not, we use the following method :
Find a whole number nearly greater than the square root of p. Let k > *jp. Test whether p is divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p is prime.
e.g,,We have to find whether 191 is a prime number or not. Now, 14 > V191.
Prime numbers less than 14 are 2, 3, 5, 7, 11, 13.
191 is not divisible by any of them. So, 191 is a prime number.
7.Composite Numbers : Numbers greater than 1 which are not prime, are known as composite numbers, e.g., 4, 6, 8, 9, 10, 12.
Note : (i) 1 is neither prime nor composite.
(ii) 2 is the only even number which is prime.
(iii) There are 25 prime numbers between 1 and 100.
8. Co-primes : Two numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g., (2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes,
V.TESTS OF DIVISIBILITY
1. Divisibility By 2 : A number is divisible by 2, if its unit's digit is any of 0, 2, 4, 6, 8.
Ex. 84932 is divisible by 2, while 65935 is not.
2. Divisibility By 3 : A number is divisible by 3, if the sum of its digits is divisible by 3.
Ex.592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which is divisible by 3.
But, 864329 is not divisible by 3, since sum of its digits =(8 + 6 + 4 + 3 + 2 + 9) = 32, which is not divisible by 3.
3. Divisibility By 4 : A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
Ex. 892648 is divisible by 4, since the number formed by the last two digits is
48, which is divisible by 4.
But, 749282 is not divisible by 4, since the number formed by the last tv/o digits is 82, which is not divisible by 4.
4. Divisibility By 5 : A number is divisible by 5, if its unit's digit is either 0 or 5. Thus, 20820 and 50345 are divisible by 5, while 30934 and 40946 are not.
5. Divisibility By 6 : A number is divisible by 6, if it is divisible by both 2 and 3. Ex. The number 35256 is clearly divisible by 2.
Sum of its digits = (3 + 5 + 2 + 5 + 6) = 21, which is divisible by 3. Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is divisible by 6.
6. Divisibility By 8 : A number is divisible by 8, if the number formed by the last
three digits of the given number is divisible by 8.
Ex. 953360 is divisible by 8, since the number formed by last three digits is 360, which is divisible by 8.
But, 529418 is not divisible by 8, since the number formed by last three digits is 418, which is not divisible by 8.
7. Divisibility By 9 : A number is divisible by 9, if the sum of its digits is divisible
by 9.
Ex. 60732 is divisible by 9, since sum of digits * (6 + 0 + 7 + 3 + 2) = 18, which is divisible by 9.
But, 68956 is not divisible by 9, since sum of digits = (6 + 8 + 9 + 5 + 6) = 34, which is not divisible by 9.
8. Divisibility By 10 : A number is divisible by 10, if it ends with 0.
Ex. 96410, 10480 are divisible by 10, while 96375 is not.
9. Divisibility By 11 : A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.
Ex. The number 4832718 is divisible by 11, since :
(sum of digits at odd places) - (sum of digits at even places)
- (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.
10. Divisibility By 12 ; A number is divisible by 12, if it is divisible by both 4 and
3.
Ex. Consider the number 34632.
(i) The number formed by last two digits is 32, which is divisible by 4,
(ii) Sum of digits = (3 + 4 + 6 + 3 + 2) = 18, which is divisible by 3. Thus, 34632 is divisible by 4 as well as 3. Hence, 34632 is divisible by 12.
11. Divisibility By 14 : A number is divisible by 14, if it is divisible by 2 as well as 7.
12. Divisibility By 15 : A number is divisible by 15, if it is divisible by both 3 and 5.
13. Divisibility By 16 : A number is divisible by 16, if the number formed by the last4 digits is divisible by 16.
Ex.7957536 is divisible by 16, since the number formed by the last four digits is 7536, which is divisible by 16.
14. Divisibility By 24 : A given number is divisible by 24, if it is divisible by both3 and 8.
15. Divisibility By 40 : A given number is divisible by 40, if it is divisible by both
5 and 8.
16. Divisibility By 80 : A given number is divisible by 80, if it is divisible by both 5 and 16.
Note : If a number is divisible by p as well as q, where p and q are co-primes, then the given number is divisible by pq.
If p arid q are not co-primes, then the given number need not be divisible by pq,
even when it is divisible by both p and q.
Ex. 36 is divisible by both 4 and 6, but it is not divisible by (4x6) = 24, since
4 and 6 are not co-primes.

VI MULTIPLICATION BY SHORT CUT METHODS
1. Multiplication By Distributive Law :
(i) a x (b + c) = a x b + a x c (ii) ax(b-c) = a x b-a x c.
Ex. (i) 567958 x 99999 = 567958 x (100000 - 1)
= 567958 x 100000 - 567958 x 1 = (56795800000 - 567958) = 56795232042. (ii) 978 x 184 + 978 x 816 = 978 x (184 + 816) = 978 x 1000 = 978000.
2. Multiplication of a Number By 5n : Put n zeros to the right of the multiplicand and divide the number so formed by 2n
Ex. 975436 x 625 = 975436 x 54= 9754360000 = 609647600
16
VII. BASIC FORMULAE
1. (a + b)2 = a2 + b2 + 2ab 2. (a - b)2 = a2 + b2 - 2ab
3. (a + b)2 - (a - b)2 = 4ab 4. (a + b)2 + (a - b)2 = 2 (a2 + b2)
5. (a2 - b2) = (a + b) (a - b)
6. (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
7. (a3 + b3) = (a +b) (a2 - ab + b2) 8. (a3 - b3) = (a - b) (a2 + ab + b2)
9. (a3 + b3 + c3 -3abc) = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)
10. If a + b + c = 0, then a3 + b3 + c3 = 3abc.

VIII. DIVISION ALGORITHM OR EUCLIDEAN ALGORITHM
If we divide a given number by another number, then :
Dividend = (Divisor x Quotient) + Remainder
IX. {i) (xn - an ) is divisible by (x - a) for all values of n.
(ii) (xn - an) is divisible by (x + a) for all even values of n.
(iii) (xn + an) is divisible by (x + a) for all odd values of n.

X. PROGRESSION
A succession of numbers formed and arranged in a definite order according to certain definite rule, is called a progression.
1. Arithmetic Progression (A.P.) : If each term of a progression differs from its preceding term by a constant, then such a progression is called an arithmetical progression. This constant difference is called the common difference of the A.P.
An A.P. with first term a and common difference d is given by a, (a + d), (a + 2d),(a + 3d),.....
The nth term of this A.P. is given by Tn =a (n - 1) d.
The sum of n terms of this A.P.
Sn = n/2 [2a + (n - 1) d] = n/2 (first term + last term).

SOME IMPORTANT RESULTS :

(i) (1 + 2 + 3 +…. + n) =n(n+1)/2
(ii) (l2 + 22 + 32 + ... + n2) = n (n+1)(2n+1)/6
(iii) (13 + 23 + 33 + ... + n3) =n2(n+1)2

2. Geometrical Progression (G.P.) : A progression of numbers in which every term bears a constant ratio with its preceding term, is called a geometrical progression.
The constant ratio is called the common ratio of the G.P. A G.P. with first term a and common ratio r is :
a, ar, ar2,
In this G.P. Tn = arn-1
sum of the n terms, Sn= a(1-rn)
(1-r)

SOLVED EXAMPLES

Ex. 1. Simplify : (i) 8888 + 888 + 88 + 8
(ii) 11992 - 7823 - 456
Sol. i ) 8888 ii) 11992 - 7823 - 456 = 11992 - (7823 + 456)
888 = 11992 - 8279 = 3713-
88 7823 11992
+ 8 + 456 - 8279
9872 8279 3713

Ex. 2, What value will replace the question mark in each of the following equations ?
(i) ? - 1936248 = 1635773 (ii) 8597 - ? = 7429 - 4358

Sol. (i) Let x - 1936248=1635773.Then, x = 1635773 + 1936248=3572021. (ii) Let 8597 - x = 7429 - 4358.
Then, x = (8597 + 4358) - 7429 = 12955 - 7429 = 5526.

Ex. 3. What could be the maximum value of Q in the following equation? 5P9 + 3R7 + 2Q8 = 1114
Sol. We may analyse the given equation as shown : 1 2
Clearly, 2 + P + R + Q = ll. 5 P 9
So, the maximum value of Q can be 3 R 7
(11 - 2) i.e., 9 (when P = 0, R = 0); 2 Q 8
11 1 4

Ex. 4. Simplify : (i) 5793405 x 9999 (ii) 839478 x 625
Sol.
i)5793405x9999=5793405(10000-1)=57934050000-5793405=57928256595.b

ii) 839478 x 625 = 839478 x 54 = 8394780000 = 524673750.
16

Ex. 5. Evaluate : (i) 986 x 237 + 986 x 863 (ii) 983 x 207 - 983 x 107
Sol.
(i) 986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000.
(ii) 983 x 207 - 983 x 107 = 983 x (207 - 107) = 983 x 100 = 98300.

Ex. 6. Simplify : (i) 1605 x 1605 ii) 1398 x 1398
Sol.
i) 1605 x 1605 = (1605)2 = (1600 + 5)2 = (1600)2 + (5)2 + 2 x 1600 x 5
= 2560000 + 25 + 16000 = 2576025.
(ii) 1398 x 1398 - (1398)2 = (1400 - 2)2= (1400)2 + (2)2 - 2 x 1400 x 2
=1960000 + 4 - 5600 = 1954404.

Ex. 7. Evaluate : (313 x 313 + 287 x 287).
Sol.
(a2 + b2) = 1/2 [(a + b)2 + (a- b)2]
(313)2 + (287)2 = 1/2 [(313 + 287)2 + (313 - 287)2] = ½[(600)2 + (26)2]
= 1/2 (360000 + 676) = 180338.

Ex. 8. Which of the following are prime numbers ?
(i) 241 (ii) 337 (Hi) 391 (iv) 571
Sol.
(i) Clearly, 16 > Ö241. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13.
241 is not divisible by any one of them.
241 is a prime number.

(ii) Clearly, 19>Ö337. Prime numbers less than 19 are 2, 3, 5, 7, 11,13,17.
337 is not divisible by any one of them.
337 is a prime number.
(iii) Clearly, 20 > Ö39l". Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19.
We find that 391 is divisible by 17.
391 is not prime.
(iv) Clearly, 24 > Ö57T. Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23.
571 is not divisible by any one of them.
571 is a prime number.

Ex. 9. Find the unit's digit in the product (2467)163 x (341)72.
Sol. Clearly, unit's digit in the given product = unit's digit in 7153 x 172.
Now, 74 gives unit digit 1.
7152 gives unit digit 1,
 7153 gives unit digit (l x 7) = 7. Also, 172 gives unit digit 1.
Hence, unit's digit in the product = (7 x 1) = 7.

Ex. 10. Find the unit's digit in (264)102 + (264)103
Sol. Required unit's digit = unit's digit in (4)102 + (4)103.
Now, 42 gives unit digit 6.
(4)102 gives unjt digit 6.
(4)103 gives unit digit of the product (6 x 4) i.e., 4.
Hence, unit's digit in (264)m + (264)103 = unit's digit in (6 + 4) = 0.

Ex. 11. Find the total number of prime factors in the expression (4)11 x (7)5 x (11)2.
Sol. (4)11x (7)5 x (11)2 = (2 x 2)11 x (7)5 x (11)2 = 211 x 211 x75x 112 = 222 x 75 x112
Total number of prime factors = (22 + 5 + 2) = 29.

Ex.12. Simplify : (i) 896 x 896 - 204 x 204
(ii) 387 x 387 + 114 x 114 + 2 x 387 x 114
(iii) 81 X 81 + 68 X 68-2 x 81 X 68.
Sol.
(i) Given exp = (896)2 - (204)2 = (896 + 204) (896 - 204) = 1100 x 692 = 761200.
(ii) Given exp = (387)2+ (114)2+ (2 x 387x 114)
= a2 + b2 + 2ab, where a = 387,b=114
= (a+b)2 = (387 + 114 )2 = (501)2 = 251001.
(iii) Given exp = (81)2 + (68)2 – 2x 81 x 68 = a2 + b2 – 2ab,Where a =81,b=68
= (a-b)2 = (81 –68)2 = (13)2 = 169.

Ex.13. Which of the following numbers is divisible by 3 ?
(i) 541326 (ii) 5967013
Sol.
(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.
Hence, 541326 is divisible by 3.

(ii) Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3.
Hence, 5967013 is not divisible by 3.

Ex.14.What least value must be assigned to * so that the number 197*5462 is r 9 ?
Sol.
Let the missing digit be x.
Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 +»2) = (34 + x).
For (34 + x) to be divisible by 9, x must be replaced by 2 .
Hence, the digit in place of * must be 2.

Ex. 15. Which of the following numbers is divisible by 4 ?
(i) 67920594 (ii) 618703572
Sol.
(i) The number formed by the last two digits in the given number is 94, which is not divisible by 4.
Hence, 67920594 is not divisible by 4.

(ii) The number formed by the last two digits in the given number is 72, which is divisible by 4.
Hence, 618703572 is divisible by 4.

Ex. 16. Which digits should come in place of * and $ if the number 62684*$ is divisible by both 8 and 5 ?
Sol.
Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 is never divisible by 8. So, 0 will replace $.
Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if * is replaced by 4.
Hence, digits in place of * and $ are 4 and 0 respectively.

Ex. 17. Show that 4832718 is divisible by 11.
Sol. (Sum of digits at odd places) - (Sum of digits at even places)
= (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.
Hence, 4832718 is divisible by 11.

Ex. 18. Is 52563744 divisible by 24 ?
Sol. 24 = 3 x 8, where 3 and 8 are co-primes.
The sum of the digits in the given number is 36, which is divisible by 3. So, the given number is divisible by 3.
The number formed by the last 3 digits of the given number is 744, which is divisible by 8. So, the given number is divisible by 8.
Thus, the given number is divisible by both 3 and 8, where 3 and 8 are co-primes.
So, it is divisible by 3 x 8, i.e., 24.

Ex. 19. What least number must be added to 3000 to obtain a number exactly divisible by 19 ?
Sol. On dividing 3000 by 19, we get 17 as remainder.
Number to be added = (19 - 17) = 2.

Ex. 20. What least number must be subtracted from 2000 to get a number exactly divisible by 17 ?
Sol. On dividing 2000 by 17, we get 11 as remainder.
Required number to be subtracted = 11.

Ex. 21. Find the number which is nearest to 3105 and is exactly divisible by 21.
Sol. On dividing 3105 by 21, we get 18 as remainder.
Number to be added to 3105 = (21 - 18) - 3.
Hence, required number = 3105 + 3 = 3108.

Ex. 22. Find the smallest number of 6 digits which is exactly divisible by 111.
Sol. Smallest number of 6 digits is 100000.
On dividing 100000 by 111, we get 100 as remainder.
Number to be added = (111 - 100) - 11.
Hence, required number = 100011.-

Ex. 23. On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37. Find the divisor.
Dividend - Remainder 15968-37
Sol. Divisor = -------------------------- = ------------- = 179.
.Quotient 89

Ex. 24. A number when divided by 342 gives a remainder 47. When the same number ift divided by 19, what would be the remainder ?
Sol. On dividing the given number by 342, let k be the quotient and 47 as remainder.
Then, number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.
The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.

Ex. 25. A number being successively divided by 3, 5 and 8 leaves remainders 1, 4
and 7 respectively. Find the respective remainders if the order of divisors be reversed,
Sol.
3 X
5 y - 1
8 z - 4
1 - 7
z = (8 x 1 + 7) = 15; y = {5z + 4) = (5 x 15 + 4) = 79; x = (3y + 1) = (3 x 79 + 1) = 238.
Now,
8 238
5 29 - 6
3 5 - 4
1 - 9,
Respective remainders are 6, 4, 2.

Ex. 26. Find the remainder when 231 is divided by 5.
Sol. 210 = 1024. Unit digit of 210 x 210 x 210 is 4 [as 4 x 4 x 4 gives unit digit 4].
Unit digit of 231 is 8.
Now, 8 when divided by 5, gives 3 as remainder.
Hence, 231 when divided by 5, gives 3 as remainder.

Ex. 27. How many numbers between 11 and 90 are divisible by 7 ?
Sol. The required numbers are 14, 21, 28, 35, .... 77, 84.
This is an A.P. with a = 14 and d = (21 - 14) = 7.
Let it contain n terms.
Then, Tn = 84 => a + (n - 1) d = 84
=> 14 + (n - 1) x 7 = 84 or n = 11.
Required number of terms = 11.

Ex. 28. Find the sum of all odd numbers upto 100.
Sol. The given numbers are 1, 3, 5, 7, ..., 99.
This is an A.P. with a = 1 and d = 2.
Let it contain n terms. Then,
1 + (n - 1) x 2 = 99 or n = 50.
Required sum = n (first term + last term)
2
= 50 (1 + 99) = 2500.
2

Ex. 29. Find the sum of all 2 digit numbers divisible by 3.
Sol. All 2 digit numbers divisible by 3 are :
12, 51, 18, 21, ..., 99.
This is an A.P. with a = 12 and d = 3.
Let it contain n terms. Then,
12 + (n - 1) x 3 = 99 or n = 30.
Required sum = 30 x (12+99) = 1665.
2

Ex.30.How many terms are there in 2,4,8,16……1024?
Sol.Clearly 2,4,8,16……..1024 form a GP. With a=2 and r = 4/2 =2.
Let the number of terms be n . Then
2 x 2n-1 =1024 or 2n-1 =512 = 29.
n-1=9 or n=10.

Ex. 31. 2 + 22 + 23 + ... + 28 = ?
Sol. Given series is a G.P. with a = 2, r = 2 and n = 8.
sum = a(rn-1) = 2 x (28 –1) = (2 x 255) =510
(r-1) (2-1)
2. H.C.F. AND L.C.M. OF NUMBERS

IMPORTANT FACTS AND FORMULAE

I. Factors and Multiples : If a number a divides another number b exactly, we say that a is a factor of b. In this case, b is called a multiple of a.
II. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.): The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly.
There are two methods of finding the H.C.F. of a given set of numbers :
1. Factorization Method : Express each one of the given numbers as the product of prime factors.The product of least powers of common prime factors gives H.C.F.
2. Division Method: Suppose we have to find the H.C.F. of two given numbers. Divide the larger number by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is the required H.C.F.
Finding the H.C.F. of more than two numbers : Suppose we have to find the H.C.F. of three numbers. Then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given numbers.
Similarly, the H.C.F. of more than three numbers may be obtained.
III. Least Common Multiple (L.C.M.) : The least number which is exactly divisible by each one of the given numbers is called their L.C.M.
1. Factorization Method of Finding L.C.M.: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors,
2. Common Division Method {Short-cut Method) of Finding L.C.M.: Arrange the given numbers in a row in any order. Divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers,
IV. Product of two numbers =Product of their H.C.F. and L.C.M.
V. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.
VI. H.C.F. and L.C.M. of Fractions:
1.H C F= H.C.F. of Numerators 2.L C M = L.C.M of Numerators__
L.C.M. of Denominators H.C.F. of Denominators
VII. H.C.F. and L.C.M. of Decimal Fractions: In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.
VIII. Comparison of Fractions: Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M. as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

SOLVED EXAMPLES

Ex. 1. Find the H.C.F. of 23 X 32 X 5 X 74, 22 X 35 X 52 X 73,23 X 53 X 72
Sol. The prime numbers common to given numbers are 2,5 and 7.
H.C.F. = 22 x 5 x72 = 980.

Ex. 2. Find the H.C.F. of 108, 288 and 360.
Sol. 108 = 22 x 33, 288 = 25 x 32 and 360 = 23 x 5 x 32.
H.C.F. = 22 x 32 = 36.

Ex. 3. Find the H.C.F. of 513, 1134 and 1215.
Sol.
1134 ) 1215 ( 1
1134
81 ) 1134 ( 14
81
324
324
x
H.C.F. of 1134 and 1215 is 81.
So, Required H.C.F. = H.C.F. of 513 and 81.
______
81 ) 513 ( 6
__486____
27) 81 ( 3
81
0

H.C.F. of given numbers = 27.

Ex. 4. Reduce 391 to lowest terms .
667
to lowest terms.

Sol. H.C.F. of 391 and 667 is 23.
On dividing the numerator and denominator by 23, we get :
391 = 391  23 = 17
667 667 23 29

Ex.5.Find the L.C.M. of 22 x 33 x 5 x 72 , 23 x 32 x 52 x 74 , 2 x 3 x 53 x 7 x 11.
Sol. L.C.M. = Product of highest powers of 2, 3, 5, 7 and 11 = 23 x 33 x 53 x 74 x 11

Ex.6. Find the L.C.M. of 72, 108 and 2100.
Sol. 72 = 23 x 32, 108 = 33 x 22, 2100 = 22 x 52 x 3 x 7.
L.C.M. = 23 x 33 x 52 x 7 = 37800.

Ex.7.Find the L.C.M. of 16, 24, 36 and 54.
Sol.
2 16 - 24 - 36 - 54
2 8 - 12 - 18 - 27
2 4 - 6 - 9 - 27
3 2 - 3 - 9 - 27
3 2 - 1 - 3 - 9
2 - 1 - 1 - 3

 L.C.M. = 2 x 2 x 2 x 3 x 3 x 2 x 3 = 432.

Ex. 8. Find the H.C.F. and L.C.M. of 2 , 8 , 16 and 10.
3 9 81 27
Sol. H.C.F. of given fractions = H.C.F. of 2,8,16,10 = 2_
L.C.M. of 3,9,81,27 81

L.C.M of given fractions = L.C.M. of 2,8,16,10 = 80_
H.C.F. of 3,9,81,27 3

Ex. 9. Find the H.C.F. and L.C.M. of 0.63, 1.05 and 2.1.
Sol. Making the same number of decimal places, the given numbers are 0.63, 1.05 and 2.10.
Without decimal places, these numbers are 63, 105 and 210.
Now, H.C.F. of 63, 105 and 210 is 21.
H.C.F. of 0.63, 1.05 and 2.1 is 0.21.
L.C.M. of 63, 105 and 210 is 630.
L.C.M. of 0.63, 1.05 and 2.1 is 6.30.

Ex. 10. Two numbers are in the ratio of 15:11. If their H.C.F. is 13, find the numbers.
Sol. Let the required numbers be 15.x and llx.
Then, their H.C.F. is x. So, x = 13.
The numbers are (15 x 13 and 11 x 13) i.e., 195 and 143.

Ex. 11. TheH.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the numbers is 77,find the other.
Sol. Other number = 11 X 693 = 99
77

Ex. 12. Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.
Sol. Required length = H.C.F. of 495 cm, 900 cm and 1665 cm.
495 = 32 x 5 x 11, 900 = 22 x 32 x 52, 1665 = 32 x 5 x 37.
H.C.F. = 32 x 5 = 45.
Hence, required length = 45 cm.

Ex. 13. Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
Sol. Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032
_______
1651 ) 2032 ( 1 1651
1651_______
381 ) 1651 ( 4
1524_________
127 ) 381 ( 3
381
0
Required number = 127.

Ex. 14. Find the largest number which divides 62, 132 and 237 to leave the same remainder in each case.
Sol . Required number = H.C.F. of (132 - 62), (237 - 132) and (237 - 62)
= H.C.F. of 70, 105 and 175 = 35.

Ex.15.Find the least number exactly divisible by 12,15,20,27.
Sol.

3 12 - 15 - 20 - 27
4 4 - 5 - 20 - 9
5 1 - 5 - 5 - 9
1 - 1 - 1 - 9

Ex.16.Find the least number which when divided by 6,7,8,9, and 12 leave the same remainder 1 each case
Sol. Required number = (L.C.M OF 6,7,8,9,12) + 1

3 6 - 7 - 8 - 9 - 12
4 2 - 7 - 8 - 3 - 4
5 1 - 7 - 4 - 3 - 2
1 - 7 - 2 - 3 - 1

L.C.M = 3 X 2 X 2 X 7 X 2 X 3 = 504.
Hence required number = (504 +1) = 505.

Ex.17. Find the largest number of four digits exactly divisible by 12,15,18 and 27.
Sol. The Largest number of four digits is 9999.
Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.
On dividing 9999 by 540,we get 279 as remainder .
Required number = (9999-279) = 9720.

Ex.18.Find the smallest number of five digits exactly divisible by 16,24,36 and 54.
Sol. Smallest number of five digits is 10000.
Required number must be divisible by L.C.M. of 16,24,36,54 i.e 432,
On dividing 10000 by 432,we get 64 as remainder.
Required number = 10000 +( 432 – 64 ) = 10368.

Ex.19.Find the least number which when divided by 20,25,35 and 40 leaves remainders 14,19,29 and 34 respectively.
Sol. Here,(20-14) = 6,(25 – 19)=6,(35-29)=6 and (40-34)=6.
Required number = (L.C.M. of 20,25,35,40) – 6 =1394.

Ex.20.Find the least number which when divided by 5,6,7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder .
Sol. L.C.M. of 5,6,7,8 = 840.
 Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 X 2 + 3)=1683

Ex.21.The traffic lights at three different road crossings change after every 48 sec., 72 sec and 108 sec.respectively .If they all change simultaneously at 8:20:00 hours,then at what time they again change simultaneously .
Sol. Interval of change = (L.C.M of 48,72,108)sec.=432sec.
So, the lights will agin change simultaneously after every 432 seconds i.e,7 min.12sec
Hence , next simultaneous change will take place at 8:27:12 hrs.

Ex.22.Arrange the fractions 17 , 31, 43, 59 in the ascending order.
18 36 45 60
Sol.L.C.M. of 18,36,45 and 60 = 180.
Now, 17 = 17 X 10 = 170 ; 31 = 31 X 5 = 155 ;
18 18 X 10 180 36 36 X 5 180

43 = 43 X 4 = 172 ; 59 = 59 X 3 = 177 ;
45 45 X 4 180 60 60 X 3 180

Since, 155<170<172<177, so, 155 < 170 < 172 < 177 180 180 180 180Hence, 31 < 17 < 43 < 59 36 18 45 60 3. DECIMAL FRACTIONSIMPORTANT FACTS AND FORMULAEI. Decimal Fractions : Fractions in which denominators are powers of 10 are known as decimal fractions. Thus ,1/10=1 tenth=.1;1/100=1 hundredth =.01; 99/100=99 hundreths=.99;7/1000=7 thousandths=.007,etcII. Conversion of a Decimal Into Vulgar Fraction : Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms.Thus, 0.25=25/100=1/4;2.008=2008/1000=251/125. III. 1. Annexing zeros to the extreme right of a decimal fraction does not change its value Thus, 0.8 = 0.80 = 0.800, etc.2. If numerator and denominator of a fraction contain the same number of decimal places, then we remove the decimal sign. Thus, 1.84/2.99 = 184/299 = 8/13; 0.365/0.584 = 365/584=5IV. Operations on Decimal Fractions :1. Addition and Subtraction of Decimal Fractions : The given numbers are soplaced under each other that the decimal points lie in one column. The numbersso arranged can now be added or subtracted in the usual way.2. Multiplication of a Decimal Fraction By a Power of 10 : Shift the decimalpoint to the right by as many places as is the power of 10.Thus, 5.9632 x 100 = 596,32; 0.073 x 10000 = 0.0730 x 10000 = 730.3.Multiplication of Decimal Fractions : Multiply the given numbers consideringthem without the decimal point. Now, in the product, the decimal point is markedoff to obtain as many places of decimal as is the sum of the number of decimalplaces in the given numbers.Suppose we have to find the product (.2 x .02 x .002). Now, 2x2x2 = 8. Sum of decimal places = (1 + 2 + 3) = 6. .2 x .02 x .002 = .000008.4.Dividing a Decimal Fraction By a Counting Number : Divide the givennumber without considering the decimal point, by the given counting number.Now, in the quotient, put the decimal point to give as many places of decimal asthere are in the dividend.Suppose we have to find the quotient (0.0204 + 17). Now, 204 ^ 17 = 12. Dividend contains 4 places of decimal. So, 0.0204 + 17 = 0.0012.5. Dividing a Decimal Fraction By a Decimal Fraction : Multiply both the dividend and the divisor by a suitable power of 10 to make divisor a whole number. Now, proceed as above.Thus, 0.00066/0.11 = (0.00066*100)/(0.11*100) = (0.066/11) = 0.006V V. Comparison of Fractions : Suppose some fractions are to be arranged in ascending or descending order of magnitude. Then, convert each one of the given fractions in the decimal form, and arrange them accordingly.Suppose, we have to arrange the fractions 3/5, 6/7 and 7/9 in descending order. now, 3/5=0.6,6/7 = 0.857,7/9 = 0.777.... since 0.857>0.777...>0.6, so 6/7>7/9>3/5

VI. Recurring Decimal : If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called a recurring decimal.
In a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it. If a set of figures is repeated, it is expressed by putting a bar on the set
______
Thus 1/3 = 0.3333….= 0.3; 22 /7 = 3.142857142857.....= 3.142857
Pure Recurring Decimal: A decimal fraction in which all the figures after the decimal point are repeated, is called a pure recurring decimal.

Converting a Pure Recurring Decimal Into Vulgar Fraction : Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures.

thus ,0.5 = 5/9; 0.53 = 53/59 ;0.067 = 67/999;etc...

Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some of them are repeated, is called a mixed recurring decimal.
e.g., 0.17333 = 0.173.

Converting a Mixed Recurring Decimal Into Vulgar Fraction : In the numerator, take the difference between the number formed by all the digits after decimal point (taking repeated digits only once) and that formed by the digits which are not repeated, In the denominator, take the number formed by as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits.

Thus 0.16 = (16-1) / 90 = 15/19 = 1/6;
____
0.2273 = (2273 – 22)/9900 = 2251/9900

VII. Some Basic Formulae :

1. (a + b)(a- b) = (a2 - b2).
2. (a + b)2 = (a2 + b2 + 2ab).
3. (a - b)2 = (a2 + b2 - 2ab).
4. (a + b+c)2 = a2 + b2 + c2+2(ab+bc+ca)
5. (a3 + b3) = (a + b) (a2 - ab + b2)
6. (a3 - b3) = (a - b) (a2 + ab + b2).
7. (a3 + b3 + c3 - 3abc) = (a + b + c) (a2 + b2 + c2-ab-bc-ca)
8. When a + b + c = 0, then a3 + b3+ c3 = 3abc

SOLVED EXAMPLES

Ex. 1. Convert the following into vulgar fraction:
(i) 0.75 (ii) 3.004 (iii) 0.0056

Sol. (i). 0.75 = 75/100 = 3/4 (ii) 3.004 = 3004/1000 = 751/250 (iii) 0.0056 = 56/10000 = 7/1250

Ex. 2. Arrange the fractions 5/8, 7/12, 13/16, 16/29 and 3/4 in ascending order of magnitude.

Sol. Converting each of the given fractions into decimal form, we get :
5/8 = 0.624, 7/12 = 0.8125, 16/29 = 0.5517, and 3/4 = 0.75
Now, 0.5517<0.5833<0.625<0.75<0.8125  16/29 < 7/12 < 5/8 < 3/4 < 13/16Ex. 3. arrange the fractions 3/5, 4/7, 8/9, and 9/11 in their descending order.Sol. Clearly, 3/5 = 0.6, 4/7 = 0.571, 8/9 = 0.88, 9/111 = 0.818. Now, 0.88 > 0.818 > 0.6 > 0.571
 8/9 > 9/11 > 3/4 > 13/ 16

Ex. 4. Evaluate : (i) 6202.5 + 620.25 + 62.025 + 6.2025 + 0.62025
(ii) 5.064 + 3.98 + 0.7036 + 7.6 + 0.3 + 2

Sol. (i) 6202.5 (ii) 5.064
620.25 3.98
62.025 0.7036
6.2025 7.6
+ __ 0.62025 0.3
6891.59775 _2.0___
19.6476

Ex. 5. Evaluate : (i) 31.004 – 17.2368 (ii) 13 – 5.1967

Sol. (i) 31.0040 (ii) 31.0000
– 17.2386 – _5.1967
13.7654 7.8033

Ex. 6. What value will replace the question mark in the following equations ?
(i) 5172.49 + 378.352 + ? = 9318.678
(ii) ? – 7328.96 + 5169.38

Sol. (i) Let 5172.49 + 378.352 + x = 9318.678
Then , x = 9318.678 – (5172.49 + 378.352) = 9318.678 – 5550.842 = 3767.836
(ii) Let x – 7328.96 = 5169.38. Then, x = 5169.38 + 7328.96 = 12498.34.

Ex. 7. Find the products: (i) 6.3204 * 100 (ii) 0.069 * 10000

Sol. (i) 6.3204 * 1000 = 632.04 (ii) 0.069 * 10000 = 0.0690 * 10000 = 690

Ex. 8. Find the product:
(i) 2.61 * 1.3 (ii) 2.1693 * 1.4 (iii) 0.4 * 0.04 * 0.004 * 40

Sol. (i) 261 8 13 = 3393. Sum of decimal places of given numbers = (2+1) = 3.
2.61 * 1.3 = 3.393.
(ii) 21693 * 14 = 303702. Sum of decimal places = (4+1) = 5
2.1693 * 1.4 = 3.03702.
(iii) 4 * 4 * 4 * 40 = 2560. Sum of decimal places = (1 + 2+ 3) = 6
0.4 * 0.04 * 0.004 * 40 = 0.002560.

Ex. 9. Given that 268 * 74 = 19832, find the values of 2.68 * 0.74.

Sol. Sum of decimal places = (2 + 2) = 4
2.68 * 0.74 = 1.9832.

Ex. 10. Find the quotient:
(i) 0.63 / 9 (ii) 0.0204 / 17 (iii) 3.1603 / 13

Sol. (i) 63 / 9 = 7. Dividend contains 2 places decimal.
0.63 / 9 = 0.7.
(ii) 204 / 17 = 12. Dividend contains 4 places of decimal.
0.2040 / 17 = 0.0012.
(iii) 31603 / 13 = 2431. Dividend contains 4 places of decimal.
3.1603 / 13 = 0.2431.

Ex. 11. Evaluate :
(i) 35 + 0.07 (ii) 2.5 + 0.0005
(iii) 136.09 + 43.9

Sol. (i) 35/0.07 = ( 35*100) / (0.07*100) = (3500 / 7) = 500
(ii) 25/0.0005 = (25*10000) / (0.0005*10000) = 25000 / 5 = 5000
(iii) 136.09/43.9 = (136.09*10) / (43.9*10) = 1360.9 / 439 = 3.1

Ex. 12. What value will come in place of question mark in the following equation?
(i) 0.006 +? = 0.6 (ii) ? + 0.025 = 80

Sol. (i) Let 0.006 / x = 0.6, Then, x = (0.006 / 0.6) = (0.006*10) / (0.6*10) = 0.06/6 = 0.01
(ii) Let x / 0.025 = 80, Then, x = 80 * 0.025 = 2

Ex. 13. If (1 / 3.718) = 0.2689, Then find the value of (1 / 0.0003718).

Sol. (1 / 0.0003718 ) = ( 10000 / 3.718 ) = 10000 * (1 / 3.718) = 10000 * 0.2689 = 2689.

___ ______
Ex. 14. Express as vulgar fractions : (i) 0.37 (ii) 0.053 (iii) 3.142857
__ ___
Sol. (i) 0.37 = 37 / 99 . (ii) 0.053 = 53 / 999
______ ______
(iii) 3.142857 = 3 + 0.142857 = 3 + (142857 / 999999) = 3 (142857/999999)
_ __ _
Ex. 15. Express as vulgar fractions : (i) 0.17 (ii) 0.1254 (iii) 2.536
_
Sol. (i) 0.17 = (17 – 1)/90 = 16 / 90 = 8/ 45
__
(ii) 0.1254 = (1254 – 12 )/ 9900 = 1242 / 9900 = 69 / 550

(iii) 2.536 = 2 + 0.536 = 2 + (536 – 53)/900 = 2 + (483/900) = 2 + (161/300) = 2 (161/300)

Ex. 16. Simplify: 0.05 * 0.05 * 0.05 + 0.04 * 0.04 * 0.04
0.05 * 0.05 – 0.05 * 0.04 + 0.04 * 0.04

Sol. Given expression = (a3 + b3) / (a2 – ab + b2), where a = 0.05 , b = 0.04
= (a +b ) = (0.05 +0.04 ) =0.09
4. SIMPLIFICATION

IMPORTANT CONCEPTS

I. ’BODMAS’Rule: This rule depicts the correct sequence in which the operations are to be executed,so as to find out the value of a given expression.

Here, ‘B’ stands for ’bracket’ ,’O’for ‘of’ , ‘D’ for’ division’ and ‘M’ for ‘multiplication’, ‘A’ for ‘addition’ and ‘S’ for ‘subtraction’.
Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order(), {} and [].

After removing the brackets, we must use the following operations strictly in the order:
(1)of (2)division (3) multiplication (4)addition (5)subtraction.

II. Modulus of a real number : Modulus of a real number a is defined as
|a| ={a, if a>0
-a, if a<0 Thus, |5|=5 and |-5|=-(-5)=5.III. Virnaculum (or bar): When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum. SOLVED EXAMPLESEx. 1. Simplify: (i)5005-5000+10 (ii) 18800+470+20 Sol. (i)5005-5000+10=5005-(5000/10)=5005-500=4505. (ii)18800+470+20=(18800/470)+20=40/20=2.Ex. 2. Simplify: b-[b-(a+b)-{b-(b-a-b)}+2a] Sol. Given expression=b-[b-(a+b)-{b-(b-a+b)}+2a] =b-[b-a-b-{b-2b+a}+2a] =b-[-a-{b-2b+a+2a}] =b-[-a-{-b+3a}]=b-[-a+b-3a] =b-[-4a+b]=b+4a-b=4a.Ex. 3. What value will replace the question mark in the following equation? 4 1+3 1+?+2 1=13 2. 2 6 3 5 Sol. Let 9/2+19/6+x+7/3=67/5 Then x=(67/5)-(9/2+19/6+7/3)x=(67/5)-((27+19+14)/6)=((67/5)-(60/6) x=((67/5)-10)=17/5=3 2 5Hence, missing fractions =3 2 5Ex.4. 4/15 of 5/7 of a number is greater than 4/9 of 2/5 of the same number by 8.What is half of that number?Sol. Let the number be x. then 4/15 of 5/7 of x-4/9 of 2/5 of x=84/21x-8/45x=8 (4/21-8/45)x=8(60-56)/315x=84/315x=8 x=(8*315)/4=6301/2x=315Hence required number = 315.Ex. 5. Simplify: 3 1{1 1-1/2(2 1-1/4-1/6)}] 4 4 2Sol. Given exp. =[13/4{5/4-1/2(5/2-(3-2)/12)}]=[13/4{5/4-1/2(5/2-1/12)}] =[13/4{5/4-1/2((30-1)/12)}]=[13/4{5/4-29/24}] =[13/4{(30-39)/24}]=[13/41/24]=[(13/4)*24]=78Ex. 6. Simplify: 10836of 1+2*3 1 4 5 4Sol. Given exp.= 1089+2*13 =108 +13 =12+13 =133 = 13 3 5 4 9 10 10 10 10Ex.7 Simplify: (7/2)(5/2)*(3/2)  5.25 (7/2)(5/2)of (3/2)sol.Given exp. (7/2)(2/5)(3/2)  5.25=(21/10)(525/100)=(21/10)(15/14) (7/2)(15/4)Ex. 8. Simplify: (i) 12.05*5.4+0.6 (ii) 0.6*0.6+0.6*0.6 ( Bank P.O 2003)Sol. (i) Given exp. = 12.05*(5.4/0.6) = (12.05*9) = 108.45 (ii) Given exp. = 0.6*0.6+(0.6*6) = 0.36+0.1 = 0.46Ex. 9. Find the value of x in each of the following equation:(i) [(17.28/x) / (3.6*0.2)] = 2(ii) 3648.24 + 364.824 + x – 36.4824 = 3794.1696(iii) 8.5 – { 5 ½ – [7 ½ + 2.8]/x}*4.25/(0.2)2 = 306 (Hotel Management,1997)Sol. (i) (17.28/x) = 2*3.6*0.2  x = (17.28/1.44) = (1728/14) = 12. (ii) (364.824/x) = (3794.1696 + 36.4824) – 3648.24 = 3830.652 – 3648.24 = 182.412.  x = (364.824/182.412) =2.(iii) 8.5-{5.5-(7.5+(2.8/x))}*(4.25/0.04) = 306  8.5-{5.5-{(7.5x+2.8)/x)}*(425/4) = 306  8.5-{(5.5x-7.5x-2.8)/x}*(425/4) = 306  8.5-{(-2x-2.8)/x}*106.25 = 306  8.5-{(-212.5x-297.5)/x} = 306  (306-221)x = 297.5  x =(297.5/85) = 3.5.Ex. 10. If (x/y)=(6/5), find the value (x2+y2)/(x2-y2) Sol. (x2+y2)/(x2-y2) = ( x2 /y2+ 1)/ ( x2 /y2-1) = [(6/5)2+1] / [(6/5)2-1] = [(36/25)+1] / [(36/25)-1] = (61*25)/(25*11) = 61/11 Ex. 11. Find the value of 4 - _____5_________ 1 + ___1___ __ 3 + __1___ 2 + _1_ 4Sol. Given exp. = 4 - _____5_______ = 4 - ____5________ = 4 - ____5_____ 1 + ___1__ 1 + _____1____ 1 + ___1__ 3 + __1___ 3 + __4__ (31/9) 2 + _1_ 9 4 = 4 - __5____ = 4 - __5___ = 4 – (5*31)/ 40 = 4 – (31/8) = 1/8 1 + _9¬_ (40/31) 31 Ex. 12. If _____2x______ = 1 ., then find the value of x . 1 + ___1___ __ 1+ __x__ 1 - xSol. We have : _____2x______ _ = 1  _____2x_____ = 1  __2x____ = 1 1 + ___1_____ 1 + ___1____ 1+ (1 – x) _(1 – x) – x [1/(1- x)] 1 - x  2x = 2-x  3x = 2  x = (2/3). Ex.13.(i)If a/b=3/4 and 8a+5b=22,then find the value of a.(ii)if x/4-x-3/6=1,then find the value of x.Sol. (i) (a/b)=3/4 b=(4/3) a. 8a+5b=22  8a+5*(4/3) a=22  8a+(20/3) a=22 44a = 66  a=(66/44)=3/2(ii) (x /4)-((x-3)/6)=1 (3x-2(x-3) )/12 = 1  3x-2x+6=12  x=6.Ex.14.If 2x+3y=34 and ((x + y)/y)=13/8,then find the value of 5y+7x.Sol. The given equations are:2x+3y=34 …(i) and, ((x + y) /y)=13/8  8x+8y=13y  8x-5y=0 …(ii)Multiplying (i) by 5,(ii) by 3 and adding, we get : 34x=170 or x=5.Putting x=5 in (i), we get: y=8. 5y+7x=((5*8)+(7*5))=40+35=75Ex.15.If 2x+3y+z=55,x-y=4 and y - x + z=12,then what are the values of x , y and z? Sol. The given equations are:2x+3y+z=55 …(i); x + z - y=4 …(ii); y -x + z =12 …(iii)Subtracting (ii) from (i), we get: x+4y=51 …(iv)Subtracting (iii) from (i), we get: 3x+2y=43 …(v) Multiplying (v) by 2 and subtracting (iv) from it, we get: 5x=35 or x=7.Putting x=7 in (iv), we get: 4y=44 or y=11.Putting x=7,y=11 in (i), we get: z=8.Ex.16.Find the value of (1-(1/3))(1-(1/4))(1-(1/5))….(1-(1/100)). Sol. Given expression = (2/3)*(3/4)*(4/5) *…….* (99/100) = 2/100 = 1/50.Ex.17. Find the value of (1/(2*3))+(1/(3*4))+(1/(4*5))+(1/(5*6))+…..+ ((1/(9*10)).Sol. Given expression=((1/2)-(1/3))+((1/3)-(1/4))+((1/4)-(1/5))+((1/5)-(1/6))+….+ ((1/9)-(1/10))=((1/2)-(1/10))=4/10 = 2/5.Ex.18.Simplify: 9948/49 * 245.Sol. Given expression = (100-1/49) * 245=(4899/49) * 245 = 4899 * 5=24495.Ex.19.A board 7ft. 9 inches long is divided into 3 equal parts . What is the length of each part?Sol. Length of board=7ft. 9 inches=(7*12+9)inches=93 inches.Length of each part = (93/3) inches = 31 inches = 2ft. 7 inches 20.A man divides Rs. Among 5 sons,4daughters and 2 nephews .If each daughter receives four times as much as each nephews and each son receives five times as much as each nephews ,how much does each daughter receive?Let the share of each nephews be Rs.x.Then,share of each daughter=rs4x;share of each son=Rs.5x;So,5*5x+4*4x+2*x=860025x+16x+2x=8600=43x=8600x=200;21.A man spends 2/5 of his salary on house rent,3/10 of his salary on food and 1/8 of his salary on conveyence.if he has Rs.1400 left with him,find his expenditure on food and conveyence.Part of salary left=1-(2/5+3/10+1/8)Let the monthly salary be Rs.xThen, 7/40 of x=1400X=(1400*40/7)=8600Expenditure on food=Rs.(3/10*800)=Rs.2400Expenditure on conveyence=Rs.(1/8*8000)=Rs.100022.A third of Arun’s marks in mathematics exeeds a half of his marks in english by 80.if he got 240 marks In two subjects together how many marks did he got inh english?Let Arun’s marks in mathematics and english be x and yThen 1/3x-1/2y=302x-3y=180……>(1)
x+y=240…….>(2)
solving (1) and (2)
x=180
and y=60

23.A tin of oil was 4/5full.when 6 bottles of oil were taken out and four bottles of oil were poured into it, it was ¾ full. how many bottles of oil can the tin contain?
Suppose x bottles can fill the tin completely
Then4/5x-3/4x=6-4
X/20=2
X=40
Therefore required no of bottles =40

24.if 1/8 of a pencil is black ½ of the remaining is white and the remaining 3 ½ is blue find the total length of the pencil?
Let the total length be xm
Then black part =x/8cm
The remaining part=(x-x/8)cm=7x/8cm
White part=(1/2 *7x/8)=7x/16 cm
Remaining part=(7x/8-7x/16)=7x/16cm
7x/16=7/2
x=8cm

25.in a certain office 1/3 of the workers are women ½ of the women are married and 1/3 of the married women have children if ¾ of the men are married and 2/3 of the married men have children what part of workers are without children?
Let the total no of workers be x
No of women =x/3
No of men =x-(x/3)=2x/3
No of women having children =1/3 of ½ ofx/3=x/18
No of men having children=2/3 of ¾ of2x/3=x/3
No of workers having children = x/8 +x/3=7x/18
Workers having no children=x-7x/18=11x/18=11/18 of all wprkers

26.a crate of mangoes contains one bruised mango for every thirty mango in the crate. If three out of every four bruised mango are considerably unsaleble and there are 12 unsaleable mangoes in the crate then how msny mango are there in the crate?

Let the total no of mangoes in the crate be x
Then the no of bruised mango = 1/30 x
Let the no of unsalable mangoes =3/4 (1/30 x)
1/40 x =12
x=480

27. a train starts full of passengers at the first station it drops 1/3 of the passengers and takes 280more at the second station it drops one half the new total and takes twelve more .on arriving at the third station it is found to have 248 passengers. Find the no of passengers in the beginning?
Let no of passengers in the beginning be x
After first station no passengers=(x-x/3)+280=2x/3 +280
After second station no passengers =1/2(2x/3+280)+12
½(2x/3+280)+12=248
2x/3+280=2*236
2x/3=192
x=288

28.if a2+b2=177and ab=54 then find the value of a+b/a-b?
(a+b)2=a2+b2+2ab=117+2*24=225
a+b=15
(a-b)2=a2+b2-2ab=117-2*54
a-b=3
a+b/a-b=15/3=5

29.find the value of (75983*75983- 45983*45983/30000)
Given expression=(75983)2-(45983)2/(75983-45983)
=(a-b)2/(a-b)
=(a+b)(a-b)/(a-b)
=(a+b)
=75983+45983
=121966

30.find the value of 343*343*343-113*113*113
343*343+343*113+113*113

Given expression= (a3-b3)
a2+ab+b2
=(a-b)
=(343-113)
.=230

31.Village X has a population of 68000,which is decreasing at the rate of 1200 per year.VillagyY has a population of 42000,which is increasing
at the rate of 800 per year .in how many years will the population of the two villages be equal?

Let the population of two villages be equal after p years
Then,68000-1200p=42000+800p
2000p=26000
p=13

32.From a group of boys and girls,15 girls leave.There are then left 2 boys for each girl.After this,45 boys leave.There are then 5 girls for each boy.Find
the number of girls in the beginning?
Let at present there be x boys.
Then,no of girls at present=5x
Before the boys had left:no of boys=x+45
And no of girls=5x
X+45=2*5x
9x=45
x=5
no of girls in the beginning=25+15=40

33.An employer pays Rs.20 for each day a worker works and for feits Rs.3 for each day is ideal at the end of sixty days a worker gets Rs.280 . for how many days did the worker remain ideal?
Suppose a worker remained ideal for x days then he worked for 60-x days
20*(60-x)-3x=280
1200-23x=280
23x=920
x=40

Ex 34.kiran had 85 currency notes in all , some of which were of Rs.100 denaomination and the remaining of Rs.50 denomination the total amount of all these currency note was Rs.5000.how much amount did she have in the denomination of Rs.50?
Let the no of fifty rupee notes be x
Then,no of 100 rupee notes =(85-x)
50x+100(85-x)=5000
x+2(85-x)=100
x=70
so,,required amount=Rs.(50*70)= Rs.3500

Ex. 35. When an amount was distributed among 14 boys, each of them got rs 80 more than the amount received by each boy when the same amount is distributed equally among 18 boys. What was the amount?
Sol. Let the total amount be Rs. X the,
x - x = 80  2 x = 80  x =63 x 80 = 5040.
14 18 126 63
Hence the total amount is 5040.

Ex. 36. Mr. Bhaskar is on tour and he has Rs. 360 for his expenses. If he exceeds his tour by 4 days, he must cut down his daily expenses by Rs. 3. for how many days is Mr. Bhaskar on tour?

Sol. Suppose Mr. Bhaskar is on tour for x days. Then,
360 - 360 = 3  1 - 1 = 1  x(x+4) =4 x 120 =480
x x+4 x x+4 120

 x2 +4x –480 = 0  (x+24) (x-20) = 0  x =20.
Hence Mr. Bhaskar is on tour for 20 days.

Ex. 37. Two pens and three pencils cost Rs 86. four Pens and a pencil cost Rs. 112. find the cost of a pen and that of a pencil.

Sol. Let the cost of a pen ands a pencil be Rs. X and Rs. Y respectively.
Then, 2x + 3y = 86 ….(i) and 4x + y =112.
Solving (i) and (ii), we get: x = 25 and y = 12.
Cost of a pen =Rs. 25 and the cost of a pencil =Rs. 12.

Ex. 38. Arjun and Sajal are friends . each has some money. If Arun gives Rs. 30 to Sajal, the Sajal will have twice the money left with Arjun. But, if Sajal gives Rs. 10 to Arjun, the Arjun will have thrice as much as is left with Sajal. How much money does each have?
Sol. Suppose Arun has Rs. X and Sjal has Rs. Y. then,
2(x-30)= y+30 => 2x-y =90 …(i)
and x +10 =3(y-10) => x-3y = - 40 …(ii)
Solving (i) and (ii), we get x =62 and y =34.
Arun has Rs. 62 and Sajal has Rs. 34.

Ex. 39. In a caravan, in addition to 50 hens there are 45 goats and 8 camels with some keepers. If the total number of feet be 224 more than the number of heads, find the number of keepers.
Sol. Let the number of keepers be x then,
Total number of heads =(50 + 45 + 8 + x)= (103 + x).
Total number of feet = (45 + 8) x 4 + (50 + x) x 2 =(312 +2x).
(312 + 2x)-(103 + x) =224 x =15.
Hence, number of keepers =15.

SQUARE ROOTS AND CUBE ROOTS
IMPORTANT FACTS AND FORMULAE

Square Root: If x2 = y, we say that the square root of y is x and we write, √y = x.
Thus, √4 = 2, √9 = 3, √196 = 14.
Cube Root: The cube root of a given number x is the number whose cube is x. We denote the cube root of x by 3√x.
Thus, 3√8 = 3√2 x 2 x 2 = 2, 3√343 = 3√7 x 7 x 7 = 7 etc.
Note:
1.√xy = √x * √y 2. √(x/y) = √x / √y = (√x / √y) * (√y / √y) = √xy / y

SOLVED EXAMPLES

Ex. 1. Evaluate √6084 by factorization method .
Sol. Method: Express the given number as the product of prime factors. 2 6084
Now, take the product of these prime factors choosing one out of 2 3042
every pair of the same primes. This product gives the square root 3 1521
of the given number. 3 507
Thus, resolving 6084 into prime factors, we get: 13 169
6084 = 22 x 32 x 132 13  √6084 = (2 x 3 x 13) = 78.

Ex. 2. Find the square root of 1471369.
Sol. Explanation: In the given number, mark off the digits 1 1471369 (1213
in pairs starting from the unit's digit. Each pair and 1
the remaining one digit is called a period. 22 47
Now, 12 = 1. On subtracting, we get 0 as remainder. 44
Now, bring down the next period i.e., 47. 241 313
Now, trial divisor is 1 x 2 = 2 and trial dividend is 47. 241
So, we take 22 as divisor and put 2 as quotient. 2423 7269
The remainder is 3. 7269
Next, we bring down the next period which is 13. x
Now, trial divisor is 12 x 2 = 24 and trial dividend is
313. So, we take 241 as dividend and 1 as quotient.
The remainder is 72. ¬
Bring down the next period i.e., 69.
Now, the trial divisor is 121 x 2 = 242 and the trial
dividend is 7269. So, we take 3as quotient and 2423
as divisor. The remainder is then zero.
Hence, √1471369 = 1213.

Ex. 3. Evaluate: √248 + √51 + √ 169 .
Sol. Given expression = √248 + √51 + 13 = √248 + √64 = √ 248 + 8 = √256 = 16.

Ex. 4. If a * b * c = √(a + 2)(b + 3) / (c + 1), find the value of 6 * 15 * 3.
Sol. 6 * 15 * 3 = √(6 + 2)(15 + 3) / (3 + 1) = √8 * 18 / 4 = √144 / 4 = 12 / 4 = 3.
Ex. 5. Find the value of √25/16.
Sol. √ 25 / 16 = √ 25 / √ 16 = 5 / 4

Ex. 6. What is the square root of 0.0009?
Sol. √0.0009= √ 9 / 1000 = 3 / 100 = 0.03.

Ex. 7. Evaluate √175.2976.
Sol. Method: We make even number of decimal places 1 175.2976 (13.24
by affixing a zero, if necessary. Now, we mark off 1
periods and extract the square root as shown. 23 75
69
√175.2976 = 13.24 262 629
524
2644 10576
10576
x

Ex. 8. What will come in place of question mark in each of the following questions?
(i) √32.4 / ? = 2 (ii) √86.49 + √ 5 + ( ? )2 = 12.3.
Sol. (i) Let √32.4 / x = 2. Then, 32.4/x = 4 <=> 4x = 32.4 <=> x = 8.1.

(ii) Let √86.49 + √5 + x2 = 12.3.
Then, 9.3 + √5+x2 = 12.3 <=> √5+x2 = 12.3 - 9.3 = 3
<=> 5 + x2 = 9 <=> x2 = 9 - 5= 4 <=> x = √4 = 2.

Ex.9. Find the value of √ 0.289 / 0.00121.

Sol. √0.289 / 0.00121 = √0.28900/0.00121 = √28900/121 = 170 / 11.

Ex.10. If √1 + (x / 144) = 13 / 12, the find the value of x.

Sol. √1 + (x / 144) = 13 / 12  ( 1 + (x / 144)) = (13 / 12 )2 = 169 / 144
x / 144 = (169 / 144) - 1
x / 144 = 25/144  x = 25.

Ex. 11. Find the value of √3 up to three places of decimal.
Sol.
1 3.000000 (1.732
1
27 200
189
343 1100
1029
3462 7100
6924 √3 = 1.732.

Ex. 12. If √3 = 1.732, find the value of √192 - 1 √48 - √75 correct to 3 places
2
of decimal. (S.S.C. 2004)
Sol. √192 - (1 / 2)√48 - √75 = √64 * 3 - (1/2) √ 16 * 3 - √ 25 * 3
=8√3 - (1/2) * 4√3 - 5√3
=3√3 - 2√3 = √3 = 1.732

Ex. 13. Evaluate: √(9.5 * 0.0085 * 18.9) / (0.0017 * 1.9 * 0.021)
Sol. Given exp. = √(9.5 * 0.0085 * 18.9) / (0.0017 * 1.9 * 0.021)
Now, since the sum of decimal places in the numerator and denominator under the radical sign is the same, we remove the decimal.
 Given exp = √(95 * 85 * 18900) / (17 * 19 * 21) = √ 5 * 5 * 900 = 5 * 30 = 150.

Ex. 14. Simplify: √ [( 12.1 )2 - (8.1)2] / [(0.25)2 + (0.25)(19.95)]
Sol. Given exp. = √ [(12.1 + 8.1)(12.1 - 8.1)] / [(0.25)(0.25 + 19.95)]

=√ (20.2 * 4) /( 0.25 * 20.2) = √ 4 / 0.25 = √400 / 25 = √16 = 4.
Ex. 15. If x = 1 + √2 and y = 1 - √2, find the value of (x2 + y2).
Sol. x2 + y2 = (1 + √2)2 + (1 - √2)2 = 2[(1)2 + (√2)2] = 2 * 3 = 6.

Ex. 16. Evaluate: √0.9 up to 3 places of decimal.
Sol.
9 0.900000(0.948
81
184 900
736
1888 16400
15104 √0.9 = 0.948

Ex.17. If √15 = 3.88, find the value of √ (5/3).
Sol. √ (5/3) = √(5 * 3) / (3 * 3) = √15 / 3 = 3.88 / 3 = 1.2933…. = 1.293.

Ex. 18. Find the least square number which is exactly divisible by 10,12,15 and 18.
Sol. L.C.M. of 10, 12, 15, 18 = 180. Now, 180 = 2 * 2 * 3 * 3 *5 = 22 * 32 * 5.
To make it a perfect square, it must be multiplied by 5.
 Required number = (22 * 32 * 52) = 900.

Ex. 19. Find the greatest number of five digits which is a perfect square.
(R.R.B. 1998)
Sol. Greatest number of 5 digits is 99999.
3 99999(316
9
61 99
61
626 3899
3756
143
 Required number == (99999 - 143) = 99856.

Ex. 20. Find the smallest number that must be added to 1780 to make it a perfect
square.
Sol.
4 1780 (42
16
82 180
164
16

 Number to be added = (43)2 - 1780 = 1849 - 1780 = 69.

Ex. 21. √2 = 1.4142, find the value of √2 / (2 + √2).
Sol. √2 / (2 + √2) = √2 / (2 + √2) * (2 - √2) / (2 - √2) = (2√2 – 2) / (4 – 2)
= 2(√2 – 1) / 2 = √2 – 1 = 0.4142.

22. If x = (√5 + √3) / (√5 - √3) and y = (√5 - √3) / (√5 + √3), find the value of (x2 + y2).
Sol.
x = [(√5 + √3) / (√5 - √3)] * [(√5 + √3) / (√5 + √3)] = (√5 + √3)2 / (5 - 3)
=(5 + 3 + 2√15) / 2 = 4 + √15.
y = [(√5 - √3) / (√5 + √3)] * [(√5 - √3) / (√5 - √3)] = (√5 - √3)2 / (5 - 3)
=(5 + 3 - 2√15) / 2 = 4 - √15.
 x2 + y2 = (4 + √15)2 + (4 - √15)2 = 2[(4)2 + (√15)2] = 2 * 31 = 62.

Ex. 23. Find the cube root of 2744.
Sol. Method: Resolve the given number as the product 2 2744
of prime factors and take the product of prime 2 1372
factors, choosing one out of three of the same 2 686
prime factors. Resolving 2744 as the product of 7 343
prime factors, we get: 7 ¬ 49
7
2744 = 23 x 73.
 3√2744= 2 x 7 = 14.

Ex. 24. By what least number 4320 be multiplied to obtain a number which is a perfect cube?
Sol. Clearly, 4320 = 23 * 33 * 22 * 5.
To make it a perfect cube, it must be multiplied by 2 * 52 i.e,50.
6.AVERAGE

Ex.1:Find the average of all prime numbers between 30 and 50?
Sol: there are five prime numbers between 30 and 50.
They are 31,37,41,43 and 47.
Therefore the required average=(31+37+41+43+47)/5 199/5  39.8.

Ex.2. find the average of first 40 natural numbers?
Sol: sum of first n natural numbers=n(n+1)/2;
So,sum of 40 natural numbers=(40*41)/2 820.
Therefore the required average=(820/40) 20.5.

Ex.3. find the average of first 20 multiples of 7?
Sol: Required average =7(1+2+3+…….+20)/20 (7*20*21)/(20*2) (147/2)=73.5.

Ex.4. the average of four consecutive even numbers is 27. find the largest of these numbers?
Sol: let the numbers be x,x+2,x+4 andx+6. then,
(x+(x+2)+(x+4)+(x+6))/4) = 27
(4x+12)/4 = 27
x+3=27  x=24.
Therefore the largest number=(x+6)=24+6=30.

Ex.5. there are two sections A and B of a class consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class?
Sol: total weight of(36+44) students=(36*40+44*35)kg =2980kg.
Therefore weight of the total class=(2980/80)kg =37.25kg.

Ex:6.nine persons went to a hotel for taking their meals 8 of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine.What was the total money spent by them?
Sol: Let the average expenditure of all nine be Rs.x
Then 12*8+(x+8)=9x or 8x=104 or x=13.
Total money spent = 9x=Rs.(9*13)=Rs.117.

Ex.7: Of the three numbers, second is twice the first and is also thrice the third.If the average of the three numbers is 44.Find the largest number.
Sol: Let the third number be x.
Then second number = 3x.
First number=3x/2.
Therefore x+3x+(3x/2)=(44*3) or x=24
So largest number= 2nd number=3x=72.

Ex.8:The average of 25 result is 18.The average of 1st 12 of them is 14 & that of last 12 is 17.Find the 13th result.
Sol: Clearly 13th result=(sum of 25 results)-(sum of 24 results)
=(18*25)-(14*12)+(17*12)
=450-(168+204)
=450-372
=78.

Ex.9:The Average of 11 results is 16, if the average of the 1st 6 results is 58 & that of the last 63. Find the 6th result.
Sol: 6th result = (58*6+63*6-60*11)=66

Ex.10:The average waight of A,B,C is 45 Kg. The avg wgt of A & B be 40Kg & that of B,C be 43Kg. Find the wgt of B.
Sol. Let A,B,c represent their individual wgts.
Then,
A+B+C=(45*3)Kg=135Kg
A+B=(40*2)Kg=80Kg & B+C=(43*2)Kg=86Kg
B=(A+B)+(B+C)-(A+B+C)
=(80+86-135)Kg
=31Kg.

Ex. 11. The average age of a class of 39 students is 15 years. If the age of the teacher be included, then the average increases by3 months. Find the age of the teacher.
Sol. Total age of 39 persons = (39 x 15) years
= 585 years.
Average age of 40 persons= 15 yrs 3 months
= 61/4 years.
Total age of 40 persons = (_(61/4 )x 40) years= 610 years.
:. Age of the teacher = (610 - 585) years=25 years.

Ex. 12. The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53 kg is replaced by a new man. Find the weight of the new
man.
Sol. Total weight increased =(1.8 x 10) kg =18 kg.
:. Weight of the new man =(53 + 18) kg =71 kg.

Ex. 13. There were 35 students in a hostel. Due to the admission of 7 new students, ;he expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Rs 1. Wbat was the original expenditure of the mess?
Sol. Let the original average expenditure be Rs. x. Then,
42 (x - 1) - 35x=42  7x= 84  x =12.
Original expenditure = Rs. (35 x 12) =Rs. 420. .

14. A batsman makes a score of 87 runs in the 17th inning and thus increases his avg by 3. Find his average after 17th inning.
Sol. Let the average after 17th inning = x.
Then, average after 16th inning = (x - 3).
:. 16 (x - 3) + 87 = 17x or x = (87 - 48) = 39.

Ex.15. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km perhour. Find the average speed of the train during the whole journey.
Sol. Required average speed = ((2xy)/(x+y)) km / hr
=(2 x 84 x 56)/(84+56)km/hr
= (2*84*56)/140 km/hr
=67.2 km/hr.

7. PROBLEMS ON NUMBERS

In this section, questions involving a set of numbers are put in the form of a puzzle. You have to analyze the given conditions, assume the unknown numbers and form equations accordingly, which on solving yield the unknown numbers.

SOLVED EXAMPLES

Ex.1. A number is as much greater than 36 as is less than 86. Find the number.
Sol. Let the number be x. Then, x - 36 = 86 - x => 2x = 86 + 36 = 122 => x = 61. Hence, the required number is 61.

Ex. 2. Find a number such that when 15 is subtracted from 7 times the number, the
Result is 10 more than twice the number. (Hotel Management, 2002)
Sol. Let the number be x. Then, 7x - 15 = 2x + 10 => 5x = 25 =>x = 5.
Hence, the required number is 5.
Ex. 3. The sum of a rational number and its reciprocal is 13/6. Find the number.
(S.S.C. 2000)
Sol. Let the number be x.
Then, x + (1/x) = 13/6 => (x2 + 1)/x = 13/6 => 6x2 – 13x + 6 = 0
=> 6x2 – 9x – 4x + 6 = 0 => (3x – 2) (2x – 3) = 0
 x = 2/3 or x = 3/2
Hence the required number is 2/3 or 3/2.

Ex. 4. The sum of two numbers is 184. If one-third of the one exceeds one-seventh
of the other by 8, find the smaller number.
Sol. Let the numbers be x and (184 - x). Then,
(X/3) - ((184 – x)/7) = 8 => 7x – 3(184 – x) = 168 => 10x = 720 => x = 72.
So, the numbers are 72 and 112. Hence, smaller number = 72.

Ex. 5. The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers.
Sol. Let the number be x and y. Then,
x – y = 11 ----(i) and 1/5 (x + y) = 9 => x + y = 45 ----(ii)
Adding (i) and (ii), we get: 2x = 56 or x = 28. Putting x = 28 in (i), we get: y = 17.
Hence, the numbers are 28 and 17.

Ex. 6. If the sum of two numbers is 42 and their product is 437, then find the
absolute difference between the numbers. (S.S.C. 2003)
Sol. Let the numbers be x and y. Then, x + y = 42 and xy = 437
x - y = sqrt[(x + y)2 - 4xy] = sqrt[(42)2 - 4 x 437 ] = sqrt[1764 – 1748] = sqrt[16] = 4.
Required difference = 4.
Ex. 7. The sum of two numbers is 16 and the sum of their squares is 113. Find the
numbers.
Sol. Let the numbers be x and (15 - x).
Then, x2 + (15 - x)2 = 113 => x2 + 225 + X2 - 30x = 113
=> 2x2 - 30x + 112 = 0 => x2 - 15x + 56 = 0
=> (x - 7) (x - 8) = 0 => x = 7 or x = 8.
So, the numbers are 7 and 8.

Ex. 8. The average of four consecutive even numbers is 27. Find the largest of these
numbers.
Sol. Let the four consecutive even numbers be x, x + 2, x + 4 and x + 6.
Then, sum of these numbers = (27 x 4) = 108.
So, x + (x + 2) + (x + 4) + (x + 6) = 108 or 4x = 96 or x = 24.
:. Largest number = (x + 6) = 30.

Ex. 9. The sum of the squares of three consecutive odd numbers is 2531.Find the
numbers.
Sol. Let the numbers be x, x + 2 and x + 4.
Then, X2 + (x + 2)2 + (x + 4)2 = 2531 => 3x2 + 12x - 2511 = 0
=> X2 + 4x - 837 = 0 => (x - 27) (x + 31) = 0 => x = 27.
Hence, the required numbers are 27, 29 and 31.

Ex. 10. Of two numbers, 4 times the smaller one is less then 3 times the 1arger one by 5. If the sum of the numbers is larger than 6 times their difference by 6, find the two numbers.
Sol. Let the numbers be x and y, such that x > y
Then, 3x - 4y = 5 ...(i) and (x + y) - 6 (x - y) = 6 => -5x + 7y = 6 …(ii)
Solving (i) and (ii), we get: x = 59 and y = 43.
Hence, the required numbers are 59 and 43.

Ex. 11. The ratio between a two-digit number and the sum of the digits of that
number is 4 : 1.If the digit in the unit's place is 3 more than the digit in the ten’s place, what is the number?
Sol. Let the ten's digit be x. Then, unit's digit = (x + 3).
Sum of the digits = x + (x + 3) = 2x + 3. Number = l0x + (x + 3) = llx + 3.
11x+3 / 2x + 3 = 4 / 1 => 1lx + 3 = 4 (2x + 3) => 3x = 9 => x = 3.
Hence, required number = 11x + 3 = 36.

Ex. 12. A number consists of two digits. The sum of the digits is 9. If 63 is subtracted
from the number, its digits are interchanged. Find the number.
Sol. Let the ten's digit be x. Then, unit's digit = (9 - x).
Number = l0x + (9 - x) = 9x + 9.
Number obtained by reversing the digits = 10 (9 - x) + x = 90 - 9x.
therefore, (9x + 9) - 63 = 90 - 9x => 18x = 144 => x = 8.
So, ten's digit = 8 and unit's digit = 1.
Hence, the required number is 81.
Ex. 13. A fraction becomes 2/3 when 1 is added to both, its numerator and denominator.
And ,it becomes 1/2 when 1 is subtracted from both the numerator and denominator. Find the fraction.
Sol. Let the required fraction be x/y. Then,
x+1 / y+1 = 2 / 3 => 3x – 2y = - 1 …(i) and x – 1 / y – 1 = 1 / 2
 2x – y = 1 …(ii)
Solving (i) and (ii), we get : x = 3 , y = 5
therefore, Required fraction= 3 / 5.

Ex. 14. 50 is divided into two parts such that the sum of their reciprocals is 1/ 12.Find the two parts.
Sol. Let the two parts be x and (50 - x).
Then, 1 / x + 1 / (50 – x) = 1 / 12 => (50 – x + x) / x ( 50 – x) = 1 / 12
=> x2 – 50x + 600 = 0 => (x – 30) ( x – 20) = 0 => x = 30 or x = 20.
So, the parts are 30 and 20.

Ex. 15. If three numbers are added in pairs, the sums equal 10, 19 and 21. Find the
numbers )
Sol. Let the numbers be x, y and z. Then,
x+ y = 10 ...(i) y + z = 19 ...(ii) x + z = 21 …(iii)
Adding (i) ,(ii) and (iii), we get: 2 (x + y + z ) = 50 or (x + y + z) = 25.
Thus, x= (25 - 19) = 6; y = (25 - 21) = 4; z = (25 - 10) = 15.
Hence, the required numbers are 6, 4 and 15.
7. PROBLEMS ON NUMBERS

In this section, questions involving a set of numbers are put in the form of a puzzle. You have to analyze the given conditions, assume the unknown numbers and form equations accordingly, which on solving yield the unknown numbers.

SOLVED EXAMPLES

Ex.1. A number is as much greater than 36 as is less than 86. Find the number.
Sol. Let the number be x. Then, x - 36 = 86 - x => 2x = 86 + 36 = 122 => x = 61. Hence, the required number is 61.

Ex. 2. Find a number such that when 15 is subtracted from 7 times the number, the
Result is 10 more than twice the number. (Hotel Management, 2002)
Sol. Let the number be x. Then, 7x - 15 = 2x + 10 => 5x = 25 =>x = 5.
Hence, the required number is 5.
Ex. 3. The sum of a rational number and its reciprocal is 13/6. Find the number.
(S.S.C. 2000)
Sol. Let the number be x.
Then, x + (1/x) = 13/6 => (x2 + 1)/x = 13/6 => 6x2 – 13x + 6 = 0
=> 6x2 – 9x – 4x + 6 = 0 => (3x – 2) (2x – 3) = 0
 x = 2/3 or x = 3/2
Hence the required number is 2/3 or 3/2.

Ex. 4. The sum of two numbers is 184. If one-third of the one exceeds one-seventh
of the other by 8, find the smaller number.
Sol. Let the numbers be x and (184 - x). Then,
(X/3) - ((184 – x)/7) = 8 => 7x – 3(184 – x) = 168 => 10x = 720 => x = 72.
So, the numbers are 72 and 112. Hence, smaller number = 72.

Ex. 5. The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers.
Sol. Let the number be x and y. Then,
x – y = 11 ----(i) and 1/5 (x + y) = 9 => x + y = 45 ----(ii)
Adding (i) and (ii), we get: 2x = 56 or x = 28. Putting x = 28 in (i), we get: y = 17.
Hence, the numbers are 28 and 17.

Ex. 6. If the sum of two numbers is 42 and their product is 437, then find the
absolute difference between the numbers. (S.S.C. 2003)
Sol. Let the numbers be x and y. Then, x + y = 42 and xy = 437
x - y = sqrt[(x + y)2 - 4xy] = sqrt[(42)2 - 4 x 437 ] = sqrt[1764 – 1748] = sqrt[16] = 4.
Required difference = 4.
Ex. 7. The sum of two numbers is 16 and the sum of their squares is 113. Find the
numbers.
Sol. Let the numbers be x and (15 - x).
Then, x2 + (15 - x)2 = 113 => x2 + 225 + X2 - 30x = 113
=> 2x2 - 30x + 112 = 0 => x2 - 15x + 56 = 0
=> (x - 7) (x - 8) = 0 => x = 7 or x = 8.
So, the numbers are 7 and 8.

Ex. 8. The average of four consecutive even numbers is 27. Find the largest of these
numbers.
Sol. Let the four consecutive even numbers be x, x + 2, x + 4 and x + 6.
Then, sum of these numbers = (27 x 4) = 108.
So, x + (x + 2) + (x + 4) + (x + 6) = 108 or 4x = 96 or x = 24.
:. Largest number = (x + 6) = 30.

Ex. 9. The sum of the squares of three consecutive odd numbers is 2531.Find the
numbers.
Sol. Let the numbers be x, x + 2 and x + 4.
Then, X2 + (x + 2)2 + (x + 4)2 = 2531 => 3x2 + 12x - 2511 = 0
=> X2 + 4x - 837 = 0 => (x - 27) (x + 31) = 0 => x = 27.
Hence, the required numbers are 27, 29 and 31.

Ex. 10. Of two numbers, 4 times the smaller one is less then 3 times the 1arger one by 5. If the sum of the numbers is larger than 6 times their difference by 6, find the two numbers.
Sol. Let the numbers be x and y, such that x > y
Then, 3x - 4y = 5 ...(i) and (x + y) - 6 (x - y) = 6 => -5x + 7y = 6 …(ii)
Solving (i) and (ii), we get: x = 59 and y = 43.
Hence, the required numbers are 59 and 43.

Ex. 11. The ratio between a two-digit number and the sum of the digits of that
number is 4 : 1.If the digit in the unit's place is 3 more than the digit in the ten’s place, what is the number?
Sol. Let the ten's digit be x. Then, unit's digit = (x + 3).
Sum of the digits = x + (x + 3) = 2x + 3. Number = l0x + (x + 3) = llx + 3.
11x+3 / 2x + 3 = 4 / 1 => 1lx + 3 = 4 (2x + 3) => 3x = 9 => x = 3.
Hence, required number = 11x + 3 = 36.

Ex. 12. A number consists of two digits. The sum of the digits is 9. If 63 is subtracted
from the number, its digits are interchanged. Find the number.
Sol. Let the ten's digit be x. Then, unit's digit = (9 - x).
Number = l0x + (9 - x) = 9x + 9.
Number obtained by reversing the digits = 10 (9 - x) + x = 90 - 9x.
therefore, (9x + 9) - 63 = 90 - 9x => 18x = 144 => x = 8.
So, ten's digit = 8 and unit's digit = 1.
Hence, the required number is 81.
Ex. 13. A fraction becomes 2/3 when 1 is added to both, its numerator and denominator.
And ,it becomes 1/2 when 1 is subtracted from both the numerator and denominator. Find the fraction.
Sol. Let the required fraction be x/y. Then,
x+1 / y+1 = 2 / 3 => 3x – 2y = - 1 …(i) and x – 1 / y – 1 = 1 / 2
 2x – y = 1 …(ii)
Solving (i) and (ii), we get : x = 3 , y = 5
therefore, Required fraction= 3 / 5.

Ex. 14. 50 is divided into two parts such that the sum of their reciprocals is 1/ 12.Find the two parts.
Sol. Let the two parts be x and (50 - x).
Then, 1 / x + 1 / (50 – x) = 1 / 12 => (50 – x + x) / x ( 50 – x) = 1 / 12
=> x2 – 50x + 600 = 0 => (x – 30) ( x – 20) = 0 => x = 30 or x = 20.
So, the parts are 30 and 20.

Ex. 15. If three numbers are added in pairs, the sums equal 10, 19 and 21. Find the
numbers )
Sol. Let the numbers be x, y and z. Then,
x+ y = 10 ...(i) y + z = 19 ...(ii) x + z = 21 …(iii)
Adding (i) ,(ii) and (iii), we get: 2 (x + y + z ) = 50 or (x + y + z) = 25.
Thus, x= (25 - 19) = 6; y = (25 - 21) = 4; z = (25 - 10) = 15.
Hence, the required numbers are 6, 4 and 15.

8. PROBLEMS ON AGES

Ex. 1. Rajeev's age after 15 years will be 5 times his age 5 years back. What is the
present age of Rajeev ? (Hotel Management,2002)

Sol. Let Rajeev's present age be x years. Then,
Rajeev's age after 15 years = (x + 15) years.
Rajeev's age 5 years back = (x - 5) years.
:. x + 15 = 5 (x - 5) x + 15 = 5x - 25  4x = 40  x = 10.
Hence, Rajeev's present age = 10 years.

Ex. 2. The ages of two persons differ by 16 years. If 6 years ago, the elder one be
3 times as old as the younger one, find their present ages. (A.A.O. Exam,2003)
Sol. Let the age of the younger person be x years.
Then, age of the elder person = (x + 16) years.
:. 3 (x - 6) = (x + 16 - 6)  3x -18 = x + 10  2x = 28  x = 14.
Hence, their present ages are 14 years and 30 years.

Ex. 3. The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita
is more than Ankit's age by 4 years, what is Nikita's age? (S.B.I.P.O,1999)
Sol. Let Ankit's age be x years. Then, Nikita's age = 240/xyears.
 2  (240 /x ) – x = 4  480 – x2 = 4x  x2 + 4x – 480 = 0
 ( x+24)(x-20) = 0  x = 20.
Hence, Nikita's age = (22_0) years = 12 years. 1
Ex. 4. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father. . (S.S.C, 2003)
Sol. Let the son's present age be x years. Then, father's present age = (3x + 3) years
 (3x + 3 + 3) = 2 (x + 3) + 10  3x + 6 = 2x + 16  x = 10.
Hence, father's present age = (3x + 3) = ((3  10) + 3) years = 33 years.

Ex. 5. Rohit was 4 times as old as his son 8 years ago. After 8 years, Rohit will be
twice as old as his son. What are their present ages?
Sol. Let son's age 8 years ago be x years. Then, Rohit's age 8 years ago = 4x years.
Son's age after 8 years = (x + 8) + 8 = (x + 16) years.
Rohit's age after 8 years = (4x + 8) + 8 = (4x+ 16) years.
 2 (x + 16) = 4x + 16  2x = 16  x = 8.
Hence, son's 'present age = (x + 8) = 16 years.
Rohit's present age = (4x + 8) = 40 years.

Ex. 6. One year ago, the ratio of Gaurav’s and Sachin’s age was 6: 7 respectively.
Four years hence, this ratio would become 7: 8. How old is Sa chin ?
(NABARD, 2002)

Sol:
. Let Gaurav's and Sachin's ages one year ago be 6x and 7x years respectively. Then, Gaurav's age
4 years hence = (6x + 1) + 4 = (6x + 5) years.
Sachin's age 4 years hence = (7x + 1) + 4 = (7x + 5) years.

6x+5 = 7  8(6x+5) = 7 (7x + 5)  48x + 40 = 49x + 35  x = 5.
7x+5 8
Hence, Sachin's present age = (7x + 1) = 36 years.

,7. Abhay’s age after six years will be three-seventh of his fathers age. Ten years ago the ratio of their ages was 1 : 5. What is Abhay’s father's age at present?

Sol. Let the ages of Abhay and his father 10 years ago be x and 5x years respectively. Then,
Abhay's age after 6 years = (x + 10) + 6 = (x + 16) years.
Father's age after 6 years = (5x + 10) + 6 = (5x + 16) years.

\x + 16) = 3 (5x + 16)  7 (x + 16) = 3 (5x + 16)  7x + 112 = 15x + 48
7
 8x = 64  x = 8.
Hence, Abhay's father's present age = (5x + 10) = 50 years.

8. PROBLEMS ON AGES

Ex. 1. Rajeev's age after 15 years will be 5 times his age 5 years back. What is the
present age of Rajeev ? (Hotel Management,2002)

Sol. Let Rajeev's present age be x years. Then,
Rajeev's age after 15 years = (x + 15) years.
Rajeev's age 5 years back = (x - 5) years.
:. x + 15 = 5 (x - 5) x + 15 = 5x - 25  4x = 40  x = 10.
Hence, Rajeev's present age = 10 years.

Ex. 2. The ages of two persons differ by 16 years. If 6 years ago, the elder one be
3 times as old as the younger one, find their present ages. (A.A.O. Exam,2003)
Sol. Let the age of the younger person be x years.
Then, age of the elder person = (x + 16) years.
:. 3 (x - 6) = (x + 16 - 6)  3x -18 = x + 10  2x = 28  x = 14.
Hence, their present ages are 14 years and 30 years.

Ex. 3. The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita
is more than Ankit's age by 4 years, what is Nikita's age? (S.B.I.P.O,1999)
Sol. Let Ankit's age be x years. Then, Nikita's age = 240/xyears.
 2  (240 /x ) – x = 4  480 – x2 = 4x  x2 + 4x – 480 = 0
 ( x+24)(x-20) = 0  x = 20.
Hence, Nikita's age = (22_0) years = 12 years. 1
Ex. 4. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father. . (S.S.C, 2003)
Sol. Let the son's present age be x years. Then, father's present age = (3x + 3) years
 (3x + 3 + 3) = 2 (x + 3) + 10  3x + 6 = 2x + 16  x = 10.
Hence, father's present age = (3x + 3) = ((3  10) + 3) years = 33 years.

Ex. 5. Rohit was 4 times as old as his son 8 years ago. After 8 years, Rohit will be
twice as old as his son. What are their present ages?
Sol. Let son's age 8 years ago be x years. Then, Rohit's age 8 years ago = 4x years.
Son's age after 8 years = (x + 8) + 8 = (x + 16) years.
Rohit's age after 8 years = (4x + 8) + 8 = (4x+ 16) years.
 2 (x + 16) = 4x + 16  2x = 16  x = 8.
Hence, son's 'present age = (x + 8) = 16 years.
Rohit's present age = (4x + 8) = 40 years.

Ex. 6. One year ago, the ratio of Gaurav’s and Sachin’s age was 6: 7 respectively.
Four years hence, this ratio would become 7: 8. How old is Sa chin ?
(NABARD, 2002)

Sol:
. Let Gaurav's and Sachin's ages one year ago be 6x and 7x years respectively. Then, Gaurav's age
4 years hence = (6x + 1) + 4 = (6x + 5) years.
Sachin's age 4 years hence = (7x + 1) + 4 = (7x + 5) years.

6x+5 = 7  8(6x+5) = 7 (7x + 5)  48x + 40 = 49x + 35  x = 5.
7x+5 8
Hence, Sachin's present age = (7x + 1) = 36 years.

,7. Abhay’s age after six years will be three-seventh of his fathers age. Ten years ago the ratio of their ages was 1 : 5. What is Abhay’s father's age at present?

Sol. Let the ages of Abhay and his father 10 years ago be x and 5x years respectively. Then,
Abhay's age after 6 years = (x + 10) + 6 = (x + 16) years.
Father's age after 6 years = (5x + 10) + 6 = (5x + 16) years.

\x + 16) = 3 (5x + 16)  7 (x + 16) = 3 (5x + 16)  7x + 112 = 15x + 48
7
 8x = 64  x = 8.
Hence, Abhay's father's present age = (5x + 10) = 50 years.

8. PROBLEMS ON AGES

Ex. 1. Rajeev's age after 15 years will be 5 times his age 5 years back. What is the
present age of Rajeev ? (Hotel Management,2002)

Sol. Let Rajeev's present age be x years. Then,
Rajeev's age after 15 years = (x + 15) years.
Rajeev's age 5 years back = (x - 5) years.
:. x + 15 = 5 (x - 5) x + 15 = 5x - 25  4x = 40  x = 10.
Hence, Rajeev's present age = 10 years.

Ex. 2. The ages of two persons differ by 16 years. If 6 years ago, the elder one be
3 times as old as the younger one, find their present ages. (A.A.O. Exam,2003)
Sol. Let the age of the younger person be x years.
Then, age of the elder person = (x + 16) years.
:. 3 (x - 6) = (x + 16 - 6)  3x -18 = x + 10  2x = 28  x = 14.
Hence, their present ages are 14 years and 30 years.

Ex. 3. The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita
is more than Ankit's age by 4 years, what is Nikita's age? (S.B.I.P.O,1999)
Sol. Let Ankit's age be x years. Then, Nikita's age = 240/xyears.
 2  (240 /x ) – x = 4  480 – x2 = 4x  x2 + 4x – 480 = 0
 ( x+24)(x-20) = 0  x = 20.
Hence, Nikita's age = (22_0) years = 12 years. 1
Ex. 4. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father. . (S.S.C, 2003)
Sol. Let the son's present age be x years. Then, father's present age = (3x + 3) years
 (3x + 3 + 3) = 2 (x + 3) + 10  3x + 6 = 2x + 16  x = 10.
Hence, father's present age = (3x + 3) = ((3  10) + 3) years = 33 years.

Ex. 5. Rohit was 4 times as old as his son 8 years ago. After 8 years, Rohit will be
twice as old as his son. What are their present ages?
Sol. Let son's age 8 years ago be x years. Then, Rohit's age 8 years ago = 4x years.
Son's age after 8 years = (x + 8) + 8 = (x + 16) years.
Rohit's age after 8 years = (4x + 8) + 8 = (4x+ 16) years.
 2 (x + 16) = 4x + 16  2x = 16  x = 8.
Hence, son's 'present age = (x + 8) = 16 years.
Rohit's present age = (4x + 8) = 40 years.

Ex. 6. One year ago, the ratio of Gaurav’s and Sachin’s age was 6: 7 respectively.
Four years hence, this ratio would become 7: 8. How old is Sa chin ?
(NABARD, 2002)

Sol:
. Let Gaurav's and Sachin's ages one year ago be 6x and 7x years respectively. Then, Gaurav's age
4 years hence = (6x + 1) + 4 = (6x + 5) years.
Sachin's age 4 years hence = (7x + 1) + 4 = (7x + 5) years.

6x+5 = 7  8(6x+5) = 7 (7x + 5)  48x + 40 = 49x + 35  x = 5.
7x+5 8
Hence, Sachin's present age = (7x + 1) = 36 years.

,7. Abhay’s age after six years will be three-seventh of his fathers age. Ten years ago the ratio of their ages was 1 : 5. What is Abhay’s father's age at present?

Sol. Let the ages of Abhay and his father 10 years ago be x and 5x years respectively. Then,
Abhay's age after 6 years = (x + 10) + 6 = (x + 16) years.
Father's age after 6 years = (5x + 10) + 6 = (5x + 16) years.

\x + 16) = 3 (5x + 16)  7 (x + 16) = 3 (5x + 16)  7x + 112 = 15x + 48
7
 8x = 64  x = 8.
Hence, Abhay's father's present age = (5x + 10) = 50 years.

8. PROBLEMS ON AGES

Ex. 1. Rajeev's age after 15 years will be 5 times his age 5 years back. What is the
present age of Rajeev ? (Hotel Management,2002)

Sol. Let Rajeev's present age be x years. Then,
Rajeev's age after 15 years = (x + 15) years.
Rajeev's age 5 years back = (x - 5) years.
:. x + 15 = 5 (x - 5) x + 15 = 5x - 25  4x = 40  x = 10.
Hence, Rajeev's present age = 10 years.

Ex. 2. The ages of two persons differ by 16 years. If 6 years ago, the elder one be
3 times as old as the younger one, find their present ages. (A.A.O. Exam,2003)
Sol. Let the age of the younger person be x years.
Then, age of the elder person = (x + 16) years.
:. 3 (x - 6) = (x + 16 - 6)  3x -18 = x + 10  2x = 28  x = 14.
Hence, their present ages are 14 years and 30 years.

Ex. 3. The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita
is more than Ankit's age by 4 years, what is Nikita's age? (S.B.I.P.O,1999)
Sol. Let Ankit's age be x years. Then, Nikita's age = 240/xyears.
 2  (240 /x ) – x = 4  480 – x2 = 4x  x2 + 4x – 480 = 0
 ( x+24)(x-20) = 0  x = 20.
Hence, Nikita's age = (22_0) years = 12 years. 1
Ex. 4. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father. . (S.S.C, 2003)
Sol. Let the son's present age be x years. Then, father's present age = (3x + 3) years
 (3x + 3 + 3) = 2 (x + 3) + 10  3x + 6 = 2x + 16  x = 10.
Hence, father's present age = (3x + 3) = ((3  10) + 3) years = 33 years.

Ex. 5. Rohit was 4 times as old as his son 8 years ago. After 8 years, Rohit will be
twice as old as his son. What are their present ages?
Sol. Let son's age 8 years ago be x years. Then, Rohit's age 8 years ago = 4x years.
Son's age after 8 years = (x + 8) + 8 = (x + 16) years.
Rohit's age after 8 years = (4x + 8) + 8 = (4x+ 16) years.
 2 (x + 16) = 4x + 16  2x = 16  x = 8.
Hence, son's 'present age = (x + 8) = 16 years.
Rohit's present age = (4x + 8) = 40 years.

Ex. 6. One year ago, the ratio of Gaurav’s and Sachin’s age was 6: 7 respectively.
Four years hence, this ratio would become 7: 8. How old is Sa chin ?
(NABARD, 2002)

Sol:
. Let Gaurav's and Sachin's ages one year ago be 6x and 7x years respectively. Then, Gaurav's age
4 years hence = (6x + 1) + 4 = (6x + 5) years.
Sachin's age 4 years hence = (7x + 1) + 4 = (7x + 5) years.

6x+5 = 7  8(6x+5) = 7 (7x + 5)  48x + 40 = 49x + 35  x = 5.
7x+5 8
Hence, Sachin's present age = (7x + 1) = 36 years.

,7. Abhay’s age after six years will be three-seventh of his fathers age. Ten years ago the ratio of their ages was 1 : 5. What is Abhay’s father's age at present?

Sol. Let the ages of Abhay and his father 10 years ago be x and 5x years respectively. Then,
Abhay's age after 6 years = (x + 10) + 6 = (x + 16) years.
Father's age after 6 years = (5x + 10) + 6 = (5x + 16) years.

\x + 16) = 3 (5x + 16)  7 (x + 16) = 3 (5x + 16)  7x + 112 = 15x + 48
7
 8x = 64  x = 8.
Hence, Abhay's father's present age = (5x + 10) = 50 years.

8. PROBLEMS ON AGES

Ex. 1. Rajeev's age after 15 years will be 5 times his age 5 years back. What is the
present age of Rajeev ? (Hotel Management,2002)

Sol. Let Rajeev's present age be x years. Then,
Rajeev's age after 15 years = (x + 15) years.
Rajeev's age 5 years back = (x - 5) years.
:. x + 15 = 5 (x - 5) x + 15 = 5x - 25  4x = 40  x = 10.
Hence, Rajeev's present age = 10 years.

Ex. 2. The ages of two persons differ by 16 years. If 6 years ago, the elder one be
3 times as old as the younger one, find their present ages. (A.A.O. Exam,2003)
Sol. Let the age of the younger person be x years.
Then, age of the elder person = (x + 16) years.
:. 3 (x - 6) = (x + 16 - 6)  3x -18 = x + 10  2x = 28  x = 14.
Hence, their present ages are 14 years and 30 years.

Ex. 3. The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita
is more than Ankit's age by 4 years, what is Nikita's age? (S.B.I.P.O,1999)
Sol. Let Ankit's age be x years. Then, Nikita's age = 240/xyears.
 2  (240 /x ) – x = 4  480 – x2 = 4x  x2 + 4x – 480 = 0
 ( x+24)(x-20) = 0  x = 20.
Hence, Nikita's age = (22_0) years = 12 years. 1
Ex. 4. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father. . (S.S.C, 2003)
Sol. Let the son's present age be x years. Then, father's present age = (3x + 3) years
 (3x + 3 + 3) = 2 (x + 3) + 10  3x + 6 = 2x + 16  x = 10.
Hence, father's present age = (3x + 3) = ((3  10) + 3) years = 33 years.

Ex. 5. Rohit was 4 times as old as his son 8 years ago. After 8 years, Rohit will be
twice as old as his son. What are their present ages?
Sol. Let son's age 8 years ago be x years. Then, Rohit's age 8 years ago = 4x years.
Son's age after 8 years = (x + 8) + 8 = (x + 16) years.
Rohit's age after 8 years = (4x + 8) + 8 = (4x+ 16) years.
 2 (x + 16) = 4x + 16  2x = 16  x = 8.
Hence, son's 'present age = (x + 8) = 16 years.
Rohit's present age = (4x + 8) = 40 years.

Ex. 6. One year ago, the ratio of Gaurav’s and Sachin’s age was 6: 7 respectively.
Four years hence, this ratio would become 7: 8. How old is Sa chin ?
(NABARD, 2002)

Sol:
. Let Gaurav's and Sachin's ages one year ago be 6x and 7x years respectively. Then, Gaurav's age
4 years hence = (6x + 1) + 4 = (6x + 5) years.
Sachin's age 4 years hence = (7x + 1) + 4 = (7x + 5) years.

6x+5 = 7  8(6x+5) = 7 (7x + 5)  48x + 40 = 49x + 35  x = 5.
7x+5 8
Hence, Sachin's present age = (7x + 1) = 36 years.

,7. Abhay’s age after six years will be three-seventh of his fathers age. Ten years ago the ratio of their ages was 1 : 5. What is Abhay’s father's age at present?

Sol. Let the ages of Abhay and his father 10 years ago be x and 5x years respectively. Then,
Abhay's age after 6 years = (x + 10) + 6 = (x + 16) years.
Father's age after 6 years = (5x + 10) + 6 = (5x + 16) years.

\x + 16) = 3 (5x + 16)  7 (x + 16) = 3 (5x + 16)  7x + 112 = 15x + 48
7
 8x = 64  x = 8.
Hence, Abhay's father's present age = (5x + 10) = 50 years.

8. PROBLEMS ON AGES

Ex. 1. Rajeev's age after 15 years will be 5 times his age 5 years back. What is the
present age of Rajeev ? (Hotel Management,2002)

Sol. Let Rajeev's present age be x years. Then,
Rajeev's age after 15 years = (x + 15) years.
Rajeev's age 5 years back = (x - 5) years.
:. x + 15 = 5 (x - 5) x + 15 = 5x - 25  4x = 40  x = 10.
Hence, Rajeev's present age = 10 years.

Ex. 2. The ages of two persons differ by 16 years. If 6 years ago, the elder one be
3 times as old as the younger one, find their present ages. (A.A.O. Exam,2003)
Sol. Let the age of the younger person be x years.
Then, age of the elder person = (x + 16) years.
:. 3 (x - 6) = (x + 16 - 6)  3x -18 = x + 10  2x = 28  x = 14.
Hence, their present ages are 14 years and 30 years.

Ex. 3. The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita
is more than Ankit's age by 4 years, what is Nikita's age? (S.B.I.P.O,1999)
Sol. Let Ankit's age be x years. Then, Nikita's age = 240/xyears.
 2  (240 /x ) – x = 4  480 – x2 = 4x  x2 + 4x – 480 = 0
 ( x+24)(x-20) = 0  x = 20.
Hence, Nikita's age = (22_0) years = 12 years. 1
Ex. 4. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father. . (S.S.C, 2003)
Sol. Let the son's present age be x years. Then, father's present age = (3x + 3) years
 (3x + 3 + 3) = 2 (x + 3) + 10  3x + 6 = 2x + 16  x = 10.
Hence, father's present age = (3x + 3) = ((3  10) + 3) years = 33 years.

Ex. 5. Rohit was 4 times as old as his son 8 years ago. After 8 years, Rohit will be
twice as old as his son. What are their present ages?
Sol. Let son's age 8 years ago be x years. Then, Rohit's age 8 years ago = 4x years.
Son's age after 8 years = (x + 8) + 8 = (x + 16) years.
Rohit's age after 8 years = (4x + 8) + 8 = (4x+ 16) years.
 2 (x + 16) = 4x + 16  2x = 16  x = 8.
Hence, son's 'present age = (x + 8) = 16 years.
Rohit's present age = (4x + 8) = 40 years.

Ex. 6. One year ago, the ratio of Gaurav’s and Sachin’s age was 6: 7 respectively.
Four years hence, this ratio would become 7: 8. How old is Sa chin ?
(NABARD, 2002)

Sol:
. Let Gaurav's and Sachin's ages one year ago be 6x and 7x years respectively. Then, Gaurav's age
4 years hence = (6x + 1) + 4 = (6x + 5) years.
Sachin's age 4 years hence = (7x + 1) + 4 = (7x + 5) years.

6x+5 = 7  8(6x+5) = 7 (7x + 5)  48x + 40 = 49x + 35  x = 5.
7x+5 8
Hence, Sachin's present age = (7x + 1) = 36 years.

,7. Abhay’s age after six years will be three-seventh of his fathers age. Ten years ago the ratio of their ages was 1 : 5. What is Abhay’s father's age at present?

Sol. Let the ages of Abhay and his father 10 years ago be x and 5x years respectively. Then,
Abhay's age after 6 years = (x + 10) + 6 = (x + 16) years.
Father's age after 6 years = (5x + 10) + 6 = (5x + 16) years.

\x + 16) = 3 (5x + 16)  7 (x + 16) = 3 (5x + 16)  7x + 112 = 15x + 48
7
 8x = 64  x = 8.
Hence, Abhay's father's present age = (5x + 10) = 50 years.

9. SURDS AND INDICES

I IMPORTANT FACTS AND FORMULAE I
1. LAWS OF INDICES:

(i) am x an = am + n
(ii) am¬ / an = am-n
(iii) (am)n = amn
(iv) (ab)n = anbn
(v) ( a/ b )n = ( an / bn )
(vi) a0 = 1

2. SURDS: Let a be a rational number and n be a positive integer such that a1/n = nsqrt(a)
is irrational. Then nsqrt(a) is called a surd of order n.

3. LAWS OF SURDS:

(i) n√a = a1/2
(ii) n √ab = n √a * n √b
(iii) n √a/b = n √a / n √b
(iv) (n √a)n = a
(v) m√(n√(a)) = mn√(a)
(vi) (n√a)m = n√am

I SOLVED EXAMPLES

Ex. 1. Simplify : (i) (27)2/3 (ii) (1024)-4/5 (iii)( 8 / 125 )-4/3

Sol . (i) (27)2/3 = (33)2/3 = 3( 3 * ( 2/ 3)) = 32 = 9
(ii) (1024)-4/5 = (45)-4/5 = 4 { 5 * ( (-4) / 5 )} = 4-4 = 1 / 44 = 1 / 256
(iii) ( 8 / 125 )-4/3 = {(2/5)3}-4/3 = (2/5){ 3 * ( -4/3)} = ( 2 / 5 )-4 = ( 5 / 2 )4 = 54 / 24 = 625 / 16

Ex. 2. Evaluate: (i) (.00032)3/5 (ii)l (256)0.16 x (16)0.18.

Sol. (i) (0.00032)3/5 = ( 32 / 100000 )3/5. = (25 / 105)3/5 = {( 2 / 10 )5}3/5 = ( 1 / 5 )(5 * 3 / 5) = (1/5)3 = 1 / 125
(ii) (256)0. 16 * (16)0. 18 = {(16)2}0. 16 * (16)0. 18 = (16)(2 * 0. 16) * (16)0. 18
=(16)0.32 * (16)0.18 = (16)(0.32+0.18) = (16)0.5 = (16)1/2 = 4.
196

Ex. 3. What is the quotient when (x-1 - 1) is divided by (x - 1) ?

Sol. x-1 -1 = (1/x)-1 = _1 -x * 1 = -1
x - 1 x - 1 x (x - 1) x
Hence, the required quotient is -1/x

Ex. 4. If 2x - 1 + 2x + 1 = 1280, then find the value of x.
Sol. 2x - 1 + 2X+ 1 = 1280  2x-1 (1 +22) = 1280
 2x-1 = 1280 / 5 = 256 = 28  x -1 = 8  x = 9.

Hence, x = 9.
Ex. 5. Find the value of [ 5 ( 81/3 + 271/3)3]1/ 4

Sol. [ 5 ( 81/3 + 271/3)3]1/ 4 = [ 5 { (23)1/3 + (33)1/3}3]1/ 4 = [ 5 { (23 * 1/3)1/3 + (33 *1/3 )1/3}3]1/ 4
= {5(2+3)3}1/4 = (5 * 53)1/ 4 =5(4 * 1/ 4) = 51 = 5.

Ex. 6. Find the Value of {(16)3/2 + (16)-3/2}

Sol. [(16)3/2 +(16)-3/2 = (42)3/2 +(42)-3/2 = 4(2 * 3/2) + 4{ 2* (-3/2)}
= 43 + 4-3 = 43 + (1/43) = ( 64 + ( 1/64)) = 4097/64.

Ex. 7. If (1/5)3y = 0.008, then find the value of(0.25)y.

Sol. (1/5)3y = 0.008 = 8/1000 = 1/125 = (1/5)3  3y = 3  Y = 1.
 (0.25)y = (0.25)1 = 0.25.

Ex. 8. Find the value of (243)n/5  32n + 1
9n  3n -1 .

Sol. (243)n/5 x32n+l = 3 (5 * n/5)  32n+l _ = 3n 32n+1
(32)n  3n - 1 32n  3n - 1 32n  3n-l

= 3n + (2n + 1) = 3(3n+1) = 3(3n+l)-(3n-l) = 32 = 9.
32n+n-1 3(3n-1)

Ex. 9. Find the value Of (21/4-1)(23/4+21/2+21/4+1)

Sol.
Putting 21/4 = x, we get :

(21/4-1) (23/4+21/2+21/4+1)=(x-1)(x3+x2+x+1) , where x = 21/4
=(x-1)[x2(x+1)+(x+1)]
=(x-1)(x+1)(x2+1) = (x2-1)(x2+1)
=(x4-1) = [(21/4)4-1] = [2(1/4*4) –1] = (2-1) = 1.

Ex. 10. Find the value of 62/3  3√67
3√66

Sol. 62/3  3√67 = 62/3  (67)1/3 = 62/3  6(7 * 1/3) = 62/3  6(7/3)
3√66 (66)1/3 6(6 * 1/3) 62

=62/3  6((7/3)-2) = 62/3  61/3 = 61 = 6.
Ex. 11. If x= ya, y=zb and z=xc,then find the value of abc.

Sol. z1= xc =(ya)c [since x= ya]
=y(ac) = (zb)ac [since y=zb]
=zb(ac)= zabc
 abc = 1.
= 24
Ex. 12. Simplify [(xa / xb)^(a2+b2+ab)] * [(xb / xc )^ b2+c2+bc)] * [(xc/xa)^(c2+a2+ca)]
Sol.
Given Expression
= [{x(o - b)}^(a2 + b2 + ob)].['(x(b - c)}^ (b2 + c2 + bc)].['(x(c - a)}^(c2 + a2 + ca])
= [x(a - b)(a2 + b2 + ab) . x(b - c) (b2 +c2+ bc).x(c- a) (c2 + a2 + ca)]
= [x^(a3-b3)].[x^(b3-e3)].[x^(c3-a3)] = x^(a3-b3+b3-c3+c3-a3) = x0 = 1.

Ex. 13. Which is larger √2 or 3√3 ?

Sol. Given surds are of order 2 and 3. Their L.C.M. is 6. Changing each to a surd of order 6, we get:
√2 = 21/2 = 2((1/2)*(3/2)) =23/6 = 81/6 = 6√8
3√3= 31/3 = 3((1/3)*(2/2)) = 32/6 = (32)1/6 = (9)1/6 = 6√9.
Clearly, 6√9 > 6√8 and hence 3√3 > √2.

Ex. 14. Find the largest from among 4√6, √2 and 3√4.
Sol. Given surds are of order 4, 2 and 3 respectively. Their L.C,M, is 12, Changing each to a surd of order 12, we get:

4√6 = 61/4 = 6((1/4)*(3/3)) = 63/12 = (63)1/12 = (216)1/12.
√2 = 21/2 = 2((1/2)*(6/6)) = 26/12 = (26)1/12 = (64)1/12.
3√4 = 41/3 = 4((1/3)*(4/4)) = 44/12 = (44)1/12 = (256)1/12.

Clearly, (256)1/12 > (216)1/12 > (64)1/12

Largest one is (256)1/12. i.e. 3√4 .
.
10.PERCENTAGE

IMPORTANT FACTS AND FORMULAE

1. Concept of Percentage : By a certain percent ,we mean that many hundredths. Thus x percent means x hundredths, written as x%.

To express x% as a fraction : We have , x% = x/100.

Thus, 20% =20/100 =1/5; 48% =48/100 =12/25, etc.

To express a/b as a percent : We have, a/b =((a/b)*100)%.

Thus, ¼ =[(1/4)*100] = 25%; 0.6 =6/10 =3/5 =[(3/5)*100]% =60%.

2. If the price of a commodity increases by R%, then the reduction in consumption so asnot to increase the expenditure is
[R/(100+R))*100]%.
If the price of the commodity decreases by R%,then the increase in consumption so as to decrease the expenditure is
[(R/(100-R)*100]%.

3. Results on Population : Let the population of the town be P now and suppose it increases at the rate of
R% per annum, then :

1. Population after nyeras = P [1+(R/100)]^n.
2. Population n years ago = P /[1+(R/100)]^n.

4. Results on Depreciation : Let the present value of a machine be P. Suppose it depreciates at the rate
R% per annum. Then,

1. Value of the machine after n years = P[1-(R/100)]n.
2. Value of the machine n years ago = P/[1-(R/100)]n.

5. If A is R% more than B, then B is less than A by

[(R/(100+R))*100]%.

If A is R% less than B , then B is more than A by

[(R/(100-R))*100]%.

SOLVED EXAMPLES

Ex. 1. Express each of the following as a fraction :

(i) 56% (ii) 4% (iii) 0.6% (iv) 0.008%

sol. (i) 56% = 56/100= 14/25. (ii) 4% =4/100 =1/25.
(iii) 0.6 =6/1000 = 3/500. (iv) 0.008 = 8/100 = 1/1250.

Ex. 2. Express each of the following as a Decimal :

(i) 6% (ii)28% (iii) 0.2% (iv) 0.04%

Sol. (i) 6% = 6/100 =0.06. (ii) 28% = 28/100 =0.28.
(iii) 0.2% =0.2/100 = 0.002. (iv) 0.04%= 0.04/100 =0.004.

Ex. 3. Express each of the following as rate percent :

(i) 23/36 (ii) 6 ¾ (iii) 0.004

Sol. (i) 23/36 = [(23/36)*100]% = [575/9]% = 63 8/9%.

(ii) 0.004 = [(4/1000)*100]% = 0.4%.

(iii) 6 ¾ =27/4 =[(27/4)*100]% = 675%.

Ex. 4. Evaluate :

(i) 28% of 450+ 45% of 280
(ii) 16 2/3% of 600 gm- 33 1/3% of 180 gm

Sol. (i) 28% of 450 + 45% of 280 =[(28/100)*450 + (45/100)*280] = (126+126) =252.

(iii) 16 2/3% of 600 gm –33 1/3% of 180 gm = [ ((50/3)*(1/100)*600) – ((100/3)*(1/3)*280)]gm = (100-60) gm = 40gm.

Ex. 5.
(i) 2 is what percent of 50 ?
(ii) ½ is what percent of 1/3 ?
(iii)What percent of 8 is 64 ?
(iv)What percent of 2 metric tones is 40 quintals ?
(v)What percent of 6.5 litres is 130 ml?

Sol.

(i) Required Percentage = [(2/50)*100]% = 4%.

(ii) Required Percentage = [ (1/2)*(3/1)*100]% = 150%.

(iii)Required Percentage = [(84/7)*100]% = 1200%.

(iv) 1 metric tonne = 10 quintals.

Required percentage = [ (40/(2 * 10)) * 100]% = 200%.

(v) Required Percentage = [ (130/(6.5 * 1000)) * 100]% = 2%.

Ex. 6.
Find the missing figures :

(i) ?% of 25 = 20125 (ii) 9% of ? = 63 (iii) 0.25% of ? = 0.04

Sol.

(i) Let x% of 25 = 2.125. Then , (x/100)*25 = 2.125
X = (2.125 * 4) = 8.5.

(ii) Let 9% of x =6.3. Then , 9*x/100 = 6.3
X = [(6.3*100)/9] =70.

(iii) Let 0.25% of x = 0.04. Then , 0.25*x/100 = 0.04
X= [(0.04*100)/0.25] = 16.

Ex. 7.
Which is greatest in 16 ( 2/3) %, 2/5 and 0.17 ?

Sol. 16 (2/3)% =[ (50/3)* )1/100)] = 1/6 = 0.166, 2/15 = 0.133. Clearly, 0.17 is the greatest.

Ex. 8.
If the sales tax reduced from 3 1/2 % to 3 1/3%, then what difference does it make to a person who purchases an article with market price of Rs. 8400 ?

Sol. Required difference = [3 ½ % of Rs.8400] – [3 1/3 % of Rs.8400]
= [(7/20-(10/3)]% of Rs.8400 =1/6 % of Rs.8400
= Rs. [(1/6)8(1/100)*8400] = Rs. 14.

Ex. 9. An inspector rejects 0.08% of the meters as defective. How many will be examine to project ?

Sol. Let the number of meters to be examined be x.
Then, 0.08% of x =2
[(8/100)*(1/100)*x] = 2
x = [(2*100*100)/8] = 2500.

Ex. 10. Sixty five percent of a number is 21 less than four fifth of that number. What is the number ?

Sol. Let the number be x.
Then, 4*x/5 –(65% of x) = 21
4x/5 –65x/100 = 21
5 x = 2100
x = 140.

Ex.11. Difference of two numbers is 1660. If 7.5% of the number is 12.5% of the other number , find the number ?
Sol. Let the numbers be x and y. Then , 7.5 % of x =12.5% of y
X = 125*y/75 = 5*y/3.
Now, x-y =1660
5*y/3 –y =1660
2*y/3= 1660
y =[ (1660*3)/2] =2490.
One number = 2490, Second number =5*y/3 =4150.

Ex. 12.
In expressing a length 810472 km as nearly as possible with three significant digits , find the percentage error.
Sol. Error = (81.5 – 81.472)km = 0.028.
Required percentage = [(0.028/81.472)*100]% = 0.034%.

Ex. 13.
In an election between two candidates, 75% of the voters cast thier thier votes, out of which 2% of the votes were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. Find the total number of votes enrolled in that election.
Sol.
Let the number of votes enrolled be x. Then ,
Number of votes cast =75% of x. Valid votes = 98% of (75% of x).
75% of (98% of (75%of x)) =9261.
[(75/100)*(98/100)*(75/100)*x] =9261.
X = [(9261*100*100*100)/(75*98*75)] =16800.

Ex.14. Shobha’s mathematics test had 75 problems i.e.10 arithmetic, 30 algebra and 35 geometry problems. Although she answered 70% of the arithmetic ,40% of the algebra, and 60% of the geometry problems correctly. she did not pass the test because she got less than 60% of the problems right. How many more questions she would have to answer correctly to earn 60% of the passing grade?

Sol. Number of questions attempted correctly=(70% of 10 + 40% of 30 + 60% 0f 35)
=7 + 12+21= 45
questions to be answered correctly for 60% grade=60% of 75 = 45

therefore required number of questions= (45-40) = 5.

Ex.15. if 50% of (x-y) = 30% of (x+y) then what percent of x is y?

Sol.50% of (x-y)=30% of(x+y)  (50/100)(x-y)=(30/100)(x+y)
5(x-y)=3(x+y)  2x=8y  x=4y
therefore required percentage =((y/x) X 100)% = ((y/4y) X 100) =25%

Ex.16. Mr.Jones gave 40% of the money he had to his wife. he also gave 20% of the remaining amount to his 3 sons. half of the amount now left was spent on miscellaneous items and the remaining amount of Rs.12000 was deposited in the bank. how much money did Mr.jones have initially?

Sol. Let the initial amount with Mr.jones be Rs.x then,
Money given to wife= Rs.(40/100)x=Rs.2x/5.Balance=Rs(x-(2x/5)=Rs.3x/5.
Money given to 3 sons= Rs(3X((20/200) X (3x/5)) = Rs.9x/5.
Balance = Rs.((3x/5) – (9x/25))=Rs.6x/25.
Amount deposited in bank= Rs(1/2 X 6x/25)=Rs.3x/25.

Therefore 3x/25=12000  x= ((12000 x 35)/3)=100000

So Mr.Jones initially had Rs.1,00,000 with him.

Short-cut Method : Let the initial amount with Mr.Jones be Rs.x
Then,(1/2)[100-(3*20)]% of x=12000

 (1/2)*(40/100)*(60/100)*x=12000
x=((12000*25)/3)=100000

Ex 17 10% of the inhabitants of village having died of cholera.,a panic set in , during which 25% of the remaining inhabitants left the village. The population is then reduced to 4050. Find the number of original inhabitants.
Sol:
Let the total number of orginal inhabitants be x.
((75/100))*(90/100)*x)=4050  (27/40)*x=4050

x=((4050*40)/27)=6000.

Ex.18 A salesman`s commission is 5% on all sales upto Rs.10,000 and 4% on all sales exceeding this.He remits Rs.31,100 to his parent company after deducing his commission . Find the total sales.
Sol:
Let his total sales be Rs.x.Now(Total sales) – (Commission )=Rs.31,100
x-[(5% of 10000 + 4% of (x-10000)]=31,100
x-[((5/100)*10000 + (4/100)*(x-10000)]=31,100
x-500-((x-10000)/25)=31,100
x-(x/25)=31200  24x/25=31200x=[(31200*25)/24)=32,500.
Total sales=Rs.32,500

Ex .19 Raman`s salary was decreased by 50% and subsequently increased by 50%.How much percent does he lose?
Sol:
Let the original salary = Rs.100
New final salary=150% of (50% of Rs.100)=
Rs.((150/100)*(50/100)*100)=Rs.75.
Decrease = 25%

Ex.20 Paulson spends 75% of his income. His income is increased by 20% and he increased his expenditure by 10%.Find the percentage increase in his savings .
Sol:
Let the original income=Rs.100 . Then , expenditure=Rs.75 and savings =Rs.25
New income =Rs.120 , New expenditure =
Rs.((110/100)*75)=Rs.165/2
New savings = Rs.(120-(165/2)) = Rs.75/2
Increase in savings = Rs.((75/2)-25)=Rs.25/2
Increase %= ((25/2)*(1/25)*100)% = 50%.

Ex21. The salary of a person was reduced by 10% .By what percent should his reduced salary be raised so as to bring it at par with his original salary ?
Sol:
Let the original salary be Rs.100 . New salary = Rs.90.
Increase on 90=10 , Increase on 100=((10/90)*100)%
= (100/9)%

Ex.22 When the price fo a product was decreased by 10% , the number sold increased by 30%. What was the effect on the total revenue ?
Sol:
Let the price of the product be Rs.100 and let original sale be 100 pieces.
Then , Total Revenue = Rs.(100*100)=Rs.10000.
New revenue = Rs.(90*130)=Rs.11700.
Increase in revenue = ((1700/10000)*100)%=17%.

Ex 23 . If the numerator of a fraction be increased by 15% and its denominator be diminished by 8% , the value of the fraction is 15/16. Find the original fraction.
Sol:
Let the original fraction be x/y.
Then (115%of x)/(92% of y)=15/16 => (115x/92y)=15/16
 ((15/16)*(92/115))=3/4

Ex.24 In the new budget , the price of kerosene oil rose by 25%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase ?
Sol:
Reduction in consumption = [((R/(100+R))*100]%
 [(25/125)*100]%=20%.

Ex.25 The population of a town is 1,76,400 . If it increases at the rate of 5% per annum , what will be its population 2 years hence ? What was it 2 years ago ?
Sol:
Population after 2 years = 176400*[1+(5/100)]^2
=[176400*(21/20)*(21/40)]
= 194481.
Population 2 years ago = 176400/[1+(5/100)]^2
=[716400*(20/21)*(20/21)]= 160000.

Ex.26 The value of a machine depreiates at the rate of 10% per annum. If its present is Rs.1,62,000 what will be its worth after 2 years ? What was the value of the machine 2 years ago ?
Sol.
Value of the machine after 2 years
=Rs.[162000*(1-(10/100))^2] = Rs.[162000*(9/10)*(9/10)]
=Rs. 131220
Value of the machine 2 years ago
= Rs.[162000/(1-(10/100)^2)]=Rs.[162000*(10/9)*(10/9)]=Rs.200000

Ex27. During one year, the population of town increased by 5% . If the total population is 9975 at the end of the second year , then what was the population size in the beginning of the first year ?
Sol:
Population in the beginning of the first year
= 9975/[1+(5/100)]*[1-(5/100)] = [9975*(20/21)*(20/19)]=10000.

Ex.28 If A earns 99/3% more than B,how much percent does B earn less then A ?
Sol:
Required Percentage = [((100/3)*100)/[100+(100/3)]]%
=[(100/400)*100]%=25%

Ex. 29 If A`s salary is 20% less then B`s salary , by how much percent is B`s salary more than A`s ?
Sol:
Required percentage = [(20*100)/(100-20)]%=25%.

Ex30 .How many kg of pure salt must be added to 30kg of 2% solution of salt and water to increase it to 10% solution ?
Sol:
Amount of salt in 30kg solution = [(20/100)*30]kg=0.6kg
Let x kg of pure salt be added
Then , (0.6+x)/(30+x)=10/10060+100x=300+10x
90x=240  x=8/3.

Ex 31. Due to reduction of 25/4% in the price of sugar , a man is able to buy 1kg more for Rs.120. Find the original and reduced rate of sugar.
Sol:
Let the original rate be Rs.x per kg.
Reduced rate = Rs.[(100-(25/4))*(1/100)*x}]=Rs.15x/16per kg
120/(15x/16)-(120/x)=1  (128/x)-(120/x)=1
 x=8.
So, the original rate = Rs.8 per kg
Reduce rate = Rs.[(15/16)*8]per kg = Rs.7.50 per kg

Ex.32 In an examination , 35% of total students failed in Hindi , 45% failed in English and 20% in both . Find the percentage of those who passed in both subjects .
Sol:
Let A and B be the sets of students who failed in Hindi and English respectively .
Then , n(A) = 35 , n(B)=45 , n(AB)=20.
So , n(AB)=n(A)+n(B)- n(AB)=35+45-20=60.
Percentage failed in Hindi and English or both=60%
Hence , percentage passed = (100-60)%=40%

Ex33. In an examination , 80% of the students passed in English , 85% in Mathematics and 75% in both English and Mathematics. If 40 students failed in both the subjects , find the total number of students.
Sol:
Let the total number of students be x .
Let A and B represent the sets of students who passed in English and Mathematics respectively .
Then , number of students passed in one or both the subjects
= n(AB)=n(A)+n(B)- n(AB)=80% of x + 85% of x –75% of x
=[(80/100)x+(85/100)x-(75/100)x]=(90/100)x=(9/10)x
Students who failed in both the subjects = [x-(9x/10)]=x/10.
So, x/10=40 of x=400 .
Hence ,total number of students = 400.

uml tutorial

2010-08-14T01:35:00.000-07:00

ans for c

2010-08-14T01:30:00.001-07:00

What is C language?
The C programming language is a standardized programming language
developed in the early 1970s by Ken Thompson and Dennis Ritchie for
use on the UNIX operating system. It has since spread to many other
operating systems, and is one of the most widely used programming
languages. C is prized for its efficiency, and is the most popular
programming language for writing system software, though it is also
used for writing applications.
printf() Function
What is the output of printf("%d")?
1. When we write printf("%d",x); this means compiler will print the
value of x. But as here, there is nothing after %d so compiler will show
in output window garbage value.
2. When we use %d the compiler internally uses it to access the
argument in the stack (argument stack). Ideally compiler determines
the offset of the data variable depending on the format specification
string. Now when we write printf("%d",a) then compiler first accesses
the top most element in the argument stack of the printf which is %d
and depending on the format string it calculated to offset to the actual
data variable in the memory which is to be printed. Now when only %d
will be present in the printf then compiler will calculate the correct
offset (which will be the offset to access the integer variable) but as
the actual data object is to be printed is not present at that memory
location so it will print what ever will be the contents of that memory
location.
3. Some compilers check the format string and will generate an error
without the proper number and type of arguments for things like
printf(...) and scanf(...).
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malloc() Function- What is the difference between "calloc(...)" and
"malloc(...)"?
1. calloc(...) allocates a block of memory for an array of elements of a
certain size. By default the block is initialized to 0. The total number of
memory allocated will be (number_of_elements * size).
malloc(...) takes in only a single argument which is the memory
required in bytes. malloc(...) allocated bytes of memory and not blocks
of memory like calloc(...).
2. malloc(...) allocates memory blocks and returns a void pointer to the
allocated space, or NULL if there is insufficient memory available.
calloc(...) allocates an array in memory with elements initialized to 0
and returns a pointer to the allocated space. calloc(...) calls malloc(...)
in order to use the C++ _set_new_mode function to set the new
handler mode.
printf() Function- What is the difference between "printf(...)" and
"sprintf(...)"?
sprintf(...) writes data to the character array whereas printf(...) writes data to the
standard output device.
Compilation How to reduce a final size of executable?
Size of the final executable can be reduced using dynamic linking for
libraries.
Linked Lists -- Can you tell me how to check whether a linked list is circular?
Create two pointers, and set both to the start of the list. Update each
as follows:
while (pointer1) {
pointer1 = pointer1->next;
pointer2 = pointer2->next;
if (pointer2) pointer2=pointer2->next;
if (pointer1 == pointer2) {
print ("circular");
}}
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If a list is circular, at some point pointer2 will wrap around and be
either at the item just before pointer1, or the item before that. Either
way, its either 1 or 2 jumps until they meet.
"union" Data Type What is the output of the following program? Why?
#include
main() {
typedef union {
int a;
char b[10];
float c;
}
Union;
Union x,y = {100};
x.a = 50;
strcpy(x.b,"hello");
x.c = 21.50;
printf("Union x : %d %s %f n",x.a,x.b,x.c);
printf("Union y : %d %s %f n",y.a,y.b,y.c);
}
What does static variable mean?
there are 3 main uses for the static.
1. If you declare within a function:
It retains the value between function calls
2.If it is declared for a function name:
By default function is extern..so it will be visible from other files if the
function declaration is as static..it is invisible for the outer files
3. Static for global variables:
By default we can use the global variables from outside files If it is
static global..that variable is limited to with in the file
Advantages of a macro over a function?
Macro gets to see the Compilation environment, so it can expand __
__TIME__ __FILE__ #defines. It is expanded by the preprocessor.
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For example, you can’t do this without macros
#define PRINT(EXPR) printf( #EXPR “=%d\n”, EXPR)
PRINT( 5+6*7 ) // expands into printf(”5+6*7=%d”, 5+6*7 );
You can define your mini language with macros:
#define strequal(A,B) (!strcmp(A,B))
Macros are a necessary evils of life. The purists don’t like them, but
without it no real work gets done.
What are the differences between malloc() and calloc()?
There are 2 differences.
First, is in the number of arguments. malloc() takes a single
argument(memory required in bytes), while calloc() needs 2
arguments(number of variables to allocate memory, size in bytes of a
single variable).
Secondly, malloc() does not initialize the memory allocated, while
calloc() initializes the allocated memory to ZERO.
What are the different storage classes in C?
C has three types of storage: automatic, static and allocated.
Variable having block scope and without static specifier have
automatic storage duration.
Variables with block scope, and with static specifier have static scope.
Global variables (i.e, file scope) with or without the static specifier also
have static scope.
Memory obtained from calls to malloc(), alloc() or realloc() belongs to
allocated storage class.
What is the difference between strings and character arrays?
A major difference is: string will have static storage duration, whereas
as a character array will not, unless it is explicity specified by using the
static keyword.
Actually, a string is a character array with following properties:
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* the multibyte character sequence, to which we generally call string,
is used to initialize an array of static storage duration. The size of this
array is just sufficient to contain these characters plus the terminating
NUL character.
* it not specified what happens if this array, i.e., string, is modified.
* Two strings of same value[1] may share same memory area. For
example, in the following declarations:
char *s1 = “Calvin and Hobbes”;
char *s2 = “Calvin and Hobbes”;
the strings pointed by s1 and s2 may reside in the same memory
location. But, it is not true for the following:
char ca1[] = “Calvin and Hobbes”;
char ca2[] = “Calvin and Hobbes”;
[1] The value of a string is the sequence of the values of the contained
characters, in order.
Difference between const char* p and char const* p
In const char* p, the character pointed by ‘p’ is constant, so u cant
change the value of character pointed by p but u can make ‘p’ refer to
some other location.
in char const* p, the ptr ‘p’ is constant not the character referenced by
it, so u cant make ‘p’ to reference to any other location but u can
change the value of the char pointed by ‘p’.
What is hashing?
To hash means to grind up, and that’s essentially what hashing is all
about. The heart of a hashing algorithm is a hash function that takes
your nice, neat data and grinds it into some random-looking integer.
The idea behind hashing is that some data either has no inherent
ordering (such as images) or is expensive to compare (such as
images). If the data has no inherent ordering, you can’t perform
comparison searches.
If the data is expensive to compare, the number of comparisons used
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even by a binary search might be too many. So instead of looking at
the data themselves, you’ll condense (hash) the data to an integer (its
hash value) and keep all the data with the same hash value in the
same place. This task is carried out by using the hash value as an
index into an array.
To search for an item, you simply hash it and look at all the data whose
hash values match that of the data you’re looking for. This technique
greatly lessens the number of items you have to look at. If the
parameters are set up with care and enough storage is available for
the hash table, the number of comparisons needed to find an item can
be made arbitrarily close to one.
One aspect that affects the efficiency of a hashing implementation is
the hash function itself. It should ideally distribute data randomly
throughout the entire hash table, to reduce the likelihood of collisions.
Collisions occur when two different keys have the same hash value.
There are two ways to resolve this problem. In open addressing, the
collision is resolved by the choosing of another position in the hash
table for the element inserted later. When the hash table is searched, if
the entry is not found at its hashed position in the table, the search
continues checking until either the element is found or an empty
position in the table is found.
The second method of resolving a hash collision is called chaining. In
this method, a bucket or linked list holds all the elements whose keys
hash to the same value. When the hash table is searched, the list must
be searched linearly.
How can you determine the size of an allocated portion of memory?
You can’t, really. free() can , but there’s no way for your program to
know the trick free() uses. Even if you disassemble the library and
discover the trick, there’s no guarantee the trick won’t change with the
next release of the compiler.
Can static variables be declared in a header file?
You can’t declare a static variable without defining it as well (this is
because the storage class modifiers static and extern are mutually
exclusive). A static variable can be defined in a header file, but this
would cause each source file that included the header file to have its
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own private copy of the variable, which is probably not what was
intended.
Can a variable be both const and volatile?
Yes. The const modifier means that this code cannot change the value
of the variable, but that does not mean that the value cannot be
changed by means outside this code. For instance, in the example in
FAQ 8, the timer structure was accessed through a volatile const
pointer. The function itself did not change the value of the timer, so it
was declared const. However, the value was changed by hardware on
the computer, so it was declared volatile. If a variable is both const and
volatile, the two modifiers can appear in either order.
Can include files be nested?
Yes. Include files can be nested any number of times. As long as you
use precautionary measures , you can avoid including the same file
twice. In the past, nesting header files was seen as bad programming
practice, because it complicates the dependency tracking function of
the MAKE program and thus slows down compilation. Many of today’s
popular compilers make up for this difficulty by implementing a
concept called precompiled headers, in which all headers and
associated dependencies are stored in a precompiled state.
Many programmers like to create a custom header file that has
#include statements for every header needed for each module. This is
perfectly acceptable and can help avoid potential problems relating to
#include files, such as accidentally omitting an #include file in a
module.
When does the compiler not implicitly generate the address of the first
element of an array?
Whenever an array name appears in an expression such as
- array as an operand of the sizeof operator
- array as an operand of & operator
- array as a string literal initializer for a character array
Then the compiler does not implicitly generate the address of the
address of the first element of an array.
What is a null pointer?
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There are times when it’s necessary to have a pointer that doesn’t
point to anything. The macro NULL, defined in , has a value that’s
guaranteed to be different from any valid pointer. NULL is a literal zero,
possibly cast to void* or char*. Some people, notably C++
programmers, prefer to use 0 rather than NULL.
The null pointer is used in three ways:
1) To stop indirection in a recursive data structure
2) As an error value
3) As a sentinel value
What is the difference between text and binary modes?
Streams can be classified into two types: text streams and binary
streams. Text streams are interpreted, with a maximum length of 255
characters. With text streams, carriage return/line feed combinations
are translated to the newline n character and vice versa. Binary
streams are uninterrupted and are treated one byte at a time with no
translation of characters. Typically, a text stream would be used for
reading and writing standard text files, printing output to the screen or
printer, or receiving input from the keyboard.
A binary text stream would typically be used for reading and writing
binary files such as graphics or word processing documents, reading
mouse input, or reading and writing to the modem.
What is static memory allocation and dynamic memory allocation?
Static memory allocation: The compiler allocates the required memory
space for a declared variable.By using the address of operator,the
reserved address is obtained and this address may be assigned to a
pointer variable.Since most of the declared variable have static
memory,this way of assigning pointer value to a pointer variable is
known as static memory allocation. memory is assigned during
compilation time.
Dynamic memory allocation: It uses functions such as malloc( ) or
calloc( ) to get memory dynamically.If these functions are used to get
memory dynamically and the values returned by these functions are
assingned to pointer variables, such assignments are known as
dynamic memory allocation.memory is assined during run time.
When should a far pointer be used?
Sometimes you can get away with using a small memory model in
most of a given program. There might be just a few things that don’t fit
in your small data and code segments. When that happens, you can
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use explicit far pointers and function declarations to get at the rest of
memory. A far function can be outside the 64KB segment most
functions are shoehorned into for a small-code model. (Often, libraries
are declared explicitly far, so they’ll work no matter what code model
the program uses.) A far pointer can refer to information outside the
64KB data segment. Typically, such pointers are used with farmalloc()
and such, to manage a heap separate from where all the rest of the
data lives. If you use a small-data, large-code model, you should
explicitly make your function pointers far.
How are pointer variables initialized?
Pointer variable are initialized by one of the following two ways
- Static memory allocation
- Dynamic memory allocation
Difference between arrays and pointers?
- Pointers are used to manipulate data using the address. Pointers use
* operator to access the data pointed to by them
- Arrays use subscripted variables to access and manipulate data.
Array variables can be equivalently written using pointer expression.
Is using exit() the same as using return?
No. The exit() function is used to exit your program and return control
to the operating system. The return statement is used to return from a
function and return control to the calling function. If you issue a return
from the main() function, you are essentially returning control to the
calling function, which is the operating system. In this case, the return
statement and exit() function are similar.
What is a method?
Method is a way of doing something, especially a systematic way;
implies an orderly logical arrangement (usually in steps).
What is indirection?
If you declare a variable, its name is a direct reference to its value. If
you have a pointer to a variable, or any other object in memory, you
have an indirect reference to its value.
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What is modular programming?
If a program is large, it is subdivided into a number of smaller
programs that are called modules or subprograms. If a complex
problem is solved using more modules, this approach is known as
modular programming.
How many levels deep can include files be nested?
Even though there is no limit to the number of levels of nested include
files you can have, your compiler might run out of stack space while
trying to include an inordinately high number of files. This number
varies according to your hardware configuration and possibly your
compiler.
What is the difference between declaring a variable and defining a variable?
Declaring a variable means describing its type to the compiler but not
allocating any space for it. Defining a variable means declaring it and
also allocating space to hold the variable. You can also initialize a
variable at the time it is defined.
What is an lvalue?
An lvalue is an expression to which a value can be assigned. The lvalue
expression is located on the left side of an assignment statement,
whereas an rvalue is located on the right side of an assignment
statement. Each assignment statement must have an lvalue and an
rvalue. The lvalue expression must reference a storable variable in
memory. It cannot be a constant.
Differentiate between an internal static and external static variable?
An internal static variable is declared inside a block with static storage
class whereas an external static variable is declared outside all the
blocks in a file.An internal static variable has persistent storage,block
scope and no linkage.An external static variable has permanent
storage,file scope and internal linkage.
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What is the difference between a string and an array?
An array is an array of anything. A string is a specific kind of an array
with a well-known convention to determine its length.
There are two kinds of programming languages: those in which a string
is just an array of characters, and those in which it’s a special type. In
C, a string is just an array of characters (type char), with one wrinkle: a
C string always ends with a NUL character.
The “value” of an array is the same as the address of (or a pointer to)
the first element; so, frequently, a C string and a pointer to char are
used to mean the same thing.
An array can be any length. If it’s passed to a function, there’s no way
the function can tell how long the array is supposed to be, unless some
convention is used. The convention for strings is NUL termination; the
last character is an ASCII NUL (‘’) character.
What is an argument? Differentiate between formal arguments and actual
arguments?
An argument is an entity used to pass the data from calling function to
the called function. Formal arguments are the arguments available in
the function definition. They are preceded by their own data types.
Actual arguments are available in the function call.
What are advantages and disadvantages of external storage class?
Advantages of external storage class
1)Persistent storage of a variable retains the latest value
2)The value is globally available
Disadvantages of external storage class
1)The storage for an external variable exists even when the variable is
not needed
2)The side effect may produce surprising output
3)Modification of the program is difficult
4)Generality of a program is affected
What is a void pointer?
A void pointer is a C convention for a raw address. The compiler has no
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idea what type of object a void Pointer really points to. If you write
int *ip;
ip points to an int. If you write
void *p;
p doesn’t point to a void!
In C and C++, any time you need a void pointer, you can use another
pointer type. For example, if you have a char*, you can pass it to a
function that expects a void*. You don’t even need to cast it. In C (but
not in C++), you can use a void* any time you need any kind of
pointer, without casting. (In C++, you need to cast it).
A void pointer is used for working with raw memory or for passing a
pointer to an unspecified type.
Some C code operates on raw memory. When C was first invented,
character pointers (char *) were used for that. Then people started
getting confused about when a character pointer was a string, when it
was a character array, and when it was raw memory.
When should a type cast not be used?
A type cast should not be used to override a const or volatile
declaration. Overriding these type modifiers can cause the program to
fail to run correctly.
A type cast should not be used to turn a pointer to one type of
structure or data type into another. In the rare events in which this
action is beneficial, using a union to hold the values makes the
programmer’s intentions clearer.
When is a switch statement better than multiple if statements?
A switch statement is generally best to use when you have more than
two conditional expressions based on a single variable of numeric type.
What is a static function?
A static function is a function whose scope is limited to the current
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source file. Scope refers to the visibility of a function or variable. If the
function or variable is visible outside of the current source file, it is said
to have global, or external, scope. If the function or variable is not
visible outside of the current source file, it is said to have local, or
static, scope.
What is a pointer variable?
A pointer variable is a variable that may contain the address of another
variable or any valid address in the memory.
What is a pointer value and address?
A pointer value is a data object that refers to a memory location. Each
memory location is numbered in the memory. The number attached to
a memory location is called the address of the location.
What is a modulus operator? What are the restrictions of a modulus
operator?
A Modulus operator gives the remainder value. The result of x%y is
obtained by (x-(x/y)*y). This operator is applied only to integral
operands and cannot be applied to float or double.
Differentiate between a linker and linkage?
A linker converts an object code into an executable code by linking
together the necessary build in functions. The form and place of
declaration where the variable is declared in a program determine the
linkage of variable.
What is a function and built-in function?
A large program is subdivided into a number of smaller programs or
subprograms. Each subprogram specifies one or more actions to be
performed for a large program. such subprograms are functions.
The function supports only static and extern storage classes. By
default, function assumes extern storage class. functions have global
scope. Only register or auto storage class is allowed in the function
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parameters. Built-in functions that predefined and supplied along with
the compiler are known as built-in functions. They are also known as
library functions.
Why should I prototype a function?
A function prototype tells the compiler what kind of arguments a
function is looking to receive and what kind of return value a function
is going to give back. This approach helps the compiler ensure that
calls to a function are made correctly and that no erroneous type
conversions are taking place.
What is Polymorphism ?
'Polymorphism' is an object oriented term. Polymorphism may be
defined as the ability of related objects to respond to the same
message with different, but appropriate actions. In other words,
polymorphism means taking more than one form. Polymorphism leads
to two important aspects in Object Oriented terminology - Function
Overloading and Function Overriding. Overloading is the practice of
supplying more than one definition for a given function name in the
same scope. The compiler is left to pick the appropriate version of the
function or operator based on the arguments with which it is called.
Overriding refers to the modifications made in the sub class to the
inherited methods from the base class to change their behavior.
What is Operator overloading ?
When an operator is overloaded, it takes on an additional meaning
relative to a certain class. But it can still retain all of its old meanings.
Examples:
1) The operators >> and << may be used for I/O operations because
in the header, they are overloaded.
2) In a stack class it is possible to overload the + operator so that it
appends the contents of one stack to the contents of another. But the
+ operator still retains its original meaning relative to other types of
data.
What is the difference between goto and longjmp() and setjmp()?
A goto statement implements a local jump of program execution, and
the longjmp() and setjmp() functions implement a nonlocal, or far,
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jump of program execution.
Generally, a jump in execution of any kind should be avoided because
it is not considered good programming practice to use such statements
as goto and longjmp in your program.
A goto statement simply bypasses code in your program and jumps to
a predefined position. To use the goto statement, you give it a labeled
position to jump to. This predefined position must be within the same
function. You cannot implement gotos between functions.
When your program calls setjmp(), the current state of your program is
saved in a structure of type jmp_buf. Later, your program can call the
longjmp() function to restore the program’s state as it was when you
called setjmp().Unlike the goto statement, the longjmp() and setjmp()
functions do not need to be implemented in the same function.
However, there is a major drawback to using these functions: your
program, when restored to its previously saved state, will lose its
references to any dynamically allocated memory between the
longjmp() and the setjmp(). This means you will waste memory for
every malloc() or calloc() you have implemented between your
longjmp() and setjmp(), and your program will be horribly inefficient.
It is highly recommended that you avoid using functions such as
longjmp() and setjmp() because they, like the goto statement, are
quite often an indication of poor programming practice.
V

c tutorials

2010-08-14T01:23:00.000-07:00

C Programming Tutorials
The C is a general-purpose, procedural, imperative computer programming language developed in 1972 by Dennis Ritchie at the Bell Telephone Laboratories for use with the Unix operating system.
C is the most widely used computer language.
This tutorial should be your starting point only.

Basic of C :
Facts about C
Why to use C ?
C Program File
C Compilers

Program Structure :
Simple C Program
C Program Compilation

Basic Datatypes :
DataTypes
Modifiers
Qualifiers
Arrays

Variable Types :
Local Variable
Global Variable

Storage Classes :
auto - storage class
register - storage class
static - storage class
extern - storage class

Using Constants :
Defining Constants
The enum Data Type

Operator Types :
Arithmetic Operators
Logical Operators
Bitwise Operators
Assignment Operators
Misc Operators

Control Statements :
Branching
Looping

Input and Output :
printf() function
scanf() function

Pointing to Data :
Pointers and Arrays
Pointer Arithmetic
Using Pointer Arithmetic with Arrays

Functions :
Using Functions
Declaration and Definition

Strings :
Reading and Writing Strings
String Manipulation Function

Structured Datatypes :
Structure
Pointer to Structure

Working with Files :
Files
Basic I/O

Bits :
Bits Manipulation
Bits Field

Pre-Processors :
Pre-Processors Examples
Parameterized Macros
Macro Caveats
Useful Concepts

Built-in Library Functions :
String Manipulation Functions
Memory Management Functions
Buffer Manipulation
Character Functions
Error Handling Functions

1.Basic Introduction
C is a general-purpose high level language that was originally developed by Dennis Ritchie for the Unix operating system. It was first implemented on the Digital Equipment Corporation PDP-11 computer in 1972.

The Unix operating system and virtually all Unix applications are written in the C language. C has now become a widely used professional language for various reasons.

Easy to learn

Structured language

It produces efficient programs.

It can handle low-level activities.

It can be compiled on a variety of computers.

Facts about C

C was invented to write an operating system called UNIX.

C is a successor of B language which was introduced around 1970

The language was formalized in 1988 by the American National Standard Institute (ANSI).

By 1973 UNIX OS almost totally written in C.

Today C is the most widely used System Programming Language.

Most of the state of the art software have been implemented using C

Why to use C ?
C was initially used for system development work, in particular the programs that make-up the operating system. C was adopted as a system development language because it produces code that runs nearly as fast as code written in assembly language. Some examples of the use of C might be:

Operating Systems

Language Compilers

Assemblers

Text Editors

Print Spoolers

Network Drivers

Modern Programs

Data Bases

Language Interpreters

Utilities

C Program File
All the C programs are written into text files with extension ".c" for example hello.c. You can use "vi" editor to write your C program into a file.

This tutorial assumes that you know how to edit a text file and how to write programming instructions inside a program file.

C Compilers
When you write any program in C language then to run that program you need to compile that program using a C Compiler which converts your program into a language understandable by a computer. This is called machine language (ie. binary format). So before proceeding, make sure you have C Compiler available at your computer. It comes along with all flavors of Unix and Linux.

If you are working over Unix or Linux then you can type gcc -v or cc -v and check the result. You can ask your system administrator or you can take help from anyone to identify an available C Compiler at your computer.

If you don't have C compiler installed at your computer then you can use below given link to download a GNU C Compiler and use it
2.Program Structure
A C program basically has the following form:

Preprocessor Commands

Functions

Variables

Statements & Expressions

Comments

The following program is written in the C programming language. Open a text file hello.c using vi editor and put the following lines inside that file.

#include

int main()
{
/* My first program */
printf("Hello, TechPreparation! \n");

return 0;
}

Preprocessor Commands:
These commands tells the compiler to do preprocessing before doing actual compilation. Like #include is a preprocessor command which tells a C compiler to include stdio.h file before going to actual compilation. You will learn more about C Preprocessors in C Preprocessors session.

Functions:
Functions are main building blocks of any C Program. Every C Program will have one or more functions and there is one mandatory function which is called main() function. This function is prefixed with keyword int which means this function returns an integer value when it exits. This integer value is returned using return statement.

The C Programming language provides a set of built-in functions. In the above example printf() is a C built-in function which is used to print anything on the screen.

Variables:
Variables are used to hold numbers, strings and complex data for manipulation. You will learn in detail about variables in C Variable Types.

Statements & Expressions :
Expressions combine variables and constants to create new values. Statements are expressions, assignments, function calls, or control flow statements which make up C programs.

Comments:
Comments are used to give additional useful information inside a C Program. All the comments will be put inside /*...*/ as given in the example above. A comment can span through multiple lines.

Note the followings

C is a case sensitive programming language. It means in C printf and Printf will have different meanings.

C has a free-form line structure. End of each C statement must be marked with a semicolon.

Multiple statements can be one the same line.

White Spaces (ie tab space and space bar ) are ignored.

Statements can continue over multiple lines.

C Program Compilation
To compile a C program you would have to Compiler name and program files name. Assuming your compiler's name is cc and program file name is hello.c, give following command at Unix prompt.

$cc hello.c

This will produce a binary file called a.out and an object file hello.o in your current directory. Here a.out is your first program which you will run at Unix prompt like any other system program. If you don't like the name a.out then you can produce a binary file with your own name by using -o option while compiling C program. See an example below

$cc -o hello hello.c

Now you will get a binary with name hello. Execute this program at Unix prompt but before executing / running this program make sure that it has execute permission set. If you don't know what is execute permission then just follow these two steps

$chmod 755 hello
$./hello

This will produce following result
Hello, TechPreparation!

Congratulations!! you have written your first program in "C". Now believe me its not difficult to learn "C".

3.Basic Datatypes
C has a concept of 'data types' which are used to define a variable before its use. The definition of a variable will assign storage for the variable and define the type of data that will be held in the location.

The value of a variable can be changed any time.

C has the following basic built-in datatypes.

int

float

double

char

Please note that there is not a Boolean data type. C does not have the traditional view about logical comparison, but that's another story.

int - data type
int is used to define integer numbers.

{
int Count;
Count = 5;
}

float - data type
float is used to define floating point numbers.

{
float Miles;
Miles = 5.6;
}

double - data type
double is used to define BIG floating point numbers. It reserves twice the storage for the number. On PCs this is likely to be 8 bytes.

{
double Atoms;
Atoms = 2500000;
}

char - data type
char defines characters.

{
char Letter;
Letter = 'x';
}

Modifiers
The data types explained above have the following modifiers.

short

long

signed

unsigned

The modifiers define the amount of storage allocated to the variable. The amount of storage allocated is not cast in stone. ANSI has the following rules:

short int <= int <= long intfloat <= double <= long double What this means is that a 'short int' should assign less than or the same amount of storage as an 'int' and the 'int' should be less or the same bytes than a 'long int'. What this means in the real world is: Type Bytes Range short int 2 -32,768 -> +32,767 (32kb)
unsigned short int 2 0 -> +65,535 (64Kb)
unsigned int 4 0 -> +4,294,967,295 ( 4Gb)
int 4 -2,147,483,648 -> +2,147,483,647 ( 2Gb)
long int 4 -2,147,483,648 -> +2,147,483,647 ( 2Gb)
signed char 1 -128 -> +127
unsigned char 1 0 -> +255
float 4
double 8
long double 12

These figures only apply to today's generation of PCs. Mainframes and midrange machines could use different figures, but would still comply with the rule above.

You can find out how much storage is allocated to a data type by using the sizeof operator discussed in Operator Types Session.

Here is an example to check size of memory taken by various datatypes.

int
main()
{
printf("sizeof(char) == %d\n", sizeof(char));
printf("sizeof(short) == %d\n", sizeof(short));
printf("sizeof(int) == %d\n", sizeof(int));
printf("sizeof(long) == %d\n", sizeof(long));
printf("sizeof(float) == %d\n", sizeof(float));
printf("sizeof(double) == %d\n", sizeof(double));
printf("sizeof(long double) == %d\n", sizeof(long double));
printf("sizeof(long long) == %d\n", sizeof(long long));

return 0;
}
4.Qualifiers
A type qualifier is used to refine the declaration of a variable, a function, and parameters, by specifying whether:

The value of a variable can be changed.

The value of a variable must always be read from memory rather than from a register

Standard C language recognizes the following two qualifiers:

const

volatile

The const qualifier is used to tell C that the variable value can not change after initialization.
const float pi=3.14159;

Now pi cannot be changed at a later time within the program.

Another way to define constants is with the #define preprocessor which has the advantage that it does not use any storage

The volatile qualifier declares a data type that can have its value changed in ways outside the control or detection of the compiler (such as a variable updated by the system clock or by another program). This prevents the compiler from optimizing code referring to the object by storing the object's value in a register and re-reading it from there, rather than from memory, where it may have changed. You will use this qualifier once you will become expert in "C". So for now just proceed.

What are Arrays:
We have seen all basic data types. In C language it is possible to make arrays whose elements are basic types. Thus we can make an array of 10 integers with the declaration.

int x[10];

The square brackets mean subscripting; parentheses are used only for function references. Array indexes begin at zero, so the elements of x are:

Thus Array are special type of variables which can be used to store multiple values of same data type. Those values are stored and accessed using subscript or index.

Arrays occupy consecutive memory slots in the computer's memory.

x[0], x[1], x[2], ..., x[9]

If an array has n elements, the largest subscript is n-1.

Multiple-dimension arrays are provided. The declaration and use look like:

int name[10] [20];
n = name[i+j] [1] + name[k] [2];

Subscripts can be arbitrary integer expressions. Multi-dimension arrays are stored by row so the rightmost subscript varies fastest. In above example name has 10 rows and 20 columns.

Same way, arrays can be defined for any data type. Text is usually kept as an array of characters. By convention in C, the last character in a character array should be a `\0' because most programs that manipulate character arrays expect it. For example, printf uses the `\0' to detect the end of a character array when printing it out with a `%s'.

Here is a program which reads a line, stores it in a buffer, and prints its length (excluding the newline at the end).

main( )
{
int n, c;
char line[100];
n = 0;
while( (c=getchar( )) != '\n' )
{
if( n < 100 )line[n] = c;n++;}printf("length = %d\n", n);} Array InitializationAs with other declarations, array declarations can include an optional initializationScalar variables are initialized with a single valueArrays are initialized with a list of valuesThe list is enclosed in curly bracesint array [8] = {2, 4, 6, 8, 10, 12, 14, 16}; The number of initializes cannot be more than the number of elements in the array but it can be less in which case, the remaining elements are initialized to 0.if you like, the array size can be inferred from the number of initializes by leaving the square brackets empty so these are identical declarations:int array1 [8] = {2, 4, 6, 8, 10, 12, 14, 16};int array2 [] = {2, 4, 6, 8, 10, 12, 14, 16}; An array of characters ie string can be initialized as follows:char string[10] = "Hello"; 5.Variable TypesA variable is just a named area of storage that can hold a single value (numeric or character). The C language demands that you declare the name of each variable that you are going to use and its type, or class, before you actually try to do anything with it.The Programming language C has two main variable typesLocal VariablesGlobal VariablesLocal VariablesLocal variables scope is confined within the block or function where it is defined. Local variables must always be defined at the top of a block.When a local variable is defined - it is not initialized by the system, you must initialize it yourself.When execution of the block starts the variable is available, and when the block ends the variable 'dies'.Check following example's output main(){int i=4;int j=10;i++;if (j > 0)
{
/* i defined in 'main' can be seen */
printf("i is %d\n",i);
}

if (j > 0)
{
/* 'i' is defined and so local to this block */
int i=100;
printf("i is %d\n",i);
}/* 'i' (value 100) dies here */

printf("i is %d\n",i); /* 'i' (value 5) is now visable.*/
}

This will generate following output
i is 5
i is 100
i is 5

Here ++ is called incremental operator and it increase the value of any integer variable by 1. Thus i++ is equivalent to i = i + 1;

You will see -- operator also which is called decremental operator and it decrease the value of any integer variable by 1. Thus i-- is equivalent to i = i - 1;

Global Variables
Global variable is defined at the top of the program file and it can be visible and modified by any function that may reference it.

Global variables are initialized automatically by the system when you define them!

Data Type Initialser
int 0
char '\0'
float 0
pointer NULL

If same variable name is being used for global and local variable then local variable takes preference in its scope. But it is not a good practice to use global variables and local variables with the same name.

int i=4; /* Global definition */

main()
{
i++; /* Global variable */
func();
printf( "Value of i = %d -- main function\n", i );
}

func()
{
int i=10; /* Local definition */
i++; /* Local variable */
printf( "Value of i = %d -- func() function\n", i );
}

This will produce following result
Value of i = 11 -- func() function
Value of i = 5 -- main function

i in main function is global and will be incremented to 5. i in func is internal and will be incremented to 11. When control returns to main the internal variable will die and and any reference to i will be to the global.

6.Storage Classes
A storage class defines the scope (visibility) and life time of variables and/or functions within a C Program.

There are following storage classes which can be used in a C Program

auto

register

static

extern

auto - Storage Class
auto is the default storage class for all local variables.

{
int Count;
auto int Month;
}

The example above defines two variables with the same storage class. auto can only be used within functions, i.e. local variables.

register - Storage Class
register is used to define local variables that should be stored in a register instead of RAM. This means that the variable has a maximum size equal to the register size (usually one word) and cant have the unary '&' operator applied to it (as it does not have a memory location).

{
register int Miles;
}

Register should only be used for variables that require quick access - such as counters. It should also be noted that defining 'register' goes not mean that the variable will be stored in a register. It means that it MIGHT be stored in a register - depending on hardware and implementation restrictions.

static - Storage Class
static is the default storage class for global variables. The two variables below (count and road) both have a static storage class.

static int Count;
int Road;

{
printf("%d\n", Road);
}

static variables can be 'seen' within all functions in this source file. At link time, the static variables defined here will not be seen by the object modules that are brought in.

static can also be defined within a function. If this is done the variable is initialized at run time but is not reinitialized when the function is called. This inside a function static variable retains its value during various calls.

void func(void);

static count=10; /* Global variable - static is the default */

main()
{
while (count--)
{
func();
}

}

void func( void )
{
static i = 5;
i++;
printf("i is %d and count is %d\n", i, count);
}

This will produce following result

i is 6 and count is 9
i is 7 and count is 8
i is 8 and count is 7
i is 9 and count is 6
i is 10 and count is 5
i is 11 and count is 4
i is 12 and count is 3
i is 13 and count is 2
i is 14 and count is 1
i is 15 and count is 0

static variables can be 'seen' within all functions in this source file. At link time, the static variables defined here will not be seen by the object modules that are brought in.

static can also be defined within a function. If this is done the variable is initialized at run time but is not reinitialized when the function is called. This inside a function static variable retains its value during various calls.

void func(void);

static count=10; /* Global variable - static is the default */

main()
{
while (count--)
{
func();
}

}

void func( void )
{
static i = 5;
i++;
printf("i is %d and count is %d\n", i, count);
}

This will produce following result

i is 6 and count is 9
i is 7 and count is 8
i is 8 and count is 7
i is 9 and count is 6
i is 10 and count is 5
i is 11 and count is 4
i is 12 and count is 3
i is 13 and count is 2
i is 14 and count is 1
i is 15 and count is 0

NOTE : Here keyword void means function does not return anything and it does not take any parameter. You can memories void as nothing. static variables are initialized to 0 automatically.

Definition vs. Declaration :
Before proceeding, let us understand the difference between definition and declaration of a variable or function. Definition means where a variable or function is defined in reality and actual memory is allocated for variable or function. Declaration means just giving a reference of a variable and function. Through declaration we assure to the complier that this variable or function has been defined somewhere else in the program and will be provided at the time of linking. In the above examples char *func(void) has been put at the top which is a declaration of this function where as this function has been defined below to main() function.

There is one more very important use for 'static'. Consider this bit of code.

char *func(void);

main()
{
char *Text1;
Text1 = func();
}

char *func(void)
{
char Text2[10]="martin";
return(Text2);
}

Now, 'func' returns a pointer to the memory location where 'text2' starts BUT text2 has a storage class of 'auto' and will disappear when we exit the function and could be overwritten but something else. The answer is to specify

static char Text[10]="martin";

The storage assigned to 'text2' will remain reserved for the duration if the program.

extern - Storage Class
extern is used to give a reference of a global variable that is visible to ALL the program files. When you use 'extern' the variable cannot be initialized as all it does is point the variable name at a storage location that has been previously defined.

When you have multiple files and you define a global variable or function which will be used in other files also, then extern will be used in another file to give reference of defined variable or function. Just for understanding extern is used to declare a global variable or function in another files.

File 1: main.c

int count=5;

main()
{
write_extern();
}

File 2: write.c

void write_extern(void);

extern int count;

void write_extern(void)
{
printf("count is %i\n", count);
}

Here extern keyword is being used to declare count in another file.

Now compile these two files as follows

gcc main.c write.c -o write

This fill produce write program which can be executed to produce result.

Count in 'main.c' will have a value of 5. If main.c changes the value of count - write.c will see the new value

NEXT >> Using Constants

7.Using Constants
A C constant is usually just the written version of a number. For example 1, 0, 5.73, 12.5e9. We can specify our constants in octal or hexadecimal, or force them to be treated as long integers.

Octal constants are written with a leading zero - 015.

Hexadecimal constants are written with a leading 0x - 0x1ae.

Long constants are written with a trailing L - 890L.

Character constants are usually just the character enclosed in single quotes; 'a', 'b', 'c'. Some characters can't be represented in this way, so we use a 2 character sequence as follows.

'\n' newline
'\t' tab
'\\' backslash
'\'' single quote
'\0' null ( Used automatically to terminate character string )

In addition, a required bit pattern can be specified using its octal equivalent.

'\044' produces bit pattern 00100100.

Character constants are rarely used, since string constants are more convenient. A string constant is surrounded by double quotes eg "Brian and Dennis". The string is actually stored as an array of characters. The null character '\0' is automatically placed at the end of such a string to act as a string terminator.

A character is a different type to a single character string. This is important point to note.

Defining Constants
ANSI C allows you to declare constants. When you declare a constant it is a bit like a variable declaration except the value cannot be changed.

The const keyword is to declare a constant, as shown below:

int const a = 1;
const int a =2;

Note:
You can declare the const before or after the type. Choose one an stick to it.
It is usual to initialize a const with a value as it cannot get a value any other way.

The preprocessor #define is another more flexible (see Preprocessor Chapters) method to define constants in a program.

#define TRUE 1
#define FALSE 0
#define NAME_SIZE 20

Here TRUE, FALSE and NAME_SIZE are constant

You frequently see const declaration in function parameters. This says simply that the function is not going to change the value of the parameter.

The following function definition used concepts we have not met (see chapters on functions, strings, pointers, and standard libraries) but for completeness of this section it is is included here:

void strcpy(char *buffer, char const *string)

The enum Data type
enum is the abbreviation for ENUMERATE, and we can use this keyword to declare and initialize a sequence of integer constants. Here's an example:

enum colors {RED, YELLOW, GREEN, BLUE};

I've made the constant names uppercase, but you can name them which ever way you want.

Here, colors is the name given to the set of constants - the name is optional. Now, if you don't assign a value to a constant, the default value for the first one in the list - RED in our case, has the value of 0. The rest of the undefined constants have a value 1 more than the one before, so in our case, YELLOW is 1, GREEN is 2 and BLUE is 3.

But you can assign values if you wanted to:

enum colors {RED=1, YELLOW, GREEN=6, BLUE };

Now RED=1, YELLOW=2, GREEN=6 and BLUE=7.

The main advantage of enum is that if you don't initialize your constants, each one would have a unique value. The first would be zero and the rest would then count upwards.

You can name your constants in a weird order if you really wanted...

#include

int main() {
enum {RED=5, YELLOW, GREEN=4, BLUE};

printf("RED = %d\n", RED);
printf("YELLOW = %d\n", YELLOW);
printf("GREEN = %d\n", GREEN);
printf("BLUE = %d\n", BLUE);
return 0;
}

This will produce following results

RED = 5
YELLOW = 6
GREEN = 4
BLUE = 5
8.Operator Types
What is Operator? Simple answer can be given using expression 4 + 5 is equal to 9. Here 4 and 5 are called operands and + is called operator. C language supports following type of operators.

Arithmetic Operators

Logical (or Relational) Operators

Bitwise Operators

Assignment Operators

Misc Operators

Lets have a look on all operators one by one.

Arithmetic Operators:
There are following arithmetic operators supported by C language:

Assume variable A holds 10 and variable holds 20 then:
Show Examples

Operator Description Example
+ Adds two operands A + B will give 30
- Subtracts second operand from the first A - B will give -10
* Multiply both operands A * B will give 200
/ Divide numerator by denominator B / A will give 2
% Modulus Operator and remainder of after an integer division B % A will give 0
++ Increment operator, increases integer value by one A++ will give 11
-- Decrement operator, decreases integer value by one A-- will give 9

Logical (or Relational) Operators:
There are following logical operators supported by C language

Assume variable A holds 10 and variable holds 20 then:
Show Examples

Operator Description Example
== Checks if the value of two operands is equal or not, if yes then condition becomes true. (A == B) is not true.
!= Checks if the value of two operands is equal or not, if values are not equal then condition becomes true. (A != B) is true.
> Checks if the value of left operand is greater than the value of right operand, if yes then condition becomes true. (A > B) is not true.
< Checks if the value of left operand is less than the value of right operand, if yes then condition becomes true. (A < B) is true. >= Checks if the value of left operand is greater than or equal to the value of right operand, if yes then condition becomes true. (A >= B) is not true.
<= Checks if the value of left operand is less than or equal to the value of right operand, if yes then condition becomes true. (A <= B) is true. && Called Logical AND operator. If both the operands are non zero then then condition becomes true. (A && B) is true. || Called Logical OR Operator. If any of the two operands is non zero then then condition becomes true. (A || B) is true. ! Called Logical NOT Operator. Use to reverses the logical state of its operand. If a condition is true then Logical NOT operator will make false. !(A && B) is false. 8.Bitwise Operators:Bitwise operator works on bits and perform bit by bit operation.Assume if B = 60; and B = 13; Now in binary format they will be as follows:A = 0011 1100B = 0000 1101-----------------A&B = 0000 1000A|B = 0011 1101A^B = 0011 0001~A = 1100 0011Show ExamplesThere are following Bitwise operators supported by C languageOperator Description Example & Binary AND Operator copies a bit to the result if it exists in both operands. (A & B) will give 12 which is 0000 1100 | Binary OR Operator copies a bit if it exists in either operand. (A | B) will give 61 which is 0011 1101 ^ Binary XOR Operator copies the bit if it is set in one operand but not both. (A ^ B) will give 49 which is 0011 0001 ~ Binary Ones Complement Operator is unary and has the effect of 'flipping' bits. (~A ) will give -60 which is 1100 0011 << Binary Left Shift Operator. The left operands value is moved left by the number of bits specified by the right operand. A << 2 will give 240 which is 1111 0000 >> Binary Right Shift Operator. The left operands value is moved right by the number of bits specified by the right operand. A >> 2 will give 15 which is 0000 1111

Assignment Operators:
There are following assignment operators supported by C language:
Show Examples

Operator Description Example
= Simple assignment operator, Assigns values from right side operands to left side operand C = A + B will assign value of A + B into C
+= Add AND assignment operator, It adds right operand to the left operand and assign the result to left operand C += A is equivalent to C = C + A
-= Subtract AND assignment operator, It subtracts right operand from the left operand and assign the result to left operand C -= A is equivalent to C = C - A
*= Multiply AND assignment operator, It multiplies right operand with the left operand and assign the result to left operand C *= A is equivalent to C = C * A
/= Divide AND assignment operator, It divides left operand with the right operand and assign the result to left operand C /= A is equivalent to C = C / A
%= Modulus AND assignment operator, It takes modulus using two operands and assign the result to left operand C %= A is equivalent to C = C % A
<<= Left shift AND assignment operator C <<= 2 is same as C = C << 2 >>= Right shift AND assignment operator C >>= 2 is same as C = C >> 2
&= Bitwise AND assignment operator C &= 2 is same as C = C & 2
^= bitwise exclusive OR and assignment operator C ^= 2 is same as C = C ^ 2
|= bitwise inclusive OR and assignment operator C |= 2 is same as C = C | 2

Short Notes on L-VALUE and R-VALUE:
x = 1; takes the value on the right (e.g. 1) and puts it in the memory referenced by x. Here x and 1 are known as L-VALUES and R-VALUES respectively L-values can be on either side of the assignment operator where as R-values only appear on the right.

So x is an L-value because it can appear on the left as we've just seen, or on the right like this: y = x; However, constants like 1 are R-values because 1 could appear on the right, but 1 = x; is invalid.

Misc Operators
There are few other operators supported by C Language.
Show Examples

Operator Description Example
sizeof() Returns the size of an variable. sizeof(a), where a is integer, will return 4.
& Returns the address of an variable. &a; will give actual address of the variable.
* Pointer to a variable. *a; will pointer to a variable
? : Conditional Expression If Condition is true ? Then value X : Otherwise value Y

Operators Categories:
All the operators we have discussed above can be categorized into following categories:

Postfix operators, which follow a single operand.

Unary prefix operators, which precede a single operand.

Binary operators, which take two operands and perform a variety of arithmetic and logical operations.

The conditional operator (a ternary operator), which takes three operands and evaluates either the second or third expression, depending on the evaluation of the first expression.

Assignment operators, which assign a value to a variable.

The comma operator, which guarantees left-to-right evaluation of comma-separated expressions.

Precedence of C Operators:
Operator precedence determines the grouping of terms in an expression. This affects how an expression is evaluated. Certain operators have higher precedence than others; for example, the multiplication operator has higher precedence than the addition operator:

For example x = 7 + 3 * 2; Here x is assigned 13, not 20 because operator * has higher precedence than + so it first get multiplied with 3*2 and then adds into 7.

Here operators with the highest precedence appear at the top of the table, those with the lowest appear at the bottom. Within an expression, higher precedence operators will be evaluated first.

Category Operator Associativity
Postfix () [] -> . ++ - - Left to right
Unary + - ! ~ ++ - - (type) * & sizeof Right to left
Multiplicative * / % Left to right
Additive + - Left to right
Shift << >> Left to right
Relational < <= > >= Left to right
Equality == != Left to right
Bitwise AND & Left to right
Bitwise XOR ^ Left to right
Bitwise OR | Left to right
Logical AND && Left to right
Logical OR || Left to right
Conditional ?: Right to left
Assignment = += -= *= /= %= >>= <<= &= ^= |= Right to left Comma , Left to right 9.Flow Control StatementsC provides two styles of flow control:BranchingLoopingBranching is deciding what actions to take and looping is deciding how many times to take a certain action.Branching:Branching is so called because the program chooses to follow one branch or another.if statementThis is the most simple form of the branching statements.It takes an expression in parenthesis and an statement or block of statements. if the expression is true then the statement or block of statements gets executed otherwise these statements are skipped.NOTE: Expression will be assumed to be true if its evaluated values is non-zero.if statements take the following form:Show Exampleif (expression)statement;orif (expression){Block of statements;}orif (expression){Block of statements;}else{Block of statements;}orif (expression){Block of statements;}else if(expression){Block of statements;}else{Block of statements;} ? : OperatorThe ? : operator is just like an if ... else statement except that because it is an operator you can use it within expressions.? : is a ternary operator in that it takes three values, this is the only ternary operator C has.? : takes the following form:Show Exampleif condition is true ? then X return value : otherwise Y value; switch statement:The switch statement is much like a nested if .. else statement. Its mostly a matter of preference which you use, switch statement can be slightly more efficient and easier to read.Show Exampleswitch( expression ){case constant-expression1: statements1;[case constant-expression2: statements2;] [case constant-expression3: statements3;][default : statements4;]} Using break keyword:If a condition is met in switch case then execution continues on into the next case clause also if it is not explicitly specified that the execution should exit the switch statement. This is achieved by using break keyword.What is default condition:If none of the listed conditions is met then default condition executed.10.LoopingLoops provide a way to repeat commands and control how many times they are repeated. C provides a number of looping way.while loopThe most basic loop in C is the while loop. A while statement is like a repeating if statement. Like an If statement, if the test condition is true: the statements get executed. The difference is that after the statements have been executed, the test condition is checked again. If it is still true the statements get executed again. This cycle repeats until the test condition evaluates to false.Basic syntax of while loop is as follows:Show Examplewhile ( expression ){Single statement orBlock of statements;} for loopfor loop is similar to while, it's just written differently. for statements are often used to process lists such a range of numbers:Basic syntax of for loop is as follows:Show Examplefor( expression1; expression2; expression3){Single statementorBlock of statements;} In the above syntax:expression1 - Initializes variables.expression2 - Conditional expression, as long as this condition is true, loop will keep executing.expression3 - expression3 is the modifier which may be simple increment of a variable.do...while loopdo ... while is just like a while loop except that the test condition is checked at the end of the loop rather than the start. This has the effect that the content of the loop are always executed at least once.Basic syntax of do...while loop is as follows:Show Exampledo{Single statementorBlock of statements;}while(expression); break and continue statementsC provides two commands to control how we loop:break -- exit form loop or switch.continue -- skip 1 iteration of loop.You already have seen example of using break statement. Here is an example showing usage of continue statement.#include main(){int i;int j = 10;for( i = 0; i <= j; i ++ ){if( i == 5 ){continue;}printf("Hello %d\n", i );}} This will produce following output:Hello 0Hello 1Hello 2Hello 3Hello 4Hello 6Hello 7Hello 8Hello 9Hello 10 11.Input and OutputInput : In any programming language input means to feed some data into program. This can be given in the form of file or from command line. C programming language provides a set of built-in functions to read given input and feed it to the program as per requirement.Output : In any programming language output means to display some data on screen, printer or in any file. C programming language provides a set of built-in functions to output required data.printf() functionThis is one of the most frequently used functions in C for output. ( we will discuss what is function in subsequent chapter. ).Try following program to understand printf() function.#include

main()
{
int dec = 5;
char str[] = "abc";
char ch = 's';
float pi = 3.14;

printf("%d %s %f %c\n", dec, str, pi, ch);
}

The output of the above would be:

5 abc 3.140000 c

Here %d is being used to print an integer, %s is being usedto print a string, %f is being used to print a float and %c is being used to print a character.

scanf() function
This is the function which can be used to to read an input from the command line.

Try following program to understand scanf() function.

#include

main()
{
int x;
int args;

printf("Enter an integer: ");
if (( args = scanf("%d", &x)) == 0)
{
printf("Error: not an integer\n");
} else {
printf("Read in %d\n", x);
}
}

Here %d is being used to read an integer value and we are passing &x to store the vale read input. Here &indicates the address of variable x.

This program will prompt you to enter a value. Whatever value you will enter at command prompt that will be output at the screen using printf() function. If you enter a non-integer value then it will display an error message.

Enter an integer: 20
Read in 20
13.Pointing to Data
A pointer is a special kind of variable. Pointers are designed for storing memory address i.e. the address of another variable. Declaring a pointer is the same as declaring a normal variable except you stick an asterisk '*' in front of the variables identifier.

There are two new operators you will need to know to work with pointers. The "address of" operator '&' and the "dereferencing" operator '*'. Both are prefix unary operators.

When you place an ampersand in front of a variable you will get it's address, this can be stored in a pointer variable.

When you place an asterisk in front of a pointer you will get the value at the memory address pointed to.

Here is an example to understand what I have stated above.

#include

int main()
{
int my_variable = 6, other_variable = 10;
int *my_pointer;

printf("the address of my_variable is : %p\n", &my_variable);
printf("the address of other_variable is : %p\n", &other_variable);

my_pointer = &my_variable;

printf("\nafter \"my_pointer = &my_variable\":\n");
printf("\tthe value of my_pointer is %p\n", my_pointer);
printf("\tthe value at that address is %d\n", *my_pointer);

my_pointer = &other_variable;

printf("\nafter \"my_pointer = &other_variable\":\n");
printf("\tthe value of my_pointer is %p\n", my_pointer);
printf("\tthe value at that address is %d\n", *my_pointer);

return 0;
}

This will produce following result.

the address of my_variable is : 0xbfffdac4
the address of other_variable is : 0xbfffdac0

after "my_pointer = &my_variable":
the value of my_pointer is 0xbfffdac4
the value at that address is 6

after "my_pointer = &other_variable":
the value of my_pointer is 0xbfffdac0
the value at that address is 10

Pointers and Arrays
The most frequent use of pointers in C is for walking efficiently along arrays. In fact, in the implementation of an array, the array name represents the address of the zeroth element of the array, so you can't use it on the left side of an expression. For example:

char *y;
char x[100];

y is of type pointer to character (although it doesn't yet point anywhere). We can make y point to an element of x by either of

y = &x[0];
y = x;

Since x is the address of x[0] this is legal and consistent. Now `*y' gives x[0]. More importantly notice the following:

*(y+1) gives x[1]
*(y+i) gives x[i]

and the sequence

y = &x[0];
y++;

leaves y pointing at x[1].

Pointer Arithmetic:
C is one of the few languages that allows pointer arithmetic. In other words, you actually move the pointer reference by an arithmetic operation. For example:

int x = 5, *ip = &x;

ip++;

On a typical 32-bit machine, *ip would be pointing to 5 after initialization. But ip++; increments the pointer 32-bits or 4-bytes. So whatever was in the next 4-bytes, *ip would be pointing at it.

Pointer arithmetic is very useful when dealing with arrays, because arrays and pointers share a special relationship in C.

15.Using Pointer Arithmetic With Arrays:
Arrays occupy consecutive memory slots in the computer's memory. This is where pointer arithmetic comes in handy - if you create a pointer to the first element, incrementing it one step will make it point to the next element.

#include

int main() {
int *ptr;
int arrayInts[10] = {1,2,3,4,5,6,7,8,9,10};

ptr = arrayInts; /* ptr = &arrayInts[0]; is also fine */

printf("The pointer is pointing to the first ");
printf("array element, which is %d.\n", *ptr);
printf("Let's increment it.....\n");

ptr++;

printf("Now it should point to the next element,");
printf(" which is %d.\n", *ptr);
printf("But suppose we point to the 3rd and 4th: %d %d.\n",
*(ptr+1),*(ptr+2));

ptr+=2;

printf("Now skip the next 4 to point to the 8th: %d.\n",
*(ptr+=4));

ptr--;

printf("Did I miss out my lucky number %d?!\n", *(ptr++));
printf("Back to the 8th it is then..... %d.\n", *ptr);

return 0;
}

This will produce following result:

The pointer is pointing to the first array element, which is 1.
Let's increment it.....
Now it should point to the next element, which is 2.
But suppose we point to the 3rd and 4th: 3 4.
Now skip the next 4 to point to the 8th: 8.
Did I miss out my lucky number 7?!
Back to the 8th it is then..... 8.

See more examples on Pointers and Array

Modifying Variables Using Pointers:
You know how to access the value pointed to using the dereference operator, but you can also modify the content of variables. To achieve this, put the dereferenced pointer on the left of the assignment operator, as shown in this example, which uses an array:

#include

int main() {
char *ptr;
char arrayChars[8] = {'F','r','i','e','n','d','s','\0'};

ptr = arrayChars;

printf("The array reads %s.\n", arrayChars);
printf("Let's change it..... ");

*ptr = 'f'; /* ptr points to the first element */

printf(" now it reads %s.\n", arrayChars);
printf("The 3rd character of the array is %c.\n",
*(ptr+=2));
printf("Let's change it again..... ");

*(ptr - 1) = ' ';

printf("Now it reads %s.\n", arrayChars);
return 0;
}

This will produce following result:

The array reads Friends.
Let's change it..... now it reads friends.
The 3rd character of the array is i.
Let's change it again..... Now it reads f iends.

Generic Pointers: ( void Pointer )
When a variable is declared as being a pointer to type void it is known as a generic pointer. Since you cannot have a variable of type void, the pointer will not point to any data and therefore cannot be dereferenced. It is still a pointer though, to use it you just have to cast it to another kind of pointer first. Hence the term Generic pointer. This is very useful when you want a pointer to point to data of different types at different times.

Try the following code to understand Generic Pointers.

#include

int main()
{
int i;
char c;
void *the_data;

i = 6;
c = 'a';

the_data = &i;
printf("the_data points to the integer value %d\n",
*(int*) the_data);

the_data = &c;
printf("the_data now points to the character %c\n",
*(char*) the_data);

return 0;
}

NOTE-1 : Here in first print statement, the_data is prefixed by *(int*). This is called type casting in C language. Type is used to caste a variable from one data type to another datatype to make it compatible to the lvalue.

NOTE-2 : lvalue is something which is used to left side of a statement and in which we can assign some value. A constant can't be an lvalue because we can not assign any value in contact. For example x = y, here x is lvalue and y is rvalue.

However, above example will produce following result:

the_data points to the integer value 6
the_data now points to the character a

NEXT >> Using Functions

16.Using Functions
A function is a module or block of program code which deals with a particular task. Making functions is a way of isolating one block of code from other independent blocks of code.

Functions serve two purposes.

They allow a programmer to say: `this piece of code does a specific job which stands by itself and should not be mixed up with anything else',

Second they make a block of code reusable since a function can be reused in many different contexts without repeating parts of the program text.

A function can take a number of parameters, do required processing and then return a value. There may be a function which does not return any value.

You already have seen couple of built-in functions like printf(); Similar way you can define your own functions in C language.

Consider the following chunk of code

int total = 10;
printf("Hello World");
total = total + l;

To turn it into a function you simply wrap the code in a pair of curly brackets to convert it into a single compound statement and write the name that you want to give it in front of the brackets:

Demo()
{
int total = 10;
printf("Hello World");
total = total + l;
}

curved brackets after the function's name are required. You can pass one or more parameters to a function as follows:

Demo( int par1, int par2)
{
int total = 10;
printf("Hello World");
total = total + l;
}

By default function does not return anything. But you can make a function to return any value as follows:

int Demo( int par1, int par2)
{
int total = 10;
printf("Hello World");
total = total + l;

return total;
}

A return keyword is used to return a value and datatype of the returned value is specified before the name of function. In this case function returns total which is int type. If a function does not return a value then void keyword can be used as return value.

Once you have defined your function you can use it within a program:

main()
{
Demo();
}

Functions and Variables:
Each function behaves the same way as C language standard function main(). So a function will have its own local variables defined. In the above example total variable is local to the function Demo.

A global variable can be accessed in any function in similar way it is accessed in main() function
17.Declaration and Definition
When a function is defined at any place in the program then it is called function definition. At the time of definition of a function actual logic is implemented with-in the function.

A function declaration does not have any body and they just have their interfaces.

A function declaration is usually declared at the top of a C source file, or in a separate header file.

A function declaration is sometime called function prototype or function signature. For the above Demo() function which returns an integer, and takes two parameters a function declaration will be as follows:

int Demo( int par1, int par2);

Passing Parameters to a Function
There are two ways to pass parameters to a function:

Pass by Value: mechanism is used when you don't want to change the value of passed parameters. When parameters are passed by value then functions in C create copies of the passed in variables and do required processing on these copied variables.

Pass by Reference mechanism is used when you want a function to do the changes in passed parameters and reflect those changes back to the calling function. In this case only addresses of the variables are passed to a function so that function can work directly over the addresses.

Here are two programs to understand the difference: First example is for Pass by value:

#include

/* function declaration goes here.*/
void swap( int p1, int p2 );

int main()
{
int a = 10;
int b = 20;

printf("Before: Value of a = %d and value of b = %d\n", a, b );
swap( a, b );
printf("After: Value of a = %d and value of b = %d\n", a, b );
}

void swap( int p1, int p2 )
{
int t;

t = p2;
p2 = p1;
p1 = t;
printf("Value of a (p1) = %d and value of b(p2) = %d\n", p1, p2 );
}

Here is the result produced by the above example. Here the values of a and b remain unchanged before calling swap function and after calling swap function.

Before: Value of a = 10 and value of b = 20
Value of a (p1) = 20 and value of b(p2) = 10
After: Value of a = 10 and value of b = 20

Following is the example which demonstrate the concept of pass by reference

#include

/* function declaration goes here.*/
void swap( int *p1, int *p2 );

int main()
{
int a = 10;
int b = 20;

printf("Before: Value of a = %d and value of b = %d\n", a, b );
swap( &a, &b );
printf("After: Value of a = %d and value of b = %d\n", a, b );
}

void swap( int *p1, int *p2 )
{
int t;

t = *p2;
*p2 = *p1;
*p1 = t;
printf("Value of a (p1) = %d and value of b(p2) = %d\n", *p1, *p2 );
}

Here is the result produced by the above example. Here the values of a and b are changes after calling swap function.

Before: Value of a = 10 and value of b = 20
Value of a (p1) = 20 and value of b(p2) = 10
After: Value of a = 20 and value of b = 10
18.Strings

In C language Strings are defined as an array of characters or a pointer to a portion of memory containing ASCII characters. A string in C is a sequence of zero or more characters followed by a NULL '\0' character:

It is important to preserve the NULL terminating character as it is how C defines and manages variable length strings. All the C standard library functions require this for successful operation.

All the string handling functions are prototyped in: string.h or stdio.h standard header file. So while using any string related function, don't forget to include either stdio.h or string.h. May be your compiler differs so please check before going ahead.

If you were to have an array of characters WITHOUT the null character as the last element, you'd have an ordinary character array, rather than a string constant.

String constants have double quote marks around them, and can be assigned to char pointers as shown below. Alternatively, you can assign a string constant to a char array - either with no size specified, or you can specify a size, but don't forget to leave a space for the null character!

char *string_1 = "Hello";
char string_2[] = "Hello";
char string_3[6] = "Hello";

Reading and Writing Strings:
One possible way to read in a string is by using scanf. However, the problem with this, is that if you were to enter a string which contains one or more spaces, scanf would finish reading when it reaches a space, or if return is pressed. As a result, the string would get cut off. So we could use the gets function

A gets takes just one argument - a char pointer, or the name of a char array, but don't forget to declare the array / pointer variable first! What's more, is that it automatically prints out a newline character, making the output a little neater.

A puts function is similar to gets function in the way that it takes one argument - a char pointer. This also automatically adds a newline character after printing out the string. Sometimes this can be a disadvantage, so printf could be used instead.

#include

int main() {
char array1[50];
char *array2;

printf("Now enter another string less than 50");
printf(" characters with spaces: \n");
gets(array1);

printf("\nYou entered: ");
puts(array1);

printf("\nTry entering a string less than 50");
printf(" characters, with spaces: \n");
scanf("%s", array2);

printf("\nYou entered: %s\n", array2);

return 0;
}

This will produce following result:

Now enter another string less than 50 characters with spaces:
hello world

You entered: hello world

Try entering a string less than 50 characters, with spaces:
hello world

You entered: hello

String Manipulation Functions:

char *strcpy (char *dest, char *src);
Copy src string into dest string.

char *strncpy(char *string1, char *string2, int n);
Copy first n characters of string2 to stringl .

int strcmp(char *string1, char *string2);
Compare string1 and string2 to determine alphabetic order.

int strncmp(char *string1, char *string2, int n);
Compare first n characters of two strings.

int strlen(char *string);
Determine the length of a string.

char *strcat(char *dest, const char *src);
Concatenate string src to the string dest.

char *strncat(char *dest, const char *src, int n);
Concatenate n characters from string src to the string dest.

char *strchr(char *string, int c);
Find first occurrence of character c in string.

char *strrchr(char *string, int c);
Find last occurrence of character c in string.

char *strstr(char *string2, char string*1);
Find first occurrence of string string1 in string2.

char *strtok(char *s, const char *delim) ;
Parse the string s into tokens using delim as delimiter
19.Structured Datatypes

A structure in C is a collection of items of different types. You can think of a structure as a "record" is in Pascal or a class in Java without methods.

Structures, or structs, are very useful in creating data structures larger and more complex than the ones we have discussed so far.

Simply you can group various built-in data types into a structure.

Object concepts was derived from Structure concept. You can achieve few object oriented goals using C structure but it is very complex.

Following is the example how how to define a structure.

struct student
{
char firstName[20];
char lastName[20];
char SSN[9];
float gpa;
};

Now you have a new datatype called student and you can use this datatype define your variables of student type:

struct student student_a, student_b;

or an array of students as

struct student students[50];

Another way to declare the same thing is:

struct {
char firstName[20];
char lastName[20];
char SSN[10];
float gpa;
} student_a, student_b;

All the variables inside an structure will be accessed using these values as student_a.firstName will give value of firstName variable. Similarly we can access other variables.

Structure Example:
Try out following example to understand the concept:

#include
struct student {
char firstName[20];
char lastName[20];
char SSN[10];
float gpa;
};

main()
{
struct student student_a;

strcpy(student_a.firstName, "Deo");
strcpy(student_a.lastName, "Dum");
strcpy(student_a.SSN, "2333234" );
student_a.gpa = 2009.20;

printf( "First Name: %s\n", student_a.firstName );
printf( "Last Name: %s\n", student_a.lastName );
printf( "SNN : %s\n", student_a.SSN );
printf( "GPA : %f\n", student_a.gpa );
}

This will produce following results:

First Name: Deo
Last Name: Dum
SSN : 2333234
GPA : 2009.

Pointers to Structs:
Sometimes it is useful to assign pointers to structures (this will be evident in the next section with self-referential structures). Declaring pointers to structures is basically the same as declaring a normal pointer:

struct student *student_a;

To dereference, you can use the infix operator: ->.

printf("%s\n", student_a->SSN);

typedef Keyword
There is an easier way to define structs or you could "alias" types you create. For example:

typedef struct{
char firstName[20];
char lastName[20];
char SSN[10];
float gpa;
}student;

Now you can use student directly to define variables of student type without using struct keyword. Following is the example:

student student_a;

You can use typedef for non-structs:

typedef long int *pint32;

pint32 x, y, z;

x, y and z are all pointers to long ints

Unions Datatype
Unions are declared in the same fashion as structs, but have a fundamental difference. Only one item within the union can be used at any time, because the memory allocated for each item inside the union is in a shared memory location.

Here is how we define a Union

union Shape
{
int circle;
int triangle;
int ovel;
};

We use union in such case where only one condition will be applied and only one variable will be used.

Conclusion:
You can create arrays of structs.
Structs can be copied or assigned.
The & operator may be used with structs to show addresses.
Structs can be passed into functions. Structs can also be returned from functions.
Structs cannot be compared!
Structures can store non-homogenous data types into a single collection, much like an array does for common data (except it isn't accessed in the same manner).
Pointers to structs have a special infix operator: -> for dereferencing the pointer.
typedef can help you clear your code up and can help save some keystrokes
20.Working with Files
When accessing files through C, the first necessity is to have a way to access the files. For C File I/O you need to use a FILE pointer, which will let the program keep track of the file being accessed. For Example:

FILE *fp;

To open a file you need to use the fopen function, which returns a FILE pointer. Once you've opened a file, you can use the FILE pointer to let the compiler perform input and output functions on the file.

FILE *fopen(const char *filename, const char *mode);

Here filename is string literal which you will use to name your file and mode can have one of the following values

w - open for writing (file need not exist)
a - open for appending (file need not exist)
r+ - open for reading and writing, start at beginning
w+ - open for reading and writing (overwrite file)
a+ - open for reading and writing (append if file exists)

Note that it's possible for fopen to fail even if your program is perfectly correct: you might try to open a file specified by the user, and that file might not exist (or it might be write-protected). In those cases, fopen will return 0, the NULL pointer.

Here's a simple example of using fopen:

FILE *fp;

fp=fopen("/home/techpreparation/test.txt", "r");

This code will open test.txt for reading in text mode. To open a file in a binary mode you must add a b to the end of the mode string; for example, "rb" (for the reading and writing modes, you can add the b either after the plus sign - "r+b" - or before - "rb+")

To close a function you can use the function:

int fclose(FILE *a_file);

fclose returns zero if the file is closed successfully.

An example of fclose is:

fclose(fp);

To work with text input and output, you use fprintf and fscanf, both of which are similar to their friends printf and scanf except that you must pass the FILE pointer as first argument.

Try out following example:

#include

main()
{
FILE *fp;

fp = fopen("/tmp/test.txt", "w");
fprintf(fp, "This is testing...\n");
fclose(fp;);
}

This will create a file test.txt in /tmp directory and will write This is testing in that file.

Here is an example which will be used to read lines from a file:

#include

main()
{
FILE *fp;
char buffer[20];

fp = fopen("/tmp/test.txt", "r");
fscanf(fp, "%s", buffer);
printf("Read Buffer: %s\n", %buffer );
flcose(fp;);

}

It is also possible to read (or write) a single character at a time--this can be useful if you wish to perform character-by-character input. The fgetc function, which takes a file pointer, and returns an int, will let you read a single character from a file:

int fgetc (FILE *fp);

The fgetc returns an int. What this actually means is that when it reads a normal character in the file, it will return a value suitable for storing in an unsigned char (basically, a number in the range 0 to 255). On the other hand, when you're at the very end of the file, you can't get a character value--in this case, fgetc will return "EOF", which is a constant that indicates that you've reached the end of the file.

The fputc function allows you to write a character at a time--you might find this useful if you wanted to copy a file character by character. It looks like this:

int fputc( int c, FILE *fp );

Note that the first argument should be in the range of an unsigned char so that it is a valid character. The second argument is the file to write to. On success, fputc will return the value c, and on failure, it will return EOF.

Binary I/O
There are following two functions which will be used for binary input and output:

size_t fread(void *ptr, size_t size_of_elements,
size_t number_of_elements, FILE *a_file);

size_t fwrite(const void *ptr, size_t size_of_elements,
size_t number_of_elements, FILE *a_file);

Both of these functions deal with blocks of memories - usually arrays. Because they accept pointers, you can also use these functions with other data structures; you can even write structs to a file or a read struct into memory.

21.Bits Manipulation
Bit manipulation is the act of algorithmically manipulating bits or other pieces of data shorter than a byte. C language is very efficient in manipulating bits.

Here are following operators to perform bits manipulation:
Bitwise Operators:

Bitwise operator works on bits and perform bit by bit operation.

Assume if B = 60; and B = 13; Now in binary format they will be as follows:

A = 0011 1100

B = 0000 1101

-----------------

A&B = 0000 1000

A|B = 0011 1101

A^B = 0011 0001

~A = 1100 0011

Show Examples

There are following Bitwise operators supported by C language

Operator Description Example
& Binary AND Operator copies a bit to the result if it exists in both operands. (A & B) will give 12 which is 0000 1100
| Binary OR Operator copies a bit if it exists in either operand. (A | B) will give 61 which is 0011 1101
^ Binary XOR Operator copies the bit if it is set in one operand but not both. (A ^ B) will give 49 which is 0011 0001
~ Binary Ones Complement Operator is unary and has the effect of 'flipping' bits. (~A ) will give -60 which is 1100 0011
<< Binary Left Shift Operator. The left operands value is moved left by the number of bits specified by the right operand. A << 2 will give 240 which is 1111 0000 >> Binary Right Shift Operator. The left operands value is moved right by the number of bits specified by the right operand. A >> 2 will give 15 which is 0000 1111

The shift operators perform appropriate shift by operator on the right to the operator on the left. The right operator must be positive. The vacated bits are filled with zero.

For example: x << 2 shifts the bits in x by 2 places to the left.if x = 00000010 (binary) or 2 (decimal)then: x >>= 2 => x = 00000000 or just 0 (decimal)

Also: if x = 00000010 (binary) or 2 (decimal)
then
x <<= 2 => x = 00001000 or 8 (decimal)

Therefore a shift left is equivalent to a multiplication by 2. Similarly a shift right is equal to division by 2. Shifting is much faster than actual multiplication (*) or division (/) by 2. So if you want fast multiplications or division by 2 use shifts.

To illustrate many points of bitwise operators let us write a function, Bitcount, that counts bits set to 1 in an 8 bit number (unsigned char) passed as an argument to the function.

int bitcount(unsigned char x)
{
int count;

for ( count=0; x != 0; x>>=1);
{
if ( x & 01)
count++;
}

return count;
}

This function illustrates many C program points:

for loop not used for simple counting operation.

x >>= 1 => x = x>> 1;

for loop will repeatedly shift right x until x becomes 0

use expression evaluation of x & 01 to control if

x & 01 masks of 1st bit of x if this is 1 then count++

Bit Fields
Bit Fields allow the packing of data in a structure. This is especially useful when memory or data storage is at a premium. Typical examples:

Packing several objects into a machine word. e.g. 1 bit flags can be compacted.

Reading external file formats -- non-standard file formats could be read in. E.g. 9 bit integers.

C allows us do this in a structure definition by putting :bit length after the variable. For example:

struct packed_struct {
unsigned int f1:1;
unsigned int f2:1;
unsigned int f3:1;
unsigned int f4:1;
unsigned int type:4;
unsigned int my_int:9;
} pack;

Here the packed_struct contains 6 members: Four 1 bit flags f1..f3, a 4 bit type and a 9 bit my_int.

C automatically packs the above bit fields as compactly as possible, provided that the maximum length of the field is less than or equal to the integer word length of the computer. If this is not the case then some compilers may allow memory overlap for the fields whilst other would store the next field in the next word
22.Pre-Processors
The C Preprocessor is not part of the compiler, but is a separate step in the compilation process. In simplistic terms, a C Preprocessor is just a text substitution tool. We'll refer to the C Preprocessor as the CPP.

All preprocessor lines begin with #

The unconditional directives are:
#include - Inserts a particular header from another file
#define - Defines a preprocessor macro
#undef - Undefines a preprocessor macro

The conditional directives are:
#ifdef - If this macro is defined
#ifndef - If this macro is not defined
#if - Test if a compile time condition is true
#else - The alternative for #if
#elif - #else an #if in one statement
#endif - End preprocessor conditional

Other directives include:
# - Stringization, replaces a macro parameter with a string constant
## - Token merge, creates a single token from two adjacent ones

Pre-Processors Examples:
Analyze following examples to understand various directives

#define MAX_ARRAY_LENGTH 20

Tells the CPP to replace instances of MAX_ARRAY_LENGTH with 20. Use #define for constants to increase readability.

#include
#include "myheader.h"

Tells the CPP to get stdio.h from System Libraries and add the text to this file. The next line tells CPP to get myheader.h from the local directory and add the text to the file.

#undef FILE_SIZE
#define FILE_SIZE 42

Tells the CPP to undefined FILE_SIZE and define it for 42.

#ifndef MESSAGE
#define MESSAGE "You wish!"
#endif

Tells the CPP to define MESSAGE only if MESSAGE isn't defined already.

#ifdef DEBUG
/* Your debugging statements here */
#endif

Tells the CPP to do the following statements if DEBUG is defined. This is useful if you pass the -DDEBUG flag to gcc. This will define DEBUG, so you can turn debugging on and off on the fly!

Stringize (#):
The stringize or number-sign operator ('#'), when used within a macro definition, converts a macro parameter into a string constant. This operator may be used only in a macro that has a specified argument or parameter list.

When the stringize operator immediately precedes the name of one of the macro parameters, the parameter passed to the macro is enclosed within quotation marks and is treated as a string literal. For example:

#include

#define message_for(a, b) \
printf(#a " and " #b ": We love you!\n")

int main(void)
{
message_for(Carole, Debra);
return 0;
}

This will produce following result using stringization macro message_for

Carole and Debra: We love you!

Token Pasting (##):
The token-pasting operator (##) within a macro definition combines two arguments. It permits two separate tokens in the macro definition to be joined into a single token.

If the name of a macro parameter used in the macro definition is immediately preceded or followed by the token-pasting operator, the macro parameter and the token-pasting operator are replaced by the value of the passed parameter. Text that is adjacent to the token-pasting operator that is not the name of a macro parameter is not affected. For example:

#define tokenpaster(n) printf ("token" #n " = %d", token##n)

tokenpaster(34);

This example results in the following actual output from the preprocessor:

printf ("token34 = %d", token34);

This example shows the concatenation of token##n into token34. Both the stringize and the token-pasting operators are used in this example.

Parameterized Macros:
One of the powerful functions of the CPP is the ability to simulate functions using parameterized macros. For example, we might have some code to square a number:

int square(int x)
{
return x * x;
}

We can instead rewrite this using a macro:

#define square(x) ((x) * (x))

Macros with arguments must be defined using the #define directive before they can be used. The argument list is enclosed in parentheses and must immediately follow the macro name. Spaces are not allowed between and macro name and open parenthesis. For example:

#define MAX(x,y) ((x) > (y) ? (x) : (y))

Macro Caveats:

Macro definitions are not stored in the object file. They are only active for the duration of a single source file starting when they are defined and ending when they are undefined (using #undef), redefined, or when the end of the source file is found.

Macro definitions you wish to use in multiple source files may be defined in an include file which may be included in each source file where the macros are required.

When a macro with arguments is invoked, the macro processor substitutes the arguments into the macro body and then processes the results again for additional macro calls. This makes it possible, but confusing, to piece together a macro call from the macro body and from the macro arguments.

Most experienced C programmers enclose macro arguments in parentheses when they are used in the macro body. This technique prevents undesired grouping of compound expressions used as arguments and helps avoid operator precedence rules overriding the intended meaning of a macro.

While a macro may contain references to other macros, references to itself are not expanded. Self-referencing macros are a special feature of ANSI Standard C in that the self-reference is not interpreted as a macro call. This special rule also applies to indirectly self-referencing macros (or macros that reference themselves through another macro).

23.Useful Concepts
Error Reporting:
Many times it is useful to report errors in a C program. The standard library perror() is an easy to use and convenient function. It is used in conjunction with errno and frequently on encountering an error you may wish to terminate your program early. We will meet these concepts in other parts of the function reference chapter also.

void perror(const char *message) - produces a message on standard error output describing the last error encountered.

errno: - is a special system variable that is set if a system call cannot perform its set task. It is defined in #include .

Predefined Streams:
UNIX defines 3 predefined streams ie. virtual files

stdin, stdout, stderr

They all use text a the method of I/O. stdin and stdout can be used with files, programs, I/O devices such as keyboard, console, etc.. stderr always goes to the console or screen.

The console is the default for stdout and stderr. The keyboard is the default for stdin.

Dynamic Memory Allocation:
Dynamic allocation is a pretty unique feature to C. It enables us to create data types and structures of any size and length to suit our programs need within the program. We use dynamic memory allocation concept when we don't know how in advance about memory requirement.

There are following functions to use for dynamic memory manipulation:

void *calloc(size_t num elems, size_t elem_size) - Allocate an array and initialise all elements to zero .

void free(void *mem address) - Free a block of memory.

void *malloc(size_t num bytes) - Allocate a block of memory.

void *realloc(void *mem address, size_t newsize) - Reallocate (adjust size) a block of memory.

Command Line Arguments:
It is possible to pass arguments to C programs when they are executed. The brackets which follow main are used for this purpose. argc refers to the number of arguments passed, and argv[] is a pointer array which points to each argument which is passed to mainA simple example follows, which checks to see if a single argument is supplied on the command line when the program is invoked.

#include h
main( int argc, char *argv[] )
{
if( argc == 2 )
printf("The argument supplied is %s\n", argv[1]);
else if( argc > 2 )
printf("Too many arguments supplied.\n");
else
printf("One argument expected.\n");
}

Note that *argv[0] is the name of the program invoked, which means that *argv[1] is a pointer to the first argument supplied, and *argv[n] is the last argument. If no arguments are supplied, argc will be one. Thus for n arguments, argc will be equal to n + 1. The program is called by the command line:

$myprog argument1

More clearly, Suppose a program is compiled to an executable program myecho and that the program is executed with the following command.

$myeprog aaa bbb ccc

When this command is executed, the command interpreter calls the main() function of the myprog program with 4 passed as the argc argument and an array of 4 strings as the argv argument.

argv[0] - "myprog"
argv[1] - "aaa"
argv[2] - "bbb"
argv[3] - "ccc"

Multidimensional Arrays:
The array we used in the last example was a one dimensional array. Arrays can have more than one dimension, these arrays-of-arrays are called multidimensional arrays. They are very similar to standard arrays with the exception that they have multiple sets of square brackets after the array identifier. A two dimensional array can be though of as a grid of rows and columns.

#include

const int num_rows = 7;
const int num_columns = 5;

int
main()
{
int box[num_rows][num_columns];
int row, column;

for(row = 0; row < num_rows; row++)for(column = 0; column < num_columns; column++)box[row][column] = column + (row * num_columns);for(row = 0; row < num_rows; row++){for(column = 0; column < num_columns; column++){printf("%4d", box[row][column]);}printf("\n");}return 0;} This will produce following result: 0 1 2 3 45 6 7 8 910 11 12 13 1415 16 17 18 1920 21 22 23 2425 26 27 28 2930 31 32 33 34 The above array has two dimensions and can be called a doubly subscripted array. GCC allows arrays of up to 29 dimensions although actually using an array of more than three dimensions is very rare.24String Manipulation Functions:char *strcpy (char *dest, char *src); Copy src string into dest string.char *strncpy(char *string1, char *string2, int n); Copy first n characters of string2 to stringl .int strcmp(char *string1, char *string2); Compare string1 and string2 to determine alphabetic order.int strncmp(char *string1, char *string2, int n); Compare first n characters of two strings.int strlen(char *string);Determine the length of a string.char *strcat(char *dest, const char *src); Concatenate string src to the string dest.char *strncat(char *dest, const char *src, int n); Concatenate n characters from string src to the string dest.char *strchr(char *string, int c); Find first occurrence of character c in string.char *strrchr(char *string, int c); Find last occurrence of character c in string.char *strstr(char *string2, char string*1); Find first occurrence of string string1 in string2.char *strtok(char *s, const char *delim) ;Parse the string s into tokens using delim as delimiter.Memory Management Functions:void *calloc(int num elems, int elem_size); Allocate an array and initialise all elements to zero .void free(void *mem address); Free a block of memory.void *malloc(int num bytes); Allocate a block of memory.void *realloc(void *mem address, int newsize); Reallocate (adjust size) a block of memory.Buffer Manipulation:void* memcpy(void* s, const void* ct, int n); Copies n characters from ct to s and returns s. s may be corrupted if objects overlap.int memcmp(const void* cs, const void* ct, int n); Compares at most (the first) n characters of cs and ct, returning negative value if csct.

void* memchr(const void* cs, int c, int n);
Returns pointer to first occurrence of c in first n characters of cs, or NULL if not found.

void* memset(void* s, int c, int n);
Replaces each of the first n characters of s by c and returns s.

void* memmove(void* s, const void* ct, int n);
Copies n characters from ct to s and returns s. s will not be corrupted if objects overlap.

Character Functions:

int isalnum(int c);
The function returns nonzero if c is alphanumeric

int isalpha(int c);
The function returns nonzero if c is alphabetic only

int iscntrl(int c);
The function returns nonzero if c is a control character

int isdigit(int c);
The function returns nonzero if c is a numeric digit

int isgraph(int c);
The function returns nonzero if c is any character for which either isalnum or ispunct returns nonzero.

int islower(int c);
The function returns nonzero if c is a lower case character.

int isprint(int c);
The function returns nonzero if c is space or a character for which isgraph returns nonzero.

int ispunct(int c);
The function returns nonzero if c is punctuation

int isspace(int c);
The function returns nonzero if c is space character

int isupper(int c);
The function returns nonzero if c is upper case character

int isxdigit(int c);
The function returns nonzero if c is hexa digit

int tolower(int c);
The function returns the corresponding lowercase letter if one exists and if isupper(c); otherwise, it returns c.

int toupper(int c);
The function returns the corresponding uppercase letter if one exists and if islower(c); otherwise, it returns c.

Error Handling Functions:

void perror(const char *s);
produces a message on standard error output describing the last error encountered.

char *strerror(int errnum );
returns a string describing the error code passed in the argument errnum.

25.

c apptitude in prog....

2010-08-14T01:02:00.000-07:00

Predict the output or error(s) for the following:
1. void main()
{
int const * p=5;
printf("%d",++(*p));
}

Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}

Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}

Answer:
I hate U

Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}

Answer:
5 4 3 2 1

Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) { printf(" %d ",*c); ++q; } for(j=0;j<5;j++){printf(" %d ",*p);++p; }} Answer: 2 2 2 2 2 2 3 4 6 5Predict the output or error(s) for the following:7. main(){ int i=-1,j=-1,k=0,l=2,m; m=i++&&j++&&k++||l++; printf("%d %d %d %d %d",i,j,k,l,m);}Answer: 0 0 1 3 1Explanation :Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.8. main(){ char *p; printf("%d %d ",sizeof(*p),sizeof(p));} Answer: 1 2Explanation:The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.9. main(){ int i=3; switch(i) { default:printf("zero"); case 1: printf("one"); break; case 2:printf("two"); break; case 3: printf("three"); break; } }Answer :threeExplanation :The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.10. main(){ printf("%x",-1<<4);}Answer: fff0Explanation :-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.11. main(){ char string[]="Hello World"; display(string);}void display(char *string){ printf("%s",string);}Answer:Compiler Error : Type mismatch in redeclaration of function display Explanation :In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.12. main(){ int c=- -2; printf("c=%d",c);}Answer: c=2;Explanation:Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.Note: However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.13. #define int charmain(){ int i=65; printf("sizeof(i)=%d",sizeof(i));}Answer: sizeof(i)=1Explanation:Since the #define replaces the string int by the macro char 14. main(){int i=10;i=!i>14;
Printf ("i=%d",i);
}

Answer:
i=0

Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

Predict the output or error(s) for the following:
15. #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

Answer:
77

Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).

16. #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}

Answer:
SomeGarbageValue---1

Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

17. #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}

Answer:
Compiler Error

Explanation:
You should not initialize variables in declaration

18. #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}

Answer:
Compiler Error

Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}

Answer:
hai

Explanation:
\n - newline
\b - backspace
\r - linefeed

20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}

Answer:
45545

Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}

Answer:
64

Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

22. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}

Answer:
ibj!gsjfoet

Explanation:
++*p++ will be parse in the given order
Ø *p that is value at the location currently pointed by p will be taken
Ø ++*p the retrieved value will be incremented
Ø when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.

23. #include
#define a 10
main()
{
#define a 50
printf("%d",a);
}

Answer:
50

Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}

Answer:
100

Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any problem
Predict the output or error(s) for the following:
25. main()
{
printf("%p",main);
}

Answer:
Some address will be printed.

Explanation:
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

26. main()
{
clrscr();
}
clrscr();

Answer:
No output/error

Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

27. enum colors {BLACK,BLUE,GREEN}
main()
{

printf("%d..%d..%d",BLACK,BLUE,GREEN);

return(1);
}

Answer:
0..1..2

Explanation:
enum assigns numbers starting from 0, if not explicitly defined.

28. void main()
{
char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}

Answer:
4..2

Explanation:
the second pointer is of char type and not a far pointer

29. main()
{
int i=400,j=300;
printf("%d..%d");
}

Answer:
400..300

Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program, then printf will take garbage values.

30. main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}

Answer:
H

Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

31. main()
{
int i=1;
while (i<=5) { printf("%d",i); if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}

Answer:
Compiler error: Undefined label 'here' in function main

Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

32. main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++) printf("%s",names[i]);}Answer:Compiler error: Lvalue required in function mainExplanation:Array names are pointer constants. So it cannot be modified.33. void main(){ int i=5; printf("%d",i++ + ++i);}Answer:Output Cannot be predicted exactly.Explanation:Side effects are involved in the evaluation of i34. void main(){ int i=5; printf("%d",i+++++i);}Answer:Compiler Error Explanation:The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators. 35. #includemain(){int i=1,j=2;switch(i) { case 1: printf("GOOD"); break; case j: printf("BAD"); break; }}Answer:Compiler Error: Constant expression required in function main.Explanation:The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error). Note:Enumerated types can be used in case statementsPredict the output or error(s) for the following:36. main(){int i;printf("%d",scanf("%d",&i)); // value 10 is given as input here}Answer:1Explanation:Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1. 37. #define f(g,g2) g##g2main(){int var12=100;printf("%d",f(var,12)); }Answer:100 38. main(){int i=0; for(;i++;printf("%d",i)) ;printf("%d",i);}Answer: 1Explanation:before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).39. #includemain(){ char s[]={'a','b','c','\n','c','\0'}; char *p,*str,*str1; p=&s[3]; str=p; str1=s; printf("%d",++*p + ++*str1-32);}Answer:MExplanation:p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32. i.e. (11+98-32)=77("M");40. #includemain(){ struct xx { int x=3; char name[]="hello"; };struct xx *s=malloc(sizeof(struct xx));printf("%d",s->x);
printf("%s",s->name);
}

Answer:
Compiler Error

Explanation:
Initialization should not be done for structure members inside the structure declaration

41. #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}

Answer:
Compiler Error

Explanation:
in the end of nested structure yy a member have to be declared

42. main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}

Answer:
Linker error: undefined symbol '_i'.

Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

43. main()
{
printf("%d", out);
}

int out=100;
Answer:
Compiler error: undefined symbol out in function main.

Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

Predict the output or error(s) for the following:
44. main()
{
extern out;
printf("%d", out);
}
int out=100;

Answer:
100

Explanation:
This is the correct way of writing the previous program.

45. main()
{
show();
}
void show()
{
printf("I'm the greatest");
}

Answer:
Compier error: Type mismatch in redeclaration of show.

Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

46. main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}

Answer:
100, 100, 100, 2
114, 104, 102, 3

47. main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}

Answer:
Compiler error: lvalue required.

Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

48. main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}

Answer:
111
222
333
344

49. main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}

Answer:
g20fy

Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

50. main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}

Answer:
ck

Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

51. main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i {
printf(“%s\n”,x);
x++;
}
}

Answer:
(blank space)
irl
rl
l

Explanation:
Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates
52. int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}

Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), ,

Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.

53. main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}

Answer:
i = -1, +i = -1

Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

54. What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).

55. what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);

Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position
of the file.

56. main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it reads all character upto another quotation mark.

57. What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.

58. main()
{
main();
}

Answer:
Runtime error : Stack overflow.

Explanation:
main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.

59. main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}

Answer:
Compiler error (at line number 4): size of v is Unknown.

Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

60. main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}

Answer:
2 5 5

Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

61. main()
{
char not;
not=!2;
printf("%d",not);
}

Answer:
0

Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

62. #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}

Answer:
TRUE

Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed
63. main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}

Answer:
1==1 is TRUE

Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

64. main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}

Answer:
2000 is a leap year

Explanation:
An ordinary program to check if leap year or not.

65. #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}

Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.

66. int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}

Answer:
30,20,10

Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

67. main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}

Answer:
10

Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

68. main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}

Answer:
i = -1, -i = 1

Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

69. #include
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}

Answer:
Compiler error

Explanation:
i is a constant. you cannot change the value of constant

70. #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}

Answer:
garbagevalue..1

Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.

71. #include
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}

Answer:
hello 5

Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.

72. main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}

Answer:
11

Explanation:
the expression i+++j is treated as (i++ + j)

Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

6. main()
{
extern int i;
i=20;
printf("%d",i);
}

Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

c intreview quesitions

2010-08-13T23:50:00.001-07:00

What is C language ?

What does static variable mean?

What are the different storage classes in C ?

What is hashing ?

Can static variables be declared in a header file ?

Can a variable be both constant and volatile ?

Can include files be nested?

What is a null pointer ?

What is the output of printf("%d") ?

What is the difference between calloc() and malloc() ?

What is the difference between printf() and sprintf() ?

How to reduce a final size of executable ?

Can you tell me how to check whether a linked list is circular ?

Advantages of a macro over a function ?

What is the difference between strings and character arrays ?

Write down the equivalent pointer expression for referring the same element a[i][j][k][l] ?

Which bit wise operator is suitable for checking whether a particular bit is on or off ?

Which bit wise operator is suitable for turning off a particular bit in a number ?

Which bit wise operator is suitable for putting on a particular bit in a number ?

Does there exist any other function which can be used to convert an integer or a float to a string ?

Why does malloc(0) return valid memory address ? What's the use ?

Difference between const char* p and char const* p

What is the result of using Option Explicit ?

What is the benefit of using an enum rather than a #define constant ?

What is the quickest sorting method to use ?

When should the volatile modifier be used ?

When should the register modifier be used? Does it really help ?

How can you determine the size of an allocated portion of memory ?

What is page thrashing ?

When does the compiler not implicitly generate the address of the first element of an array ?

What is the benefit of using #define to declare a constant ?

How can I search for data in a linked list ?

Why should we assign NULL to the elements (pointer) after freeing them ?

What is a null pointer assignment error ? What are bus errors, memory faults, and core dumps ?

When should a type cast be used ?

What is the difference between a string copy (strcpy) and a memory copy (memcpy)? When should each be used?

How can I convert a string to a number ?

How can I convert a number to a string ?

Is it possible to execute code even after the program exits the main() function?

What is the stack ?

How do you print an address ?

Can a file other than a .h file be included with #include ?

What is Preprocessor ?

How can you restore a redirected standard stream ?

What is the purpose of realloc( ) ?

What is the heap ?

How do you use a pointer to a function ?

What is the purpose of main( ) function ?

Why n++ executes faster than n+1 ?

What will the preprocessor do for a program ?

What is the benefit of using const for declaring constants ?

What is the easiest sorting method to use ?

Is it better to use a macro or a function ?

What are the standard predefined macros ?

Have a Question ? post your questions here. It will be answered as soon as possible.

tcs4

2010-08-13T02:48:00.000-07:00

Test Paper :19

Paper Type : Whole Testpaper
Test Date : 21 January 2010
Test Location : UIT RGPV, Bhopal
Posted By : Brashaket Pratap Singh

TCS PAPER ON 21st JANUARY AT BHOPAL

Hello Guys This is Brashaket Pratap Singh Bais one of the student of the Technocrats Institute of Technology, Bhopal Pursuing MCA.
TCS on January 21st , 2010 at RGPV BHOPAL (in the UIT Bhopal).
It was an open campus. All the reputed Colleges of Bhopal appear in the online test.
Nearly 360 students From my college attended for online test, out of which 138 were short listed for interview. Out of the 360 there r 81 students from the MCA, and only 7 students including me are able to clear the online test and short listed for the Interview. All the remaining students are from the B.E.

Complete tcs selection procedure consists of :
1) Online Test 2) Technical Interview 3) M.R. Round 4) H.R. Round

(PAPER) TCS Online Test

The test was divided in three sections as:
1. Verbal reasoning (20 min - 32 Qs )
2. Quantitative aptitude (40 min – 36 Qs)
3. Critical reasoning (30 min - 3 passages - 12 Qs)

No Sectional Cut off.
No Negative Marks.
No Psychometric Test was there for us.

1. Verbal reasoning: (20 min 32 Qs)

Antonyms and Synonyms: (20) (I don’t remember which were for antonyms and which were for synonym)\
1. Effrontery -- extreme rudeness and lack of ability to understand that your behavior is not acceptable to other people.

2. ERRATIC(SYNONYM)
a. Unromantic
b. Free
c. Popular
d. Steady
e. Unknown(answer)

3.what is the synonyms of RAPT
Ans:- concealed

4. SYNONYMS OF tranquil
Ans:- serene

5. Awry – skewed, crooked, wrong

6. Repellent x attractive(antonym)

7. Raucous – rough, wild, hoarse, guttering(synonym)

8. Benign – kind, benevolent, compassionate

9. Pristine x sullied

10. Florid= Ornate, Showy(SYN)

11. Servility= Surrender(SYN)

12. Chide * praise(ANT)

13. Stilted * natural(ANT)

I don’t Know all the words, but for scoring good marks in this section Refer Barrons 750 word list. As Well as refer FreshersWorld.com mug up all the synonyms and antonyms of approx last 50-70 papers.
Definitely you will Score good marks here, Attempt all the question as there is no negative marking.
I Think my score is good in this section.

There was paragraph making which i cud not do all the questions of that as it was very tough, so i guessed the answers and hoped atleast few hits the target the. RC had a passage from barons, i had practiced all the barrons RC and it helped .
For better score refer Barron’s for this section.

Quantitative aptitude: -
FOR THE QUANT SECTION I WILL SUGGEST U TO SOLVE THE R.S.AGRAWAL AND A LOT BETTER TO SOLVE THE LAST 3-4 YEARS QUESTION PAPERS FROM THE FreshersWorld.com. It Will really Help You.

Sorry I don’t remember the questions. But I am able to correctly solve the 33 questions and remaining r the targeted hits.
I m Sending U some sample questions with their solutions, questions on my paper r like these,

1) Complete the series 7, 9 , 13 , __ , 27, 37
Ans: 19

2) What is the largest prime number that can be stored in an 8-bit memory?
Ans: 127

3) Number of faces, vertices and edges of a cube
Ans : 6,8,12(Sometimes they change the sequence as on my paper it comes as number of vertices ,faces and edges , so first make a very close look to the question and decide the sequence of values that matching the one of the options).

4).In a triangle which one is not possible. Sides are (5 ,5 ,5.), (5 ,4 ,5 ), (4 ,4, 9 ), (3,4 ,5,)?
Ans: (4,4,9). (HINT : for a triangle sum of smaller 2 sides should be greater than the other one which is larger)

5) Match the following:
1. Male - Boy a. A type of
2. Square - Polygon b. A part of
3. Roof - Building c. Not a type of
4. Mushroom – Vegetables d. A superset of
Ans: 1- d, 2- a, 3- b, 4- c

6) Given $ means Tripling and % means change of sign then find the value of $%$6-%$%6 ?
Ans : -72

8) In a two-dimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of X (8, 5).
Ans : 3212
(Hint : Every 1 deg longitude is equal to 4 minutes . If west to east add time else subtract time)

9) Select the odd one out. a. Java b. Lisp c. Smalltalk d. Eiffel.
Ans: LISP

10) In which of the system, decimal number 194 is equal to 1234?
Ans: 5

11) A power unit is there by the bank of the river of 750 meters width. A cable is made from power unit to power a plant opposite to that of the river and 1500mts away from the power unit. The cost of the cable below water is Rs. 15/- per meter and cost of cable on the bank is Rs.12/- per meter. Find the total of laying the cable.
Ans: 20250

12) Which of the following are orthogonal pairs? a. 3i+2j b. i+j c. 2i-3j d. -7i+j
Ans : (A) & (C).

13) If VXUPLVH is written as SURMISE, what is SHDVD?
Ans : PEASA

14) If A, B and C are the mechanisms used separately to reduce the wastage of fuel by 30%, 20% and 10%. What will be the fuel economy if they were used combined.
Ans : 20%

15) A can copy 50 papers in 10 hours while both A & B can copy 70 papers in 10 hours. Then for how many hours required for B to copy 26 papers?
Ans : 13

16) Find the singularity matrix from a given set of matrices? (Hint determinant(A)=0)

17) units of basic quantities :
(energy * time * time)/(mass * dist) = distance

18) For a round trip, a car used 4 1/2 gallons of gasoline. If it used 1/4 more gasoline going than coming back, how much gasoline was used coming back?
Ans: 2 gallons

19)A shopkeeper bought a watch for Rs.400 and sold it for Rs.500.What is his profit percentage?
Ans. 25%

20)g[0]=1,g[1]=-1,g[n]=2*g[n-1]-3*g[n-2] then calculate g[4]=

21) .The ratio of boys to girls is 6:4.60% of the boys and 40% of girls take lunch in the canteen.What % of class takes lunch?
Ans) 52%

22) Which of the following set of numbers has the highest Standard deviation?
a)1,0,1,0,1,0 b) -1, -1, -1, -1, -1, -1
c)1,1,1,1,1,1 d) -1,1,-1, 1,-1, 1
Ans : D

23) (momentum * velocity)/(force * time) = velocity

24-27) there were 3 questions on venn diagrams (like 20 spek eng,30 speak hindi 3 speak both and so on….)

28). The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied?
Ans. 0.5%.

29).Odd one out:
a. Linux b.windows NT c. SQL server d. Unix

30). In the word ORGANISATIONAL, if the first and second, third and forth, forth and fifth, fifth and sixth words are interchanged up to the last letter, what would be the tenth letter from right?

31). If TAFJHH is coded as RBEKGI then RBDJK can be coded as?

32). FIND THE MISSING NO. IN SERIES ?
9 10 11 13 15 ? 21 28
Ans: 19

33). In the word CHAMPIONSHIP, if the first and second, third and forth, fifth and sixth, etc words are interchanged, what would be the 8th letter from right? (Ans: ‘I’)

34). If A, B, C are the mechanisms used separately to reduce the wastage of fuel by 30%, 40%, 10%.What will be the fuel economy if they were used combine?
1.68.4 2.62.2 3.58 4.27
(Ans- 62.2) sol: (70/100)*(60/100)*(90/100)*100=37.8 Eco = (100-37.8) =62.2

35). What is the value of the following expression M(373,5)+T(7.7)+R(4.4)-T(3.6)
Where M- MODULAS R- ROUNDOFF T- TRUNCATE
(Ans : 11) Sol: 3 + 7+ 4 - 3 ==11

36). Match the following (this type of question but not same)
A B
1. Mammal, cow ---> a. A type of
2. Snake reptile ---> b. A part of
3. Roof - Building ---> c. Not a type of
4. Mushroom - Vegetables ---> d. A superset of
(Ans: 1-d, 2-c, 3-b, 4-a)

37). If G(0) = -1 G(1)= 1 and G(N)=G(N-1) - G(N-2) then
What is the value of G(6)?(Ans: -3)

Sol: g(2)=g(1)-g(0) ==1-(-1)=2, similarly g(3),…g(6);
ONCE DO R.S.AGGARWAL (QUANT) (arithmetic section)
GOOD FOR U.

38). Which set of data exhibits a higher Standard Deviation?
(a) 9, 0, -9, 9, 0, -9 (b) 9, 9, 9, 0, 9, 9 (c) -9, -9, -9, -9, -9, -9 (d) 9, 9, 9, 9, 9, 9 (e) 9, -9, 9, -9, 9, -9
Ans: e……….take mean(sumof all/no.of elmnts),take difference of all frm mean,now apply formula (s.d)^2=(sum of (diff)^2/n)

39). In Chennai , temperature at noon varies according to -t^2/2 + 8t + 3, where t is elapsed time. Find how much temperature more or less in 4pm to 9pm.
Ans. (put 9 & 4in eq. n subtract )

40). Largest prime no 3 digit
Ans: 997

41). Find the value of @@+25 - ++@16, where @ denotes "square" and + denotes "square root".
Ans. 621

42). (Momentum*Velocity)/(Acceleration * distance ) find units.
Ans. Mass

43). In a two-dimensional array, X(9,7), with each element occupying 2 bytes of memory, with the address of the element X(1,1) is 3000,
find the address of X(8,5)?
Ans: 3106

44). Find the fourth row, having the bit pattern as an integer in an 8-bit computer, and express the answer in its decimal value.
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
C 0 1 0 1 0 1 0 1
(A OR(B AND C)) ?
Ans: B AND C
1 AND 1=1 else 0
1 OR 0/1=1
So B AND C--à 00010001 This OR A->00011111

45). An aircraft takes off from A (89o N Lat, 20o E ) at 6.00 AM local time to B (50o S , 40o W ). If the flying time is 10 hours what is the local time of landing at B?
Ans:12:00pm
don’t take care of N & S.E & W matters only.
1o change == 4mins
Here 20 o+40 o=60 o
So 60*4=240 mins……4hrs
If time change not considered then After 10 hrs it il reach at the same palce after 6:00+10:00=4:00pm
Going frm east to west so substract 4hrs frm 4:00pm…
4:00pm-4hrs=12:00pm
(if plane is going frm west to east then then Add those hrs)

46). Select the odd one out
a. SMTP b.WAP c. SAP d.ARP
Ans : c

47). select odd one- sql,db2,sybase,http {ans-htttp}

48). select odd one-sybase,db2,oracle,unix {ans-unix}

49). A can copy 50 papers in 10 hrs while A &B together can copy 70 papers in 10 hrs.how many hrs will be required for B to copy 26 papers?
Ans: 13hrs

50). Find eqn of line having intercepts (0,3) & (-2,0)
Ans Y=(3/2)X + 3
Formula x/a + y/b = 1

51). ) if n=68*12*51 which one is not int?
a>n/136 b>n/244 c> n/72
Ans: c

52). Which shape will be obtained by using the following values for X and Y
X 0 10 100 1000 9999
Y 0.00001 1.02 1.72 3.00 4.72
Ans: Y= log10(X)

53). If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ?
Sol) log 0.317=0.3332 and log 0.318=0.3364, then
log 0.319=log0.318+(log(0.318-0.317)) = 0.3396

54). A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ?
Sol) x= 2 kg Packs
y= 1 kg packs
x + y = 150 .......... Eqn 1
2x + y = 264 .......... Eqn 2
Solve the Simultaneous equation; x = 114
so, y = 36
Ans : Number of 2 kg Packs = 114.

55). A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time ?
Sol) Since it is moving from east to west longitide we need to add both
ie,40+40=80
multiply the ans by 4
=>80*4=320min
convert this min to hours ie, 5hrs 20min
It takes 8hrs totally . So 8-5hr 20 min=2hr 40min
So the ans is 10am+2hr 30 min
=>ans is 12:30 it will reach

56). A fisherman's day is rated as good if he catches 9 fishes, fair if 7 fishes and bad if 5 fishes. He catches 53 fishes in a week n had all good, fair n bad days in the week. So how many good, fair n bad days did the fisher man had in the week
Ans:4 good, 1 fair n 2 bad days
Sol) Go to river catch fish
4*9=36
7*1=7
2*5=10
36+7+10=53...
take what is given 53
good days means --- 9 fishes so 53/9=4(remainder=17) if u assume 5 then there is no chance for bad days.
fair days means ----- 7 fishes so remaining 17 --- 17/7=1(remainder=10) if u assume 2 then there is no chance for bad days.
bad days means -------5 fishes so remaining 10---10/5=2days.
Ans: 4 good, 1 fair, 2bad. ==== total 7 days.
x+y+z=7--------- eq1
9*x+7*y+5*z=53 -------eq2
multiply eq 1 by 9,
9*x+9*y+9*z=35 -------------eq3
from eq2 and eq3
2*y+4*z=10-----eq4
since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z = 2 or 1 from eq 4
for first y=1,z=2 then from eq1 x= 4
so 9*4+1*7+2*5=53.... satisfied
now for second y=3 z=1 then from eq1 x=3
so 9*3+3*7+1*5=53 ......satisfied
so finally there are two solution of this question
(x,y,z)=(4,1,2) and (3,3,1)...

57) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X?
Sol) Let no. of fish x catches=p
no. caught by y =r
r=5p.
r+p=42
then p=7,r=35

58) Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings?
suppose total income is 100
so amount x is getting is 80
y is 70
z =60
total=210
but total money is 300
300-210=90
so they are getting 90 rs less
90 is 30% of 300 so they r getting 30% discount

59) The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income?
Sol) incomes:3:4
expenditures:4:5
3x-4y=1/4(3x)
12x-16y=3x
9x=16y
y=9x/16
(3x-4(9x/16))/((4x-5(9x/16)))
Ans:12/19

60).The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied ?
Sol) M=sqrt(100N)
N is increased by 1%
therefore new value of N=N + (N/100)
=101N/100
M=sqrt(100 * (101N/100) )
Hence, we get M=sqrt(101 * N).

61). A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53 and the product went up by 540. What was the raised product?
a) 780 b) 1040 c) 1590 d) 1720
Sol) x*53-x*35=540=> x=30 therefore, 53*30=1590 Ans

Just Go through the previous year question papers and solve all the questions. The pattern of the questions r same only the values are changed, so go through and solve those questions.
Wishing U all the best.

Critical reasoning (30 min - 3 passages - 12 Qs)
For the Critical reasoning section refer barons 12th edition critical reasoning and solve all the model question papers that are given there. As try to mug up the answers.

Out of the 12 questions 8 are from the model test papers and other 4s are easily solvable.
As I think all the 12 answers of this section is of mine are correct.

RESULT OF THE ONLINE TEST DECLARED AT EVENING 09:50 Approx.

I am one of the candidates who was short listed for the Technical Interview.

As it is was very late. So they Call Us at the next day at 1:00 PM sharp.

Technical Interview
TechnAs the technical round was supposed to take place at 1:00 PM, but unfortunately the technical interviews for the our college starts at 07:00 pm, My Technical Interview takes place at 08:30 Pm.
I have to wait for the long time at the door of the Interviewer cabin as the Interviewer is writing something on the paper.
As I entered the room.
ME:-Good Evening Sir.
SIR: Have seat.
ME: Thank you very much sir, and takes the seat.
SIR: Tell me about yourself.
Me: I said my name, my college name, my degree, about my family and my hobbies, some of my positive qualities, my role model.
SIR: Can U write the program of string palindrome in C++.
Me: Yes Sir I Can. He gave me a paper , And Then I wrote The whole Code for the String palindrome using pointers. Please Be specific and don’t waste time , if u don’t know the code, just say sorry sir/mam.
SIR: Can U write the program
1 4.
2 3
3 2
4 1.
Me: Yes Sir I Can. And then I wrote the code , It just take two for loops(nested) one move from 1 to 4 and other from the 4 to 1.
Sir: What are your core (favorite) subjects in MCA?
Me: I said Computer Networks sir.
Sir:-Oh Really Computer Networks.
Me:-Yes Sir, I think as I said computer networks that was out of the leak no MCA students before me told that answer(As I think), But I gave that answer because I m having very good knowledge about the Computer networks both theoretically as well as practically. (Please don’t bluff).
SIR:- OK, Brashaket Tell me What Is WINC?
ME:- It was an blunder for me as said Computer networks , and I don’t know the answer of that question . But I m still confident and simply say very politely sorry sir.
SIR:-OK, What is URL?
Me- URL Means Uniform Resource Locator. And then told all the methods and the whole working of URL.
SIR:- What is DNS.
ME:-DNS means (Domain name systems) and then told whole working of the DNS.
SIR:-Can U tell me what is SMTP?
ME:-It means Simple Mail Transfer Protocol. And told the working of the SMTP.
SIR:-What is Cryptography?
ME:-Told.
SIR:- Sir: What are the different types of Topologies?, what are merits &demerits of each?
Me:-BUS,STAR,MESH,RING,TREE,PARTIAL MESH topology. And told the merits and demerits of all.
Sir: what is the Topology used in Ur college?
ME:- I said it is star topology.
SIR:-What r the different networking devices?
ME:- Sir, HUB, Switch, Routers, Bridges, etc.
SIR:-What is the difference between the switch, routers and bridges?
Me:-I Told.
SIR:- Then He made a very close look to my resume. (Be prepare with each and every aspect of your resume and don’t write any single thing that u don’t know.). Then He asked me About IBM DB2, as I have wrote IBM DB2 in the technical skills section in my resume.
ME:-Told.(As I did the training of DB2 in our College).
SIR:-Ok Brashaket. Then He asked me about my projects? Please explain?
Me:-As I wrote the 2 of my projects their. I Explained. At that time when I was explaining I asked him for a blank paper , and I explained that on a paper. My Project is on Data Dictionary of Oracle, SQL-Server and MS-ACCESS. It was titled Data dictionary management system. It eliminate the writing of complex queries.
I told Each and every aspect of the project from the technologies to the reports.(the explanation took approx 15-20 minutes.
SIR:- He said Very good Brashaket. And then he asks me that R U really interested in development ?
Me: I told yes very much sir.(I don’t know why he asks that question).
SIR:-Ok Brashaket You may leave now . Then He Offer shake hands to me ?
Me:-Shake hand with him and Said thank you sir? Have a good day. And leave the room.
My technical round took about 40-45 minutes .
At 09:30 PM the results of the technical round was declared , My name is in the list of the selected candidates . and then they gave us a form to fill. And as it was very late they told us to come tomorrow at 09:00 AM sharp.
As 138 students of my college appeared for the tech. round . out of 138 , 70 are short listed for the Managerial reasoning (M.R.) round.

Speak very polite and be confident in the interview. Don’t try answering the question that you really don’t know. Be very careful with the resumes keep it as simple as you can .

MR Interview
At 09:AM sharp the M.R. interviews of the students of my college had started.
My Number came at approx 11:00 AM.
Me:- I have to go to the panel number 9.
Me:- Knocked the door. And said Sir, May I come in ? There is an gentle man of age approx 50.
Sir:-Please come in?
Sir:- Have Seat?
Me:- I took and told thank u very much sir.
Sir:- He asked me what is the correct pronunciation of your name?
Me:- I told.
Sir:- Then he asked me that what is the meaning of ur name?
Me:-Told (It is one the name of the lord Shiva).
Sir:- Ok Brashaket . Tell me about yourself?
Me:- I told blah blah ?
Sir:- Tell me something about the family background?
Me:- Told.
Sir:- Then He gave me a puzzle to solve.
Me:- I am able to solve that one perfectly.
Sir:- What do u know about the new technology?
Me:-I told abt the nano technology.
Sir:-Then He asked me about the criteria behind launching the new OS?
ME:- I Told. That For the better appearance and for the better GUI and For the Better Security the new OS is lunched.
Sir:-he seem to looked that he is not satisfied with the answer and then he told me that on the basis of these thing u will launch a new OS.
Me:-yes Sir, Try to explain more, but he is not satisfied, as I think he is checking my patient . Again he was tried to frustrate me but I said sir in my opinion the new OS are launched on these basis. And the other third party s/w can easily be installed.
Sir:- Ok , Any Questions Brashaket?
Me:-Sir what kind of preparation I have to do before joining the TCS training program?
Sir:- Just learn new and more languages and technologies.
Sir:- Thank U very much Brashaket?
Me:- Thank U very much sir , It was really nice Meeting U?
Sir:- Ok, You may leave now?
Me:- Leave and told again thank u sir and have a good day sir. and left.
My M.R. Interview lasting approx for about 35 minutes.
I m very scared as the persons whose name is called think that he has cleared the MR round but some of the candidates r called the interviewers who took the MR interviews shake hands with them and then said “You may leave for home now”. It was a blunder for the person that think that he is selected but the result is that he is rejected.
After 2 hour I was Called for the HR round. Luckily I m selected for the HR Round.

HR Interview
Again I have to wait for a long time as the interviewer is writing something. But he was as decent personality. He said sorry for that, that u have to wait for a long time, I told Its OK, No problem sir.
SIR:- May I know Ur good name please?
ME:-Told.
SIR:-What is the meaning of the name?
Me:-Told
Sir:- How was the Day Brashaket ?
Me:- It was Nice Sir.
Sir:- How was Previous two interviews?
Me: Both of the interviews are Very nice sir.
SIR:-Tell me something abt ur family background?
Me:-Told
SIR:- How many siblings do u have?
Me:- Told
Sir:- what Ur father is doing?
Me:- Told( As my father is working in a cement industry In Maihar)
Sir:- He asked me abt the other cement plants there?
Me:-Told.
Sir:- He told that satna is known for the cement industries naa , can u tell me the names of the cement industries situated their?
Me:-Told.
Sir:- Ok Brashaket Suppose U have nine dots and u have to connect these dots with the help of the four line and u don’t have to lift ur pen. Can U?
Me:- I said I will Try.

He gave me the pattern like this

. . .

. . .

. . .

I m able to solve that one , just put ur pen in the middle dot of the second row and then make to diagonals line and to perpendicular lines. like that:-

. . .

. . .

. . .

Sir:- OK , Suppose that U have a pencil and U have to cross a river with the help of that pencil. How you will cross?

Me:- I gave the answer like this
Sir there might be the probability that there is no water is available on the river. Or we can map the depth of the river with the help of that pencil by dipping it in the river and if the river is not so deep then we can cross the river easily.
Sir:-Oh, really?
Me:-Yes sir , As I think.
Sir:- Then He gave me a situation , that suppose u r appointed in a team in a s/w company. The team consist of the 4 members. He said U belongs to a very low ranked institute like TIT suppose one of the member of the team is from IIT kanpur, one from IIT Bombay and one from the Other IITs and they think that u belongs to a very low standard institute so they r not coordination with u , in that situation what u will do?
Me:- Told , I am trying to satisfy him and I am able to satisfy him with my answer.
Sir:- Is their any pan for the higher studies?
Me:- No Sir, As I m going to complete My PG program and Now I wanted do job now.
Sir:- Tell me something about ur weaknesses?
Me:-Told.(BE Prepare and don’t bluff).
Sir:-Asked some more questions abt the my strengths and weaknesses?
Me:-I told.
Sir:- Is there any location constraints?
Me:- No Sir, Yoy can through me in any part of the the world.
Sir:-laughed. Are U ready to sign the bond ?
Me:-Yes sir.
Sir:- All The Best Brashaket.
Me:-Thank U Sir.

Sir:-Ok Brashaket , Thanks and U may leave Now.
Me:-Thank U very much Sir and have a good day sir and left.

My H.R. Interview lasting approx for about 45 minutes.
After Waiting for a long time the final result of the selection process is declared at 10:00 PM . And Bhagwan ki Kripa aur maa & papa ke ashirwaad aur meri mehnat ke karan I Was One of the candidates who is declared as a part of TCS.
Finally 56 Students r selected from our college, but only 4 r from the MCA. Remaining 52 r from B.E..
It was the happiest moment in my life.
Friends any one can easiliy grab TCS if you are technically strong in 3 of your department paper and with 100% confidence and positive attitude. GOOD LUCK. Meet you in TCS.
Regards
Brashaket Pratap Singh Bais

(TIT Bhopal)

android os

2010-08-13T00:32:00.000-07:00

ANDROID TUTORIAL

------------------------

Google has recently released the Android platform for developing Mobile applications. The language used for developing Android programs, is Java. But, it is not j2me.

No wireless application developer can ignore Android. Google is the best known brand name, among the users of the web and Android comes from Google..

I am presenting this hands-on tutorial, as a sequel to my j2me series. Adequate knowledge of core-java , especially, Event-handling , Swing and inner-classes is assumed. Though Android does not make use of Swing, it uses similar ideas.

We can develop Android lessons and applications in Eclipse environment. Google have provided an Eclipse-plugin for Android. This is the popular method. Google have not given direct support to Netbeans. But, some Netbeans users have developed a method for running Android in Netbeans .It is available at http://undroid.nolimit.cz/. You can find more screenshots and guidance in:

http://eppleton.com/blog/.

But, we can develop Android lessons without using either Eclipse or Netbeans. The necessary command-line tools have been provided by Google. I found that using these command-line tools is easier than either Eclipse or Netbeans method. So, I am basing all the following lessons on these tools. I think, most readers will agree with my view, after they try this method as well as Eclipse method. The Android site at 'code.google.com/android' has already given step-by-step instructions about Android in Eclipse. You can also get more details with screen shots from a wonderful website

at www.tanguay.info/web/welcome.php"

titled 'Edward's Web Developer site'. He gives excellent guidance with plenty of screen shots

But, the Android site lacks clarity, about the command-line method. Hence, I think I am adding something useful by writing on command-line method instead of simply repeating the material ,in Android site.

------------------------------------------

Let us begin from the beginning. The first step is downloading the Android SDK from

code.google.com/android/download

(version m5-rc14, Feb-12, 2008). Android was released in November, 2007 . It has been revised in the Feb-2008 version.Some of the earlier examples may not work in the newer version.

---

I am working in Windows-2000 and so I downloaded the windows version. The supported platform in Windows is either Windows-XP or Vista.( Mac OS 10 & Ubuntu Linux are the other platforms mentioned). However, it works well in my Win-2000. It is advisable to have atleast 512MB memory. The android SDK is a zip file. I unzipped it to C:\unzipped\android and later, I copied that folder to d:\android. If you want, you can simply copy it to another drive like g:\android also. In the following lessons

d:\android. is used.

If you want to develop using Eclipse, you must have installed either Eclipse3.2 or Eclipse3.3(Europa). I have tested with Eclipse3.2. No problem.It works. But, we require ADT (ie) Android Development Tools plugin for Eclipse, if you are using Eclipse.You can get this plugin from code.google.com/android/adt_download.You have to be careful about the ADT version number.It is ADT-0.3.3.

--

As my present focus is on command-line method, let me begin straight away and give a simple demo.

The procedure given here is common for all our experiments and so I will not be repeating it in each demo. So, please note it down carefully.

In my first demo, I will have a customary button and textbox ( called EditField in Android). When I click the button, I want the message "SUCCESS!" to be displayed in textbox. Just as an exercise, I am using two buttons and two textboxes.

-------------------------------------

The first step is to start the Emulator

cd to d:\android\tools

d:\android\tools>emulator

It will take a long time to get started. Do not be in a hurry. Wait till it gets fully started. And do not close that window carelessly by mistake. In that case, you will have to start it again and wait for a long time again. Finally, we get the emulator screen( starting)..

< xml="true" ns="urn:schemas-microsoft-com:vml" prefix="v" namespace="">

The second step is to give the following command., from another command window.

d:\android\tools>

activityCreator --out demo mypack.mydemos.demo

This means that my project is 'demo' and my package is 'mypack.mydemos'. Clear enough.

A number of folders are created automatically by this command. such as

tools\demo\src, tools\demo\bin ,

tools\demo\res.

We need to note the src and res folders carefully. We will place the java source file in src folder and main.xml file in res\layout, overwriting any files that are generated automatically. For the moment, we can think of the res\layout folder as the one which decides the gui design. As in asp.net, flex etc, the gui details are specified in xml file. But how shall we write the XML file? by hand? Not too difficult .But....Luckily, there is an open-source gui designer named 'DroidDraw' available in www.droiddraw.org .It is a nice tool and if you like it, you can send your appreciation to brendan.d.burns@gmail.com. He has given a simple tutorial too, on how to use this gui tool.

I downloaded this software from the above site. I unzipped it. ( any folder). When we click on the icon, we get the screen as given below.(droid-draw)

( drawing canvas area)

======================================

( toolbox & blank area)

====================================

Thus we get a window, showing the drawing canvas on leftside and toolbox and a blank area in the rightside. ( for printing purpose, I have split them into two screens) as above.

From the toolbox, we learn that we are having controls like button,check,radio,spinner,edittext(textbox) and textview (label) etc. There is also a combo to choose layout, in the canvas screen. . I am choosing 'absolute layout'. I simply drag and drop a button and an editview on the canvas.( drag and drop..do not click and drop! It won't work!).You will notice a number of tabs in toolbox. Select 'properties' tab.. After clicking the button on the canvas, give the id property of button as @id/button1. Similarly, for editview as @id/text1. Also, button2 and text2 After this, just click the 'generate' button at the bottom of the blank window. We get the following XML file(main.xml) automatically generated.

============================================

// main.xml

********************************************

***********************************************

// d:\android\mydemos\ex4\demo.java ( checkbox)

(to be copied to tools\demo\mypack\mydemos)

package mypack.mydemos;

import android.app.Activity;

import android.os.Bundle;

import android.view.View;

import android.widget.*;

public class demo extends Activity

{

Button button1;

CheckBox check1,check2;

EditText text1;

@Override

public void onCreate(Bundle icicle)

{ super.onCreate(icicle);

setContentView(R.layout.main); text1=(EditText)this.findViewById(R.id.text1); check1=(CheckBox) findViewById(R.id.check1); check2=(CheckBox) findViewById(R.id.check2);

button1 =(Button) findViewById(R.id.button1);

button1.setOnClickListener(new clicker());

} class clicker implements Button.OnClickListener

{ public void onClick(View v)

{

String r = "";

if(check1.isChecked())

{

r = r+"java"+"\n";

}

if(check2.isChecked())

{

r = r+"c#";

}

text1.setText(r);

}

}

}

******************************************

This is just the usual Java code and needs very little explanation. The only difference is the way , the controls are defined ( through res\layout\xml file).

********************************************

DEMO-5 ( RADIO BUTTONS)

The next standard control is the radiobutton, within a radiogroup.

//d:\android\mydemos\ex5\main.xml

(to be placed in tools\demo\res\layout)