habbul telescope photos

jntu(k)2008 5th sem quesition papers

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?.


*************Related Posts by Categories****

jntu(k)2009 5th sem quesition papers

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)

images

java

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

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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linux commands

•
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

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 < 177 180 180 180 180 Hence, 31 < 17 < 43 < 59 36 18 45 60 3. DECIMAL FRACTIONS IMPORTANT FACTS AND FORMULAE I. 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,etc II. 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=5 IV. Operations on Decimal Fractions : 1. Addition and Subtraction of Decimal Fractions : The given numbers are so placed under each other that the decimal points lie in one column. The numbers so arranged can now be added or subtracted in the usual way. 2. Multiplication of a Decimal Fraction By a Power of 10 : Shift the decimal point 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 considering them without the decimal point. Now, in the product, the decimal point is marked off to obtain as many places of decimal as is the sum of the number of decimal places 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 given number 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 as there 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 < 13/16 Ex. 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 EXAMPLES Ex. 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 5 Hence, missing fractions =3 2 5 Ex.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=315 Hence required number = 315. Ex. 5. Simplify: 3 1{1 1-1/2(2 1-1/4-1/6)}] 4 4 2 Sol. 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]=78 Ex. 6. Simplify: 10836of 1+2*3 1 4 5 4 Sol. Given exp.= 1089+2*13 =108 +13 =12+13 =133 = 13 3 5 4 9 10 10 10 10 Ex.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 = 0.46 Ex. 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_ 4 Sol. 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 - x Sol. 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=75 Ex.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=8600 25x+16x+2x=8600 =43x=8600 x=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.x Then, 7/40 of x=1400 X=(1400*40/7) =8600 Expenditure on food=Rs.(3/10*800)=Rs.2400 Expenditure on conveyence=Rs.(1/8*8000)=Rs.1000 22.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 y Then 1/3x-1/2y=30 2x-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.