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CAMPUS RECRUITMENT COMPLETE REFERENCE

Indian Airlines Colony,

- 09246844022 Email: [email protected] Website: www
- campusrecruitment

- 2010 ISBN 978-81-930685-1-9 Copyright ©2006
- -2015 Praxis Groups

All Rights Reserved All rights reserved

No part of this book may be reproduced in any form through any electronic or mechanical means (including photocopying,

- recording,

or information storage and retrieval) without prior permission in writing from the publisher,

except for reading and browsing via the World Wide Web

Information contained in this book has been published by Praxis Groups and the content has been obtained and developed by its authors from sources believed to be reliable and correct to the best of their knowledge

the Publisher and the Authors shall,

- in no event,
- be liable for any errors,

omissions or damages arising out of the use of this information and specifically disclaim any implied warranties or merchant ability or fitness for any particular use

Disputes of any are subject to Hyderabad Jurisdiction only

Institutions and other organizations wishing to make bulk purchases of any book,

published by PRAXIS GROUPS should contact at Phone: 09246844022 ,

- 07032206275,
- 040-40203585,

Email: [email protected] Address: Mig-B-35,

Hyderabad

Pin: 500003

About the Authors Ramanadham Ramesh Babu,

- of English
- - GITAM University,

Visakhapatnam This book is fortunate enough to have him as an author

He has profound experience in teaching and training the students in various national and international academic and competitive exams

his passion for English made him work on innovative research and he is going to receive his doctorate soon

- Israel Battu (MCA,

National Institute of Technology,

Calicut,

Kerala): A man with enthusiasm to contribute for the field of education which led him to establish an institute

His expertise and enthusiasm resulted in producing outstanding material in Logical Reasoning

- Akash R Bhutada (B

IMT Hyderabad): His passion for Aptitude exams and Aptitude contests produced exceptional material for Quantitative Aptitude

- Vijaya Laxmi Krishnan (M

A PGDTE) and A

Kalyani (M

coupled with their interaction with students of varying abilities has contributed significantly in the creation of this material in Verbal Ability

About 'Technical Interview' Authors

Chandra Shekhar,

Rega Rajendra,

Department of EEE,

Professors in Dept of Civil Engineering,

B-Tech,

Tech lead in Prokarma Info Tech are esteemed professors of Osmania University

- honesty and hard work

He has done remarkable work in the crucial areas Quality Analysis,

coordinating the authors/artists/data entries and finally contributing his creative thoughts

this book wouldn’t have seen the light of the day

Awez akram have played a very important role in managing entire work of the book without any confusion on the computer

They have produced a quality output despite many hurdles

BFA): Ranjith has contributed the art work for the book

QUANTITATIVE APTITUDE

Number System

Chain Rule / Variation

Percentages

Blood Relations

Syllogisms

Alphabet Test

Character Puzzle

Profit and Loss

Symbol Based Operations

Arithmetic Reasoning

Permutations and Combinations

Parts of Speech

Data Sufficiency

- • Noun

- • Pronoun

• Tabular Data Interpretation

- ▪ Articles
- • Bar Graphs
- • Verb

• Line Graphs or Cartesian Graphs

- ▪ Tenses
- • Pie Chart
- ▪ Subject–Verb Agreement
- • Venn Diagrams
- ▪ Active and Passive Voice
- • Net Diagram
- • Adverb
- • Mixed Diagrams
- • Adjective
- • Missed Data Tables
- • Preposition

Inequalities

- • Conjunction
- ▪ Reported Speech

Functions

Antonyms

Quadratic Equations

Analogy

- 53 Phrasal Verbs
- 58 Idiomatic Expressions

GROUP DISCUSSION

Introduction

Skills required for Group Discussion

Theme Detection

PLACEMENT PAPERS Placement Paper-1

44 Summary

Placement Paper-2

- 11 Placement Paper-3
- 22 Placement Paper-4
- 33 Placement Paper-5
- 43 Placement Paper-6 to 40 visit www

CampusRecruitment

- 1 Electronics and Communication Engineering
- 6 Electrical and Electronics Engineering
- 11 Mechanical Engineering
- 18 Computer Science/ Information Technology
- 22 Project Interview Questions

- 1 Tips for Interview
- 2 Do's and Dont's of Interview
- 2 Do's and Dont's of Body Language
- 3 Answering Interview Questions
- 4 Mock Interview with Body Language
- 6 Reasons for failure in Interviews

TIPS FOR SEARCHING/ APPLYING JOBS Different ways to apply for jobs

- 18 Applying jobs through companies website
- 19 Jobs at Government/PSUs
- 21 Apply Govt/PSU jobs in Naukri/Monster
- 22 Selection Procedure in PSUs
- 23 List of PSUs
- 23 Tips to use Naukri
- 26 Tips to use Monster
- 31 Career Path based on Academic Percentage

- 37 Things to avoid when writing Resume
- 37 Contents of a Good Resume
- 37 Sample Resume Format

Introduction to Campus Recruitment Procedure Every student chooses an educational institute on the basis of three important factorsA

The potential jobs that would be offered on the campus to the students who pass out of the institute

The last factor is apparently the most important criteria that would be evaluated by the students in choosing a particular academic institution

The campus recruitment procedure has become one of the most popular avenues to recruit people into companies

- delivery roles,
- software/ hardware testing,
- research and development,

application maintenance network security and support etc

Management graduates are hired for roles in business development,

IT consulting,

- business analysis,

customer relationship management,

HR roles,

- marketing,
- finance etc

Before a student braces himself to face the arduous task of appearing for the campus selection procedure,

there are a few points that he has to bear in mind – 1

Taking a cue from the seniors would effectively help a student to know the areas that would require extensive preparation and the ones that wouldn’t

This is extremely important because there may be scenarios in which a student may have more than one offer at hand

besides the CTC offered would benefit the students to narrow down their choices

culture and values of the companies that recruit college graduates

Campus Recruitment Procedure: Most colleges that offer campus recruitment facilitate the recruitment through a special department known as the placement department

The placement department is steered by a placement officer who oversees the entire recruitment process

The various stages that are involved in a typical campus recruitment program are as follows➢ Pre-placement talk ➢ Aptitude tests ➢ Group discussion ➢ Technical Interview ➢ HR Interview

Pre Placement Talk: The pre-placement talk is a presentation that is given by the recruiting company’s HR and recruiting team

Various aspects of the company such as its profile,

- history,
- milestone achievements,
- organizational goals,
- its vision,
- mission,
- the job profile,
- products,
- services,
- product lines,
- customers,
- locations,
- branches,
- organizational chart,
- senior management etc

- designation etc
- are also explained in detail

The general format of the selection process remains the same across companies that hire campus graduates

Aptitude Test: Aptitude test is conducted to evaluate how effectively a student could respond to a task or a situation and their communication skills

In short,

this area tests a candidate’s problem solving ability

- logical reasoning,

verbal ability and data sufficiency

- campusrecruitment

Quantitative Aptitude: Numerical ability entails multiple choice questions that are from the topics mostly covered in high school along with some advanced topics

Ratio and proportion,

- mixtures and allegation,
- permutations,
- combinations,
- probability etc

The purpose of this test is to assess the problem solving ability of a candidate under constraints in time

reading ability and also the grammatical knowledge of a candidate

The type of questions that may be asked in verbal ability include grammar based questions (sentence correction/ error identification),

vocabulary based questions (para jumbles,

- synonyms,
- antonyms,
- fill in the blanks,
- cloze passages),
- idioms and phrases,

reading comprehension and occasionally descriptive writing (essays,

- formal/informal letters,

analytical/ issue writing section)

It is mandatory for a candidate to have basic rules of English in place before he or she appears for the campus placement process

Analytical and Logical Reasoning: This section tests the logical reasoning and the analytical ability of a candidate

The questions are generally given in the form of puzzles and a set of questions follow the puzzle

Verbal based reasoning questions such as cause and effect,

assertion and reasons may also be asked

Data Interpretation and Data Sufficiency: Data is presented in various forms such as bar graphs,

pie charts and data should be interpreted accordingly

In data sufficiency,

a problem is presented with some data and a candidate has to determine if the given amount of data is sufficient for problem solving

The aptitude round cannot be underestimated because it is a process of eliminating candidates who do not have enough problem solving abilities,

reasoning skills or acceptable levels of communication

While a few companies may lay more emphasis on communication and numerical abilities,

a few others may stress upon analytical abilities

Regardless of how well a candidate fares academically,

he or she should prepare sufficiently for the aptitude test as this stage in an inevitable phase of any campus selection process

Group discussion is a process of selection rather than a process of elimination

The recruiting team will evaluate certain personality traits like confidence,

- communicating with the team,
- participation,

ability to present one’s views in a clear and concise manner,

- interpersonal skills,
- leadership skills etc

The main intention of group discussion is to assess the behavior of a candidate in a group

In the GD round,

there are usually a minimum of 5 and a maximum of 10 candidates

- hypothetical situations,
- problematic situations,
- abstract topics etc

have a clear thought process and are able to articulate their thoughts lead the group discussion

- students need to be positive,

confident and dynamic in their attitude in this round

They should also develop effective listening skills that would enable them to listen and understand others perspective

Students are advised to keep abreast of current affairs and are expected to familiarize themselves with the popular topics in news

They are advised to form small groups and discuss various topics which would bolster their efforts to successively participate in the group discussions

A student appearing for the technical round should be thorough with the fundamental aspects of his subject

he or she is expected to be proficient in the basic aspects of the subject and able to present the subject in a well formatted manner to his interviewers

It would greatly benefit the students if they would have completed their projects on their own rather than plagiarizing (copying) from other sources

This would exhibit the ingenuity of a student and increase his chances of clearing the technical round

Students who have interned in good organizations have an edge above the others in the technical round as company internships are greatly valued by the recruiters

- campusrecruitment

- attitude,
- confidence,
- flexibility,
- enthusiasm,
- behavioral skills etc

The company HR will market their company through the HR round

but not over confident in the HR round

He or she should be honest and polite in answering the questions and also ask questions to the company HR regarding basic policies,

- procedures and of course,
- the CTC

Many a time it is quite common for technically strong candidates to fail to get through the HR round

is overconfident or is rude or extremely timid in his or her attitude

Self-grooming is very important for a candidate to clear this round of the campus selection

- learnability,
- proper body language,
- confidence,
- clarity of thought,
- interpersonal skills,
- future goals etc

This test the student’s fundamental knowledge of the core subjects in his or her branch and also its practical use

Preparing a good resume is equally important while bracing for the campus selection process

The resume should be prepared in such a way that it reflects a candidate’s capabilities,

- his strengths,

achievements and areas of interests

Hence sufficient efforts should go into preparing a good resume

refer to ‘Resume’ section of this book

Knowledge about current affairs and extra-curricular activities carry about 5% weightage

These areas should also be concentrated upon to increase the chances of making it through the campus interviews

a thorough and systematic preparation in each of the areas mentioned above would go a long way in ensuring that a candidate gets through the right company

- campusrecruitment

- 9 called digits to represent any number

- 9 are called significant digits

• Classification of Numbers: Natural Numbers: The numbers 1,

which we use in counting are known as natural numbers

The set of all natural numbers can be represented by N = {1,

} Whole Numbers: If we include 0 among the natural numbers then the numbers 0,

- are called whole numbers

every natural number is a whole number

} Every natural number is an integer but every integer is not natural number

} is the set of all positive integers

–1} is the set of all negative integers

- 0 (zero) is neither positive nor negative

} is the set of all non negative integers

p is not divisible by q and q ≠ 0,

- are known as rational numbers

(or) Any number that can be written in fraction form is a rational number

- terminating decimals,

and repeating decimals as well as fractions

- 3 5 5 1 3 ,

: 7 2 9 2 5 The set of rational numbers is denoted by Q

- non-repeating decimals

10 22 ,

- √ 2 ,
- √ 3 ,
- : Absolute value of 3 7

Note: A terminating decimal will have a finite number of digits after the decimal point

5 25 3 =0

: 4 4 16 Repeating Decimals: A decimal number that has digits that repeat forever

- 3 repeats forever

: 3 A decimal that neither terminates nor repeats is termed as a Non–Repeating Decimal

- : √ 2=1

4142135623

Real Numbers: The rational and irrational numbers together are called real numbers

- 13 2 3 +4 ,
- : etc are real numbers
- 21 5 7 2 The set of real numbers is denoted by R

- are even numbers

Odd Numbers: An integer that can not exactly divided by 2 is an Odd number

- are odd numbers

Prime Numbers: A Prime Number can be divided evenly only by 1,

- or itself

- : Numbers 2,
- are prime

• 2 is the least and only even prime number

• 3 is the least odd prime number

• Prime numbers up to 100 are 2,

83,89,97

are known as composite numbers

Two numbers which have only 1 as the common factor are called co–primes (or) relatively prime to each other

- : 3 and 5 are co primes

Whole Numbers = 0 (Zero) + Natural Numbers

in and ask doubt with Question Id

• Test of Divisibility: Divisibility by 2: A number is divisible by 2 if the unit's digit is either zero or divisible by 2

: Units digit of 76 is 6 which is divisible by 2 hence 76 is divisible by 2

: The number 273 is divisible by 3 since 2 + 7 + 3 = 12 which is divisible by 3

Divisibility by 4: A number is divisible by 4,

if the number formed by the last two digits in it is divisible by 4,

or both the last digits are zeros

: The number 5004 is divisible by 4 since last two digits 04 is divisible by 4

: 375 is divisible by 5 as 5 is in the units place

: The number 6492 is divisible as it is even and sum of its digits 6 + 4 + 9 + 2 = 21 is divisible by 3

Divisibility by 7: If you double the last digit and subtract it from the rest of the number and the answer is,

- 0 (zero) or divisible by 7

(Note: you can apply this rule to that answer again if you want) e

- : Consider the number 10717

On doubling the unit's digit 7 we get 14

On omitting the unit digit of 10717 we get 1071

- 1071–14 = 1057 is divisible by 7
- ∴ 10717 is divisible by 7

Divisibility by 8: A number is divisible by 8,

if the number formed by last 3 digits is divisible by 8

: The number 6573392 is divisible by 8 as the last 3 digits '392' is divisible by 8

Divisibility by 9: A number is divisible by 9 if the sum of its digit is divisible by 9

: The number 15606 is divisible by 9 as the sum of the digits 1 + 5 + 6 + 0 + 6 = 18 is divisible by 9

- if it ends in zero

: The last digit of 4470 is zero

- it is divisible by 10

Divisibility by 11: A number is divisible by 11 if the difference of the sum of the digits at odd places and sum of the digits at the even places is either zero or divisible by 11

(or) Subtract the first digit from a number made by the other digits

- : In the number 9823,

the sum of the digits at odd places is 9+2=11 and the sum of the digits at even places is 8+3=11

The difference between them is 11 – 11 = 0

the given number is divisible by 11

: 14641 1464 − 1 is 1463 146 − 3 is 143 14−3 = 11,

- which is divisible by 11,
- so 14641 is divisible by 11

• When a number is divisible by another number,

then it is also divisible by each of the factors of that number

: The number 1644 is divisible by 12 as it is divisible by 3 and 4

Divisibility by 14: The number is divisible by 7 and 2

: The number 80388 is divisible by 18 as it satisfies the divisibility test of 9

Divisibility by 25: A number is divisible by 25 if the number formed by the last two digits is divisible by 25 or the last two digits are zero

: The number 7975 is divisible by 25 as the last two digits are divisible by 25

: The number 10824 is divisible by 88 as it is divisible by both 11 and 8

Divisibility by 125: A number is divisible by 125 if the number formed by last three digits is divisible by 125 or the last three digits are zero

: 43453375 is divisible by 125 as the last three digits 375 are divisible by 125

• Common Factors: A common factor of two or more numbers is a number which divides each of them exactly

: 3 is a common factor of 6 and 15

▪ Highest Common Factor (HCF): Highest common factor of two or more numbers is the greatest number that divides each of them exactly

- 12 are the factors of 12 and 36

Among them the greatest is 12

Hence the HCF of 12,

36 is 12

HCF is also called as Greatest common divisor (GCD) or Greatest Common measure (GCM)

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RATIO – PROPORTION CONCEPTS Ratio: A ratio is the relation between two quantities which is expressed by a fraction

• The ratio of the number 'a' to the number 'b' is a (or) a : b or a to b written as b e

: The ratio of 5 hours to 3 hours can be written as 5 (or) 5 : 3

- 3 • The ratio is always a comparison between the quantities of same kind or of same units

For example,

you cannot form the ratio between 5 hours and 3 days

Because the two numbers are expressed in different units

- convert 3 days to hours
- 3 days = 72 hours

Thus the proper form of this ratio 5 (or) 5 : 72

is 72 • Two quantities which are being compared (a : b) are called its terms

The first term (a) is called antecedent and second term (b) is called consequent

• The ratio of two quantities is always an abstract number (without any units)

• If the terms of a ratio are multiplied or divided by the same quantity the value of the ratio remains unaltered

: The ratio a : b is same as Ma : Mb

- a : b and c': d,

then proportion is written as,

a c'= a : b :: c': d'(or) a : b = c': d'(or) b d'Here a,

- d are called Terms

d are called Extremes (end terms) and b,

c are called Means (middle terms)

: Since the ratio 4 : 20 (or) is equal to the ratio 20 1 1 : 5 (or) we may write the proportion as 4 : 20 :: 1 : 5 5 4 1 = or 4 : 20 = 1 : 5 or 20 5 • In a proportion,

product of means (middle terms) is equal to product of extremes (end terms)

- ad= bc or =

b d'Key Notes: If a and b are two quantities,

then 1) Duplicate ratio of a : b=a 2 : b2 2) Sub-duplicate ratio of a : b=√ a: √ b 3) Triplicate ratio of a : b=a 3 : b3 3 3 4) Sub-triplicate ratio a : b= √ a: √ b

- 1 1 5) Inverse or reciprocal ratio of a : b= : a b 2 b 6) Third proportional to a and b is a 7) If a : b = x : y and b : c'= p : q,

then x×p a) a : c= y× q b) a : b : c'= px : py : qy c'e × ×

b d'f 9) The ratio in which two kinds of substances must be mixed together one at x per kg and another at y per kg,

so that the mixture may cost n per kg

The ratio is n y

x n 10) Let the incomes of two persons be in the ratio of a : b and their expenditure be in the ratio of x : y and ff the savings of each person is n then income of each is bn (y x ) an( y x ) and respectively

ay bx ay bx 11) In a mixture the ratio of milk and water is a : b

then the ratio of milk and water in the resulting mixture became a : m

the quantity of milk in the original mixture an = and the quantity of water in the original m b bn mixture = m b 12) In a mixture of n liters,

the ratio of milk and water is x : y

If another m liters of water is added to the mixture,

the ratio of milk and water in the resulting mixture = xn :( yn+mx+my) 13) If four numbers a,

c and d'are given then ad bc a) should be added to each of these (b+c) ( a+d ) numbers so that the resulting numbers may be proportional

ad bc b) should be subtracted from each of (a+d) (b+c) these numbers so that the resulting numbers may be proportional

- 8) Compound Ratio of (a : b),
- (c : d),
- (e : f) is

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- 13 13 Ask doubt with Question Id: 4988 18) I strike to II strike = 1 interval II strike to III strike = 1 interval i
- 2 strikes has one interval time 3 strikes has 2 interval time 10 strikes has 9 interval time Similarly,
- 6 strikes time = 5 interval time 9 intervals time = 27 seconds 5 ×27=15 sec ∴ 5 intervals time = 9 Ask doubt with Question Id: 4989 19)
- 8 men : 11 days 11 men : x If number of men increase,
- number of days will decrease

arrows are in opposite direction

- 8 x 11×8 ∴ = ⇒x= =8 days 11 11 11 Ask doubt with Question Id: 7676 24) 2 kg = 2000 gms
- 2000 gm : 80 rupees 750 gm : x As weight decreases,
- price decreases
- 2000 80 ∴ = ⇒ x= 30 750 x Ask doubt with Question Id: 7677 25) More shadow,
- more height

P1 P2 P3 P4 P5 ∴ Direct proportion between height and shadow

- d'd d'd 12
- 75 = ⇒ x=17
- 5 Poles = 5

- 25 Similarly,
- 18 poles has 17 intervals of distance Ask doubt with Question Id: 7678 i
- 18 poles has ‘17d’ distance 26) Out of 4 parts,
- 3 parts of the work is completed in 6 Similarly,
- 25 poles has ‘24d’ distance

6 = 2 days

For 1 part it takes 3 24 d'Hence,

to complete the last part it takes 2 more days

'24d' distance in = ×34=48 min

- 17 d'Ask doubt with Question Id: 7679 Ask doubt with Question Id: 4990 27) 60 women meal = 100 children meal

th 1 75 children have taken the meal

- of work in 3 days
- 20) P can complete 5 75 children = 45 women

Q can do the complete work in 4 × 4 = 16 days

- 25 children = 15 women
- 31 1 1 Ask doubt with Question Id: 7680 One day’s work of P and Q = = + 6 m : 22
- 8 kg 240 15 16 28) 11
- 25 m : x 240 23 P and Q can complete the work in = =7 days As length increases,
- weight also increases

31 31 6 22

- 8 Ask doubt with Question Id: 4991 = ⇒ x=42
- 75 kg 21) 4 machines produce 140 bottles/min 11
- 25 x 140 Ask doubt with Question Id: 7681 =35 bottles/min ∴ 1 machine produce = 29) Men Work Days 4 26 : 13 : 13 ∴ 1 machine/minute = 35 bottles 39 : 13 : x ∴ 9 machine/ minute = 35×9 = 315 bottles 26 13 x 26×13 Now,
- for 4 minutes,
- 35×4 = 1260 bottles × = ⇒ x= = 8
- 66 more days

- 12 15 Ask doubt using Question Id: 7682 9 x 30) Let required number of days be x

As number of days decrease,

- number of pumps increase
- 5 hens eat 4 packets in 20 days
- 15 x 20 = ⇒ 9 x=15×12 ⇒ x=20 pumps ∴1 packet= =5 days (5 hens in 5 days eat 1 packet) 9 12 4 ∴ Extra pumps = 20–12 = 8

in and ask doubt with Question Id

The word 'per cent' or 'percentage' means for every one hundred

In other words,

it gives rate of a parameter per hundred

: 30% means 30 out of one hundred or 100 Key Notes: • To convert a percent into a fraction,

- divide by 100

20 1 = e

: 20 %= 100 5 • To convert a fraction into a percent,

- multiply by 100
- 3 3 = ×100=75 % e

: 4 4 • To write a decimal as a percent we move the decimal point two places to the right and put the % sign

35 =35 % e

- 35= 100 • Conversely to write a percent as a decimal,

we drop the % sign and insert or move the decimal point two places to the left

- : 43% = 0

12% = 0

Calculating a Percentage:

- ( ) Value

Total For example,

if you obtained 18 marks out of 25 marks,

what was your percentage of marks

? Explanation: Total marks = 25

Marks obtained = 18

- 18 ×100=72 %

∴ Percentage of marks obtained = 25 Calculating Percentage Increase or Decrease: • % Increase : New value = (1+ Increase %) × (Original Value) • % Decrease : New value = (1−Decrease %) × (Original Value) e

: If a book costs 80 and few months later it was offered at a 30% discount

- ? Explanation: Percentage=

30 ×80=0

- 70×80= 56 100 • Calculating Percent Change: Percentage change refers to the relative percent change of an increase or decrease in the original amount

: If a book costs 80 and few monthes later it was offered at a price of 64

What was the discount percentage on that book

? Explanation: Change = 80–64 = 16

Original Value = 80

- 16 1 ×100= ×100= 20% Discount Percentage = 80 5 Calculating Successive Percentages: • If a number is successively increased by x% and y% then a single equivalent increase in that number will be xy x+ y+ %

: The price of an article is successivey increased by 10% and 20%

- 10% Increase
- 20% Increase 110 Overall 32% Increase
- (or) By using formula:

(10)(20)

- %=30+ 2=32
- 100 100 • If there's an increase and a decrease,
- in that case,

the decrease will be considered a negative value

: If there is an increase of 20% and then a decrease of 10% on the price of a commodity,

the successive percentage will be 20×( 10) 20+( 10)+ =20 10 2=8 % increase

- 100 • In case of discounts,

the value of discount percentages will be considered negative

: If a shop keeper give 20% and 10% discounts on a festival day,

the final discount given by shopkeeper is ( 50 )( 50) ( 50)+( 50)+ =

- -100 + 25 = 75% dicount
- 100 • If there are three discounts as x%,

y% and z% then first find the total discount of x% and y% and using it find the total discount with z%

• If the price of commodity increases by x%,

the percentage should a family reduce its consumption so as not to increase the expenditure on the comodity = x ×100

- 100+x • If the price of commodity decreases by x%,

the percentage should a family increase its consumption so as not to decrease the expenditure on the comodity = x ×100

100 x =

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TIME AND WORK CONCEPTS 1) If a person completes a piece of work in 'n' days,

then th 1 work done by that person in one day = part of the n work

th 1 2) If a person completes part of the work in one n day,

then the person will take 'n' days to complete the work

- 3) The total work to be done is usually considered as one unit
- 4) If M1 persons can do W1 work in D1 days and M2 persons can do W2 work in D2 days then M1 D1 W2 = M2 D2 W1
- 5) If M1 persons can do W1 work in D1 days working T1 hours per day and M2 persons can do W2 work in D2 days working T2 hours per day then M1 D1 T1 W2 = M2 D2 T2 W1
- 6) If A can do a piece of work in 'x' days and B can do it in 'y' days then A and B working together will do the xy days
- same work in (x+ y ) 7) If A,

B and C can do a piece of work in x,

y and z days respectively then all of them working together can xyz days

finish the work in (xy+ yz+ zx) 8) If A is trice as good a workman as B then,

- 9) If A is 'k' times efficient than B and is therefore able to finish a work in 'n' days less than B,

then a) A and B working together can finish the work in kn days

- 2 k 1 n days

b) A working alone can finish the work in k 1 kn days

c) B working alone can finish the work in k 1 10) If A,

working alone takes a days more than A and B working together

- 1) A is twice as good a workman as B and takes 10 days less to do a piece of work than B takes

Find the time in which B alone can complete the work

a) 22 days b) 25 days c) 23 days d) 20 days Explanation: Let B alone takes 'x' days to complete the work

x It means A takes days to complete the work

- 2 x From the given information we can write x – =10 2 2x – x x ⇒ =10 ⇒ =10⇒ x=20
- 2 2 Alternate Method: Using formula

k = 2 and n = 10 ∴ Time taken by B working alone to complete the work= 2×10 kn days ⇒ =20 days

k 1 2 1 Ask doubt with Question Id: 1179 2) 25 men can reap a field in 20 days

if the whole field is to be reaped in 37⅟₂ days after they leave the work

a) 5 days b) 4 days c) 3 days d) 4½ days Explanation: 25 men can reap the field in 20 days

⇒ 1 man can reap that field in 25×20 i

500 days

Let 15 men leave the work after x days so that remaining 10 men can complete the work in 37⅟₂ days

It means 25 men have worked for x days and 10 men have worked for 37⅟₂ days

- 1 ∴ 25 x+10×37 = 500⇒25 x=500 375=125(or) x= 5 2 ∴ 15 men must leave the work after 5 days

and forfeits 5 for each day he is idle

At the end of 60 days he gets 50

- he was idle for ____ days

a) 20 b) 25 c) 30 d) 50 Explanation: Suppose,

- the man was idle for x days

∴ 30(60 – x) – 5x = 50 ⇒ x = 50 Ask doubt with Question Id: 1181 4) 12 men or 15 women can do a work in 20 days

- ? a) 21

8 b) 22

8 c) 25

- 3 d) 29 Explanation: or means either men or women
- and means both men and women
- 12 men or 15 women ⇒ 12 men = 15 women ⇒ 4 men = 5 women

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12) Let,

he bought the mobile phone at x

th 1 1 x=7500 (By losing Then x on buying cost) 6 6 5 x=7500 ⇒ x = 9000

∴ 6 Ask doubt with Question Id: 7731 13) For 30000,

- the man loses 25%
- 25 x x=30000 ⇒ x = 40000

100 Now,

- the man wants gain of 25%
- ∴ 40000×
- 18) He sells 0
- 9 mt pipe at rate of 1 mt pipe
- ∴ SP of 0
- 9 mt = CP of 1 mt Let,

- 9 mt = 100,

- 11 then SP of 1 mt = 0
- 11 100 ×100=11
- 11% ∴ Profit = 100 Alternate Method: Using direct formula

100 meters

= 50000 100 Ask doubt with Question Id: 7730 14) Let the price of the article is x

- 8 108 x= x A sold to B at 8% Profit = x + 100 100 108 112 × x B sold to C at 12% Profit = 100 100 108 108 112 x : × x Ratio of the selling prices = 100 100 100 28 = 25 : 28

= 1: 25 Ask doubt with Question Id: 7732 15) Difference between selling prices = 3 In the above explanation,

ratio of selling prices = 25 : 28

The difference of these two (25 and 28) is also 3

one of the selling prices can be either 25 or 28

- ( ( ( (
- 5 50 3 20 6
- 90 meters 100 100 ×100 – 100×90 100 ×100 = 100×90 9 Ask doubt with Question Id: 7736 19) Let,

SP = 100

loss = 20 ∴ CP = (100 + 20) = 120 20 ×100%=16

- 66% ∴ Loss % = 120 Ask doubt with Question Id: 7737
- ) ) ) )

×100=10 %

×100=12

- 5 % 40 ∴ Option-(b) is best,
- as percentage is highest

Ask doubt with Question Id: 7734 x 4 17) Let,

∴ 6× x=8× y ⇒ = y 3 100 4 x y =33

- 33 ×100 = 1 ×100 = Gain % = 3 y 3 Option-(d): Profit percentage =

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INTERESTS AND DISCOUNTS CONCEPTS • The money borrowed or lent out for a certain period is called the principal or the sum

• Interest is the money paid for the use of borrowed money i

extra money paid for using others money is called interest

• Sum of interest and principal is called amount

if the interest on a certain sum borrowed is reckoned uniformly,

then it is called simple interest

SI = Simple Interest T = Time (in years) R = Rate percent per annum • Time must be expressed in the same units used for time in the Rate

: If 1000 is borrowed for 3 years at 10% simple interest,

what is the total amount after 3 yrs Explanation: Year Principal Interest (10%) Amount 1st 1000 100 1100 nd 2 1100 100 1200 3rd 1200 100 1300 1000×10×3 PRT = 300

I= = 100 100 Amount = Principal + Interest = 1000 + 300 = 1300 e

: If 1500 is invested at 15% simple annual interest,

how much interest is earned after 9 months

? Explanation: Here time is in terms of months but interest is in terms of years

- 9 months = years
- 12 1500×15×9 Now,

I = = 168

- 75 12×100 • Key Notes on Simple Interest 1) If a sum of money at simple interest amounts to A1 in T1 years and A2 in T2 years,
- then Principal=

- 3) If a sum of money at simple interest becomes n times of itself in T years then in how many years it will become m times of itself

(m 1 )×T years Required time = ( n 1) 1 th of the 4) If simple interest on a sum of money is x principal and the time T is equal to the rate percent R

- 1 then Rate = Time 100

x 5) A certain sum is at simple interest at a certain rate for T years

- then it would fetch x more

Then the Principal = T×R 1 6) The annual payment that will discharge a debt of P due in T years at the rate of interest R% per annum is 100 P

r2% per annum for next t2 years and r3% for the period beyond that

Then Principal = t 1 r1 +t 2 r 2+(t 3 t 1 t 2 )r 3 8) The simple interest on a certain sum of money at r1% per annum for t1 years= m

r 1 t1 r 2 t2 • Compound Interest: If interest as it becomes due and is not paid to the lender but is added on to the principal,

then the money is said to be lent at compound interest

Compound Interest = P 1+

(A 2 A1 )×100

Amount = P 1+ T 2 T1 A 1 T 2 A 2 T1 C×100 2) A sum of money becomes n times of itself in T years Where T = Number of years and C = Number of times at simple interest,

- then the rate of interest is,
- compounded annually
- 100(n 1) % Rate = T

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Ask doubt with Question Id: 1497 2) Let the numbers be 3x,

- 4x (3x + 4x) = 420 (or) x = 420 ÷ 7 = 60 Smaller number = 3x = 180 Ask doubt with Question Id: 1498 3) Let the other number be x

- 33 Ask doubt with Question Id: 1499 4) Let,
- first number = x

Then other number = 24 – x

- 7x + 5(24–x) = 146 ⇒ 2x = 146–120 ⇒ x = 13

Ask doubt with Question Id: 1500 5) Let the number be x y x + y = 9 and (10x+y)+27 = 10y+x ⇒ x = 3 and y = 6 Hence,

- 36 is the number

Ask doubt with Question Id: 1501 6) (10x + y) – (10y + x) = 9(x – y) This is always divisible by 9

Ask doubt with Question Id: 1502 3 7) 10y + x = of (10x + y) and x + y = 9 8 These equations when solved for x and y will give x = 7,

then y =4x and 10x+y+27 = 10y+x Solving these equations,

- we get,

Thus 14 is the required number

Ask doubt with Question Id: 1504 x +2 3 = ⇒ 5 x+10=66 3 x ⇒ x=7 9) 20 x+2 5 7 Hence the fraction is 13 Ask doubt with Question Id: 1505 x 10) If be the fraction

y x+1 1 x 1 = = then and ⇒ x = 3 and y = 8

y 2 y+1 3 Ask doubt with Question Id: 1506 11) 10x + y = 4 (x + y) and 10x + y + 18 = 10y + x ⇒ x = 2,

Ask doubt with Question Id: 1507 +10 1 12) x 9=+10× ⇒ x 9= ⇒ x2 – 9x – 10 =0 x x (x–10) (x+1) = 0 i

- x = 10,

Number is positive i

x = 10 Ask doubt with Question Id: 7859

- 13) Check with options

- option–c is correct

- let the other number is x
- 6+x = 5(6–x) (or) x+6 = 5(x–6) 6+x = 30–5x (or) x+6 = 5x–30 6x = 24 (or) 4x = 36 ⇒ x = 4 (or) x = 9 Hence,
- both 4 and 9 are correct

option–a: 14 + 41 = 55 (option–a is correct) option–b: 24+42 = 66 ≠ 55 Hence,

- only option–a is correct

- 3x + x = 28 ⇒ 4x = 28 ⇒ x = 7 Largest number = 3 x=3×7=21

Ask doubt with Question Id:7863 17) Let 5 consecutive natural numbers are: x,

(x)+( x+1)+(x +2)+(x +3)+(x +4) =4 ∴ 5 5x +10 = 20 ⇒ 5x = 10 ⇒ x = 2 ∴ Smallest number= x = 2 Ask doubt with Question Id:7864 18) Let,

- the number be x
- ∴ By given condition,

x + √ x= √ 400 x + √ x=20 Now,

by checking with options: option–a: 4 + √ 4 = 4+2 = 6 ≠ 20 option–b: 16+ √16 = 16+4 = 20 ∴ Option-b is correct

Ask doubt with Question Id: 7865 19) Let,

Salman has x then Sohail has x2

∴ x+x2 =110 ⇒ x2+x–110 =0 ⇒ x2+11x–10x–110=0 x(x+11)–10(x+11)=0 ⇒ x = 10,

x = –11 ∴ Rupee is positive,

- ∴ Sohail has x2 = 102 = 100

Ask doubt with Question Id: 7866 20) Let,

- the digits are x and y

∴ 10x + y + 9 = 10y + x 9x – 9y = – 9 ⇒ x – y = –1

- (1) By other condition,
- x + y = 3
- (2) Solving (1) and (2),

y = 2 ∴ Number = 10x + y = 10×1+2 = 12 Ask doubt with Question Id: 7867

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A 10) Centroid: The point of concurrence of the medians D'of a triangle is called Centroid and is denoted by 'G'

≈ 11) If G is the Centroid of ΔABC,

then B C E F a) AD is one of its median then G divides AD in the AB BC AC = = ratio 2 : 1

ΔABC ~ ΔDEF ↔ DE BF DF b) AB 2 + BC 2 + CA 2 = 3 (AG 2 + BG 2 + CG 2 )

Similarity: If the ratio of the corresponding sides G of two triangles are equal then they are similar

S Similarity: If the ratio of two corresponding sides A C of two triangles are equal and their included angles are 1 ΔABC

c) ΔABG = ΔBCG = ΔACG = equal then they are similar

- 3 QUADRILATERALS B A quadrilateral is a plane figure bounded by four straight lines called sides

CA and AB respectively then parallel opposite sides

- 1 a Δ ABC Perimeter = Sum of all sides and ΔAEF = ΔBDF = ΔCDE = ΔDEF = 4 ( a+b)×h Area = h 3(AB2 + BC2 + CA2) = 4(AD2 + BE2 + CF2)

- 14) In an equilateral triangle with side 'a' and altitude 2 h 3 2 3 √ √

a Area= a 'h' we have 'h' = or and 4 2 √3 1 ×base× height

- 15) The area of a ΔABC = 2 1 1 1 a b sin C = b c'sin A = a c'sin B = 2 2 2 16) If in a ΔABC,

a circle is inscribed by touching the sides at P,

b • A parallelogram is a quadrilateral with (a) two pairs of parallel opposite sides (b) two pairs of equal oppsite sides (c) two pairs equal opposite angles

• A rhombus is a parallelogram with (a) four equal sides a yo a (b) equal opposite angles d1 (c) no parallel sides

xo xo d'Perimeter = 4a 2 yo 1 a a ×d 1 ×d 2 Area = 2 In a rhombus,

- d12 + d22 = 4a2
- d1 and d2 are two diagonals

B • A square is a rhombus with R P (a) 2 pairs of parallel lines (b) 4 equal sides A C Q (c) 4 equal internal right angles

- 90o Similar triangles: Two triangles ABC,

DEF are said to Perimeter = 4a

be similar if their corresponding angles are equal (or) Area = a2

a a the ratio of their corresponding sides are equal

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PERMUTATIONS AND COMBINATIONS CONCEPTS •Fundamental Principal of Multiplication: In general if some procedure can be performed in n1 different ways,

- and if,
- following this procedure,

a second procedure can be performed in n2 different ways,

- and if,

following this second procedure,

a third procedure can be performed in n3 different ways,

and so fourth then the number of ways the procedure can be performed in the order indicated is the product n1

: A letter lock consists of 5 rings each marked with 10 different letters

What is the maximum number of unsuccessful attempts to open the lock

Hence each ring has 10 positions

the total number of attempts that can be made to open the lock is 10 x 10 × 10 × 10 × 10 = 105 Out of these,

there must be one attempt in which the lock will open

∴ Total number of unsuccessful attempts = 105

- -1 •Fundamental Principle of Addition: If there are two operations such that they can be performed independently in m and n ways respectively,

then either of the two operations can be performed in (m+n) ways •Factorial: The product of first 'n' natural numbers is called the 'n'-factorial and is denoted by n

- (n–2)
- (n–1)
- n Example: 4

4 = 24,

5 = 125,

24 = 120,

120 = 720

Note: 1) 0

!=1 2) The product of 'r' consecutive positive integers is divisible by r

- ! 3) (kn)
- ! Is divisible by (n

!)k for all k is a positive constant

- 4) The product of 2n

! consecutive positive integers is equal to 2(n

PERMUTATIONS •Permutation: An arrangement of any r ≤ n of these objects in a given order is called an r–permutation or a permutation of the 'n' objects taken 'r' at a time

Example: Consider the set of letters a,

dcba and acdb are permutations of the 4 letters taken all at time

- (ii) bad,

cbd and bca are permutations of the 4 letters taken 3 at a time

- (iii) ad,

da and bd are permutations of the 4 letters taken 2 at a time

Before we derive the general formula for P(n,

- r) we consider a special case

Find the number of permutations of 7 objects,

- g taken three at a time

find the number of 'three letter words' with distinct letters that can be formed from the above seven letters

- following this,

the second letter can be chosen in 6 different ways

the last letter can be chosen in 5 different ways

Thus by the fundamental principle of counting there are 7

- 5=210 possible three letter words without repetitions from the seven letters

(or) There are 210 permutations of 7 objects taken 3 at a time

3) = 210

r) follows the procedure in the preceding example: The first element in an r-permutation of n-objects can be chosen in 'n' different ways

- following this,

the second element in the permutation can be chosen in (n–1) ways

the third element in the permutation can be chosen in (n–2) ways

we have that the rth (last) element in the r–permutation can be chosen in n–(r–1) = n–r+1 ways

- ! Thus P(n,
- r) = n(n–1) (n–2)
- (n–r+1) = (n r)

! The second part of the formula follows from the fact that n(n–1)(n–2)

(n–r+1) = n (n 1)( n 2 )⋯⋯(n r+1)⋅( n r )

- ! = (n r)

!) ∴ A formula for the number of possible permutations of n

! 'r' objects from a set of 'n' is P(n,

- r) or n p r = (n r )

! In the special case that r = n,

- we have P(n,
- n) = n(n–1) (n–2)
- ! (in other words there are n

! permutations of 'n' objects taken all at a time)

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PROBABILITY CONCEPTS •Random Experiment: Probability is the study of random or non deterministic experiments

If the die is tossed in the air,

then it is certain than the die will come down,

- but is non certain that,
- say a 3 will appear

Definition: A random experiment is an experiment whose result would not be predicted but the list of possible outcomes are known

The result of random experiments may not be predicted exactly but the result must be with in the list of predicted outputs

- 2) Rolling a die is a random experiment,

since its results could not be predicted in any trial

- 3) Selection of a plastic component and verification of its compliance
- 4) Life time of a computer
- 5) Number of calls to a communication system during a fixed length interval of time

•Outcome: The result of a random experiment will be called an outcome

Example: 1) Tossing a coin

The result is ether Head(H) or Tail(T) 2) In an experiment of throwing a six-faced die

5 and 6

•Sample space: The set of all possible outcomes of some given experiment is called sample space

an element in in that set is ca