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CAMPUS RECRUITMENT COMPLETE REFERENCE

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Praxis Groups

PRAXIS GROUPS India’s First online educational material publishers Address: Mig-B-35,

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Third Book Edition 2015 Online Edition 2007,

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About the Authors  Ramanadham Ramesh Babu,

D): Assistant Professor,

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

He has gifted us with fabulous content for the Verbal Ability

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

University of Pune,

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

A PGDTE) and A

Kalyani (M

A English) : Their vast academic experience,

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,

Associate Professor,

Department of ECE,

Rega Rajendra,

Professor in Mechanical Engineering,

Siva Kumar,

Assistant Professor,

Department of EEE,

Srinivasa Kumar and Dr

Suresh Kumar,

Professors in Dept of Civil Engineering,

Kiran Kumar Cheriyala MS(UK),

B-Tech,

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

They have been visiting MNCs and conducting interviews for the past several years

About Quality Analyst,

Co-ordinator and Executive  Vamshi Krishna Kenche (B-Tech,

JNTU): A man of commitment,

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

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

Without Vamshi,

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

About Data Entry operators  Aruna sherpally and Md

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

About Graphic designers  Ranjith Kumar Durishetty (B-com,

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

Inside the Book Introduction to Campus Recruitment Procedure

QUANTITATIVE APTITUDE

LOGICAL REASONING

Number System

Coding–Decoding

Averages

Seating Arrangement

Ratio – Proportion

Day Sequence/ Calender

Chain Rule / Variation

Direction Sense Test

Percentages

Blood Relations

Mixtures and Allegations

Syllogisms

Time and Work

Pipes and Cisterns

Analogy – Classification

Time Speed and Distance

Alphabet Test

Character Puzzle

Profit and Loss

Symbol Based Operations

Interests and Discounts

Arithmetic Reasoning

Partnerships

Statements and Arguments

Problem Solving

Non Verbal Reasoning

Basic Geometry

Cubes and Dices

Solid Mensuration

VERBAL ABILITY

Co-ordinate Geometry

Grammar

Permutations and Combinations

Sentence

Probability

Parts of Speech

Data Sufficiency

Data Interpretation

• Tabular Data Interpretation

• Line Graphs or Cartesian Graphs

Inequalities

Functions

Sentence Correction/ Spotting Errors

Logarithms

Vocabulary

Set Theory

Antonyms

Progressions

Synonyms

Quadratic Equations

Analogy

Reading Comprehension

GROUP DISCUSSION

Closet – Fill in the blanks

Introduction

Sentence Rearrangement

Skills required for Group Discussion

Theme Detection

Steps to approach a Group Discussion

PLACEMENT PAPERS Placement Paper-1

Reasons for failure in Group Discussion

44 Summary

Placement Paper-2

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TECHNICAL INTERVIEW Civil Engineering

HR INTERVIEW What qualities does an Interviewer observe

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

RESUME PREPARATION Tips to build an effective Resume

How to know a company is genuine

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

What would be the quality of education imparted

How would the education benefit the student academically or otherwise

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

Companies hire engineering graduates into roles that range from software application development,

application maintenance network security and support etc

Management graduates are hired for roles in business development,

IT consulting,

customer relationship management,

HR roles,

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

Interact with senior students who have been hired into different organizations

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

Interact with the faculty and understand how the technical interviews could be facilitated

Map your strengths to the profile of the job

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

Understanding the profile of the job and the opportunities for growth within the organization,

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

Study and understand the philosophy,

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,

The role of the job offered as well as its description along with the selection criteria,

Generally the pre-placement talk is presented by a senior member of the delivery or the HR team

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

Minor variation may be present

Selection happens in the following stages

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

The areas that are normally tested are numerical or quantitative ability,

verbal ability and data sufficiency

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

The various topics from which questions may be asked are Number theory,

Averages,

Ratio and proportion,

Time and Distance,

Percentages,

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

This area can be effectively countered if a student prepares sufficiently beforehand

Verbal Ability: This area tests the communication skills,

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,

reading comprehension and occasionally descriptive writing (essays,

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

It is required by a candidate to rationally approach the puzzle by interpreting the logic

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: Those selected in the aptitude test will be called for group discussion

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

The recruiting team will evaluate certain personality traits like confidence,

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

These are the traits that the employers would want to see in their potential employees

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

The topic of the discussion is normally related current topics,

Students who are confident,

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

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

Technical Interview: The pre final round of the selection process is the technical interview

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

While a student may not be expected to know the entire subject inside out,

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

HR Interview: The HR interview is more of a two way process

A student is assessed for his communication,

The company HR will market their company through the HR round

A student should be confident,

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,

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

This may happen if the candidate has poor presentation skills,

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

The recruitment process weighs heavily on the attitude of a candidate

A candidate should exhibit positive thinking,

This segment has about 75% weightage

Academic knowledge has a weightage of about 10%

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

Communication skills carry a weightage of about 10%

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

The resume is the first impression a recruiter has of the candidate

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

achievements and areas of interests

Hence sufficient efforts should go into preparing a good resume

For further reading,

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

QUANTITATIVE APTITUDE

NUMBER SYSTEM CONCEPTS In Hindu–Arabic system we use ten symbols 0,

This is the decimal system where we use the digits 0 to 9

Here 0 is called insignificant digit where as 1,

• 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,

every natural number is a whole number

The set of whole numbers is represented by W

Integers: All counting numbers and their negatives including zero are known as integers

The set of integers can be represented by Z or I

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

Positive Integers: The set I + = {1,

} is the set of all positive integers

Positive integers and Natural numbers are synonyms

Negative Integers: The set I – = {

–1} is the set of all negative integers

Non Negative Integers: The set {0,

} is the set of all non negative integers

Rational Numbers: The numbers of the form q where p and q are integers,

p is not divisible by q and q ≠ 0,

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

This includes integers,

and repeating decimals as well as fractions

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

Irrational Numbers: Any real number that cannot be written in fraction form is an irrational number

These numbers include the non-terminating,

10 22 ,

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

4142135623

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

Even Numbers: All those natural numbers which are exactly divisible by 2 are called even numbers

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

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

And it must be a whole number greater than 1

All primes which are greater than 3 are of the form (6n+1) or (6n–1)

Note: • 1 is not a prime number

• 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

There are 25 prime numbers up to 100

Composite Number: Natural numbers greater than 1 which are not prime,

are known as composite numbers

The number 1 is neither prime nor composite

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

Note: Natural Numbers = Prime + Composite Numbers

Whole Numbers = 0 (Zero) + Natural Numbers

Integers = Negative Integers + 0 + Positive Integers

Real Numbers = Rational + Irrational Numbers

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• 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

Units digit of 330 is 0 so it is divisible by 2

Divisibility by 3: A number is divisible by 3 if sum of all digits in it is divisible by 3

: 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

Divisibility by 5: A number is divisible by 5 if the units digit in the number is either 0 or 5

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

Divisibility by 6: A number is divisible by 6 if it is even and sum of all digits is divisible by 3

: 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,

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

On doubling the unit's digit 7 we get 14

On omitting the unit digit of 10717 we get 1071

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

Divisibility by 10: A number is divisible by 10,

: The last digit of 4470 is zero

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

If that number is divisible by 11 then the original number is,

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,

• When a number is divisible by another number,

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

Divisibility by 12: A number is divisible by 12 if it is divisible by 3 and 4

: 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

Divisibility by 15: The number is divisible by 3 and 5

Divisibility by 16: The number is divisible by 8 and 2

Divisibility by 18: An even number satisfying the divisibility test by 9 is also divisible by 18

: 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

Divisibility by 88: A number is divisible by 88 if it divisible by 11 and 8

: 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

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

For example,

you cannot form the ratio between 5 hours and 3 days

Because the two numbers are expressed in different units

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

Proportion: Equality of two ratios is called proportion

Consider the two ratios,

then proportion is written as,

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

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)

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

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

In this mixture another n liters of water is added,

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

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L and M complete the work in

arrows are in opposite direction

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

Distance = 4d x 40

6 = 2 days

If car crosses ‘17d’ distance in 34 minutes then it crosses 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

th 1 75 children have taken the meal

P can do the complete work in 3 × 5 = 15 days

Still food is left for 25 children

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

31 31 6 22

Ask doubt with Question Id: 7674 39 13 13 39 22) Pumps Days ∴ Cannot be completed in 7 days

As number of days decrease,

Then 1 hen requires 5×5 = 25 days

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PERCENTAGES CONCEPTS A percentage is a way of expressing a number as a fraction of 100

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

In other words,

it gives rate of a parameter per hundred

It is denoted by the symbol %

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

20 1 = e

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

: 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

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

12% = 0

Calculating a Percentage:

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

∴ 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

How much does the book cost now

30 ×80=0

Change ×100 Percent= OriginalValue New 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

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

What is the overall percent increase in the price of the article

Explanation:

(10)(20)

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

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)+ =

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

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

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,

Ratio of work done by A and B = 3 : 1

Ratio of times taken by A and B to finish a work =1:3

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

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

Then the number of days taken by A and B working together to finish the job is √ ab

CONCEPTUAL EXAMPLES

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

A is twice as good workman as B

x It means A takes days to complete the work

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

When should 15 men leave the work,

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

Ask doubt with Question Id: 1180 3) A man is paid 30 for each day he works,

and forfeits 5 for each day he is idle

At the end of 60 days he gets 50

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

∴ 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

In how many days 7 men and 5 women would complete the work

8 b) 22

8 c) 25

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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,

100 Now,

CP of 1 mt = 100 ∴ If SP of 0

1×100 =111

100 meters

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

= 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

Option-d is correct choice

Ask doubt with Question Id: 7733 16) Checking from options

Calculating profit percentages

Option-(a): Profit percentage = Option-(b): Profit percentage = Option-(c): Profit percentage =

SP = 100

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

×100=10 % ×100=15 %

×100=10 %

×100=12

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

SP = x and CP = y

∴ 6× x=8× y ⇒ = y 3 100 4 x y =33

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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

Amount = Principal + Interest • Simple interest: For a certain period,

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

then it is called simple interest

Denoted by S

I= P R T 100 Amount = Principal + Simple Interest P = Principal

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

Time must be expressed in the same units used for time in the Rate

I = = 168

A1 T 2 A 2 T 1

(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

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

If it had been put at R1% higher rate,

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

RT (T 1) Annual payment 100T+ 2 7) Let the rate of interest for first t1 years is r1% per annum

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

Suppose all together the simple interest for t3 years is 'SI' 100×SI

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

The interest on the same sum for t2 years at r2% per annum = n

Then the sum = (m n)×100

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

And the total sum owed after a given time is called the amount at compound interest for that time

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,

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EXPLANATIONS 1) x2 – y2 = 1160 ⇒ x + y = 116 ⇒ x y=

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

There can be 2 possibilities ⇒ x+10 = 2(x–10) (or) x+10 = 2(10–x) ⇒ x = 30 (or) 3

Then other number = 24 – x

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,

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,

Ask doubt with Question Id: 1503 8) If the number is (xy),

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

Thus 14 is the required number

Its square is 196

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,

The two digit number = 24

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

Number is positive i

x = 10 Ask doubt with Question Id: 7859

Option-a: 52 + 72 = 74 ≠ 202

Option-b: 72 +92 = 130 ≠ 202 Option-c: 92+112 = 202 = 202

Ask doubt with Question Id::7860 14) By given condition,

Ask doubt with Question Id:7861 15) Check with options

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

Ask doubt with Question Id:7862 16) Let the number be 3x and x

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,

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,

Ask doubt with Question Id: 7866 20) Let,

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

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 corresponding angles of two B triangles are equal then they are similar

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

G D'The sum of the four angles of a quadrilateral is 360o

A C Types of Quadrilaterals: 12) In a Δ ABC if D,

F are the mid points of the sides • A trapezium is a quadrilateral with one pair of BC,

CA and AB respectively then parallel opposite sides

C D'13) If the lengths of 2 medians of a triangle are equal then it is an isosceles triangle

If the lengths of 3 medians of triangle are equal then it is an equilateral triangle

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

a circle is inscribed by touching the sides at P,

R respectively then AP+BQ+CR = PB+QC+RA = (AB+BC+CA)

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

Perimeter = 2(a+b) yo Area = a × h

• 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,

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

DEF are said to Perimeter = 4a

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

Diagonal = √ 2

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,

a second procedure can be performed in n2 different ways,

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

Explanation: Each ring is marked with 10 different letters

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

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

4 = 24,

5 = 125,

24 = 120,

120 = 720

Note: 1) 0

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

!)k for all k is a positive constant

! 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,

Then (i) bdca,

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

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

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

The number of permutations of 'n' objects taken 'r' at a time will be denoted by P(n,

Before we derive the general formula for P(n,

Find the number of permutations of 7 objects,

In other words,

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

Let the general three letters word be represented by three boxes

Now the first letter can be chosen in 7 different ways

the second letter can be chosen in 6 different ways

the last letter can be chosen in 5 different ways

Write each number in its appropriate box as follows: 7

Thus by the fundamental principle of counting there are 7

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

3) = 210

The derivation of the formula for P(n,

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

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

Continuing in this manner,

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

! 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 )

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

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

! In the special case that r = 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,

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

The unpredicted outcomes could not be taken under random experiments

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

Example: 1) Tossing a fair coin

since its results could not be predicted in any trial

•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

The possible outcomes are 1,

5 and 6

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

A particular outcome,

an element in in that set is ca

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