193243068 banking exam idea

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WWW.JAGRANJOSH.COMSTUDY MATERIAL FOR IBPS

CLERK RECRUITMENT 2012-13


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Preface

Paths can be many, but to reach the correct destination, the correct path must be chosen &

followed.

Unless there is a proper focus on your goal with correct guidance, same cannot be achieved in a

set time frame. Most people fail not because they lack knowledge but because they lack

direction.

What is the Book About?

In one sense, it is a guide book for IBPS CWE Clerical Cadre Recruitment Exam whoseSummary

book is already available on jagranjosh.com. This book is complete in itself. It contains the

ingredients, the methods that aspirants will need to follow to crack the Bank Recruitment

Exams & have an absolute understanding of the pattern of Questions in the Bank Recruitment

Exams. This Book comprehensively covers all the topics as per the format of the IBPS CWE Clerk

Recruitment Exam. This is a result of efforts made by Jagranjosh.com in providing appropriate

study material for Competitive Exams.

In this book all the sections have been made interesting for aspirants with a practical & user

friendly approach.

How to Read this Book

The concepts & exercises in this book cannot be absorbed by casual browsing. It should be read

properly & all the exercises must be practiced on a regular basis.

To strategize the preparation, aspirants must first read carefully the theory & concepts given at

the beginning of each section. Once the theory & concepts have been thoroughly understoodby an aspirant, then he or she can move on to the practice exercises.

To achieve accuracy & timeliness, aspirants must solve the practice exercises in a set

timeframe. This book has been designed keeping in view the requisites for the current pattern

of the Bank recruitment Exams.

Jagranjosh.com at a glance

Jagranjosh.com since its inception has always worked tirelessly in bringing up content for all

serious aspirants. Jagranjosh.com as a site can be accessed by students applying for admissions,

looking for government sector jobs, aspiring to excel in all fields of education especially the

competitive examinations. It expertises and is a gateway to infinite knowledge on Current

Affairs, UPSC/State PSCs examinations, Banking, SSC and also provides information on MBA,

CBSE and other admission related queries. Jagranjosh.com is more than a site it is companion

for life, for those wanting to be successful; it is the key to success.

Best of Luck!
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TABLE OF CONTENTS

Numerical Ability ............................................................................................................................. 5

Simplification ............................................................................................................................... 5

H.C.F. & L.C.M. of Numbers ....................................................................................................... 13

Average ...................................................................................................................................... 19

Surds & Indices .......................................................................................................................... 25

Percentage ................................................................................................................................. 30

Profit & Loss ............................................................................................................................... 36

Partnership ................................................................................................................................ 42

Time & Work .............................................................................................................................. 48

Time & Distance ......................................................................................................................... 53

Problems on Trains .................................................................................................................... 58

Simple Interest ........................................................................................................................... 63

Compound Interest .................................................................................................................... 70

Probability .................................................................................................................................. 75

Permutations & Combinations .................................................................................................. 81

Data Interpretation .................................................................................................................... 84

Reasoning ...................................................................................................................................... 92

Analogy ...................................................................................................................................... 92

Classification .............................................................................................................................. 98

Series ....................................................................................................................................... 104

Coding - Decoding .................................................................................................................... 111

Puzzle Test ............................................................................................................................... 116

Mathematical Puzzles .............................................................................................................. 123

Direction Sense Test ................................................................................................................ 131

Syllogism .................................................................................................................................. 137

Abstract Reasoning .................................................................................................................. 143


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English Language ......................................................................................................................... 160

Synonyms ................................................................................................................................. 160

Antonyms ................................................................................................................................. 166

Spotting the error .................................................................................................................... 171

Sentence Completion .............................................................................................................. 176

Sentences Improvement ......................................................................................................... 182

One word Substitution ............................................................................................................ 189

Spelling Test ............................................................................................................................. 193

Idiom & Phrases ....................................................................................................................... 200

Sentence Reconstruction ......................................................................................................... 204

Cloze Test ................................................................................................................................. 207

Comprehension ....................................................................................................................... 225

Computer Knowledge ................................................................................................................. 243

General Awareness (with special reference to banking Industry) ............................................. 260

Person Appointed .................................................................................................................... 260

Awards / Honours .................................................................................................................... 264

Person Died .............................................................................................................................. 266

International ............................................................................................................................ 268

National ................................................................................................................................... 273

Economy .................................................................................................................................. 276

Corporate ................................................................................................................................. 285

Sports ....................................................................................................................................... 289

One Liner (From February to August) ...................................................................................... 302

Current Affairs for Bank Exams ............................................................................................... 315


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NUMERICAL ABILITY

SIMPLIFICATION

Some Important Concepts

BODMAS Rule: This BODMAS Rule shows the correct sequence of all the operations that are

to be executed to find out the value of a given expression. In this rule B Stands for Bracket,

O stands for of, D for Division, M for Multiplication, A for Addition and S for

Subtraction.

Therefore, the correct order to simplify an expression is:

(a) ( )

(b)

{}

(c)

[]

(d)of

(e)Division

(f)

Multiplication

(g) Addition

(h)Subtraction

Modulus of a Real Number:If the real number is r, then

| r | = {r, if r0 .Solved Examples

Example1: What will be the value of x in the following equation?

5+

+ x +2

= 9

.

Solution: Simplifying the above equation

x = 9

2

3 - 5

1

3 -

1

4 2

1

5 x =

29

3 -

16

3 -

1

4 -

11

5

x =58032015132

60

x =113

60= 1

53

60.

Example2: Simplify: 6000 240 + 10.


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Solution:Given expression:6000

240+ 10

= 25 +10

= 35.

Example3: of of x is greater than of of x by 4. What is the value of .Solution:Given that,

1

5

2

7 x -

1

7

2

9 x = 4

95

315= 4

4315

= 4 x =315

Required number3=

315

3= 105.

Example 4: Simplify 120 24 of + 2 .Solution:Given expression: 120 24

1

3+

1

5

5

2

120 8 +1

2

120

8+

1

2

31

2.

Example5: If=

, then find the value of

+ .

Solution: Given expression: 3 + 3 3 3 =

(/)3 + 1(/)3 1

(3/2)3 + 1

(3/2)3 1 (x/y =3/2)

27/8 + 1

27/8 1 35/8

19/8

35

19 .

Example 5:15.04 ..= ?

Solution: Given expression: 15.04 4.8

0.6=15.04

48

6

=15.04 8 = 120.32.

Example 6: Find the value of

+

+

+

+

.

Solution: Given Expression:1

2 +

1

23+

1

34+

1

45+

1

56

= ( 1 -1

2) + (

1

2-

1

3) + (

1

3-

1

4) + (

1

4-

1

5) + (

1

5-

1

6)


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= 1 -1

6=

5

6.

Example 7: Arjun spendsof his salary on house rent,

of his salary on food and

of his

salary on conveyance. If he has Rs.2400 left with him, then find his expenditure on

conveyance.

Solution:Suppose Arjuns monthly salary is Rs. x

Then, remaining part of his salary = x- (1

5+

1

6+

1

10) x = x (

14

30) x =

16 x

30=

8 x

15.

Now,8 x

15= 2400 x =Rs. 4500.

Expenditure on Conveyance = Rs.( 110

4500 ) = Rs.450.

Exercise

1.

(-2) (-4) (5) (1

10) = ?

(a) 4

(b)

-4

(c) 2

(d)

1

2.

7 77 77077 =?

(a) 707

(b)777

(c) 7007

(d)

7707

3.

If 56 [12 {2+4 (2+?)}] = 45.What should come at the place of ? in the equation.

(a) 3

(b)

4

(c)

5

(d)6

4.5

?=

10

7

(a) 2/7


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(b)

3/7

(c) 7/3

(d)7/2

5.

Evaluate: 92[4(1+1)]2|62||36 |3 .(a)

2

(b)3

(c) 4

(d)5

6.How many1

4s are there in 49

1

2.

(a)

195

(b)196

(c) 197

(d)

198

7.The smallest number which should be subtracted from the sum of 11

5, 2

1

3, and 3

1

6 to make

the result a whole number is:

(a) 0.5

(b)

0.8

(c) 0.7

(d)0.9

8.

If x * y = 2+ y, then what is the value of p in the expression (2 * 4) * p = 100?(a) 36

(b)

74

(c)

65

(d)76

9.

5 is added to a certain number; the sum is multiplied by 3; the product is divided by 4 and 2

is subtracted from the quotient. The remainder left is 4. What will be the number?

(a)

4


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(b)

5

(c) 2

(d)3

10.

0.004 0.02 0.8 (0.2 0.40) =?

(a) 0.8

(b)0.08

(c) 0.008

(d)

0.0008

11.The value of1

3+1

3+1

313

is:

(a)

26

89

(b)27

89

(c)28

89

(d)

25

89


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12.

Which of the following values of m and n satisfy the equation I and II.

I. 2m + 4n = 8

II. 8m + 6n = 22

(a)

1,2

(b)2,1

(c) 1,1

(d)2,2

13.

How many pieces of 25 cm length can be cut from a 62.5 meters long Aluminum stick?

(a)

500

(b)230

(c) 240

(d)225

14.If one fourth of a tank holds 45 liters of milk, then the quantity of water that one third of

the tank holds is:

(a) 65L

(b)60L

(c)

62L

(d)

61L

15.A cricket team won 3 matches more than they lost. If a win gives them 2 points and loss (-

1) point, how many matches, in all, have they played if their score is 23.

(a)

34 matches

(b)

35 matches

(c) 36 matches

(d)

37 matches

Answers

1. (a)

2. (c)

3. (a)


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4.

(d)

5.

(a)

6. (d)

7.

(c)

8. (a)

9. (d)

10.(d)

11.

(b)

12.

(a)

13.(d)

14.(b)

15.

(d)

Solution and Explanation

1.

Given expression is (2 4 5 1

10) = 4.

2.

7 77 77077 =77077

11= 7007

3.56 [12 {2+4 (2+?)}] = 45 56 [12 4 ?] = 45 ? = 3.

4.5

?=

10

7 ? =

5 7

10 ? =

7

2.

5.Given Expression:92[4(1+1)]2

|62||36 |3 =7[42 ]2|4||3 |3 = =

722433=

7141 =

6

3= 2.

6.Required number:49

1

21

4

=99

2

4

1= 198.

7.

Sum of all the fractions = 11

5+ 2

1

3+ 3

1

6=

201

30= 6.7. The whole number just less than 6.7 is 6,

then the required number = 6.7- a = 6 a = 0.7.

8.

Given expression: (2 * 4) * p = 100 (4 + 4) * p = 100 64 + p = 100 p =36.

9.Suppose the required number is x then,

3 ( x+5)

4 2 = 4 3(x + 5) 8 = 16 x + 5 = 24/3 x = 3.

10.0.004 0.02 0.8

0.2 0.40 =

64

8 10000= 0.0008.


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11.

Given expression:1

3+1

3+1

313

=1

3+1

3+183

=1

3+1

3+38

=1

3+1

278

=1

3+8

27

=1

3+8

27

=1

89

27

=27

89

12.I. 2m + 4n = 8 II.8m + 6n =22.

Multiplying (I) by 4 and subtract the (II) from (I), then we get n =1.

Put n=1 in (I) then we get m=2.

13.

Number of pieces =62.5 100

25=

6250

25= 250.

14.Suppose the capacity of the tank is x litres. Then,1

4x = 45 x = 180

1

3x = 60.

15.Suppose they have lost x matches. Then, number of matches won = x+3.

2(x+3) x = 23 x =17.

Hence, total number of matches played by the team = x + (x+3) = 2x + 3 =37.


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H.C.F. & L.C.M. OF NUMBERS

If a number x divides another number y exactly, then x is said to be a factor of y. To add,

subtract, or just compare two fractions, we need to convert both fractions to a common

denominator. The least common multiple is usually used, although any common multiple would

work in this particular case. Here comes the concept of Least Common Multiple (L.C.M.).

Also to simplify a fraction, we divide the numerator and the denominator by the same number.

If we divide them by the greatest common factor, then no further simplifications are required.

That greatest common factor is called the Highest Common Factor (H.C.F.) or Greatest

Common Divisor (G.C.D.).

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.

There are two methods of finding the L.C.M .of two or more than two numbers:

1.

Factorization Method: In this method first resolve each one of the given numbers into aproduct of Prime factors. Then Least Common Multiple is the product of highest powers of

all the factors.

2. Common Division Method: In this method first arrange the set of numbers in any order.

Then take a number which divides exactly at least two of the given numbers and carry

forward the remaining numbers which are not divisible by that particular number. Repeat

the process till no two of the numbers are divisible by the same number. Then Least

Common Factor is the product of the divisors and the undivided numbers.

Highest Common Factor (H.C.F.): 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 two or more than two numbers:

Factorization Method: First resolve each one of the given numbers as the product prime

factors. Then H.C.F. is the product of least powers of common prime factors.

Division Method: If we want to find out the H.C.F. of two numbers then divide the larger

number by smaller one. Again divide the divisor by the remainder. We should repeat this

process till zero is obtained as remainder. If we have three numbers then H.C.F. of {(H.C.F. of

any two) and (the third number)} will be the H.C.F. of three given numbers.

Important Formulas of L.C.M. and H.C.F. of Numbers

Product of two numbers =Product of their H.C.F and L.C.M.

1.

H.C.F. =H.C.F .of Numerators

L.C.M .of Denominators

2.

L.C.M. =L.C.M .of Numerators

H.C.F .of Denominators


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Solved Examples

Example 1. Find the L.C.M. of 50, 35 and 70.

Solution.Let us apply the Factorization Method to find the L.C.M. of 50, 35 and 70.

Express each one of the given number as the product of prime factors.

50 = 2 52, 35 = 5 7, 70 = 2 5 7.

L.C.M. = Product of highest powers of 2, 5 and 7 = 2 52 7 = 350.

Example 2. Find the L.C.M. of 18, 20, 24, and 54.

Solution.Let us apply Common Division Method to find the L.C.M. of 18, 20, 24, 54.

Example 3. Find the H.C.F. of 5, 5 and 2 Solution. Here prime numbers common to given three numbers are 2 and 3.

H.C.F. = 2 32= 18.

Example 4.The L.C.M. and H.C.F. of two numbers are 3 and 105 respectively. If the sum ofthese numbers is 36, then what is the sum of the reciprocals of the numbers?

Solution. Suppose the first and the second numbers are x and y respectively.

Given that, L.C.M. of two numbers = 3 and H.C.F of two numbers = 105

Then,

2 18 20 24 - 54

2 9 10 12 27

3 9 5 6 27

3 3 5 2 9

2 1 5 2 3

1 5 1 3


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L.C.M. H.C.F. = 3 105

x y = 3 105 [Because we know that, If X and Y are two numbers then

(L.C.M. of x and y) (H.C.F. of x and y) = x y]

x y = 315 ______________ (1)

Also given that

x + y = 36 _______________ (2)

From equation (1) and (2)

X + Y

X Y =

36

315

X

X Y+

Y

X Y=

36

315

1Y +

1X =

435

1

X +

1

Y =

4

35

Hence, the sum of the reciprocals of two numbers x and y is4

35.

Exercise

Directions: Mark the correct Answer:

1. Find the highest common factor of 54 and 36.

(a)

4

(b)6

(c) 16

(d)

18

2.Find the lowest common multiple of 24, 20 and 30.

(a) 120

(b)

50

(c) 64

(d)24

3.The H.C.F. of2

5,

16

9,

18

25 is:

(a)

4

225


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(b)

2

225

(c)24

125

(d)2

125

4.

The L.C.M. of2

3,

16

27,

5

12is:

(a)80

3

(b)

80

9

(c)

69

5

(d)49

5

5.The ratio of two numbers is 3:4 and their H.C.F. is 4.Their L.C.M. is:

(a)

12

(b)16

(c) 24

(d)48

6.

The L.C.M. of two numbers is 189 and H.C.F. of two numbers is 9. If one of the number is 63

then other number is:

(a)

27

(b)28

(c)

29

(d)30

7.The H.C.F. of 3.6, 7.2, 2.1, 0.81 is:

(a) 0.3

(b)

0.03

(c)

0.9

(d)0.09

8.

The L.C.M. of 1.8, 3.6, and 0.72 is:

(a)

3.6


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(b)

11.8

(c) 0.36

(d)10.80

9.

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainderin each case

(a) 4

(b)7

(c) 9

(d)13

10.

Three different containers contain 204 litres, 1190 litres and 1445 litres of Pepsi, Slice and

Coca-Cola respectively. What biggest measure can be measured from all the different

quantities of cold drinks exactly?

(a) 17

(b)18

(c)

19

(d)

20

Answers

1.

(d)

2.

(a)

3. (b)

4. (a)

5. (d)

6.

(a)

7.

(b)

8. (d)

9.

(a)

10.(a)



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1.

54 = 2333, 36 = 3 232,

Therefore, H.C.F. = 233 =18.

2.24 = 22 23, 20 = 22 5 , 30= 235.

Therefore, L.C.M. = products of highest powers of 2, 3 and 5 = 23

3 5 =120.

8.L.C.M. of 180, 360,72 =1080.

Therefore, L.C.M. of 1.8, 3.6, and 0.72 is 10.80

9.Required number will be = H.C.F. of (91 -43), (183 -91) and (183 43)= H.C.F. of 48, 92, 140 =4


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AVERAGE

The term Average refers to the sum of all observations divided by the total number of

observations. Average is used quite regular in our day to day life. For example to calculate the

average marks of the students, Average height of a particular group etc. The term average is

also referred to as Mean. Basic formula to calculate the average is as follows:

Average = (Sum of all observation

Number of observation)

Solved Examples

Example1:Find the average of first 10 natural numbers.

Solution: We know that sum of first n natural numbers =n ( n+1)

2

Then, the sum of first 10 natural numbers =10 (10+1)

2=

10 11

2= 55.

The required average = 55.

Example2: If the average of four consecutive even numbers is 9, then find the smallest of

these numbers.

Solution:Suppose four consecutive even numbers are a, a+2, a+4 and a+6.

Then,a+(a+2)+(a+4)+(a+6)

4= 9

4a+12

4= 9 a+3 = 9a=6.

Hence, the smallest even number = 6.

Example3: There are two departments, Accounts and Technical in an organization, consisting

of 24 and 36 employees respectively. If the average weight of all the employees in Accounts

section is 40 kg and that of Technical department is 45 kg. Find the average weight of both

the Accounts and Technical departments.

Solution: According to the given problem,

Total weight of all the employees in Accounts Department

Number of Employees in Accounts Department= 40

Total weight of all the employees in Accounts Department = 4024 = 960 kg.

Similarly, total weight of all the employees in Technical Department = 4536 =1620 kg

Total weight of (Accounts + Technical) Department =2580 Kg

And total employees in both the Departments = 60.

Average Weight =2580

60= 43 kg


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Example4: The average of 35 numbers is 70.If the average of first 17 numbers is 60 and that of

last 17 numbers is 65, then find the 18thnumber.

Solution: Sum of first 17 numbers = 1760 = 1020

Also, sum of last 17 numbers = 1765 = 1105

18thnumber = {Sum of 35 numbers (sum of first 17 numbers + sum of last 17 numbers)}

= {3570 (1020+1105)} 2450 2125= 325.

Example5: A famous batsman makes a score of 92 runs in the 15th

inning and by this his

average increased by 5.what is his average after the 15th

inning.

Solution: Suppose his average after 15th

inning = a

Then, average after 14th

inning = (a 5)

15 a = 14 (a 5) + 92a =22.

Exercise

1. Find the average of first five multiples of 2

(a)

5.5

(b)6.5

(c) 5

(d)6

2.

If the average of 5, 3, 6, 7, 9, 2, 8, a is 12, then what is the value of a?

(a) 55.5

(b)59

(c)

56

(d)

56.5

3. In Aruns opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not

agree with Arun and he thinks that Aruns weight is greater than 60 Kg but less than 70 kg

.His mothers view is that his weight cannot be greater than 68 kg. If all of them are correctin their estimation, what is the average of different probable weights of Arun?

(S.B.I.P.O. 2000)

(a) 68 kg

(b)

67 kg


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(c)

69 kg

(d)None of these

4. A certain number of sweets were distributed among 56 children equally. After each child

getting 8 sweets, 17 sweets were left out. How many total sweets were there initially?

(Clerks Grade, 2009)

(a) 448

(b)431

(c) 475

(d)465

(e)

None of these

5.

The average of five positive integers is 385.The average of first two integers is 568.5. Theaverage of the fourth and fifth integers is 187.5. What is the third integer?


(a)

420

(b)382

(c) 415

(d)Cannot be determined

(e)

None of these

6. Find the average of the following set of numbers:

352, 283, 625, 518, 445, 700, 878

(a)

544

(b)

568

(c) 599

(d)

None of these

7. The average height of 40 boys is 1.5 m. When 10 boys leave the group then the average

height increases by 0.2 m. What is the average height of the 10 boys who leave?

(a) 0.8m

(b)0.9m


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(c)

Cannot be determined

(d)None of these

8. A car driver travels to a place 100 km away at an average speed of 60km/hr. His average

speed for the whole journey in km/hr is:

(a)

45km/hr

(b)43km/hr

(c) 46km/hr

(d)48km/h

9. The average of 5 consecutive even numbers A, B, C, D and E respectively is 74. What is the

product of C and E?

(a)

5928

(b)5616

(c) 5538

(d)

5772

(e)None of these


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Answers

1.

(d)

2. (c)

3.

(d)

4. (d)

5. (e)

6. (d)

7.

(b)

8.

(d)

9. (d)


1.

Required Average=2(1+2+3+4+5)

5= 6.

2.

(5+3+6+7+9+2+8+a)

8 = 12 a = 96 40 = 56.

3.

Suppose Aruns weight is w kg.

In Aruns opinion, 65< w < 72

According to Aruns mother, w < 68

According to Aruns brother, 60< w < 70.

There could be two values of w, 66 and 67.

4. Total number of sweets = 56 8 + 17 = 465.

5.

Third integer = 385 5 568.5 2 187.5 2 = 413.

6.

(d)

7.

Sum of the heights of 10 boys = 40 1.5 30 1.7 = 9.


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Then, Average height of 10 boys =9

10= 0.9m.

8.

Average Speed =2xy

x+ykm/hr =

2 60 40

60+40= 48 km/hr.

9.

Suppose five consecutive even numbers are x, x+2, x+4, x+6 and x+8.

Then, x + x+2 + x+4 + x+6 + x+8 = 74 5 5x +20 = 370 x =70.

A=70, C= 70+4 =74, E= 78

C E =74 78 =5772


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SURDS & INDICES

The term Index (Indices is a plural) refers to the power to which a number is raised. If a2is a

number then a is called a base2 is called the power(Index or Exponent).

For Example: 81 = 92

, then we say that 81 is equal to base 9 raised to the power 2.

The Laws on Indices:

(i) = + (ii)

( ) = (iii)

=

(iv)() = (v)

() =

(vi)

0= 1When the power is a fraction say

1

, then is a surd of order n. Here a is a rational numberand n is a positive integer.

The laws of Surds:

(i) =1 (ii)

( ) = a(iii) =

(iv) =

(v)

= (vi)( ) = ()

Solved examples

Example1. Simplify:().Solution.Given that, (64)

5

6= (26)5

6 = (2)5= 32.

Example2. Simplify:( ). ()..Solution.Given that,( 81)0.24 (9)0.02 = ( 92)0.24 (9)0.02= (9)0.48 (9)0.02 = (9)0.50

= (32)1

2= 3.

Example3.If + += 1028, then find the value of x.Solution. 42 + 4+2= 1028 42 ( 1+44) = 1028

42( 257) = 1028 42= 4 x 2 = 1 x = 3.


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Exercise

1. Find the value of (81)14.(a)

3

(b)3(c) 9(d) 33

2. ( 164

)4

3 = ?

(a)

254

(b)-254

(c) 256

(d)-256

3. 1089 + 3 = (? )2 (SBI Clerk 2012)(a)6

(b)5

(c)

3

(d)8

4. The value of [ (89)100 (89)98] is:

(a)7920

(b)7939

(c)7921

(d)7922

5. (0.25)2.5= ?(a)64

(b)32

(c)16


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(d)60

6. If (x

y)a5= ( y

x)a7 , then the value of a is:

(a)5

(b)

6

(c)7

(d)3

7. If 5a =2b =10c , then find the value of1

a+

1

b-

1

c.

(a)3

(b)2

(c)

1

(d)0

8. Find the smallest number from among 36 , 44 and33 .(a) 36 (b) 44 (c) 33 (d)

All are equal

9. Number of prime factors in66 (35)8

(14)6is:

(a)14

(b)15

(c)16

(d) 17

Answers

1.(b)

2.

(d)

3.

(a)

4.(c)


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5.

(b)

6.

(b)

7.(d)

8.

(a)

9.(c)


1. (81)14= (3 3 3 3 )12 14= (34 )12 14= (3)12=3.2. ( 1

64)

43 = ( 1

4)

3 (4)

3 =( 14)4= (4)4= 256.

3.

1089 + 3 = (? )2

332+ 3 = (? )2 33 + 3 = (? )2 (6)2=(? )2 ? = 6.

4. [ (89)100 (89)98] =(89)100

(89)98= (89)2= 7921.

5. (0.25)2.5= ( 25100

)25

10 = (1

4)

52= (4)

5

2=25= 32.

6.

(x

y)a5= ( y

x)a7 ( x

y)a5= ( x

y)7a a5 = 7 a a =6.

7. 5a =2b=10c = k 5=k1

a , 2 = k1

b and 10 = k1

c ( 1+ 1 )= k1c 1 + 1= 1c 1 + 11c = 08. L.C.M. of 6, 4 and 3 is 12.

Then, = ()= () , = ()= () , = ()= () = () , = () , = () Hence, 36 is the smallest number.

9.

()() =

= 36 58 72

Numbers of prime factors = 16.


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PERCENTAGE

In mathematics, a percentage is a way of expressing a number like Ratio, as a fraction of 100.It

is denoted by the symbol %. Percentages are used to express how large/small one quantity is,

relative to another quantity. The percent value is computed by multiplying the numeric value of

the ratio by 100.

For example 50% read as Fifty percent and it is equal to 50/100 or 0.5.

Calculation of Percentage

The Percent Value is computed by multiplying the numeric value of the ratio by 100.

For Example:

If we want to calculate the percentage of 20 oranges out of 80 oranges then,

The required percentage= (20/80) 100 = 25%

Calculation of Percentage of a Percentage

To calculate a percentage of a percentage first we should convert all the percentages to

fractions of 100.

For Example:

40% of 50% = (40/100) (50/100) = 0.40 0.50 = 0.20 = 20/100 = 20%

Percentage Increase

When the value increased by 10 percent, it becomes 1.1 times of itselfSuppose the value of X increased by 10%

X + 10% of X

X + {(10/100)X}

X + (0.1X)

(1+0.1) X

(1.1)X

Here new value is 1.1 times of its original value

Percentage Decrease

When any value decreased by 10 percent, it becomes 0.9 times of itself

Suppose the value of X decreased by 10%

X - 10% of X


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X {(10/100)X}

X (0.1X)

(1-0.1) X

(0.9)X

Here new value is 0.9 times of its original value.

Exercise

1. If x% of x is 81, then x is equal to:

(a)

30

(b)

92

(c) 90

(d)91

2.65% of a number is more than its two fifth by 140. What is 30% of that number?


(a) 186

(b)168

(c)

164

(d)

182

3.78% of a number is 2496.What is 55% of that number? (Clerks Grade 2009)

(a)

1815

(b)1650

(c) 1760

(d)1705

4.

The difference between 78% of a number and 59% of the same number is 323. What is 62%

of that number? (SBI Clerk, 2009)

(a)

1071

(b)1178

(c)

1037

(d)

1159


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(e)

None of these

5.96% of a number is 5600. What is the value of the number?

(a) 21,000

(b)

20,000

(c) 22,000

(d)24,000

6.35% of 411 - x % of 272 = 84.01. Then the value of x is:

(a)

42

(b)

36

(c) 18

(d)22

(e)None of these

7. Ishan invests Rs. 4,448.Which is 25% of his monthly income in insurance policy. What is his

monthly income? (OBC Clerk, 2009)

(a) Rs. 17, 792

(b)Rs. 16, 584

(c)

Rs. 16,442

(d)

Rs. 17,774

(e)None of these

8.

A number is increased by 10% and then again by 10%.By what % should the increased

number be reduced so as to get back the original number?

(a)

17.36%

(b)17.11%

(c)

17.23%

(d)17%

9.Milk contains 5% water. What quantity of pure milk should be added to 10 litres of milk to

reduce this to 2%? (Bank P.O., 2003)

(a) 5 litres


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(b)

7 litres

(c) 15 litres


(e)

None of these

10.If price of a T- Shirt is increased by 10% and then reduced by 10%. The final price of T-shirt

is:

(a) Decrease by 1%

(b) Increase by 1%

(c) Does not change

(d)

Cannot determined

11.

The total population of a village is 5000. The number of males and females increases by 10%and 15% respectively and consequently the population of the village becomes 5600. What

was the number of males in the village? (BANK P.O. 2003)

(a) 2000

(b)

2500

(c)

3000

(d)4000

12.

Mr. Devar spends 38% of his monthly income on food and the remaining amount of Rs.5800 he saves. What is Mr. Devars monthly income? ( SBI Clerk, 2009)

(a)

Rs. 23,200

(b)Rs. 24,200

(c) Rs. 23,800

(d)Rs. 24,400

(e)

None of these

Answers

1. (c)

2.

(b)

3.

(c)

4. (e)


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5.

(b)

6.

(d)

7. (a)

8.

(a)

9. (c)

10.(a)

11.(c)

12.

(a)


1.

x = 81

2= 8100 x = 90.

2. Suppose the number is x then

65% of x -2

5x = 140

13 8 20

= 140 x = 560.

3.78

100x = 2496 x = 3200, then 55% of x =

55

100 3200 = 1760.

4.

Suppose the number is x, then (78 59) % of x = 323 x = 1700.

62 % of 1700 = 1054.

5. 96% of x = 19,200 x= 20,000

6.35

100411 x

272

100= 84.01 5984 = 272 x x = 22.

7.

Let us suppose Ishans monthly income = x Rs.

Then,

25% of x = 4448 x = 17,792.

8.

Suppose the original number is 100

Then, 110% of 110% of 100 =110

100

110

100100 = 121.


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Decrease on 121 = 21 Decrease on 100 = (21

121100) % =17.36%.

9. Water quantity in 10 litres = 5% of 10 = 0.5 litre

Suppose x litres of pure milk is required, then

0.5

0.5+ =2

100 2x = 30 x = 15.

10.

Suppose the original price of T-Shirt is 100 Rs.

Now, new price = 90% of (110% of 100) = 99

Decrease = (100-99)% =1%

11.

Let us suppose the number of males = x

Then, number of females = (5000 x)

10% of x + 15% of (5000 x) = 5600 - 5000 5x = 15000 x = 3000.

12.

Let us suppose Devars monthly income =x

X (38

100 + 25

100 + 12

100)= 5,800 x = Rs. 23,200


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PROFIT & LOSS

Profit and loss is a very important branch of basic Mathematics. This branch deals with the

study of Profit and loss made in the business and in our daily life.

Some basic terms used in Profit and loss are:Cost price The price, at which an article is purchased, is called Cost price and it is abbreviated

by C.P.

Selling Price The price, at which an article is sold, is called its selling price and it is abbreviated

by S.P.

Profit

If S.P. > C.P., then seller is said to have a profit.

Loss

If SP < CP, Then seller is said to have incurred a loss.

Formulae

Profit or Gain = S.P. C.P.

Loss = C.P. S.P.

Gain % =Gain 100

C.P.

Loss % =Loss 100

C.P.

S.P. =(100+Gain %)

100 C.P.

S.P. =(100Loss %)

100 C.P.

C.P =100

(100+Gain %) S.P.

C.P =100

(100Loss %) S.P.

If a person sells two similar items, one at a gain of A%, and the other at a loss of A%,

then the seller always incurs a loss. This loss can be calculated by:

Loss % =(Common loss and gain %

10)2

If an article sold at two different selling pricesale price1and sale price2

respectively. On one gain1 is made and on the other gain2is made then:


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sale price 1

100+gain 1 %=

sale price 2

100+gain 2%

Solved Examples

Example1: 100 apples are bought at the rate of Rs. 500 and sold at the rate of Rs. 84 per

dozen. What will be the percentage of profit and loss?

Solution:We will solve this in steps

Step I:Given that C.P. of 100 apples = 500

Then, C.P. of 1 apple =500

100

= 5

Step II:Also given that per dozen S.P. of apples = 84

Then, S.P. of 1 apple =

84

12

=7

Step III:Now, we know that

Gain % =Gain 100

C.P.

=(S.P.of 1 appleC.P.of 1 apple )100

C.P.

=(75)100

5

=2 100

5

= 40%

Therefore, there is a profit of 40% in the whole selling process.

Example2: Kartik sold an item for Rs. 6,500 and incurred a loss of 20%.At what price should he

have sold the item to have gained a profit of 20%. (Clerks Grade, 2011)

Solution:If selling price = Rs. 6,500

Then, loss = 20%

We know the formula S.P. =(100Loss %)

100 C.P. C.P. =

S.P.100

(100Loss %) 6500100

80 Rs.8,125

Now new S.P =(100+20)

100 8125. Rs.9750

Hence, new selling Price should be Rs. 9750.

Exercise


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1.

Arjun buys an old car for Rs 1,12,000 and spends 8000 on its repairs. If he sells the Car for

Rs 1,80,000, his gain percent is:

(a)

50

(b) 51

(c)

52

(d) 53

2. Ajay sells a painting which costs him Rs 500 to Bablu at a profit of 20%. Bablu then sells it

to Chirag, making a profit of 30% on the price he pays to A. How much does Chirag pay to

Bablu?

(a) 789

(b) 790

(c) 782

(d) 780

3. When a Scooter is sold for Rs. 25,000, the owner loses 20%.At what price must the scooter

be sold in order to gain 20 %?

(a) 37,500

(b)

37,520

(c)

37,000

(d) 39,000

4.

The ratio between the sale price and the cost price of an article is 8:3.What is the ratio

between the profit and sale price of that article?

(a)

5:3

(b) 5:8

(c) 5:9

(d) 5:7

5. On selling 19 balls at Rs. 840, there is a loss equal to the cost price of 7 balls. The cost price

of a ball is:

(a) 84

(b) 83

(c)

70


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(d)

72

6. Rajesh purchased toffees at 12 for a rupee. How many for a rupee must he sell to gain of

20%?

(a) 8

(b)

9

(c) 10

(d) 11

7. A man bought some bananas at the rate of 16 for Rs.24 and sold them at the rate of 8 for

Rs 18.What is the profit percentage?

(a) 30%

(b)

60%

(c) 50%

(d) 55%

8.

If loss is 1/4 of S.P., the loss percentage is:

(a) 25%

(b) 20%

(c) 35%

(d)

40%

9. By selling an article for Rs. 150, a man gains Rs. 25.Then his gain percent is:

(a) 22%

(b)

25%

(c)

20%

(d) 30%

10.

Rahul gain 80 paisa on Rs. 40.His gain percent is:

(a) 1%

(b)2%

(c) 3%

(d)4%


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Answers

1.

(a)

2. (d)

3.

(a)

4. (b)

5. (c)

6. (c)

7.

(c)

8.

(b)

9. (c)

10.(b)


1. C.P. = Rs. (1,12,000 + 8,000) = Rs. 1,20,000.

S.P = 1,80,000, Gain % = (60,000

1,20,000100) =50%

2.

C.P. for Bablu = 120 % of 500 = Rs.(120

100 500) = Rs.600.

C.P. for Chirag = 130 % of 600 = Rs.(130

100 600) = Rs.780

3.(100+20)

New S.P.= .

(10020)25,000

New S.P. = 120 25,000

80= Rs. 37,500

4. S.P. = 8a, C.P. = 3a, Gain = 8a 3a =5a.

Ratio of between the profit and sale price =5a

8a= 5:8

5.

C.P. of 19 balls S. P. of 19 balls = C.P. of 7 balls.

C.P. of 12 balls = S.P. of 19 balls =Rs.840

C.P. of 1 ball = Rs (840

12) = Rs. 70.

8.

S.P. = a, Loss = a/4, CP. =a + a/4=5/4 a


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Loss% =(4

4

5100) % = 20%


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PARTNERSHIP

Meaning:When two or more than two persons run a business jointly, they are called partners

in that business and the deal between them is known as partnership.

There are two types of partners in the business1. Working Partner:A person who manages the business is known as working partner.

2. Sleeping Partner:A person who simply invests the money is known as sleeping partner.

Some Important Formulae:

Suppose two persons P and Q invests Rs. X and Rs. Y respectively for a year in a business, then

their share of profit or loss at the end of the year:

Share of P s profit (or loss )Share of Q sprofit (or loss ) =

X

Y

Suppose two persons P and Q invests Rs. X for m month and Rs. Y for n months respectively,

then

Share of P s profit (or loss )Share of Q sprofit (or loss )=

Xm

Yn

Solved Examples

Example1: P, Q and R started a business by investing Rs. 1,50,000, Rs. 2,50,000 and 3,50,000

respectively. At the end of the year, out of an annual profit of Rs.30,000 find the share of P, Q

and R respectively.

Solution:Ratio of shares of P, Q and R respectively = Ratio of Ratio of their investments

= 1,50,000 : 2,50,000 : 3,50,000

= 3 : 5 :7

Share of Ps profit = Rs. ( 30,000 3

15) = Rs. 6,000.


15) = Rs. 10,000.


15) = Rs. 14,000.

Example2: Ram started a business investing Rs. 50,000.After 4 months, Shyam joined him

with a capital of Rs. 30,000 .After another 2 months, Mohan joined them with a capital of Rs.

60,000.At the end of the year, they made a profit of Rs. 24,000.Find the share of profits of

Ram, Shyam and Mohan.

Solution: According to the given problem, it is clear that Ram invested his capital for 12

months, Shyam invested for 8 months and Mohan invested for 6 months.


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Then, ratio of their Capitals = (50,000 12) : (30,000 8) : (60,000 6)

= 60:24:36 = 5:2:3

Share of Rams profit = (24,000 5

10) = Rs.12,000;

Share of Shyams profit = (24,000 210

) = Rs.4,800;

Share of Mohans profit = (24,000 3

10 ) = Rs.7,200.

Example3: Three friends P, Q and R enter into partnership. P invests 2 times as much as Q

invests and Q invests one-fourth of what R invests. If the profit earned by them is Rs.63,000,

then what is the share of profit of P at the end of the year.

Solution: Let us suppose Rs Capital = Rs. X. Then, Qs Capital = Rs.1

4 X

Now, Ps capital = Rs. (2 Qs Capital) = Rs. (2

1

4 X) = Rs. (

1

2X)

Ratio of their capitals =1

2X :

1

4 X : X =2:1:4

Hence, share of Ps profit = Rs.(63,0002

7) =18,000.

Example4: P, Q and R start a business by investing Rs. 10,000.After 4 month P invests 5,000

more, Q withdraws Rs. 2,000 and R withdraws Rs. 4,000. If the total Rs.88,000 profit was

recorded at the end of the year, then find the share of Q.

Solution: Ratio of capitals of P, Q and R

= (10,000 4 + 15,000 8) : (10,000 4 + 8,000 8) : ( 10,000 4 + 6,000 8)

= (40,000+1,20,000) : (40,000 + 64,000) : (40,000 + 48,000)

= 1,60,000 : 1,04,000 : 88,000 =160 : 104 : 88 = 20 : 13 : 11

Hence, share of Q = Rs. (88,000 13

44) = Rs.26,000.


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Exercise

Directions: Select the correct option:

1.Radha and Meera started a business investing Rs. 50,000 and Rs. 65,000 respectively. In

what ratio the profit earned after 5 years be divided between Radha and Meera?

(a)

13:12

(b)13:10

(c) 10:13

(d)10:11

2.Anurag invested Rs.45,000 in a business .After few months, his friend Brijesh joined him

with Rs. 40,000. At the end of the year, the total profit was divided between them in the

ratio 3:2. After how many months did Brijesh join.

(a) 2 months

(b)3 months

(c) 4 months

(d)

5 months

3.

Shankar started a marriage burro business by investing Rs.1, 00,000. After 4 months, his

wife Geeta joined him with a capital of Rs.50,000 and after 2 years, they earned a profit of

Rs. 34,000. What was the Shankars share in the profit?

(a)

Rs.24,500

(b)Rs.23,000

(c)

Rs.24,000

(d)

Rs.22,000

4.Nitin, Pranjali and Vivek hired a car for Gwalior darshan and used it for 3, 4 and 5 hours

respectively and they paid Rs.900.Hire charges paid by Nitin were:

(a) Rs.8,000

(b)

Rs.8,562

(c) Rs.6,000

(d)Rs.6,230

5.P, Q and R subscribe Rs.89,000 for online business. P subscribes Rs.4,000 more than Q and

Q Rs.8,000 more than R. Out of a total profit of Rs, 17,800 , P receives:


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(a)

Rs.7,000

(b)Rs.7,890

(c) Rs.7,999

(d)

Rs.6,890

6.Riya and Jaya invested in a business. They divided their profit in the ratio 4:5.If Riya invested

20,000,then the amount invested by Jaya:

(a) Rs.25,000

(b)Rs.26,000

(c) Rs.20,000

(d)

Rs. 27,000

7.

Four milkmen Ram, Shyam, Mohan and John rented a pasture. Ram grazed 18 cows for 2months; Shyam 6 cows for 3 month; Mohan 3 cows for 2 month and John 15 cows for 2

months. If Rams share of rent is Rs. 540, find the total rent of the field.

(a) Rs.1100

(b)

Rs.1290

(c)

Rs.1300

(d)Rs.1350

8.

Two partners Ajay and Vijay shared the profit in a business in the ratio 3:4. They hadpartnered for 9 months and 5 months respectively. What was the ratio of their investment?

(a)15

36

(b)35

14

(c)

15

35

(d)15

31

9.

If 3 (Ps share) = 4 (Qs share) = 6 (Rs share), then out of a profit of Rs. 5,400, P will receive:

(a) Rs.2,424

(b)

Rs.2,450

(c)

Rs.2,400

(d)Rs.2,490


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10.

Kapil started a business with Rs,15,000 and after x months Sunil joined him with Rs.20,000.If

the profits at the end of the year are divided equally then the value of x:

(a)

2 months

(b)4 months

(c)

5 months

(d)3 months

Answers

1. (c)

2. (b)

3.

(c)

4.

(c)

5.

(a)

6.

(a)

7. (d)

8. (a)

9. (c)

10.

(d)


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1. Radha : Meera = 50,000 : 65,000 = 10 : 13

2. Let us suppose Brijesh joined after x months. Then his money was invested for (12-x)

months.

45,000 12

40,000 (12x)=3

2 9 = 12-x x = 3.

3. Shankar : Geeta = (1,00,000 24) : (50,000 20) = 24 : 10 = 12 : 5

Shankars share in the profit = Rs. (12

17 34,000) = Rs.24,000.

4.

Nitin : Pranjali : Vivek = 3 : 4 : 5

Hire charges paid by Nitin = Rs. (3

12 24,000) = Rs.6,000.

5. Suppose R= a. Then Q= a+8,000 and P= 4,000+8,000+a

a+a+ 8,000+4,000+8,000+a = 89,000 3a+20,000 = 89,000 3a = 69,000 a=23,000

Now, P: Q: R =35:31:23

Then Ps share = Rs.(35

89 17,800 )= Rs.7,000.

6. Suppose Jaya invested x Rs. in the business. Then,20,000

x=

4

5x= 25,000.

7.

Ram:Shyam:Mohan:John =(18 2):(63):( 32) : ( 152) = 36 : 18 : 6 : 30 = 6 : 3 : 1: 5

Suppose total rent = r Rs. Then Rams share =6

15 r.

6

15 r = 540 r = 1350.

8.

Suppose Ajay was invested Rs. x for 9 months and Vijay was invested Rs. y for 5 months.

Then,9x

5y=

3

4 36x =15y

x

y=

15

36

9. 3P: 4Q : 6R = CP =C

3, Q =

C

4, R =

C

6P : Q : R = 4 : 3 : 2

Ps share in the profit = (4

95,400) =2,400.

10. 15,000 12 = 20,000 (12 x) 15 12 = 20 (12 - x) x = 3.


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TIME & WORK

Some basic facts with formulae

1. If a person P can do a piece of work in n days, then work done by him in 1 day =1

n.

2. If a Person P can do1

nwork in 1 day, then he can finish the work in n days.

3.

If a Person P can do a piece of work in n days and another person Q can do the same

work in m days then time taken by them to complete the work isnm

n+m days when they

work together but independently.

4.

If the efficiency of P to do a work is x times than efficiency of Q to do a same work then,

Ratio of work done by P and Q = x: 1

Ratio of time taken by P and Q to complete a work = 1:3

Solved Examples

Example1. Prerna can do a piece of work in 16 days while Monami can do the same work in

20 days. How long should it take both Prerna and Monami, working together but

independently, to do the same work?

Solution. Prernas 1 days work =1

16 , Monamis 1 days work =

1

20

(Prerna + Monami) s 1 days work =1

16 +

1

20 =

5 + 4

80 =

9

80

Time taken by Prerna and Monami to finish the work =80

9days = 8

8

9 days.

Example2. Efficiency of Arun is thrice than Brijesh to do a piece of work and together they

finish a piece of work in 27 days. In how many days will Arun finish the work?

Solution. ( Aruns 1 day work ) : ( Brijeshs 1 day work ) = 3 : 1

(Arun + Brijesh )s 1 day work =1

27

Then, Aruns 1 day work =1

27

3

4=

1

36

Arun can finish the work in 36 days.

Exercise

1. Aryan takes 5 hours to do a job, and Aryaman takes 6 hours to do the same job.How long

should it take both Aryan and Aryaman, working together but independently, to do the

same job?


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(a)

3 days

(b)

28

11days

(c) 29

11days

(d)4 days

2.

X is thrice as good a workman as Y and together they finish a piece of work in 21 days. In

how many days will X alone finish the work?

(a) 29 days

(b)31 days

(c) 28 days

(d)27 days

3.

A and B can do a piece of work in 12 days, B and C can do it in 16 days, A and C can do it in

20 days. In how many days will A, B and C complete it, working together?

(a) 84

7days

(b)82

7days

(c) 31days

(d)

32 days

4.

A can do a certain work in 45 days. B is 80% more efficient than A. How many days does B

alone take to do the same work?

(a) 24 days

(b)23 days

(c)

28 days

(d)25 days

5. X and Y undertake to do a piece of work for Rs.800. X alone can do it in 6 days while Y alone

can do it in 8 days. With the help of Z, they finish it in 3 days. Then what will be the share ofX?

(a) 100

(b)220

(c) 400


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(d)

300

6. A and B working separately can do a piece of work in 12 and 36 days respectively. If they

work for a day alternately, A beginning, in how many days, the work will be completed?

(a) 9 days

(b)

10 days

(c) 18 days

(d)16 days

7. Eight men can complete a work in 10 days. They started the work and after 6 days, two men

left. In how many days will the work be completed by the remaining men?

(a) 25 days

(b)

22 days

(c) 62

5days

(d)63

5days

8.

A sum of money is sufficient to pay A salary for 12 days and Bs salary for 24days. The same

money is sufficient to pay the salaries of both for:

(a) 8 days

(b)9 days

(c) 10 days

(d)12 days

9. A takes thrice as much time as B or twice as much time as C to finish a piece of work.

Working together, they can finish the work in 4 days. C can do the work alone in:

(a) 14 days

(b)12 days

(c)

10 days

(d)8 days

10.A can finish a work in 12 days and B can do the same work in1

3the time taken by A. Then,

working together, what part of same work they can finish in a day?

(a)1

5


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(b)

1

4

(c)1

3

(d)1

2

Answers

1. (b)

2. (c)

3. (a)

4. (d)

5. (c)

6.

(c)

7. (c)

8. (a)

9. (b)

10.(c)


1.

(Aryan +Aryaman)s 1 day work = (

1

5+

1

6) =

11

30

Both together will finish the work in 28

11days.

2. ( Xs 1 day work) : (Ys 1 day work) = 3:1

(X + Y) 1 day work =1

21

Then, Xs 1 day work =1

21

3

4 =

1

28

X alone can finish the work in 28 days.

3.

(A+B)s 1 day work =1

12+

1

10

(B+C)s 1 day work =1

10+

1

20

(A+C)s 1 day work =1

12+

1

20

Then, (A+B+C)s 1 day work =1

2 (

1

12+

1

10+

1

20) =

7

60


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A, B and C can finish the work in 84

7days.

4. Ratio of time taken by A and B to do a work =180

100=

9

5.

If B takes x days to do a work then, 9:5:: 45:x

x = 25 days

5. Zs 1 day work = {1

3- (

1

6+

1

8)} =

1

24

Ratio of X, Y and Z s 1 day work = 4:3:1

Now, Xs share = Rs. (800 4

8 ) =Rs.400.

6. (A+B)s 2 day work =1

12+

1

36=

1

9

Work done by A and B in 9 pairs of days = (9 1

9) = 1.

Total days to complete the work = 18 days.


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TIME & DISTANCE

Time and Distance Formulae relates Time, Distance and Speed. These relationships have many

practical applications.

For example if you know the speed of any vehicle and the distance covered by that vehicle,then we can easily calculate the time taken in whole journey by using the formula of Time and

Distance.

Important Formulae

(i) Speed =Distance

Time,Distance= SpeedTime, Time=

Distance

Speed

(ii)

S km/hr =(s5

18)m/sec

(iii)S m/sec =(s18

5)km/hr

Example: A Taj Express travelling at5

11of its actual speed and covers 42 km in 1 hr 40 min 48

sec. Find the actual speed of the Taj Express.

Solution: Suppose the actual speed of Taj = S km/hr

Then new speed =5

11S

Time taken by Taj with new speed

= 1 hr 40 min 48 sec

= 1hr+40 160

hr+48 16060

[Because 1hr = 60 minutes,

1 minute = 60 Second,

& 1 minute =1

60hr ,

1 Second =1

60minute ]

= 1hr +2

3hr +

1

75hr

=126

75hrs

Now, according to the formula

New Speed Time taken by Taj with new speed =

Distance covered by Taj


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5

11S

126

75= 42

S =427511

1265

S =55km/hr

Hence, the actual speed of the Taj Express is 55 km/hr

Points to remember:

1. Read the units of time speed and distance carefully.

2. If the distance is given in km and the speed is in m/s then always convert the units. According

to the demand of the question you can change the kilometer in to meter or m/s in to km /hr.

Exercise

1. If Narendra runs at a speed of 10 km/hr then how many minutes does he take to cover a

distance of 200 m?

(a) 72 Second

(b)73 Second

(c) 74 Second

(d)

75 Second2.

Rahul can cover a certain distance in 1 hr 30 minutes by covering one fourth of the

distance at 5 km/he and the rest at 6 km/hr. Find the total distance covered by Rahul.

(a)

8km

(b)7km

(c)

8.57km


3.

Ram runs along the four sides of a Square at the speeds of 2, 4 , 6 and 8 km/hr. what is theaverage speed of Ram around the field?

(a) km/hr

(b)3km/hr

(c)

3.84km/hr


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(d)

3.2km/hr

4.A car covers a distance of 666 km in 9 hours. What is the speed of the car?


(a)

78km/hr

(b)76km/hr

(c) 74km/hr


(e)

None of these

5.

Gwalior and Agra are two stations 160 km apart. A train starts from Gwalior at 8 a.m. and

travels towards Agra at 60 km/hr .Another train starts from Agra at 9 a.m. and travels

towards Gwalior at 40 km/hr. At what time do they meet?

(a) 9.15a.m.

(b)9.45a.m

(c)

9.55a.m.

(d)10a.m.

6.A motor car starts with the speed of 70 km/hr with its speed increasing every two hours by

10km/h. In how many hours will it cover 345 kms? (Bank P.O. 2003)

(a)

21

4hrs

(b)4 hrs 5 min

(c)

41

2hrs

(d)

Cannot be determined

(e)

None of these

7. If Shyam walks at 10 km/hr instead of 8 km/hr, he would have walked 5 km more. The actual

distance travelled by him is:

(a) 21km

(b)

20km

(c) 19km

(d)18km


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8.

A Truck covers a distance of 330km at the speed of 30km/hr. What is the average speed of a

car which travels a distance of 110km more than the truck in the same time?

(IBPS Clerk, 2011)

(a) 42km/hr

(b)

48km/hr

(c) 39km/hr

(d)38km/hr

(e)None of these

Answers

1.

(a)

2.

(c)

3.

(c)

4.

(c)

5. (d)

6. (c)

7. (b)

8.

(e)


1. Narendras Speed = 10km/hr =10 5

18m/s =

25

9m/s

Time taken by Narendra to cover 200 m = (200 9

25

) second = 72 Second.

2. Total time taken = 1 hr 30 minutes =3

2hrs

Suppose total distance = xkm

Then,/4

5+

3/46

=3

2

20

+8=

3

2 x =

3 20

7= 8.57km


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3.

Suppose each side of the square is x km and the average speed of Ram around the field

is y km.

Then,2+

4+

6+

8=

4 y = 3.84 km/hr

4.

Speed of the Car =666

9 kmph =74kmph

5. Suppose they meet h hours after 8a.m.Then,

(Distance covered by first train in h hrs) + (Distance covered by second train in (h-1) hrs)

= 100

60h + 40(h-1) = 160 100h = 200 h = 2hrs.

Hence, they meet 2hrs after 8a.m. i.e. at 10 a.m.

6. Distance covered in 2 hrs = (70 2) km = 140km.

And distance covered in next 2 hrs = (80 2) km = 160 km

Now the remaining distance = 345-140-160 = 45km

Speed of the car in fifth hour = 90 km/hr.

Time taken to cover 45 km = 45/90 hr = hr.

Total time taken by the Car = 2+2+1/2 = 4 12

hrs.

7.

If the actual distance travelled by Shyam is x km , then

8= +510 x = 20km

8. Total time taken by a Truck =330

30= 11 hrs.

New distance of a Car = 330 + 110 =440

Average Speed = 440/11 = 40 km/hr


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PROBLEMS ON TRAINS

Some Important Facts and Formulae

1.

Speed =DistanceTime

,Distance= SpeedTime, Time= DistanceSpeed

2. S km/hr =(s5

18)m/sec

3.

S m/sec =(s18

5)km/hr

4. If the length of a train is l meters then time taken to pass a pole or a standing man or a

signal post is equal to the time taken by a train to cover l meters.

5. If the length of a train is l meters then time taken to pass a stationary object of length a

meters is equal to the time taken by a train to cover (l + a) meters.6. Suppose two trains or two objects are moving in the same direction and their respective

speeds are u m/s and v m/s. If u > v, then their relative speed is (u v) m/s.

7. Suppose two trains or two objects are moving in opposite directions and their

respective speeds are u m/s and v m/s, then their relative speed is (u + v) m/s.

8.

If two trains of length l meters and m meters are moving in opposite directions and their

respective speeds are u m/s and v m /s, then they will cross each other after(+ )(+)

seconds.

9.

If two trains of length l meters and m meters are moving in the same direction and their

respective speeds are u m/s and v m/s, then the faster train will take(+ )(+)seconds to

cross the slower train.

Exercise

1. A train of length 90 metres running at the speed of 54 km/hr crosses a telegraph post. The

time taken by the train to cross the telegraph post is

(a) 2 seconds

(b)4 seconds

(c) 6 seconds

(d)8 seconds


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2.

A train of length 150 metres running at the speed of 72 km/hr crosses a bridge of length 30

metres. The time taken by the train to cross the bridge is

(a)

9 seconds

(b)10 seconds

(c)

11 seconds

(d)12 seconds

3. Two trains, 73 metres and 57 metres in length respectively, are running in the same

direction at the speed 118 km/hr and 100 km/hr respectively. In what time will they be

completely clear of each other from the moment they meet?

(a) 25 seconds

(b)26 seconds

(c) 30 seconds

(d)31 seconds

4. Two trains, 92 metres and 83 metres in length respectively, are running in the opposite

direction at the speed 67 km/hr and 59 km/hr respectively. In what time will they be

completely clear of each other from the moment they meet?

(a) 2 seconds

(b)

3 seconds

(c)

4 seconds

(d)5 seconds

5.

A train of length 90 metres, running at 80 km/hr passes a man who is walking in the same

direction at 8 km/hr. The time taken by train to cross the man is

(a)

3.5 seconds

(b)4.5 seconds

(c) 5.5 seconds

(d)

6.5 seconds

6. A train of length 125 metres, running at 83 km/hr passes a man who is walking against the

train at 7 km/hr. The time taken by train to cross the man is

(a) 4 seconds

(b)4.5 seconds


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(c)

5 seconds

(d)5.5 seconds

7. A train of length 125 metres, running at 83 km/hr passes a man who is sitting in another

train which is running in the same direction at the speed 65 km/hr. The time taken by train

to cross the man is

(a)

20 seconds

(b)

25 seconds

(c) 30 seconds

(d)35 seconds

8. A train of length 140 metres, running at 83 km/hr passes a man who is sitting in another

train which is running in the opposite direction at the speed 43 km/hr. The time taken by

train to cross the man is

(a)

4 seconds

(b)5 seconds

(c) 6 seconds

(d)7 seconds

9.

A train of length 71 metres, running at 51 km/hr crosses another train which is running in

the opposite direction at 67 km/hr in 4.5 seconds. The length of the train which is running

at 67 km/hr is(a) 50 metres

(b)55 metres

(c) 60 metres

(d)

64 metres

10.A train of length 63 metres, running at 81 km/hr crosses another train which is running in

the same direction at 63 km/hr in 24 seconds. The length of the train which is running at 63

km/hr is

(a) 57 metres

(b)

60 metres

(c)

63 metres

(d)66 metres


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Answers

1.

(c)

2. (a)

3.

(b)

4. (d)

5. (b)

6. (c)

7.

(b)

8.

(a)

9. (d)

10.(a)


1. To cross a telegraph post the train has cover its own length which is 90 metres and its speed

is 54 km/hr = 54 5

18m/s= 15m/s

Therefore the required time =90

15 = 6 seconds.

3.

Here the faster train will cover its own length as well as the length of the other train which

is73 + 57 = 130 metres.

The relative speed of faster train = 118 km/hr 100 km/hr = 18 km/hr = 18 5

18m/s= 5

m/s

The required time =130

5= 26 seconds.

4. Here the relative speed = 67 km/hr + 59 km/hr = 126 km/hr = 126 5

18m/s = 35 m/s.

Because trains are running in opposite direction.

6.

In this question, the man is walking in the opposite direction with a speed 7 km/hr then the

relative speed of the train = 83 km/h + 7 km/hr = 90 km/hr =90 5

18m/s= 25 m/s.

The length of the train 125 metres is to be covered with the speed of 25 m/s.

Therefore the required time =125

25= 5 seconds.


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9.

In this question, the trains are running in the opposite direction.

Therefore the relative speed = 51 km/hr + 67 km/hr = 108 km/hr = 108 5

18m/s= 30 m/s.

Here the length of the second train is unknown. Let the length be x metres. Then the total

length which is to be covered = (71 + x) metres.

From t =d

s, 4.5 =

(71 + )30

4.5 30 = 71 + x 135 71 = x x = 64.

Therefore the length required = 64 metres.


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SIMPLE INTEREST

Definition:

If a person X borrows some money from another person Y for a certain period, then after that

specified period, X (borrower) has to return the borrowed money with some additional money.This additional money that X (borrower) has to pay is called Interest. The actual borrowed

money is called Principal or Sum. The Principle and interest together is called amount, and the

time for which X the borrower has been used the borrowed money is called the time. The

interest that X has to pay for every 100 rupees each year is called rate percent per annum.

If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called

Simple Interest and it is denoted byS.I.

FORMULAE:

Let Principle = P, Rate =R% per annum, and Time = T years. Then

S.I. = (P R T

100)

Or

P =(100 S.I.

R T)

or

R =(100 S.I.

P T)

or

T =(100 S.I.

P R)

Now,

Simple Interest + Principle = Amount

If we denote the amount by A, then

Simple Interest = A P

S.I. = A P =P R T

100

A=P (1+

100) = SI (1+

100

)Two different cases can be compared by using the following formula


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A1P 1A2P2

=111222

Solved Examples

Example 1:What will be the simple interest on Rs. 78,000 at 10% per annum for 9 years?

Solution:Here, given that

Principal (P) =78,000

Rate (R) = 10%

Time (T) = 9 years

Now, we know that

S.I. = (P R T

100)

S.I. = (78,000 10 9

100)

S.I. = Rs. 70,200

Therefore, the simple interest on Rs. 78,000 at 10% per annum for 9 years will be Rs. 70, 200.

Exercise

1. What will be the simple interest earned on an amount of Rs. 18,000 in 6 months at the rate

of 25% p.a.?

(a)

Rs.2250.50

(b)Rs.2350.50

(c) Rs.2, 250

(d)Rs 2,400

2. What will be the simple interest on Rs. 2,400 at 41

6% per annum for the period from 1

stFeb,

2005 to 15th

April, 2005?

(a) Rs. 20

(b)

Rs.20.5

(c)

Rs. 22

(d)Rs.25

3.

A sum of Rs. 600 amounts to Rs. 900 in 5 years at simple interest. What would be the

amount if the interest rate is increased by 5%?


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(a)

Rs.1, 000

(b)Rs 1,230

(c) Rs.1, 050

(d)

Rs.1, 125

4. Ram borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to

another person at 61

4% p.a. for 2 years. Find his gain in the transaction per year.

(a)

Rs.112.50

(b)Rs.150

(c) Rs.125

(d)Rs. 167.50

5.

How much time will it take for an amount of Rs. 400 to yield Rs. 72 as interest at 9% per

annum of simple interest?

(a) 3 year

(b)3.5 years

(c)

4 years

(d)

2 years

6. What is the present worth of Rs. 1100 due in 2 years at 5% simple interest per annum?

(a) Rs.900

(b)

Rs.1000

(c) Rs.1010

(d)Rs.1025

7. What will be the ratio of simple interest earned by certain amount at the same rate of

interest for 9 years and that for 12 years?

(a)

1

2

(b)4

5

(c)3

4

(d)

5

7


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8.

A sum of money at simple interest amounts to Rs. 750 in 3 years and to Rs. 794 in 4 years.

The sum is:

(a)

Rs.620

(b)RS.630

(c)

Rs.625

(d)Rs. 618

9. A sum of money becomes7

6of itself in 3 years at a certain rate of simple interest. The rate

per annum is:

(a) 18%

(b)25%

(c)

559%

(d)65

9%

10.The difference between the simple interest received from two different sources on Rs.1200

for 5 years is Rs. 18.60.the difference between their rates of interest is:

(a) 0.31%

(b)

0.32%

(c) 0.5%

(d)0.4%

11.Ram a moneylender finds that due to a fall in the annual rate of interest from 11% to 9%,

his yearly income diminishes by Rs. 24.50.His capital is:

(a)

Rs. 1250

(b)Rs.1290

(c) Rs.1225

(d)

Rs.120012.

A man lends Rs. 20,000 in four parts. If he gets 10% on Rs.4000, 15% on Rs. 8000 and 18%

on 2000.If his average annual interest is 20% what percent must he get for the remainder?

(a)

32%

(b)35%


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(c)

35.4%

(d)34%

13.What should be the least number of years in which the simple interest on Rs. 3200 at 51

3%

will be an exact number of rupees?(a)

6

(b)5

(c) 4

(d)3

14.

Ram takes a loan from a bank of Rs 1000 at 5% simple interest. He returns Rs. 500 at the

end of 1 year .In order to clear his dues at the end of 2 years, he would pay:

(a)

Rs.570

(b)Rs.572

(c)

Rs.575

(d)

Rs.576

15.A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same

rate of simple interest .The rate of interest per annum is:

(a) 12%

(b)

13%

(c) 14%

(d)15%

Answers

1.

(c)

2.

(a)

3. (c)

4.

(a)

5.

(d)

6. (b)

7. (c)


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8.

(d)

9.

(c)

10.(a)

11.

(c)

12.(d)

13.(d)

14.(c)

15.

(a)


1. P= 18,000, R = 25%, T = 6/12 year

S.I. =P R T

100=

18000 25 1

100 2= Rs.2, 250.

2. Principal (P) = Rs. 2,400, Rate (R) =25

6, Time (T) = (27+31+15) days =73 days =

73

365year =

1

5

year

Now, S.I. = (P R T

100) S.I. = (2,400

25

6

1

5

1

100) S.I. = 20.

Therefore, the simple interest on Rs. 2,400 at 4 1

6% per annum for the period from 1

stFeb,

2005 to 15thApril, 2005 will be Rs. 20.

NOTE: The day on which money is withdrawn is counted while the day on which money is

deposited is not counted.

3. According to the given situation a sum of Rs. 600 amounts to Rs. 950 in 5 years, then

S.I. = Rs. 900 Rs.600 = Rs.300

P = Rs. 600, T = 5 Years

Therefore,

R =S.I.100

PT

R = (300 100

6005)% = 10%

If the interest rate is increased by 10 %, then

New Rate = (10+5) % =15%, New S.I. =Rs.(600155

100)= Rs.450

Then, New Amount = Rs.600 + Rs.450 = Rs.1050


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4. Gain in 2 years = Rs. [ ( 5000 25

4

2

100) (

5000 42

100) ] = Rs.(625 400) =Rs.225

Then, gain in 1 year = Rs. (225

2) = Rs. 112.50

10. (1200 1 5

100

) - (1200 2 5

100

)= 18.60 6000(R1 -R2) = 1860 R1 R2 = 0.31%.


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COMPOUND INTEREST

When the borrower X and the lender Y agrees to fix up a certain time for example yearly, half

yearly or quarterly to settle the previous money, then the difference between the amount and

the money borrowed is said to be the Compound Interest and it denoted by C.I. In these

calculations, principal for the second unit of time is the amount of first unit of time and so on.

Some important facts and formulae

Let Principal = P, Time = n years

(i)

If interest is compounded annually, then

Amount = P (1 +R

100)n

(ii) If interest is compounded half- yearly, then

Amount = P (1+R/2

100)2n

(iii) If interest is compounded Quarterly, then

Amount = P (1+R/4

100)4n

(iv) If time is in fractions and the interest is compounded yearly, say 2 , then

Amount = P (1 +R

100)2(1+

1

2R

100)

Solved Examples

Example1. What will be the Compound Interest on Rs. 5000 at 5% per annum for 3 years,

compounded annually?

Solution. Amount = Rs. [5,000 ( 1 +5

100)3] = Rs. [5,000 (

21

20)3] = Rs. [5,000

9,261

8,000]

= Rs.5788.125

Compound Interest = Rs. (5788.125 5000) = Rs.788.125.

Example2. What will be the Compound Interest on Rs. 2000 at 10% per annum for 1 year 2months, compounded annually?

Solution.Given that principal = Rs. 2000, R = 10%, Time = 1 year 2 months = 11

6year

Amount = Rs. [2000 ( 1 +10

100) ( 1 +

1

610

100) ] = Rs. [2000 (

11

10) (

61

60) ] = Rs. 2236.67


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Compound Interest = Rs. (2236.67 2000) =Rs. 236.67.

Example3. What is the compound interest on Rs. 12000 in 4 years at 20 % per annum, the

interest being compounded half yearly.

Solution.Given that,

Principal = Rs. 12000, Rate = 20% per annum, Time = 4 years

Now, Amount = Rs.[ 12000 ( 1 +20

2

100) 2= Rs. (120 11 11) = Rs. 14520

Compound Interest = Rs.( 14520 12000) = Rs. 2520.

Example4. If the simple interest on Rs. 1000 at x % per annum for 2 years is Rs.200, then

calculate the compound interest on the same amount for the same period at the same time.

Solution.Here,

Principal = Rs.1000, Rate = x% per annum, Time = 2 years, S.I. = Rs.200

Then, S.I. = Rs.(1000 x 2

100) Rs.200 = Rs.(

1000 x 2

100) x = 10%.

Exercise

1. What will be the compound interest on a sum of Rs 8000 at the rate of 15% per annum after

2 years?

(a) Rs.2400

(b)Rs.2450

(c) Rs.2580

(d)

Rs.2650

2.

What is the difference between the compound interests on Rs. 10000 for 21

2years at 4%pe

annum yearly and half yearly?

(a) Rs.8

(b)

Rs.8.49

(c)

Rs.7

(d)

Rs. 10

3. A sum of money tripled itself at compound interest in 10 years. In how many years will it

become 27 times.


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(a)

35 years

(b)31 years

(c) 30 years

(d)

32 years

4. At what rate of compound interest per annum will a sum of Rs.10000 becomes Rs. 20736 in

4 years.

(a) 21%

(b)20%

(c) 25%

(d)

26%

5.

What will be the compound interest on a sum of Rs.7200 at 5 percent per annum in 2years? (SBI Assistance and Stenographer Exam 2012)

(a) Rs.841

(b)

Rs.738

(c) Rs.793

(d)Rs.812

(e)Rs.694

6.

There is 60% increase in an amount in 6 years at Simple Interest what will be the compound

interest of Rs. 12000 after 3 years at the same rate?

(a) Rs.2160

(b)Rs.3120

(c) Rs.3972

(d)

Rs.6240

(e)None of these

Answers

1.

(c)

2. (b)

3. (c)


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4.

(b)

5.

(b)

6. (b)


1. Amount = Rs. [8000 1 + 15100

2]Rs. [8000

23 23

20 20]= Rs. 10,580

C.I. = Rs. [10580 8000] = Rs. 2,580

2.

C.I. , When interest is compounded yearly

Amount =

10000

1 +

4

1002

1 +

1

24

100

Amount = 10000 26 2625 25

5150

= Rs.11,032.32C.I. = 11032.32 10000 = 1032.32C.I. When interest is compounded half yearly

Amount = Rs [10000 ( 1 +4

2

100)

2 5

2] = Rs. 11040.80803

C.I. = 11040.80803 10000 = 1040.808032

Diff = 8.488032 Rs. 8.9

3. P 1 + 100

10= 3P1 +

10010= 3

Suppose after n years it will become 27 times

Then,

P 1 + 100 = 27 p 1 + 100 = 331 +

100 = 1 +

10030 n = 30.

4. Let the rate of C.I. = R, then


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P 1 + 100

= 20736 10000 1 + 100

4= 20736 1 + 100

4= 1210

4 100

=12

10-

1

R = 20%5.

Amount = 7200 1 + 51002=

7200 21 21

20 20= Rs.7938

C.I. = Rs.(7938 7200) = Rs.738

6. Suppose principle = Rs. 100

Then, S.I. = 60 Rs, Time = 6 years

S.I. =

100 60 =

100 6100

R = 10%

Now , Amount = 12000 1 + 10100

= Rs. 15972C.I. = Rs.(15972 12000) = Rs.3972


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PROBABILITY

Probability: It is a mathematical measurement of uncertainty of an event. The value of

Probability varies from 0 to 1. The 0 value of an events probability indicates that the event

cannot happen. On the other side the value 1 for any event indicates that the event is certain.

Some Important Term

1.

Experiment:It is an operation which can produce some well defined outcomes.

2. Random Experiment:It is an experiment in which all the possible outcomes are known but

the exact output cannot be predicted in advance.

For example:Rolling an unbiased dice, tossing a fair coin etc.

3. Sample Space: It is a set of all possible outcomes of a random experiment.

For Example; if we toss a coin the possible outcomes are head (H) and tail (T). Therefore the

sample space for tossing a coin = {H, T}.

4.

Event: It is a subset of a sample space.

For Example; if we toss two coins, the possible outcomes are {HH, HT, TH, TT}. Here the

events can be; only H = {HH}, only one H = {HT, TH} or no H = {TT}.

5.

Probability of an Event:

Let S be the sample space and E be an Event in the given sample space.

Then, E S.

() = ()().Some Important Results

1.

P(S) = 1,

2. 0P(E)1

3.

P() = 0

4. For event A and B, P (A B) = P (A) + P (B) P (A B)


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5.

If A is (not A), then P ( A ) = 1 P (A).

Some Important Formulae

1.

n ! = n (n 1) (n 2) ......... 3.2.1

2.

0 ! = 1 = 1 !

3.

Selection of r things out of n things = n C r= !

! ()!= ( 1)( 2)

( 1)( 2) .. 3.2.1

4.

n C n=n C 0= 1

5.n

C r=n

C (n r)


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Exercise

1. Three unbiased coins are tossed. The probability of getting two heads is

(a)

1

8

(b)

3

8

(c)5

8

(d)7

8

2.

Three unbiased coins are tossed. The probability of getting one head is

(a)1

8

(b)

3

8

(c)5

8

(d)7

8

3.

Three unbiased coins are tossed. The probability of getting at least one head is

(a)

1

8

(b)

3

8

(c)5

8

(d)7

8

4. Three unbiased coins are tossed. The probability of getting at most one head is

(a)1

2

(b)3

8

(c)

58

(d)

7

8

5. Two dice are thrown simultaneously. What is the probability of obtaining a total score of

six?

(a)1

2

193243068 banking exam idea

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