ibps po quantitative aptitude

2 IBPS BANK PO EXAM 2013 : Quantitative Aptitude

Table of Contents

PREFACE ............................................................................................................................................. 6

IMPORTANT CHAPTER GUIDELINES ........................................................................................... 7

CHAPTER: SIMPLIFICATION & APPROXIMAITON ................................................................... 8

Solved Examples (Simplification & Approximation) ............................................................................................. 12

Practice Set-1 (Simplification & Approximation) .................................................................................................. 22

Practice Set-2 (Simplification & Approximaiton) .................................................................................................. 25

Simplification & Approximaiton Averages Practice Set-1 (Answers) .................................................................... 27

Simplification & Approximaiton Averages Practice Set-2 (Answers) .................................................................... 28

CHAPTER: NUMBER SERIES ......................................................................................................... 29

Solved Examples (Number Series) ........................................................................................................................ 33

Practice Set (Number Series)................................................................................................................................ 38

Number Series Practice Set (Answers) ................................................................................................................. 41

CHAPTER: NUMBER SYSTEM ....................................................................................................... 42

Solved Examples (Number System) ...................................................................................................................... 48

Practice Set-1 (Number System) .......................................................................................................................... 53

Practice Set-2 (Number System) .......................................................................................................................... 55

Number System Practice Set-1 (Answers) ............................................................................................................ 58

Number System Practice Set-2 (Answers) ............................................................................................................ 59

CHAPTER: RATIO, PROPORTION & ALLIGATION .................................................................. 60

Solved Examples (Ratio, Proportion & Alligation) ................................................................................................ 66

Practice Set-1 (Ratio, Proportion & Alligation) ..................................................................................................... 70

Practice Set-2 (Ratio, Proportion & Alligation) ..................................................................................................... 74


Ratio, Proportion & Alligation Practice Set-1 (Answers) ....................................................................................... 77

Ratio, Proportion & Alligation Practice Set-2 (Answers) ....................................................................................... 78

CHAPTER: AVERAGES ................................................................................................................... 79

Solved Examples (Averages) ................................................................................................................................ 80

Practice Set (Averages) ........................................................................................................................................ 86

Averages Practice Set (Answers) .......................................................................................................................... 90

CHAPTER: PERCENTAGES, PARTNERSHIP AND SHARE ....................................................... 91

Solved Examples (Percentages, Partnership and Share) ....................................................................................... 95

Practice Set-1 (Percentages, Partnership and Share) .......................................................................................... 101

Practice Set-2 (Percentages, Partnership and Share) .......................................................................................... 104

Percentages, Partnership and Share Practice Set-1 (Answers) ........................................................................... 106

Percentages, Partnership and Share Practice Set-2 (Answers) ........................................................................... 107

CHAPTER: PROFIT & LOSS ........................................................................................................ 108

Solved Examples (Profit & Loss) ......................................................................................................................... 112

Practice Set (Profit & Loss) ................................................................................................................................. 117

Profit & Loss Practice Set (Answers) .................................................................................................................. 121

CHAPTER: SIMPLE INTEREST & COMPOUND INTEREST .................................................. 122

Solved Examples (Simple Interest & Compound Interest) .................................................................................. 126

Practice Set (Simple Interest & Compound Interest) .......................................................................................... 132

Simple Interest & Compound Interest Practice Set (Answers) ............................................................................ 136

CHAPTER: TIME AND WORK .................................................................................................... 137

Solved Examples (Time and Work) ..................................................................................................................... 140

Practice Set (Time and Work)............................................................................................................................. 146

Averages Practice Set (Time and Work) ............................................................................................................. 151


CHAPTER: SPEED DISTANCE & TIME ..................................................................................... 152

Solved Examples (Speed Distance & Time)......................................................................................................... 156

Practice Set (Speed Distance & Time) ................................................................................................................ 162

Speed Distance & Time Practice Set (Answers) .................................................................................................. 166

CHAPTER: MENSURATION ........................................................................................................ 167

Solved Examples (Mensuration) ........................................................................................................................ 174

Practice Set (Mensuration) ................................................................................................................................ 178

Mensuration Practice Set (Answers) .................................................................................................................. 181

CHAPTER: PERMUTATIONS, COMBINATIONS & PROBABILITY ..................................... 182

Solved Examples (Permutations, Combinations & Probability) .......................................................................... 188

Practice Set-1 (Permutations, Combinations & Probability) ............................................................................... 195

Practice Set-2 (Permutations, Combinations & Probability) ............................................................................... 197

Permutations, Combinations & Probability Practice Set-1 (Answers) ................................................................. 199

Permutations, Combinations & Probability Practice Set-2 (Answers) ................................................................. 200

CHAPTER: DATA INTERPRETATION ...................................................................................... 201

CHAPTER: DATA INTERPRETATION-TABLE CHART ......................................................... 204

Solved Examples (Data Interpretation-Table Chart) ........................................................................................... 206

Practice Set (Data Interpretation-Table Chart) ................................................................................................... 213

Data Interpretation-Table Chart Practice Set (Answers) .................................................................................... 221

CHAPTER: DATA INTERPRETATION-LINE GRAPHS ........................................................... 222

Solved Examples (Data Interpretation-Line Graphs) .......................................................................................... 225

Practice Set (Data Interpretation-Line Graphs) .................................................................................................. 229

Data Interpretation-Line Graphs Practice Set (Answers) .................................................................................... 233

CHAPTER: DATA INTERPRETATION-BAR GRAPHS ............................................................ 234


Solved Examples (Data Interpretation-Bar Graphs)............................................................................................ 237

Practice Set (Data Interpretation-Bar Graphs) ................................................................................................... 243

Data Interpretation-Bar Graphs Practice Set (Answers) ..................................................................................... 247

CHAPTER: DATA INTERPRETATION-PIE DIAGRAM .......................................................... 248

Solved Examples (Data Interpretation-Pie Diagram) .......................................................................................... 250

Practice Set (Data Interpretation-Pie Diagram) .................................................................................................. 256

Data Interpretation- Pie Diagram Practice Set (Answers) ................................................................................... 262

CHAPTER: DATA INTERPRETATION-CASE LETS ................................................................ 263

Solved Examples (Data Interpretation-Caselets) ................................................................................................ 266

PracticeSet-(Data Interpretation- Caselets)........................................................................................................ 270

Data Interpretation-Caselets Practice Set (Answers).......................................................................................... 274

Solved Examples (Data Interpretation-Miscellaneous) ...................................................................................... 275

Practice Set (Data Interpretation-Miscellaneous) .............................................................................................. 285

Data Interpretation-Miscellaneous Practice Set (Answers) ................................................................................ 291

FEEDBACK ..................................................................................................................................... 292


Preface

Jagranjosh’s IBPS PO Exam 2013: Quantitative Aptitude e-Book is a one-stop solution for aspirants who endeavor to leave no stone unturned to score considerably in IBPS PO Written Examination 2013. The eBook is highly useful for all officers level Banking exams - IBPS Specialist officers Exam and SBI PO Exam. IBPS PO Exam 2013: Quantitative Aptitude e-Book is prepared by Jagranjosh’s team of subject matter

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Aptitude section of Officers level Banking Exams. This eBook is a perfect blend of chapter-wise basic

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of Officers level Banking Exams. The chapter-wise compilation of this e-Book makes the concept of

Quantitative aptitudes easy to understand for students. It further includes previous year questions and

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• Chapter wise Basic Concepts & Questions

• IBPS PO Previous Year Questions

• Chapter wise Model Practice Set

• Important Chapter Guidelines

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Important Chapter Guidelines

Important Chapter Guidelines for IBPS PO 2013 : Quantitative Aptitude

Chapter Importance

Number System Very Important

Number Series Very Important

Percentage Partnership and Share Very Important

Simplification and Approximation Very Important

Average Important

Ratio ,Proportion & Alligation Important

Time & Work Important

Profit, Loss & Discount Very Important

Simple Interest and Compound Interest Important

Time, Distance and Speed Important

Permutation and Combination & Probability Very Important

Data Interpretation-Table Charts Very Important

Data Interpretation-Bar Graph Very Important

Data Interpretation-Line Graph Very Important

Data Interpretation-Pie Charts Very Important

Data Interpretation-Caselets Very Important

Data Interpretation-Miscelleneous Graphs Very Important

Mensuration Important

Note : Prepared on the basis of questions asked in previous year papers


Chapter: Simplification & Approximaiton

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 | = .

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

5 + + x +2 = 9 .

Solution: Simplifying the above equation

x = 9 - 5 - – 2 x = - - -

x =

x = = 1 .


Example: If = , then find the value of .

Solution: Given expression: =

( x/y =3/2)

.

Example: Arjun spends of 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 Arjun’s monthly salary is Rs. x

Then, remaining part of his salary = x- ( + + ) x = x – ( ) x = = .

Now, = 2400 x =Rs. 4500.

Expenditure on Conveyance = Rs.( 4500 ) = Rs.450.


Approximation

In these types of questions there is no need to find out the exact values. The candidate is required to calculate the approximate value in the following manner:

Step I: Round off the numbers given in the question.

Step II: Simplify the value

Step III: Round off the final simplified value

Rounding off numbers:

Rounding off the numbers given in the question can be done in the following manner:

(a) Rounding off to the nearest 10: Example: Rounded off value of 56 is 60, because the digit at unit place is greater than 5, therefore, we will add 1 to the digit at tens place and replace the unit’s digit by 0. Example: Rounded off value of 54 is 50, because the digit at unit place is less than 5, therefore, the value at tens place will remains the same and unit’s digit will be replaced by 0. Example: Rounded off value of 35 is 40, because the digit at unit place is equal to 5, therefore, we will add 1 to the digit at tens place and replace the unit’s digit by 0.

(b) Rounding off to the nearest 100: Example: Rounded off value of 386 is 400, because the digit at tens place is greater than 5, therefore, we will add 1 to the digit at hundreds place and replace the digit at unit and tens places by 00. Example: Rounded off value of 741 is 700, because the digit at tens place is less than 5, therefore, the value at hundreds place will remains the same. Unit and tens digit will be replaced by 00.

(c) Rounding off to the nearest 1000 Example: Rounded off value of 1963 is 2000, because the digit at hundreds place is greater than 5, therefore, we will add 1 to the digit at thousand place and replace the ones, tens and hundreds by 000.

Although the numbers can be rounded off by the above procedures but it depends on the other numbers involved in the simplification.

Rounding off a number to a decimal place:


The candidate is required to follow the following steps to round off a number to the nth decimal place:

Step I: check the digit immediately, next right to the nth place.

Step II: Add 1 to the digit in the nth place, if the next right digit is 5 or more, otherwise the digit will remains the same.

Step III: Remove all the digits in places to the right of the nth place.

Example: Rounded off value of 4.693 to the second place is 4.69, because next right digit to the second place is 3, therefore, the value will remain the same and the digit in place to the right of the second place will be removed.


Solved Examples (Simplification & Approximation)

Directions (1-5) What will come in place of the question mark (?) in the following question? (IBPS PO/MT Exam 2012) 1. 4003 × 77 - 21015=? × 116

(1) 2477 (2) 2478 (3) 2467 (4) 2476 (5) None of these

Solution: ? × 116 = 4003 × 77 – 21015 or, ? × 116 = 308231 – 21015 = 287216 or, ? × 116 = 287216

? = = 2476

Ans: (4)

2. (1) 143

(2) (3) 134

(4) (5) None of these Solution:

= - 361 = - 361 = 504 -361 = 143 Ans: (1)


3. (4444÷40) + (645÷25) + (3991÷26) =? (1) 280.4 (2) 290.4 (3) 295.4 (4) 285.4 (5) None of these Solution: (4444÷40) + (645÷25) + (3991÷26)

= + +

= 111.1 + 25.8 + 153.5 = 290.4 Ans: (2)

4. × - (83)2 = (?)2 + (37)2 (1) 37 (2) 33 (3) 34 (4) 38 (5) None of these Solution: (?)2 + (37)2 = × - (83)2 or, (?)2 + (37) 2 = 182×51 - (83) 2 or, (?)2 + 1369 = 9282-6889 =2393 or, (?)2 = 2393 - 1369 = 1024

?= = 32 Ans: (5)

5. 5 × 4 × 11 + 2 =?

(1) 303.75 (2) 305.75

(3)

(4)

(5) None of these


Solution: ? = 5 × 4 × 11 + 2

= × × +

= × × +

= 101× 3 + = 303 + =

= = 305.75

Ans: (2) Direction (6-10): What approximate value should come in place of the question mark (?) in the following question? (Note: You are not excepted to calculate the exact value.) (IBPS PO/MT Exam 2012) 6. 8787 ÷ 343 × =? (1) 250 (2) 140 (3) 180 (4) 100 (5) 280 Solution: ? = 8787 ÷ 343 × = 25.61 × 7.07 = 181.09 Ans: (3)

7. × (303÷8) = (?)2 (1) 48 (2) 38 (3) 28 (4) 18 (5) 58

Solution: × (303÷8) = (?)2


or, 38 × 37.8 = (?)2 (37.8 ~ 38) or, 38 × 38 = (?)2

Ans: (2)

8. of 4011.33 + of 3411.22 =?

(1) 4810 (2) 4980 (3) 4890 (4) 4930 (5) 4850

Solution: ? = 4011.33 + 3411.22

= +

= 2507.08 + 2387.854 = 2507 + 2387 = 4894 Ans: (3) 9. 23% of 6783 + 57% of 8431=? (1) 6460 (2) 6420 (3) 6320 (4) 6630 (5) 6360 Solution: ?= 23% of 6783 + 57% of 8431

6783 + 8431

= 23 = 1560.09 + 4805.67 = 6365.76 Ans: (5)


10. 335.01 × 244.99÷55=? (1) 1490 (2) 1550 (3) 1420 (4) 1590 (5) 1400 Solution: ? = 335.01 × 244.99 55 = 335 × 245 55

= 335 × = = 1422.27 1490

Ans: (1)

Directions (Q.11-15): What will come in place of the question mark (?) in the following questions? (IBPS PO/MT Exam 2011)

11. 3463 × 295 - 18611 =? + 5883

(1) 997091

(2) 997071

(3) 997090

(4) 999070

(5) None of these

Solution: 3463 x 295 -18611 =? + 5883

? = 1021585 - 18611 - 5883 = 997091

Ans: (1)


12. (23.1)2 + (48.6)2 - (39.8)2 =? + 1147.69

(1) (13.6)2

(2)

(3) 163.84

(4) 12.8

(5) None of these

Solution: 533.61 + 2361.96 - 1584.04 =? + 1147.69

or, ? = 1311.53 - 1147.69 = 163.84

Ans: (3)

13. + = ?

1)

2) 0.75

3) 1

4) None of these

Solution: + = = = =

Ans: (4)


14. [(3 + ) X (8 + 7 )]- 98 =?

(1) 2

(2) 8

(3) 382 (4) 386 (5) None of these

Solution: [ (3 +1) (8 + 7) ]- 98

= [4 15 ] - 98

= [60 8] - 98 = 480 - 98 = 382

Ans: (3)

15. - (54)2 = + (74)2

(1) 3844 (2) 3721 (3) 3481 (4) 3638 (5) None of these

Solution: - (54)2 - (74)2 =

or , = - 2916 - 5476

= 8453 - 2916 - 5476 = 61


or , ? = (61)2 = 3721

Ans: (2)

Directions ( 16-20): What approximate value should come in place of question mark (?) in the following questions? (Note: You are not expected to calculate the exact value] (IBPS PO/MT Exam 2011)

16. 39.897% of 4331 + 58.779% of 5003 =?

(1) 4300

(2) 4500

(3) 4700

(4) 4900

(5) 5100

Solution: 40 + 59 = 1732 + 2950 = 4682 4700

Ans: (3) 17. 43931.03 / 2111.02 x 401.04 =?

1) 8800

2) 7600

3) 7400

4) 9000

5) 8300

Solution: 43931 2111 401 =?

or, ? = 44000 400


or, ? = 400 = 8800

Ans: (5)

18. 34.993 =?

1) 3000

2) 2800

3) 2500

4) 3300

5) 2600

Solution: 34.993 = 80 x 35 = 2800

Ans: (2)

19. +349 =? 21.003

1) 7600

2) 7650

3) 7860

4) 7560

5) 7680

Solution: 17 + 349 =? 21

or, 366 21 = ?

or, ? = 7686 7680.

Ans: (5)


20. 59.88 12.21 6.35 =?

1) 10

2) 50

3) 30

4) 70

5) 90

Solution: 60 12 6 = 30

Ans: (3)


Practice Set-1 (Simplification & Approximation)

Directions (Q. 1-5): What will come in place of question mark (?) in the following questions? (IBPS RRB Grade A Officer Exam 2012)

1.

(1)

(2) 2

(3) 16

(4) 8

(5) None of these

2. 55% of

(1) 126.5 (2) 126.6 (3) 124.6 (4) 125.4 (5) None of these

3.

(1)

(2)

(3) (4) 18 (5) 32

4.

(1) 81

(2) 64

(3) –8

(4) –7

(5) 9

5.

(1) 6 (2) 2 (3) 4 (4) 0 (5) None of these

Directions (Q. 6-10). What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value.)

6. 68% of 1288 + 26% of 734 – 215 =?

(1) 620 (2) 930 (3) 540 (4) 850 (5) 710

7.

(1) 670 (2) 530 (3) 420 (4) 780 (5) 960

8. 6578 ÷ 67 ⤬ 15 =? ⤬ 6


(1) 200 (2) 250 (3) 150 (4) 100 (5) 300

9.

(1) 540 (2) 760 (3) 800 (4) 1260 (5) 1040

10.

(1) 840 (2) 910 (3) 1320 (4) 1120 (5) 1550

Directions (Q.11 to 15): What will come in place of question mark (?) in the following questions? (RBI Grade ‘B’ Officers Exam 2011)

11. [(3024 ÷ 189)1/2 + (684 ÷ 19)2] = (?)2 + 459

(1) -27 (2) -29 (3) 31 (4) 841 (5) 1089

12. 4.4 times of 30% of 216 =?

(1) 81.9 (2) 83.7 (3) 87.3 (4) 89.1

(5) None of these

13. (0.0729 ÷ 0.1)3 ÷ (0.081 × 10)5 × (0.3 × 3)5 = (0.9)? + 3

(1) 1 (2) 2 (3) 4 (4) 7 (5) None of these

14. of × 5) = 149.8 – 112

(1) (2) 18 (3) 324 (4) 24 (5) None of these

15. (27)2 × 6 ÷ 9 + (7)3 + 71 = (?)3 – 431

(1) 11 (2) (13)3 (3) 13 (4) (11)2 (5) None of these

Directions (Q. 16-20): What will come in place of question mark (?) in the following questions? (Corporation Bank PO Exam 2011)

16. 5907 – 1296 ÷ 144 = ? ⤬ 8

(1) 726.75 (2) 767.25 (3) 737.25 (4) 676.75 (5) None of these

17.


(1)

(2) (3) 3 (4) 9 (5) -3

18.

(1) 11 (2) 5 (3) 7 (4) 9 (5) None of these

19.

(1) (2) 200

(3) (4) 100 (5) –10

20.

(1) 169 (2) 13 (3) 14 (4) 196 (5) None of these

25 IBPS PO EXAM 2013 : Quantitative Aptitude

Practice Set-2 (Simplification & Approximaiton)

Directions (Q. 1-5): What approximate value will come in place of question mark (?) in the following questions? (Corporation Bank PO Exam 2011)

(You are not expected to calculate the exact value.)

1. 3237 ÷ 31 ⤬ 15 = ? ⤬ 17

(1) 90 (2) 100 (3) 110 (4) 120 (5) 80

2.

(1) 1760 (2) 1880 (3) 1950 (4) 1720 (5) 1650

3. (34.34)2 + (5.96) 2 – (23.09) 2 =?

(1) 510 (2) 540 (3) 620 (4) 680 (5) 650

4.

(1) 200 (2) 180 (3) 120 (4) 140

(5) 160

5. 61% of 981 – 150.17 = ? – 65% of 676

(1) 760 (2) 780 (3) 830 (4) 860 (5) 890

Directions-(Q. 6-10): What will come in place of question-mark (?) in the following questions? (Allahabad Bank Probationary Officers Exam 2011)

6. of 30% of 3420 = (?)2 x 2

(1) (81)2 (2) 7 (3) 9 (4) 81 (5) 49

7. 1898 ÷ 73 x 72 = (?)2 x 13

(1) - 256 (2) 256 (3) 12 (4) 144 (5) -16

8. =?

(1) 42 (2) 1024 (3) 1764 (4) (1024)2 (5) 32


9. (0·81)2+ (0·729)3 x (0·9)2 = (0.9)?-3

(1) 6 (2) 2 (3) 4 (4) 0 (5) None of these

10. 65% of x 5 =? + 154

(1) 56 (2) 28 (3) 35 (4) 32 (5) None of these

Directions-(Q. 11-15) What approximate value will come in place of question-mark (?) in the following questions? (Allahabad Bank Probationary Officers Exam 2011)

(You are not expected to calculate the exact value.)

11. =? ÷ 8

(1) 620 (2) 670 (3) 770 (4) 750 (5) 700

12. 89·988% of 699·9 + 50·002% of 999·99 - 170·015 =?

(1) 990 (2) 900 (3) 920 (4) 960 (5) 860

13. ÷ =?

(1) 760 (2) 800 (3) 690 (4) 870 (5) 780

14. 6999 ÷70·005 x 94·998 =? x 19·999

(1) 475 (2) 420 (3) 320 (4) 540 (5) 525

15. (49·99)2 – (8·9)2 – (15·9)2 =?

(1) 2165 (2) 2000 (3) 1965 (4) 1920 (5) 1885


Simplification & Approximaiton Averages Practice Set-1

(Answers)

1) 5

2) 1

3) 1

4) 4

5) 3

6) 4

7) 2

8) 2

9) 5

10) 3

11) 2

12) 4

13) 1

14) 5

15) 1

16) 3

17) 2

18) 4

19) 5

20) 2


Simplification & Approximaiton Averages Practice Set-2

(Answers)

1) 1

2) 1

3) 5

4) 3

5) 5

6) 3

7) 3

8) 5

9) 4

10) 2

11) 2

12) 4

13) 4

14) 1

15) 1


Chapter: Number Series

In Number Series, questions are asked on the basis of relation between numbers given in a series. The questions asked can be divided into different types:

Type I: In this type of questions, a series of numbers is given with one number missing represented by a question mark. Candidate has to select from the options choices to correct option in place of the question mark. The given sequences of numbers will be such that each number follows its predecessor in the same way, i.e., according to a particular pattern. Candidates are required to find out the correct ways in which the sequence is formed and there after find out the number to complete the series.

1. 30, 34, 43, 59, 84, 120,?

(1) 169

(2) 148

(3) 153

(4) 176

(5) None of these

Solution: (1) The given pattern is:

+22, 32, +42, + 62, +72

So, missing term is 169=120 +72

2. 40, 54, 82,?, 180,250

(1) 142

(2) 124

(3) 136

(4) 163

(5) None of these

Solution: (2) The pattern is: +14, + 28, + 42, + 52, + 70


So, missing term is 82 + 42=124

Type II: Here, we are given a sequence of number. Whole sequence, except the odd number follow a certain rule. You have to find that number which does not follow the rule.

1. 0, 1,3,8,18,35,264

(1) 62

(2) 35

(3) 18

(4) 8

(5) None of these

Solution: (1) The pattern is +(02+1), +(12+1), + (22+1) ,+ (32+1), + (42+1), + (52+1)

So, 62 is wrong and must be replaced by 35 + (52+1) = 62

2. 1, 9, 125, 49, 729, 121, 2147

(1) 2147

(2) 729

(3) 125

(4) 1

(5) None of these

Solution: (1)

Type III: In this type of questions, a number series is given. After the series is over, in the next line, a number is followed by (A), (B), (C), (D) and (E). The candidates have to complete the series starting with the number given following the sequence of the given series.

1. 2, 4, 9, 20, 43, 90

3 (A) (B) (C) (D) (E)

Which number will come in place of (D)?

(1) 58

(2) 99


(3) 48

(4) 59

(5) None of these

Solution: (4)

Similarly,

Hence, the number 59 will come in place of D.

2. 3, 4, 10, 33, 136,

3 (A) (B) (C) (D) (E)

Which number will come in place of (E)?

(1) 1035

(2) 1165

(3) 1039

(4) 891

(5) None of these

Solution: (2)


Similarly,

Hence, the number 1165 will come in place of E.


Solved Examples (Number Series)

Directions: (1-5) In each of these questions a number series is given. In each series only one number is wrong. Find out the wrong number. (IBPS CWE PO MT 2012) 1. 5531 5506 5425 5304 5135 4910 4621 (1) 5531 (2) 5425 (3) 4621 (4) 5135 (5) 5506 Solution: The number should be 5555 in place of 5531. -72, -92, -112, -132, -152, -172... Ans: (1) 2. 6 7 9 13 26 37 69 (1) 7 (2) 26 (3) 69 (4) 37 (5) 9 Solution: The number should be 21 in place of 26. +1, +2, +4, +8, +16, +32 Ans: (2) 3. 1 3 10 36 152 760 4632 (1) 3 (2) 36 (3) 4632 (4) 760


(5) 152 Solution: The number should be 770 In place of 760. ×1 +2, ×2 +4, ×3 +6, ×4 + 8, ×5 +10, ×6 + 12, ... Ans: (4) 4. 4 3 9 34 96 219 435 (1) 4 (2) 9 (3) 34 (4) 435 (5) 219 Solution: The series is 02+ 4, 12+2, 32+0, 62-2, 102-4, 152- 6,212 - 8... Hence, 435 should be replaced with 433 Ans: (1) 5. 157.5 45 15 6 3 2 1 (1) 1 (2) 2 (3) 6 (4) 157.5 (5) 45 Solution: The number should be 2 in place of 1.

3.5, 3, 2.5, 2, 1.5, 1, ... Ans: (1)

Directions (6-10): In the following number series only one number is wrong.

Find out the wrong number.

6. 7 12 40 222 1742 17390 208608

(1) 7

(2) 12

(3) 40


(4) 1742

(5) 208608

Solution: The pattern of number series is as follows:

7 2 - 2 = 12

12 4 - (2 + 6) = 48 - 8 = 40

40 6 - (8 + 10) = 240 - 18 = 222

222 8 - (18 + 14) = 1776 - 32 = 1744 1742

1744 10 - (32 + 18) = 17440 - 50 = 17390

Ans: (4)

7. 6 91 584 2935 11756 35277 70558

(1) 91

(2) 70558

(3) 584

(4) 2935

(5) 35277


6 7 + 72 = 42 + 49 = 91

91 6 + 62 = 546 + 36 = 582 584

582 5+52 =2910 + 25=2935

2935 4 + 42 = 11740 + 16 = 11756

11756 x 3 + 32 = 35268 + 9 = 35277

Ans: (3)


8. 9050 5675 3478 2147 1418 1077 950

(1) 3478

(2) 1418

(3) 5675

(4) 2147

(5) 1077


9050 – I53 = 9050 - 3375 = 5675

5675 - 133 = 5675 - 2197 = 3478

3478 - 113 = 3478 - 1331 = 2147

2147 - 93 = 2147 - 729 = 1418

1418 - 73 = 1418 - 343 = 1075 1077

Ans: (5)

9. 1 4 25 256 3125 46656 823543

(1)3125

(2) 823543

(3) 46656

(4) 25

(5) 256


11 = 1;-22 = 4; 33 = 27 25; 44 = 256; 55 = 3125; 66 = 46656;

77 = 823543

Ans: (4)


10. 8424 4212 2106 1051 526.5 263.25 131.625

(1) 131.625

(2) 1051

(3) 4212

(4) 8424

(5) 263.25


8424 2 = 4212

4212 2 = 2106

2106 2 = 1053 1051

1053 2 = 526.5

526.5 ÷ 2 = 263.25 263.25 ÷ 2 = 13 1.625

Ans: (2)


Practice Set (Number Series)

Directions (Q. 1-3): What will come in place of question mark (?) in the following number series? (IBPS RRB Grade Officer Exam 2012)

1. 987 587 331 187 123 (?)

(1) 104

(2) 113

(3) 107

(4) 114

(5) None of these

2. 125 171 263 401 585 (?)

(1) 835

(2) 815

(3) 792

(4) 788

(5) None of these

3. 121 132 167 226 309 (?)

(1) 424

(2) 413

(3) 427

(4) 416

(5) None of these

Directions (Q. 4-5): In the following number series, only one is wrong. Find out the wrong number.

4. 454 327 648 524 842 713 1036

(1) 327

(2) 648

(3) 521

(4) 842

(5) 713

5. 72.5 86 113 168 275 491 923

(1) 86

(2)113

(3)168

(4)275

(5)491

Directions (Q. 6-10): What will come in place of question mark (?) in the following number series? (RBI Grade’B’ Officer’s Exam 2011)

6. 17 19 33 (?) 129 227

(1) 64 (2) 73 (3) 67 (4) 72 (5) None of these

7. 35 256 451 620 763 (?)


(1) 680 (2) 893 (3) 633 (4) 880 (5) None of these

8. 18 139 868 917 (?) 1051

(1) 1042 (2) 1036 (3) 942 (4) 996 (5) None of these

9. 2890 (?) 1162 874 730 658

(1) 1684 (2) 1738 (3) 1784 (4) 1672 (5) None of these

10. 14 1004 1202 1251.5 1268 (?)

(1) 1267.5 (2) 1276.25 (3) 1324.5 (4) 1367.25 (5) None of these

Directions (Q. 11-15): What will come in place of question mark (?) in the following number series? (Corporation Bank PO 2011)

11. 8 11 20 47 128 (?)

(1) 483 (2) 488 (3) 397 (4) 371 (5) None of these

12. 71 78 99 134 183 (?)

(1) 253 (2) 239 (3) 246 (4) 253 (5) None of these

13. 342 337.5 328.5 315 297 (?)

265.5

274.5

270

260

None of these

14. 161 164 179 242 497 (?)

(1) 1540 (2) 1480 (3) 1520 (4) 1440 (5) None of these

15. 239 254 284 344 464 (?)

(1) 726 (2) 716 (3) 724 (4) 714 (5) None of these

Directions-(Q. 16-20) What will come in place of question-mark (?) in the following number series? (Allahabad Bank Probationary Officers Exam 2011)

16. 958 833 733 658 608, (?)

(1) 577 (2) 583 (3) 567 (4) 573 (5) None of these


17. 11 10 18 51 200, (?)

(1) 885 (2) 1025 (3) 865 (4) 995 (5) None of these

18. 25 48 94 186 370 (?)

(1) 738 (2) 744 (3) 746 (4) 724 (5) None of these

19. 14 24 43 71 108 (?)

(1) 194 (2) 154 (3) 3) 145 (4) 4) 155 (5) 5) None of these

20. 144 173 140 169 136 (?)

(1) 157 (2) 148 (3) 164 (4) 132 (5) None of these


Number Series Practice Set (Answers)

1) 3

2) 2

3) 4

4) 5

5) 3

6) 3

7) 4

8) 1

9) 2

10) 2

11) 4

12) 3

13) 2

14) 3

15) 5

16) 2

17) 4

18) 1

19) 2

20) 5


Chapter: Number System

Introduction: Number is a symbol which represents quantity. There are three types of numbers:

1. Real Numbers: Real numbers are those numbers which can be easily indentify and quantify. For example: -10, -7.33, -1, 0, 1, 2, 5.77 etc.

2. Imaginary Numbers: Imaginary numbers are those numbers which we can just imagine but cannot physically perceive.

For example: , etc, is represented by i . Square root of all negative numbers are imaginary.

3. Complex Number: Combination of Real and Imaginary number is called complex numbers. For example: (2+5i) , (1+3i) etc .

Here we will only discuss about the real numbers . Types of Real Numbers: there are two types of real numbers

1. Rational numbers: - All that numbers which can be expressed in the form of where p

& q are integers and q 0 are called rational numbers.

For example: - -1, 2, , 0, 1 , 2.7 etc. , here -1 = =

Again Rational numbers are classified as:

(a) Integers: All rational numbers which do not have decimal or fractional parts are called integers. For example: -3, -1, 0 , 1 , 2 etc .

Integers are of two type whole numbers and natural numbers. All the non negative integers are whole numbers , for example 0 , 1 , 2 , 3 etc and all the whole numbers except 0 are natural numbers , for example 1, 2 , 3 etc


(b) Fractions: All rational numbers which are in the form of where p & q are integers and

q 0 and p is not a multiple of q are called fractions.

For example: - 1.2, , , 1.7, .3 etc.

Fractions are of three types:

Proper

Improper

Mixed fractions

A proper fraction is a fraction whose numerator is smaller than denominator, for example

, etc. An improper fraction is a fraction whose numerator is equal to or greater than its

denominator for example , etc. and a mixed fraction is an integer plus a fraction, for

example , etc.

2. Irrational Numbers:- All that numbers which cannot be expressed in the form of are

called irrational numbers. They have non-terminating and non-recurring decimal parts.

For example: , , etc.

On the basis of origin:

1. Prime Numbers: All the natural numbers greater than 1 which are only divisible by 1 and the number itself are called prime numbers. For example: 2, 3, 5, 7, 11, 13, 17 etc.

2. Composite Numbers: All the natural numbers greater than 1 which are divisible by at least one more number other than 1 and the number itself are called composite numbers. For example: 4, 6, 8, 9, 10, 12 etc. Note: 1 is neither a prime number nor a composite.

On basis of divisor:

1. Even Numbers: All the natural numbers which are multiple of 2 are called even numbers. For example: 2, 4, 6, 8, 10 etc.


Odd Numbers: All the natural numbers which are not a multiple of 2 are called odd

numbers. They are denoted as 2k 1, where k is a natural number. For example: 3, 5, 7, 9, 11 etc.

To find whether a number is prime or not. Step 1: Find the approximate square root of a number . Step 2: Check if any prime number from 2 to that square root divides that number or not. Step 3: If none of those prime number divides the number than the number must be prime number. Example: Take 631 , the approx square root of 631 is 25 , now from 2 to 25 there are 2 , 3 ,5 , 7 , 11 , 13 , 17 , 19 and 23 prime number . Since none of these divides 631, so 631 must be a prime number.

Conversion of recurring decimal into fraction.

The form of purely recurring number =

Let x = 0.77777…. 10x = 7.77777……

If x = 0.27272727 …. 100 x = 27.272727 ……

The form of purely recurring number =

Let x= 0.143333333…….. 100x = 14.333333……. 1000x = 143.3333….

= 129


Quotient and remainder:

Dividend = (Divisor Quotient) + remainder.

For example: If dividend = 15968, Quotient = 89 and remainder = 37 then Divisor is?

Divisor =

=


Divisibility Rules:

For Number

Divisibility Rule Example Note

2 If the last digit of a number is 0,2,4,6,8 , then the number is divisible by 2

742 is divisible by 2 but 743 is not.

3 If the sum of all the digits of a number is divisible by 3 , then the number is divisible by 3

1458 (sum of digits = 18) is divisible by 3, but 766 (sum of digits =19) is not divisible by 3.

766 (sum of digits =19) the remainder when 19 is divided by 3 i.e. 1 will also be the remainder when 766 is divided by3

4 If the last two digits of a number are divisible by 4, then the number is also divisible by 4

6732 is divisible by 4 as 32 is divisible by 4, but 2142 is not divisible by 4.

Similarly the remainder when 42 is divided by 4 i.e 2 will also be the remainder when 2142 is divided by 4.

5 If the last digits of a number are 0 and 5, then the number is divisible by 5.

1465, 1320 are divisible by 5 as their last digit is 5 and 0 respectively.

6 If the number is divisible by 2 and 3 both, then it is also divisible by 6.

1452 is divisible by both 2 and 3 so it is divisible by 6 also, but 3362 is not divisible by 6 as it is not divisible by 3.

If the number is divisible by 4 and 6 both, then it is not necessary that it is

divisible by 24 (6 4).

8 If the last three digit of a number are divisible by 8 or are 000, then the number is divisible by 8 .

43102 and 13000 are divisible by 8 since 102 is divisible by 8 and 13000 have 000 as last three digits, but 2148 is not as

The remainder when 148 is divided by 8 i.e. 4 will also be the remainder of 2148


148 is not divisible by 8 when divided by 8.

9 If the sum of all digits of a number is divisible by 9 , then the number is also divisible by 9.

25344 (sum of digits = 18) is divisible by 9 , 764 (sum of digits =17) is not.

The remainder when 17 is divided by 9 i.e 8 will also be the remainder when 764 is divided by 9.

11 If the difference between the sum of the digits in the even places and the sum of the digits in the odd places is either 0 or is divisible by 11 , then the number is also divisible by 11.

9415956 is divisible by 11 as the difference of 9+1+9+6 =25 and 4+5+5 = 14 is 11, but 31872 is not as the sum of even places = 13 and sum of odd is 8 their difference is neither 0 nor 11.


Solved Examples (Number System)

1. When X is subtracted from the numbers 9, 15 and 27, the remainders are in continued proportion. What is the value of X? (IBPS CWE PO MT 2012)

(1) 8 (2) 6 (3) 4 (4) 5 (5) None of these Solution: Let be subtracted from the numbers 9, 15 and 27 we get continue proportion. Now, 9 – : 15 – : 27 –

b2 = ac (15 – )2 = (9 – ) (27 – ) or, 225 – 30 + 2 = 243 + 2 – 36 or, 6 = 243 – 225 = 18

X = 3 Hence number become 9 - x = 9-3=6 15 – x = 15 -3 = 12 And 27-x = 27 -3 = 24

6 : 12 : 24 = 1: 2: 4 Thus 1: 2: 4 is continued proportion Ans: (5) 2. Sum of three consecutive numbers is 2262. What is 41% of the highest number? (IBPS CWE PO MT 2012)

(1) 301.5.1 (2) 303.14 (3) 308.73 (4) 306.35 (5) 309.55 Solution: Let the three consecutive number be x, x + 1 and x + 2. Then, x + x + 1 + x + 2 =2262 or, 3x = 2262 – 3 = 2259

x = = 753


The Numbers are 753, 754, 755. The highest number is 755 41% of 755 = 755 = 41 7.55 = 309.55

Ans: (5)

3. Rachita enters a shop to buy ice-creams, cookies and pastries. She has to buy at least 9 units of each. She buys more cookies than ice-creams and more pastries than cookies. She picks up a total of 32 items. How many cookies does she buy? (IBPS CWE PO MT 2012) (1) Either 12 or 13 (2) Either 11 or 12 (3) Either 10 or 11 (4) Either 9 or 11 (5) Either 9 or 10 Solution: Total number of items = 32 Maximum number of ice creams = 9

pastries> cookies> ice cream So, 13 10 9 12 11 9 Hence number of cookies is either 10 or 11. Number of pastries is either 13 or 12. Ans: (3) 4. The fare of a bus is Rs. x for the first five kilometres and Rs. 13/- per kilometre thereafter. If a passenger pays Rs. 2402/- for a journey of 187 kilometres, what is the value of X? (IBPS CWE PO MT 2012) (1) Rs. 29 (2) Rs. 39 (3) Rs. 36 (4) Rs. 31 (5) None of these Solution: Let the fare of first five kilometres be Rs. x. Total distance = 187 km Remaining distance = 187 - 5 = 182 km Now, x + 182 13 = 2402


x = 2402 - 2366= Rs.36 Ans: (3) 5. The product of three consecutive even numbers is 4032. The product of the first and the third number is 252. What is five times the second number? (1) 80 (2) 100 (3) 60 (4) 70 (5) 90 Solution: Let the three consecutive even numbers be 2x, 2x + 2 and 2x+4. Then, (2x) (2x + 2) (2x + 4) = 4032 ... (I) Again, product of first and third number 2x (2x + 4) = 252 ... (II) Putting the values of the product of first and third number in eqn (I), we have (2x + 2) 252 = 4032 or, 2x + 2 = = 16

x = 7 Hence, first number = 7 8 = 14 Second number = 7 x 2 + 2 = 16 And third number = 7 x 2 + 4 = 18 Five times of second number = 5 x 16 = 80 Ans: (1)

6. Rubina could get equal number of Rs. 55, Rs. 85 and Rs. 105 tickets for a movie. She spends Rs. 2,940 for all the tickets. How many of each did she buy?

(1) 12 (2) 14 (3) 16 (4) Cannot be determined (5) None of these

Solution: Let she buy x tickets. Then total money spent = 55x + 85x + 105x or, 245x = 2940 or, x = 12

Ans: (1)


7. Seema bought 20 pens, 8 packets of wax colours, 6 calculators and 7 pencil boxes. The price of one pen is Rs. 7, one packet of wax colour is for Rs. 22, one calculator is for Rs. 175 and one pencil box costs Rs. 14 more than the combined price of one pen and one packet of wax colours. How much amount did Seema pay to the shopkeeper?

(1) Rs. 1,491 (2) Rs. 1,725 (3) Rs. 1,667 (4) Rs. 1,527 (5) None of these

Solution: Price of one pencil box = 14 + (Price of one pen + Price of one packet of wax colours)= 14+ (7+22) = Rs. 43 Total amount paid by Seema = {(20 × 7) + (8 × 22) + (6 × 175) + (7 × 43)} = Rs 1667

Ans: (3)

8. In a cricket match, Sachin and Viru scored a century each (more than 100 runs) and Yuvi and Gauti scored a half century each. Gauti scored 76 runs and Yuvi scored 12 runs less than Gauti. Viru scored 102 runs. The sum of the scores of all the four players is 442. How many runs did Sachin score? (Corporation Bank PO 2011)

(1) 200 (2) 210 (3) 180 (4) 160 (5) None of these

Solution: Sachin's score = 442 – 76 – (76 – 12) – 102 = 200

Ans: (1)

9. The sum of seven consecutive even numbers of Set A is 378. What is the sum of a different set of four consecutive numbers whose lowest number is 23 less than the mean of Set A?

(1) 132

(2) 146

(3) 156

(4) 124

(5) None of these


Solution: Mean of Set A

Lowest number of Set B = (54 – 23) = 31 Sum of the four numbers of Set B = 31 + 32 + 33 + 34 = 130

Ans: (5)

10. The sum of six consecutive even numbers of Set-A is 402. What is the sum of another Set-B of four consecutive numbers whose lowest number is 15 less than double the lowest number of set- A? (Allahabad Bank Probationary Officers Exam 2011)

(1) 444 (2) 442 (3) 440 (4) 446 (5) None of these

Solution: Third even number = = 67 - 1 = 66

smallest even number = 62

smallest number of set B = 2 x 62 - 15 = 109

required sum = 109 + 110 + 111+ 112 = 442

Ans: (2)


Practice Set-1 (Number System)

1. A person on tour has Rs 360 for his daily expense. He decides to extend his tour programme by 4 days which leads to cutting down daily expense by Rs 3 a day. The number of days of his tour programme is

(1) 15 (2) 20 (3) 18 (4) 16

2. Two times a two-digit number is 9 times the number obtained by reversing the digits and sum of the digits is 9. The number is

(1) 72 (2) 54 (3) 63 (4) 81

3. If 5 students utilize 18 pencils in 9 days, how long at the same rate will 66 pencils last for 15 students?

(1) 10 days (2) 12 days (3) 11 days (4) None of these

4. A man has certain number of small boxes to pack into parcels. If he packs 3, 4, 5 or 6 in a parcel, he is left with one over; if he packs 7 in a parcel, none is left over. What is the number of boxes he may have to pack?

(1) 106 (2) 301 (3) 309 (4) 400

5. Out of a group of swans, 7/2 times the square foot of the number are playing on the shore of the pond. The two remaining are inside the pond. What is the total number of swans?

(1) 10 (2) 14 (3) 12


(4) 16

6. In a family, each daughter has the same number of brothers as she has sisters and each son has twice as many sisters as he has brothers. How many sons are there in the family?

(1) 4 (2) 3 (3) 2 (4) 5

7. A yearly payment to a servant is Rs. 90 plus one turban. The servant leaves the job after 9 months and received 65 and a turban. Then find the price of the turban.

(1) Rs 10 (2) Rs 15 (3) Rs 7.50 (4) Cannot be determined

8. Mr. Mukherjee is 5 yr older to his wife and they have one son and one daughter. If the daughter is 23 yr younger to his mother, what is the difference in the ages of the brother and the sister?

(1) 7 yr (2) 12 yr (3) 4 yr (4) 2 yr

9. The expenses of a hotel consist of two parts. The first one varies with the number of inmates, while the second one is fixed. When the numbers of inmates are 275 and 350, the expenses are Rs 1600 and Rs 1900 respectively. The expenses when the number of inmates is 315, are

(1) Rs 1760 (2) Rs 1680 (3) Rs 1780 (4) Rs 1660

10. If the numerator of a fraction is doubled and the denominator is increased by 3, the new fraction is 3/5 what is the original fraction, if its denominator is more than twice the numerator by 1?

(1) 3/7 (2) 6/13 (3) 1/3 (4) 5/11


Practice Set-2 (Number System)

1. The sum and product of two numbers are 5 and 6, respectively. The sum of reciprocals of their squares is

(1)

(2)

(3)

(4)

2. The sum of the squares of 3 consecutive positive numbers is 365. The sum of the numbers is

(1) 30 (2) 33 (3) 36 (4) 45

3. The sum of a natural number and its square equals the product of the first three prime numbers. The number is

(1) 2 (2) 3 (3) 5 (4) 6

4. 47 is added to the product of 71 and an unknown number. The new number is divisible by 7, giving the quotient 98. The unknown number is a multiple of

(1) 2 (2) 7 (3) 5 (4) 3


5. Two natural numbers are in the ratio 3: 5 and their product is 2160. The smaller of the number is

(1) 36 (2) 24 (3) 18 (4) 12

6. The sum of two numbers is 10. Their product is 20. Find the sum of the reciprocals of the two numbers.

(1) 1

(2)

(3)

(4)

7. The sum of the digits of a two digit numbers is 10. The number formed by reversing the digits is 18 less than the original number. Find the original number.

(1) 81 (2) 46 (3) 64 (4) 60

8. The least number to be subtracted from 36798 to get a number which is exactly divisible by 78 is

(1) 18 (2) 60 (3) 38 (4) 68

9. The sum of first sixty number from one to sixty is divisible by

(1) 13 (2) 59 (3) 60 (4) 61


10. Eight consecutive number are given. If the average of the two numbers that appear in the middle is 6, then the sum of the eight given numbers is

(1) 36 (2) 48 (3) 54 (4) 64


Number System Practice Set-1 (Answers)

1 (2)

2 (4)

3 (3)

4 (2)

5 (4)

6 (2)

7 (1)

8 (4)

9 (1)

10 (1)


Number System Practice Set-2 (Answers)

1 (1)

2 (2)

3 (3)

4 (4)

5 (1)

6 (3)

7 (3)

8 (2)

9 (4)

10 (2)


Chapter: Ratio, Proportion & Alligation

Introduction:-

Ratio is the relation which one quantity bears to another of the same kind. The ratio of two

quantities a and b is the fraction and we write it as a: b.

In the ratio a: b, we call a as the first term or antecedent and b, the second term or consequent.

Note: The multiplication or division of each term of a ratio by the same non- zero number does not affect the ratio.

Compound Ratio: - It is obtained by multiplying together the numerators for new numerator and denominators for new denominator.

Example 1. If the ratios are 4:3, 15:20, 2:6 and 3:5 find the compound ratio?

Sol 1. Required ratio =

Duplicate ratio of a: b =

Triplicate ratio of a: b = etc.

Example2. If we divide 4185 into two parts such that they are in ratio 7:2, then find the values of both the parts?

Sol 2. Let the actual variable be 7x and 2x.

Then 7x+2x = 4185

So, the 1st part = 7

The 2nd part = 2

Note:


The ratio of first , second and third quantities is given by

ac : bc : bd

If the ratio between first and second quantity is a:b and third and fourth is c:d.

a: b

c: d

ac : bc : bd

Similarly, the ratio of first, second, third and fourth quantities is given by

ace : bce : bde : bdf

If the ratio between first and second quantity is a: b and third and fourth is c:d .

a : b

c : d

e : f

ace : bce : bde : bdf

Example 3. If Savita has Rs 1880. How much money does Ravina have if the ratio of money with savita and Rita is 15: 7 and that with Rita and Ravina 16: 7?

Solution3:

Savita : Rita : Ravina

15 : 7

16 : 7

15

240 : 112 : 14

The ratio of money with Savita , Rita and Ravina is 240 : 112 : 14.

We see that 240 x = 1880

Hence 14


Proportion

Introduction:-

Four quantities are said to be proportional if the two ratios are equal i.e. the A, B, C and

D are proportion. It is denoted by “::” it is written as A : B : C : D where A and D are extremes and B and C are called means .

Product of the extreme = Product of the means

Direct proportion: - The two given quantities are so related that if one quantity increases (or decreases) then the other quantity also increases (or decreases).

Example 1. If 5 pens cost Rs 10 then 15 pen cost?

Sol 1. It is seen that if number of pens increases then cost also increases. So,

5 pens: 15 pens:: Rs 10 : required cost

Required cost =

Inverse proportion: - The two given quantities are so related that if one quantity increases (or decreases) then the other quantity also decreases (or increases).

Example 2.If 10 men can do a work in 20 days then in how many days 20 men can do that work?

Sol 2. Here if men increase then days should decrease, so this is a case of inverse proportion, so

10 men: 20 men :: required days : 20 days

Required days =

Rule of three: It Is the method of finding 4th term of a proportion if all the other three are given, if ratio is a:b :: c:d then ,

d =

Compound proportion: - Lets take an example to explain this.


Example3. If 9 men can do a piece of work in 40 days of 10 hours each, how many men will it take to do 12 times the amount of work if they work for 30 days of 9 hours?

Solution 3:

Step 1. Days: ……. : …….

Hours: ……. : …….. :: 9: required no. of men

Work: ……. : ……..

Step 2.

1. Compare days with men : to do the work in less days we will need more men , so it is the

case of inverse proportion hence ,

30 : 40 :: 9 : required no. of men

2. Compare hours with men : to do the work in less hours we will need more men , so it is

the case of inverse proportion hence ,

9 : 10 :: 9 : required no. of men

3. Compare work with men : to do the more work we will need more men , so it is the

case of direct proportion hence ,

1: 12:: 9: required no. of men

Put all the values in step 1,

30: 40

9: 10 :: 9: required no. of men

1: 12

Now,


ALLIGATION

Introduction:-

The word allegation means linking. It is used to find:

1. The proportion in which the ingredients of given price are mixed to produce a new

mixture at a given price.

2. The mean or average value of mixture when the price of the two or more ingredients

and the proportion in which they are mixed are given.

Mathematical Formula:

For two ingredient:-

CP of unit quantity of cheaper (c) CP of unit quantity of dearer (d)

Mean Price

(d-m) (m-c)

Then (cheaper quantity): (dearer quantity) = (d-m): (m-c)

Example 1: If the rice at Rs 3.20 per kg and the rice at Rs 3.50 per kg be mixed then what should be their proportion so that the new mixture be worth Rs 3.35 per kg ?

Sol 1: CP of 1 kg of cheaper rice CP of 1 kg of dearer rice

320 paisa 350 paisa

Mean Price (m)

335 paisa

15 15


By allegation:

Hence they must be mixed in equal proportion i.e. 1:1

For three ingredient:-

1. The price of the ingredient should be reduced to one denomination and then place

them in ascending order under one another.

2. After that place the mean prices to the left of all the price

3. Then pair the price so that price less than and greater than the mean prices go together.

4. Then find out the difference between mean price and each price and place it opposite to

the price with which it is linked.

5. These difference will help to find out the given answer and similarly it will work for four

ingredients .

Example 2: Find out the ratio of new mixture so that it will cost Rs 1.40 per kg from the given three kinds of rice costing Rs 1.20, Rs 1.45 and Rs 1.74?

Sol 2: 1st rice cost = 120, 2nd rice cost = 145 and 3rd rice cost = 174 paisa.

From the above rule: we have,

120 5+34 [(145-140) + (174-140)]

140 145 20 (140-120)

174 20 (140-120)

Therefore, three rice must be mixed in 39: 20: 20 ratios to have a new mixture of rice.


Solved Examples (Ratio, Proportion & Alligation)

1. A certain amount was to be distributed among A, B and C in the ratio 2: 3: 4 respectively, but was erroneously distributed in the ratio 7: 2: 5 respectively. As a result of this, B got Rs 40 less. What is the amount? (IBPS CWE PO MT 2012) (1) Rs. 210 (2) Rs. 270 (3) Rs. 230 (4) Rs. 280 (5) None of these Solution: Let the amount be x. B's share =

Due to error B's share =

Difference in B's share due to error = 40

- = 40

or, = 40

or, 24x = 40 126

x = = Rs. 210

Ans: (1) 2. Rs.73,689/- are divided between A and B in the ratio 4: 7. What is the difference between thrice the share of A and twice the share of B? (IBPS CWE PO MT 2012) (1) Rs. 36,699 (2) Rs. 46,893 (3) Rs. 20,097 (4) Rs. 26,796 (5) Rs. 13,398 Solution: Let A’s share be 4x and B’s share be 7x.

4x +7x = 73689 or, 11x = 73689


x = = 6699

A’s share = 6699 4 = 26796 B’s share = 6699 7 = 44893 Thrice the share of A = 26796 3 = 80388 Twice the share of B = 46893 2= 93786 Difference = 93786 - 80388 = Rs. 13398 Ans: (5)

3. The ratio of the present age of Manisha and Deepali is 5:X. Manisha is 9 years younger than Parineeta. Parineeta’s age after 9 years will be 33 years. The difference between Deepali's and Manisha's age is the same as the present age of Parineeta. What should come in place of X?

(1) 23 (2) 39 (3) 15 (4) Cannot be determined (5) None of these Solution: Parineeta’s present age = (33 - 9 =) 24 yrs.

Manisha's present age = (24 - 9 =) 15 yrs.

Deepali’s present age = IS + 24 = 39 yrs.

Ratio of the present age of Manisha and Deepali

= 15 : 39 = 5 : 13 X = 13

Ans: (5)

4. The ratio between the three angles of a quadrilateral is 3 : 5 : 9. The value of the fourth angle of the quadrilateral is 71˚. What is the difference between the largest and the smallest angles of the quadrilateral? (IBPS RRB Group ‘A’ Officers Exam 2012)

(1) 82˚ (2) 106˚ (3) 102˚ (4) 92˚ (5) None of these


Solution. Let the quadrilateral angles be 3x, 5x, 9x and 71˚. Total sum of angles = 3x + 5x + 9x + 71˚ = 360˚ or, 17x = 360 – 71 = 289 ∴ x = 17° Hence angles are 51˚, 85˚, 153°, and 71˚. ∴ Difference = 153 – 51 = 102˚.

Ans: (3)

5. The second largest and the smallest angles of a triangle are in the ratio of 6 : 5. The difference between the second largest angle and the smallest angle 'of the triangle is equal to 9°. What is the difference between the smallest and the largest angles of the triangle? (IBPS RRB Group ‘A’ Officers Exam 2012)

(1) 36° (2) 24° (3) 12° (4) 18° (4) None of these

Solution. Let the second largest angle of the triangle be 6x and the smallest angle 5x.

Now, 6x - 5x = 9°

or, x = 9°

Second largest angle = 54°

Smallest angle = 45°

Sum of angles of a triangle = 180°

∴ largest angle = 180 - 99 = 810

∴ Difference = 81 - 45 = 36°.

Ans: 1

6. The ratio between the speed of a bus and train is 15 : 27 respectively. Also, a car covered a distance of 720 km in 9 hours. The speed of the bus is three- fourth the speed of the car. How much distance will the train cover in 7 hours? (Allahabad Bank Probationary Officers Exam 2011)

(1) 760 km (2) 756 km (3) 740 km


(4) Cannot be determined (5) None of these

Solution: Speed of the car = = = 80hm/hr

Speed of the bus = 60km/hr

Speed of the train = 108km/hr

Distance covered by train in 7 hours = (7 x 108 =) 756 km

Ans: (2)


Practice Set-1 (Ratio, Proportion & Alligation)

1. A container has 80 L of milk. From

this container 8 L of milk was taken

out and replaced by water. The

process was further repeated twice.

The volume of milk in the container

after that is [SSC Quantitative

Aptitude (Arihant)]

a) 58.23 L

b) 85.23 L

c) 58.32 L

d) 85.32 L

2. A can contains a mixture of two

liquids A and B in the ratio 7 : 5.

When 9 L of mixture is drawn off

and the can is filled with B, the

ratio of A and B becomes 7 : 9.

Litres of liquid A contained by the

can initially was

a) 10

b) 20

c) 21

d) 25

3. What number should be added to

or subtracted from each term of the

ratio 17 : 24 so that it becomes

equal to 1 : 2?

a) 5 is subtracted

b) 10 is added

c) 7 is added

d) 10 is subtracted

4. The ratio of weekly incomes of A

and B is 9 : 7 and the ratio of their

expenditures is 4 : 3. If each saves

Rs. 200 per week, then the sum of

their weekly incomes is

a) Rs. 3200

b) Rs. 4200

c) Rs. 4800

d) Rs. 5600

5. The ratio of alcohol and water in 40

L of mixture is 5 : 3.8 L of the

mixture is removed and replaced

with water, Now, the ratio of the

alcohol and water in the resultant

mixture is

a) 1 : 2

b) 1 : 1

c) 2 : 1

d) 1 : 3

6. Rama's expenditure and savings are

in the ratio 3 : 2. His income

increases by 10%. His expenditure

also increases by 12%. His saving

increases by

a) 7%

b) 10%

c) 9%

d) 13%

7. Three numbers are in the ratio 3 : 4

: 5. The sum of the largest and the

smallest equals the sum of the

second and 52. The smallest

number is

a) 20


b) 27

c) 39

d) 52

8. The ratio of the ages of Ram and

Rahim 10 yr ago was 1 : 3. The ratio

of their ages 5 yr hence will be 2 : 3.

Then, the ratio of their present ages

is

a) 1 : 2

b) 3 : 5

c) 3 : 4

d) 2 : 5

9. The ratio of milk and water in

mixtures of four container are 5 : 3,

2 : 1, 3 : 2 and 7 : 4, respectively. In

which container is the quantity of

milk, relative to water, minimum?

a) First

b) Second

c) Third

d) Fourth

10. In a mixture of 25 L, the ratio of

acid to water is 4 : 1. Another 3 L of

water is added to the mixture. The

ratio of acid to water in the new

mixture is

a) 5 : 2

b) 2 : 5

c) 3 : 5

d) 5 : 3

11. A shopkeeper buys two varieties of

tea, the price of the first being

twice the second. He sells the

mixture at Rs 36 per kilogram and

makes a profit of 20%. If the ratio of

quantities of the first and second

variety in these mixture is 3 : 4,

then what is the cost price of each

variety of tea ? [Mission MBA MAT

(Arihant)]

a) Rs 21, 42

b) Rs 15, 30

c) Rs 16.5, 33

d) Rs 17, 34

12. Two liquids are mixed in the ratio 3

: 5 and the mixture is sold at Rs 120

with a profit of 20%. If the first

liquid is costlier than the second by

Rs 2 per litre, find the cost of the

costlier liquid per litre.

a) Rs 92.30

b) Rs 74.10

c) Rs 101.25

d) Rs 99.25

13. A grocer buys two kinds of rice at

Rs 1.80 and Rs 1.20 per kg

respectively. In what proportion

should these be mixed, so that by

selling the mixture at Rs 1.75 per

kg, 25% may be gained?

a) 2 : 1

b) 3 : 2

c) 3 : 4

d) 1 : 2

14. A jar full of whisky contains 40% of

alcohol. A part of this whisky is

replaced by another containing 19%

alcohol and now the percentage of

alcohol was found to be 26. The

quantity of whisky replaced is

a)


b)

c)

d)

15. A container contains 240 L of wine.

80 L is taken out of the container

everyday and an equal quantity of

water is put into it. Find the

quantity of the wine that remains in

the container at the end of the

fourth day.

a) 39.2 L

b) 32 L

c) 42.5 L

d) 47.40 L

16. A tea trader mixed two varieties of

tea, one costing Rs 3.50 per kg and

the other costing Rs 4 per kg and

sells 40 kg of the mixture to a

vendor at Rs 4.50 per kg and makes

a profit of 20%. How much of each

variety did the vendor mix?

a) 30 kg, 10 kg

b) 20 kg, 20 kg

c) 10 kg, 30 kg

d) None of these

17. A vessel contains 50 L milk. The

milkman delivers 10 L to the first

house and adds an equal quantity

of water. He does exactly the same

at the second and third house.

What is the ratio of milk and water

when he has finished delivering at

the third house?

a) 61 : 64

b) 27 : 37

c) 16 : 19

d) None of these

18. Prabhu purchased 30 kg of rice at

the rate of Rs 17.50 per kg and

another 30 kg rice at a certain rate.

He mixed the two rice and sold the

entire quantity at the rate of Rs

18.60 per kg and made 20% over all

profit. At what price per kg did he

purchase the lot of another 30 kg

rice?

a) Rs 14.50

b) Rs 12.50

c) Rs 15.50

d) Rs 13.50

19. A person has a chemical of Rs 50

per litre. In what ratio should water

be mixed in that chemical so that

after selling the mixture at Rs 40

per litre he may get a profit of 50%.

a) 8 : 7

b) 9 : 8

c) 10 : 7

d) 4 : 3

20. A trader has 50 kg of pulses, part of

which he sells at 8% profit and the

rest at 18% profit. He gains 14% on

the whole. What is the quantity

sold at 18% profit?

a) 30 kg

b) 35 kg

c) 40 kg

d) None of these


21. A container of capacity 120 L is

filled with milk and water. 80% of

milk and 40% of water is taken out

of vessel. It is found that the vessel

is vacated by 65%. What is the ratio

of milk to water?

a) 5 : 3

b) 6 : 5

c) 3 : 5

d) 4 : 3

22. How much water should be added

to 60 L of milk at L for Rs 10 so

as to have a mixture worth Rs

per litre?

a) 16 L

b) 15 L

c) 18 L

d) 20 L


Practice Set-2 (Ratio, Proportion & Alligation)

1. An employer reduces the number

of employees in the ratio 8 : 5 and

increases their wages in the ratio 7 :

9. As a result, the overall wages bill

is

a) increased in the ratio 56 : 69

b) decreased in the ratio 56 : 45

c) increased in the ratio 13 : 17

d) decreased in the ratio 17 : 13

2. A can contains a mixture of two

liquids A and B in the ratio 7 : 5.

When 9 L of mixture is drawn off

and the can is filled with B, the

ratio of A and B becomes 7 : 9.

Litres of liquid A contained by the

can initially was

e) 10

f) 20

g) 21

h) 25

3. What number should be added to

or subtracted from each term of the

ratio 17 : 24 so that it becomes

equal to 1 : 2?

e) 5 is subtracted

f) 10 is added

g) 7 is added

h) 10 is subtracted

4. The ratio of weekly incomes of A

and B is 9 : 7 and the ratio of their

expenditures is 4 : 3. If each saves

Rs. 200 per week, then the sum of

their weekly incomes is

e) Rs. 3200

f) Rs. 4200

g) Rs. 4800

h) Rs. 5600

5. Rama's expenditure and savings are

in the ratio 3 : 2. His income

increases by 10%. His expenditure

also increases by 12%. His saving

increases by

e) 7%

f) 10%

g) 9%

h) 13%

6. If A : B is 2 : 3, B : C is 6 : 11, then A :

B : C is

a) 2 : 3 : 11

b) 4 : 6 : 22


c) 4 : 6 : 11

d) 2 : 6 : 11

7. The ratio of the ages of Ram and

Rahim 10 yr ago was 1 : 3. The ratio

of their ages 5 yr hence will be 2 : 3.

Then, the ratio of their present ages

is

e) 1 : 2

f) 3 : 5

g) 3 : 4

h) 2 : 5

8. If the annual income of A, B and C

are in the ratio 1 : 3 : 7 and the total

annual income of A and C is Rs.

800000, then the monthly salary of

B is

a) Rs. 20000

b) Rs. 25000

c) Rs. 30000

d) Rs. 15000

9. Two numbers are such that the

ratio between them is 4 : 7. If each

is increased by 4 the ratio becomes

3 : 5. The larger number is

a) 36

b) 48

c) 56

d) 64

10. Acid and water are mixed in a

vessel A in the ratio of 5 : 2 and in

the vessel B in the ratio of 8 : 5. In

what proportion should quantities

be taken out from the two vessels,

so as to form a mixture in which the

acid and water will be in the ratio

of 9 : 4?

a) 7 : 2

b) 2 : 7

c) 7 : 4

d) 2 : 3

11. In an alloy, Zinc and Copper are in

the ratio 1 : 2. In the second alloy,

the same elements are in the ratio

2 : 3. If these two alloys be mixed to

form a new alloy in which two

elements are in the ratio 5 : 8, the

ratio of these two alloys in the new

alloy is

a) 3 : 10

b) 3 : 7

c) 10 : 3

d) 7 : 3

12. A boy has a few coins of

denominations 50 paise, 25 paise

and 10 paise in the ratio 1 : 2 : 3. If

the total amount of the coins is Rs.


6.50, the number of 10 paise coins

is

a) 5

b) 10

c) 15

d) 20

13. The sum of the ages of a father and

his son is 100 yr now. 5 yr ago their

ages were in the ratio of 2 : 1. The

ratio of the ages of father and son

after 10 yr will be

a) 5 : 3

b) 4 : 3

c) 10 : 7

d) 3 : 5

14. In a school having roll strength 286,

the ratio of boys and girls is 8 : 5. If

22 more girls get admitted into the

school, the ratio of boys and girls

becomes

a) 12 : 7

b) 10 : 7

c) 8 : 7

d) 4 : 3

15. A box contains Rs. 1,50 paise and

25 paise coins in the ratio 8 : 5 : 3. If

the total amount of money in the

box is Rs. 112.50, the number of 50

paise coins is

a) 80

b) 50

c) 30

d) 42


Ratio, Proportion & Alligation Practice Set-1 (Answers)

1) c

2) c

3) d

4) a

5) b

6) a

7) c

8) b

9) c

10) a

11) a

12) c

13) d

14) c

15) d

16) b

17) a

18) d

19) a

20) a

21) a

22) b


Ratio, Proportion & Alligation Practice Set-2 (Answers)

1) b

2) c

3) d

4) a

5) a

6) c

7) b

8) b

9) c

10) a

11) a

12) c

13) a

14) d

15) b


Chapter: Averages

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 = ( )

Example. What is the average of First 10 Prime numbers?

Solution: First 10 Prime number are 2,3,5,7,11,13,17,19,23,29.

Hence, Average = {2+3+5+7+11+13+17+19+23+29} / 10

= 129 / 10

= 12.90

So, Average of First 10 Prime numbers is 12.90.

Example. The total number of sales visits made by a Salesman in the month of June is 90. What is the Average visit he makes per day?

Solution: Number of days in the month of June are 30

Hence, Average Visit per day = Number of total visits / Number of total days

= 90 / 30

= 3

So, the salesman makes 3 visits per day.


Solved Examples (Averages)

1. Ramola's monthly income is three times Ravina's monthly income. Ravina's monthly income is fifteen per cent more than Ruchira's monthly income. Ruchira's monthly income is Rs. 32,000. What is Ramola's annual income?

(1) Rs. 1,10.400 (2) Rs. 13,24,800 (3) Rs. 36,800 (4) Rs. 52,200 (5) None of these

Solution: Ravina's monthly income = 32000 = Rs. 36800

Ramola's monthly income = 3 36800 = Rs. 110400

Ramola's annual income = 12 110400 = Rs. 1324800

Ans: (2)

2. The average marks in English of a class of 24 students is 56. If the marks of three students were misread as 44, 45 and 61 in lieu of the actual marks 48, 59 and 67 respectively, then what would be the correct average?

(1) 56.5 (2) 59 (3) 57.5 (4) 58 (5) None of these

Solution: Total marks = 24 × 56 = 1344

Total of actual marks

= 1344 - (44 + 45 + 61) + (48 + 59 + 67) = 1368

Actual Average = = 57


Ans: (5)

3. In a test, a candidate secured 468 marks out of maximum marks' A:. Had the maximum marks' A' converted to 700, he would have secured 336 marks. What was the maximum marks of the test?

(1) 775 (2) 875 (3) 975 (4) 1075 (5) None of these

Solution: Converted maximum marks = 700

Converted marks = 336

% marks = x 100= 48%

468 is 48 % of maximum marks 'A'

A= × 100 =975

Ans : (3)

4. The ratio between the speed of a truck, car and train is 3: 8: 12. The car moved uniformly and covered a distance of 1040 km in 13 hours. What is the average speed of the truck and the train together? (IBPS RRB Group ‘A’ Officers Exam 2012)

(1) 75km/hr

(2) 60 km/hr

(3) 48 km/hr

(4) Cannot be determined

(5) None of these

Solution: Speed of car

Ratio of speed of truck, car and train = 3 : 8 : 9

Now 8x = 80


∴ x = 10

Hence truck = 30 kmph.

Train = 90 kmph.

∴ Average speed of truck and train together

Ans: 2

5. The circumference of a circle is twice the perimeter of a rectangle. The area of the circle is 5544 sq cm. What is the area of the rectangle if the length of the rectangle is 40cm? (IBPS RRB Group ‘A’ Officers Exam 2012)

(1) 1120 sq cm

(2) 1020 sq cm

(3) 1140 sq cm

(4) 1040 sq cm

(5) None of these

Solution: Area of circle

∴ r = 42

Circumference of circle = 2 ⤬ perimeter of rectangle

Or,

Or, perimeter of rectangle = 132 cm

Or, 2(l + b) = 132

∴ l + b = 66

∴ b = 66 – 40 = 26


Area of rectangle = 40 ⤬ 26 = 1040 cm2 = 1040 sq.cm.

Ans: 4

6. Among three numbers the first is twice the second thrice the third. If the average of the three numbers is 49.5, then the difference between the first and the third number is

(1) 54

(2) 28

(3) 39.5

(4) 41.5

Solution: Let first number = x

Then, second number =

and third number =

According to given condition,

X + + = 3 × 49.5

= 148.5

11x = 6 × 148.5

x = = 81

First number = 81

Third Number = = 27

Required difference = 81-27=54

Ans: (1)


7. 10 yrs ago, the average age of P and Q was 20 yrs. Average age of P, Q is 30 yrs now. After 10 yr, the age of R will be

(1) 35 yr (2) 40 yr (3) 30 yr (4) 45 yr

Solution: 10 yr ago, total age P and q

= 20 × 2 = 40 yr

Present total age of P and Q = 40+2×10= 60 yr

Present total age of P and Q and R=30×3= 90 yr

R’s age = 90-60 = 30 yr

After 10 yr R’s age = (30+10) yr = 40 yr

Ans: (2)

8. In an examination, the average of marks was found to be 50. For 100 students, marks was computed wrongly as 90 instead of 60.For deducting marks for computational errors, the average of marks came down to 45. The total number of candidate, who appeared at the examination, was

(1) 600 (2) 300 (3) 200 (4) 150

Solution: Let the total number of candidate be x.

By given condition.

50x- 90x100+ 60x100= 45x

50x – 9000+6000 = 45x

50x – 45x = 3000

x = = 600.

Ans: (1)


9. The average of 25 observations 13. It was later found that an observation 73 was wrongly entered as 48. The new average is

(1) 12.6 (2) 14 (3) 15 (4) 13.8

Solution: Required new average

=

= = = 14

Ans: (2)

10. A tabulator while calculating the average marks of 100 students of an examination, by mistake enters 68, instead of 86 and obtained the average marks of those students are

(1) 58.8 (2) 57.82 (3) 58.81 (4) 57.28

Solution: Actual total number of 100 students

= 5800 + (86 - 68)

= 5818

Required actual average = = 58.18

Ans: (1)


Practice Set (Averages)

1. The average age of a jury of 5 is 40.

If a member aged 35 resigns and a

managed 25 becomes a member,

then the average age of the new

jury is

a) 30 yr

b) 38 yr

c) 40 yr

d) 42 yr

2. The average of the runs made by

Raju, Shyam and Hari is 7 less than

that made by Shyam, Hari and

Kishore. If the number of Kishore's

run is 35, what is Raju's run?

a) 21

b) 35

c) 7

d) 14

3. The mean of 50 numbers is 30.

Later it was discovered that two

entries were wrongly entered as 82

and 13 instead of 28 and 31. Find

the correct mean.

a) 36.12

b) 30.66

c) 29.28

d) 38.21

4. One third of a certain journey is

covered at the rate of 25 km/h,

one-fourth at the rate of 30 km/h

and the rest at 50 km/h. The

average speed for the whole

journey is

a) 35 km/h

b)

c) 30 km/h

d)

5. The average of 5 numbers is 140. If

one number is excluded, the

average of the remaining 4

numbers is 130. The excluded

number is

a) 135

b) 134

c) 180

d) 150


6. The average weight of 5 persons

sitting in a boat is 38 kg. The

average weight of the boat and the

persons sitting in the boat is 52 kg.

What is the weight of the boat?

a) 228 kg

b) 122 kg

c) 232 kg

d) 242 kg

7. There are 50 students in a class.

Their average weight is 45 kg.

When one student leaves the class

the average weight reduces by 100

g. What is the weight of the student

who left the class?

a) 45 kg

b) 47.9 kg

c) 49.9 kg

d) 50.1 kg

8. Out of 4 numbers, whose average is

60, the first one is one fourth of the

sum of the last three. The first

number is

a) 15

b) 45

c) 48

d) 60

9. In a class, the average score of girls

in an examination is 73 and that of

boys is 71. The average score for

the whole class is 71.8. Find the

percentage of girls

a) 40%

b) 50%

c) 55%

d) 60%

10. The average of the first 100 positive

integers is

a) 100

b) 51

c) 50.5

d) 49.5

11. Out of the three numbers, the first

number is twice of the second and

the second is thrice of the third

number. If the average of these 3

numbers is 20, then the sum of the

largest and smallest numbers is

a) 24

b) 42

c) 54

d) 60


12. The average age of 40 students of a

class is 18 yr. When 20 new

students are admitted to the same

class, the average age of the

students of the class is increased by

6 months. The average age of newly

admitted students is

a) 19 yr

b) 19 yr 6 months

c) 20 yr

d) 20 yr 6 months

13. The average age of 40 students of a

class is 15 yr. When 10 new

students are admitted, the average

age is increased by 0.2 yr. The

average age of new students is

a) 15.2 yr

b) 16 yr

c) 16.2.yr

d) 16.4 yr

14. The average of 6 observations is

45.5. If one new observation is

added to the previous observation,

then the new average becomes 47.

The new observation is

a) 58

b) 56

c) 50

d) 46

15. The average of marks scored by the

students of a class is 68. The

average of marks of the girls in the

class is 80 and that of boys is 60.

What is the percentage of boys in

the class?

a) 40

b) 60

c) 65

d) 70

16. The average age of 30 boys in a

class is 15 yr. One boy aged 20 yr,

left the class but two new boys

came in his place whose ages differ

by 5 yr. If the average age of all the

boys now in the class still remains

15 yr, the age of the younger

newcomer is

a) 20 yr

b) 15 yr

c) 10 yr

d) 8 yr


17. Out of 10 teachers of a school, one

teacher retires and in his place a

new teacher of age 25 yr joins. As a

result of it, the average age of the

teachers is reduced by 3 yr. The age

of the retired teacher is

a) 60 yr

b) 58 yr

c) 56 yr

d) 55 yr

18. The mean weight of 34 students of

a school is 42 kg. If the weight of

the teacher be included, the mean

rises by 400 g. Find the weight of

the teacher (in kg).

a) 66

b) 56

c) 55

d) 57

19. Ram aims to score an average of 80

marks in quarterly and half yearly

exams. But his average in quarterly

is 3 marks less than his target and

that in half yearly is 2 marks more

than his aim. The difference

between the total marks scored in

both the exams is 25. Total marks

aimed by Ram is

a) 380

b) 400

c) 410

d) 430


Averages Practice Set (Answers)

1) b

2) d

3) c

4) b

5) c

6) b

7) c

8) c

9) a

10) c

11) b

12) b

13) b

14) b

15) b

16) b

17) d

18) b

19) a


Chapter: Percentages, Partnership and Share

INTRODUCTION

The word per cent means per hundred. Thus 19 parts out of 100 parts. This can also be written

as .

Fraction Equivalents of important Percentages.

PER CENT TO FRACTION

To convert a per cent to a fraction, divide it by 100 and delete the % sign.

Example: 2% can be converted to a fraction as

PER CENT OF A NUMBER

Per cent of a number is the product of equivalent fraction (of rate per cent) and the number.

⥤

Example: To find out 25% of 500


Solution: Required value = 25% of 500

equivalent fraction for 25%

= 125

Example: 9% of what number is 36?

Solution: the required number (base number)

= 400

Example: If 30% of a number is 48, then what is 40% of the number?

Solution: Here, unitary method can be used to save the time.

30% → 48

⥤ 1% →

⥤ 40% →

Hence, the required value is 64

Calculating % EXCESS OR % SHORTNESS

When a number A exceeds the another number B by x %, then % shortness of

It implies that B is less than A by .

Similarly, if a number A is short of (or less than) B by x%, then % excess of B


i.e. B is more than A by

Example: If the income of Ram is more than that of Mohan by 25% then by much percentage

Mohan's income is than that of Ram?

Solution: Required % shortness (less) income of Mohan

= 20%

Therefore, income of Mohan is 20% less than that of Ram.

Partnership and Shares

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 business

1. 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:

=

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

=

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 their investments


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

= 3: 5:7

Share of P’s profit = Rs. (30,000 × ) = Rs. 6,000.

Share of Q’s profit = Rs. (30,000 × ) = Rs. 10,000.

Share of R’s profit = Rs. (30,000 × ) = 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.

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

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

Share of Ram’s profit = (24,000 × ) = Rs.12,000;

Share of Shyam’s profit = (24,000 × ) = Rs.4,800;

Share of Mohan’s profit = (24,000 × ) = Rs.7,200.


Solved Examples (Percentages, Partnership and Share)

1. Akash scored 73 marks in subject A. He scored 56% marks in subject B and X marks in subject C. Maximum marks in each subject were 150. The overall percentage marks obtained by Akash in all the three subjects together were 54%. How many marks did he score in subject C? (IBPS CWE PO MT 2012) (1) 84 (2) 86 (3) 79 (4) 73 (5) None of these Solution: Akash scored in subject A = 73 marks Subject B = = 84 marks

Total marks Akash got in all the three subjects together = 54 4.5 = 243 marks

Let Akash's marks in subject C be X. A + B + C = 243 or, A + B + X = 243 or, X = 243- (84 +73) = 243 -157= 86 marks Ans: (2)

2. An HR Company employs 4800 persons, out of which 45 per cent are males and 60 per cent of the males are either 25 years or older. How many males are employed in that HR Company who are younger than 25 years?

(1) 2640

(2) 2160

(3) 1296

(4) 864

(5) None of these

Solution: Total number of persons = 4800


Number of males = 45% of 4800 = = 2160

Now, according to the question,

Number of males who are younger than 25 years

= (100-60 =) 40% of 2160

= 864

Ans: (4)

3. Six-elevenths of a number is equal to 22 per cent of the second number. The second number is equal to one- fourth of the third number. The value of the third number is 2400. What is 45% of the first number?

(1) 109.8 (2) 111.7 (3) 117.6 (4) 123.4 (5) None of these

Solution: According to the question,

x First number = 22% of second number

Second number = x Third number

or, Second number = x 2400 = 600

or, First number = ×

= × 242

required answer = 45% of 242 = = 108.9


Ans: (5)

4. In an entrance examination, Ritu scored 56 per cent marks, Smita scored 92 per cent marks and Rina scored 634 marks. The maximum marks of the examination is 875. What is the average marks scored by all the three girls together?

(1) 1929 (2) 815 (3) 690 (4) 643 (5) None of these

Solution: Ritu's marks = 875 x = 490

Smita's, marks = 875 x = 805

Rina's marks = 634

Total marks = 490 + 80S + 634 = 1929

Average = = 643

Ans: (4)

5. If twenty five per cent of three-sevenths of twenty six per cent of a number is 136.5, what is the number? (IBPS RRB Group ‘A’ Officers Exam 2012)

(1) 6300 (2) 5600 (3) 4800 (4) 4900 (5) None of these

Solution: Let the number be x.

Then,

Ans: (4)


6. Two-thirds of Ranjit’s monthly salary is equal to Raman’s monthly salary. Raman’s monthly salary id thirty per cent more than pawan’s monthly salary. Pawan’s monthly salary is Rs. 32000. What is Ranjit’s monthly salary? (IBPS RRB Group ‘A’ Officers Exam 2012)

(1) Rs. 64200 (2) Rs. 62500 (3) Rs. 64500 (4) Rs. 62400 (5) None of these

Solution: Pawan’s monthly salary = Rs. 32,000

Raman’s monthly salary

Ranjit’s monthly salary

Ans: (4)

7. In a class there are 60 students, out of whom 15 per cent are girls. Each girl’s monthly fee is Rs. 250 and each boy’s monthly fee is 34 per cent more than a girl. What is the total monthly fees of girls and boys together? (IBPS RRB Group ‘A’ Officers Exam 2012)

(1) Rs. 19335 (2) Rs. 18435 (3) Rs. 19345 (4) Rs. 19435 (5) None of these

Solution: Number of girls

Total monthly fee of girls = 250 ⤬ 9 = Rs. 2250

Number of boys = 60 – 9 = 51

Monthly fee of one boy

Total monthly fees of boys = 51 ⤬ 335 = Rs. 17085

∴ Sum = 17085 + 2250 = Rs. 19,235

Ans: (1)


8. Ravi scored 225 marks in a test and failed by 15 marks. If the passing percentage of the test

is 25 per cent, what is the maximum marks of the test?

(1) 860

(2) 840

(3) 920

(4) 960

(5) None of these

Solution: Passing marks = 225 + 15 = 240

Maximum marks

Ans: (4)

9. Ravi scored 225 marks in a test and failed by 15 marks. If the passing percentage of the test is 25 per cent, what is the maximum marks of the test?

(1) 860

(2) 840

(3) 920

(4) 960

(5) None of these

Solution: Passing marks = 225 + 15 = 240

Maximum marks

Ans: (4)

10. In a school, there are 800 students out of whom 12 per cent are girls. Each boy's monthly fee is Rs. 220 and each girl's monthly fee is 25 per cent less than that of a boy. What is the total monthly fee of all the girls and boys together? (Corporation Bank PO 2011)

(1) Rs. 1,72,020

(2) Rs. 1 ,80,780

(3) Rs. 1,70,720

(4) Rs. 1,80,600

(5) None of these


Solution: Total number of girls

Total number of boys = 800 – 96 = 704

Each girl’s monthly fee

Total monthly fee of girls and boys together

= 96 ⤬ 165 + 704 ⤬ 220

= 15840 + 154880 = Rs. 170720

Ans: (3)


Practice Set-1 (Percentages, Partnership and Share)

1. The number of seats in an

auditorium is increased by 25%. The

price of a ticket is also increased by

12%. Then, the increase in revenue

collection will be

a) 40%

b) 35%

c) 45%

d) 48%

2. Two numbers are 30% and 40%

more than the third number,

respectively. The first number is x %

of the second. Then, x is equal to

a)

b)

c)

d)

3. The price of cooking oil has

increased by 25%. The percentage

of reduction that a family should

effect in the use of cooking oil, so

as not to increase the expenditure

on this account is

a) 15%

b) 20%

c) 25%

d) 30%

4. In an examination, a student had to

obtain 33% of the maximum marks

to pass. He got 125 marks and

failed by 40 marks. The maximum

marks were

a) 500

b) 600

c) 800

d) 1000

5. In an office 40% of the staff is

female, 40% of the female and 60%

of the male voted for me. The

percentage of votes I got was

a) 24%

b) 42%

c) 50%

d) 52%

6. If A's income is 50% less than that

of B's, then B's income is what per

cent more than that of A?

a) 125%

b) 100%

c) 75%

d) 50%

7. In an examination, 35% of the

candidates failed in Mathematics

and 25% in English. If 10% failed in

both Mathematics and English,

then how much per cent passed in

both the subjects?

a) 50%

b) 55%

c) 57%

d) 60%


8. The price of sugar rise by 25%. If a

family wants to keep their expenses

on sugar the same as earlier, the

family will have to decrease its

consumption of sugar by

a) 25%

b) 20%

c) 80%

d) 75%

9. If the numerator of a fraction is

increased by 20% and the

denominator is decreased by 5%,

the value of the new fraction

becomes . The original fraction is

a)

b)

c)

d)

10. If the price of tea is increased by

20%, by how much per cent the

consumption of tea be reduced, so

that there is no increase in the

expenditure on it?

a)

b) 20%

c)

d)

11. A number reduced by 25% becomes

225. What per cent should it be

increased, so that it becomes 375?

a) 25%

b) 30%

c) 35%

d) 75%

12. 25% of the candidates who

appeared in an examination failed

to qualify and only 450 candidates

qualified. The number of

candidates, who appeared in the

examination was

a) 700

b) 600

c) 550

d) 500

13. A worker suffers a 20% cut in his

wages. He may regain his original

wages by obtaining a rise of

a) 27.5%

b) 25.0%

c) 22.5%

d) 20.0%

14. A certain company has 80

engineers. If the engineers

constitute 40% of its workers, then

the number of people employed in

the company is

a) 150

b) 800

c) 200

d) 3200


15. A saves 20% of his monthly salary.

If his monthly expenditure is Rs.

6000. Then, his monthly saving is

a) Rs. 1200

b) Rs. 4800

c) Rs. 1500

d) Rs. 1800


Practice Set-2 (Percentages, Partnership and Share)

1. A starts business with Rs. 7000 and after 5 months, B joined as a partner. After a year the profit is divided in the ratios 2: 3. The capital of B is

(a) Rs 10000

(b) Rs 6500

(c) Rs 18000

(d) Rs 9000

2. A, Band C started a business by investing Rs 40500, Rs 45000 and Rs 60000, respectively. After 6 months C withdrew Rs 15000 while A invested Rs 4500 more. In annual profit of Rs 56100 the share of C will exceed that of A by

(a) Rs 900

(b) Rs 1100

(c) Rs 3000

(d) Rs 3900

3. A, B and C entered into partnership in a

business. A got of the profit and B and C

distributed the remaining profit equally. If C got Rs 400 less than A, the total profit was

(a) Rs 1600

(b) Rs 1200

(c) Rs 1000

(d) Rs 800

4. A began a business with Rs 2250 and was joined afterwards by B with Rs 2700. If the profits at the end of the year were divided in the ratio of 2: 1, after how much time B joined the business?

(a) 5 months

(b) 6 months

(c) 3 months

(d) 7 months

5. Amit and Brijesh started a business with initial investments in the ratio of 12: 11 and their annual profits were in the ratio of 4: 1. If Amit invested the money for 11 months, then for what time Brijesh invested the money?

(a) 9 months

(b) 3 months

(c) 5 months

(d) 10 months

6. A began a business with Rs 10500 and is joined afterwards by B with Rs 18000. After how many months did B join, if the profit at the end of year is divided equally?

(a) 5 months

(b) 15 months

(c) 10 months


(d) 9 months

7. A and B started a business by investing Rs 35000 and Rs 20000, respectively. After 5 months B left the business and C joined the business with a sum of Rs 15000. The profit earned at the end of year is Rs 84125. What is the share of B in profit?

(a) Rs 14133

(b) Rs 15000

(c) Rs 13460

(d) Cannot be determined


Percentages, Partnership and Share Practice Set-1

(Answers)

1) a

2) d

3) b

4) a

5) d

6) b

7) a

8) b

9) c

10) c

11) a

12) b

13) b

14) c

15) c


Percentages, Partnership and Share Practice Set-2

(Answers)

1) c

2) d

3) c

4) d

5) b

6) a

7) c


Chapter: Profit & Loss

Profit and Loss is an extension of the chapter of percentages. It is a very important branch of basic Mathematics. This branch deals with the study of Profit and loss made in any commercial transaction. The entire economy and the concept of capitalism is based on the so called “Profit Motive”.

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 % =

Loss % =

S.P. = × C.P.

S.P. = × C.P.

C.P = × S.P.


C.P = × S.P.

Example: 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 =

= 5 Step II: Also given that per dozen S.P. of apples = 84

Then, S.P. of 1 apple =

=7 Step III: Now, we know that

Gain % =

=

=

=

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

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 % =


Example : A man sold two plots for Rs. 15, 00,000 each. On one he gains 25% while on the other he loses 25%. How much does he gain or loss in the whole transaction.

Solution: In such a case there is always a loss

Loss % = = =6.25%.

If an article sold at two different selling price

. On one is made and on the other is made then:

=

Profit calculation on the basis of equating the Amount Spent and the Amount Earned:

If the person is going through the transaction has got back all the money that he has spent, but has ended up with some amount of goods left over after the transaction.

% Profit = 100

Example: A fruit vendor recovers the cost of 15 oranges by selling 10 oranges. Find his percentage profit.

Solution: Here the money spent is equal to the money earned the percentage profit is given by

% Profit = 100 = 5 100/10 =50%.

Discount:

Discount is the reduction offered amount on the market price.

Therefore,

SP = MP

Where,

MP = Market Price of the product


d = discount in percentage on the market price

Equivalent Single discount for successive discounts a% and b% = (a+b - ) %

Example: What are the successive discounts of 10 %, 12 % and 15% amount to a single discount?

Solution: Suppose the marked price = Rs. 100

The, S.P. = 85% of 88% of 90% of Rs. 100 = Rs. 67.32.

Therefore, the single discount = (100 – 67.32) % = 32.68 %.


Solved Examples (Profit & Loss)

1. An article was purchased for Rs. 78,350/-. Its price was marked up by 30%. It was sold at a discount of 20% on the marked up price. What was the profit percent on the cost price? (IBPS CWE PO MT 2012) (1) 4 (2) 7 (3) 5 (4) 3 (5) 6 Solution: Cost price = Rs. 78350 Marked price = 78350 = Rs. 101855

Selling price = 101855 = Rs. 81484

Profit = 81484 -78350 = 3134

Reqd % profit = 100 = 4%

Ans: (1)

2. The cost of 8 kg of almonds is equal to the cost of 50 kg of apples. The cost of 19 kg of mangoes is Rs. 456. The cost of 1 kg of apples is twice the cost of 2 kg of mangoes. What is the total cost of 3 kg of almonds and 4 kg of apples together? (1) Rs. 2,168

(2) Rs. 2,248

(3) Rs. 2,184

(4) Rs. 2,264

(5) None of these

Solution: Cost of 1 kg of mangoes

Cost of 1 kg of apples = 2 ⤬ 48 = Rs. 96

Cost of 1 kg of almonds


Cost of 3 kg of almonds and 4 kg of apples

= 3 ⤬ 600 + 4 ⤬ 96 = Rs. 2184

Ans: (3)

3. Meena purchases 1500 ml of milk every day. If the cost of one liter of milk is Rs. 44, how much amount will she pay in 20 days? (Corporation Bank PO 2011)

(1) Rs. 1,340

(2) Rs. 1,320

(3) Rs. 1,280

(4) Rs. 1,260

(5) None of these

Solution: Amount paid in 20 days = 1.5 ⤬ 20 ⤬ 44= Rs. 1320

Ans: (2)

4. Kamya purchased an item of Rs. 46,000 and sold it at loss of 12 per cent. With that amount she purchased another item and sold it at a gain of 12 per cent. What was her overall gain/loss? (Allahabad Bank Probationary Officers Exam 2011)

(1) Loss of Rs 662.40 (2) Profit of Rs 662.40 (3) Loss of Rs 642.80 (4) Profit Rs 642.80 (5) None of these

Solution: First S.P. = = Rs 40480

Second S.P = = Rs 45337.6

Loss = Rs (46000-45337.6) = RS 662.4

Ans: (1)

5. A grain dealer cheats to the extent of 10% while buying as well as selling by using false weights. His total profit percentage is

(1) 21%

(2) 23%


(3) 25%

(4) 20%

Ans: (1)

Solution: Here, x = 10%, y = 10%

Total profit percentage

= = 21%

6. When the price of sugar decreases by 10%, a man could buy 1 kg more for Rs 270. Then, the original price of sugar per kg is

(1) Rs 25

(2) Rs 30

(3) Rs 27

(4) Rs 32

Ans: (2)

Solution: Let original CP of sugar = Rs x per kg

New CP of sugar = X x = Rs

According to the give condition,

- = 1

= 1

= 1

x = = Rs 30 per kg


7. For a certain article, if discount is 25% then profit is 25%. If the discount is 10% then the profit is

(1) 50%

(2) 40%

(3) 30%

(4) %

Ans: (1)

Solution: Suppose cost price of the article = Rs x

Then, X × = Rs 510

x = = Rs

If discount is 10% then profit

= × = 150

Profit price of the gift item 50%


8. A man bought orange at the rate of 8 for Rs 34 and sold them at the rate of 12 for Rs 57. How many orange should be sold to earn a net profit of RS 45?

(1) 90

(2) 100

(3) 135

(4) 150

Ans: (1)

Solution: Cost price of one orange =

and selling price of one orange =

Profit of one orange = -

= = =

On selling one orange, we get Rs

On getting profit of Rs 45, we well

= 45 × 2 = 90 orange


Practice Set (Profit & Loss)

1. A man makes a profit of 20% on the sale by selling 20 articles for Rs 1, the number of articles he bought by Rs 1 is

(a) 20

(b) 24

(c) 25

(d) 30

2. A man buys one table and one chair for Rs 500. He sells the table at a loss of 10% and the chair at a gain of 10%. He still gains Rs 10 on the whole. The cost price of the chair is

(a) Rs 300

(b) Rs 350

(c) Rs 200

(d) Rs 250

3. What single discount is equivalent to two successive discounts of 20% and 15%?

(a) 35%

(b) 32%

(c) 34%

(d) 30%

4. If the selling price of 10 articles is equal to the cost price of 11 articles, then the gain per cent is

(a) 10%

(b) 11%

(c) 15%

(d) 25%

5. Krishna purchased a number of articles at Rs 10 for each and the same number for Rs 14 each. He mixed them together and sold them for Rs 13 each. Then, his gain or loss per cent is

(a) loss 8 %

(b) gain 8 %

(c) loss 8 %

(d) gain 8 %

6. A shopkeeper makes a profit of 20% even after giving a discount of 10% on the marked price of an article. If marked price is Rs 500, then the cost price of the article is

(a) Rs 350

(b) Rs 375

(c) Rs 425

(d) Rs 475


7. A fruit seller bought 240 bananas at the

rate of Rs 48 per dozen. He sells of them

at the rate of Rs 5 per banana. th of the

remaining are 6 found to be rotten. The price per banana at which he has to sell the remaining bananas to get a profit of 25% on his entire investment is

(a) Rs 5.5

(b) Rs 6.0

(c) Rs 5.0

(d) Rs 6.5

8. The difference between the selling price and cost price of an article is Rs 210. If the profit per cent is 25, then the selling price of the article is

(a) Rs 950

(b) Rs 1050

(c) Rs 1150

(d) Rs 1250

9. A shopkeeper allows 23% commission on his advertised price and still makes a profit of 10%. If he gains Rs 56 on one item, his advertised price of the item, is

(a) Rs 820

(b) Rs 780

(c) Rs 790

(d) Rs 8OO

10. By selling an article for Rs 144, a person gained such that the percentage gain

equals the cost price of the article. The cost price of the article is

(a) Rs 90

(b) Rs 8O

(c) Rs 75

(d) Rs 60

11. When the price of cloth was reduced by 25%, the quantity of cloth sold increased by 20%. What was the effect on gross receipt of the shop?

(a) 5% increase

(b) 5% decrease

(c) 10% increase

(d) 10% decrease

12. If an article is sold at 200% profit, then the ratio of its cost price to its selling price will be

(a) 1:2

(b) 2:1

(c) 1:3

(d) 3:1


13. If an electricity bill is paid before due date one gets a reduction of 4% on the amount of the bill. By paying the bill before due date a person got a reduction of Rs 13. The amount of his electricity bill was

(a) Rs 125

(b) Rs 225

(c) Rs 325

(d) Rs 425

14. A shopkeeper marks an article at (. 60 and sells it at a discount of 15%. He also gives a gift worth Rs 3. If he still makes 20% profit, the cost price, (in Rs) is

(a) 22

(b) 32

(c) 40

(d) 42

15. A shopkeeper earns a profit of 12% on selling a book at 10% discount on the printed price. The ratio of the cost price and the printed price of the book is

(a) 45:56

(b) 45:51

(c) 47: 56

(d) 47: 51

16. A reduction of 10% in the price of sugar enables a housewife to buy 6.2 kg more for Rs 1116. The reduced price per kg is

(a) Rs 12

(b) Rs 14

(c) Rs 16

(d) Rs 18

17. A man buys a certain number of oranges at 20 for Rs 60 and an equal number at 30 for Rs 60. He mixes them and sells them at 25 for Rs 60. What is gain or loss per cent

(a) Gain of 4%

(b) Loss of 4%

(c) Neither gain nor loss

(d) Loss of 5%

18. The marked price of a shirt and trousers are in the ratio 1:2. The shopkeeper gives 40% discount on the shirt. If the total discount on the set of the shirt and trousers is 30%, the discount offered on the trousers is

(a) 15%

(b) 20%

(c) 25%

(d) 30%


19. The percentage of loss when an article is sold at Rs 50 is the same as that of the profit when it is sold at Rs 70. The above mentioned percentage of profit or loss on the article is

(a) 10%

(b) 16 %

(c) 20%

(d) 22 %

20. I purchased 120 exercise books at the rate of Rs 3 each and sold of them at the rate of Rs

4 each, of them at the rate of Rs 5 each and the rest at the cost price. My profit per cent was

(a) 44 %

(b) 44 %

(c) 44 %

(d) 45%


Profit & Loss Practice Set (Answers)

1) b

2) a

3) b

4) a

5) d

6) b

7) b

8) b

9) d

10) b

11) d

12) c

13) c

14) c

15) a

16) d

17) b

18) c

19) b

20) b


Chapter: Simple Interest & Compound 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 by S.I.

FORMULAE:

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

S.I. = (

Or

P = )

or

R = )

or

T =

Now,

Simple Interest + Principle = Amount

If we denote the amount by ’A’, then


Simple Interest = A – P

S.I. = A – P =

A=P (1+ ) = SI (1+ )

Two different cases can be compared by using the following formula

=

Example: 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. = (

S.I. = (

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.


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

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

Amount = P

(iii) If interest is compounded Quarterly, then

Amount = P

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

Amount = P (1+ )

Example: What will be the Compound Interest on Rs. 5000 at 5% per annum for 3 years, compounded annually?

Solution: Amount = Rs. [5,000 × ] = Rs. [5,000 × ] = Rs. [5,000 × ]

= Rs.5788.125

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


Example: 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 = Rs. (120 × 11× 11) = Rs. 14520

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


Solved Examples (Simple Interest & Compound Interest)

1. What is the difference between the simple and compound interest on Rs. 7,300/- at the rate of 6 p.c.p.a. in 2 years? (IBPS CWE PO MT 2012) (1) Rs. 29.37 (2) Rs. 26.28 (3) Rs. 31.41 (4) Rs. 23.22 (5) Rs. 21.34

Solution: SI = = = 876

CI = 7300 [(1 )2= -1] = 7300 [( )2 - 1]

7300 [( )] = 7300 ) = 902.28

Difference = 902.28 – 876 = 26.28

Quick Method:

CI = - (6 + 6)

= 12.36 -12 = 0.36% = 0.36 per cent of 7300 = 26.28 Ans: (2)

2. The simple interest accrued on an amount of Rs. 22,500 at the end of four years is Rs. 10,800. What would be the compound interest accrued on the same amount at the same rate of interest at the end of two years?

(1) 16,908 (2) 5,724 (3) 28,224 (4) 8,586 (5) None of these


Solution: r = = 12%

CI= 22500 2 – 22500

= 22500 - 22500 = 28224 – 22500 = 5724

Ans: (2)

3. The simple interest accrued on a sum of a certain principal is Rs. 35,6727 in seven years at the rate of 8 pcpa. What would be the compound interest accrued on that principal at the rate of 2 pcpa in 2 years?

(A) Rs. 2573.48

(B) Rs. 2564.86

(C) Rs. 2753.86

(D) Rs. 2654.48

(E) None of these

Solution. Principal

CI

Quicker Method:

Ans: (1)


4. 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. = (

S.I. = (

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

5. 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

Solution : P= 18,000, R = 25%, T = 6/12 year

S.I. = = = Rs.2, 250.

6. What will be the simple interest on Rs. 2,400 at 4 % per annum for the period from 1st

Feb, 2005 to 15th April, 2005? (a) Rs. 20 (b) Rs.20.5 (c) Rs. 22 (d) Rs.25

Solution: Principal (P) = Rs. 2,400, Rate (R) = , Time (T) = (27+31+15) days =73 days = year

= year


Now, S.I. = ( S.I. = (2,400 S.I. = 20.

Therefore, the simple interest on Rs. 2,400 at 4 % per annum for the period from 1st Feb,

2005 to 15th April, 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.

7. 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%?

(a) Rs.1, 000 (b) Rs 1,230 (c) Rs.1, 050 (d) Rs.1, 125

Solution : 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 =

R = % = 10%

If the interest rate is increased by 10 %, then

New Rate = (10+5) % =15%, New S.I. =Rs. = Rs.450

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

8. 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


Solution : Amount = Rs. [8000 ]

Rs. [8000 = Rs. 10,580

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

9. What is the difference between the compound interests on Rs. 10000 for 2 years at 4%pe

annum yearly and half yearly? (a) Rs.8 (b) Rs.8.49 (c) Rs.7 (d) Rs. 10

Solution: C.I. , When interest is compounded yearly

Amount =

Amount = = Rs.11,032.32

∴ C.I. = 11032.32 10000 = 1032.32 C.I. When interest is compounded half yearly

Amount = Rs [10000 ] = Rs. 11040.80803

∴ C.I. = 11040.80803 10000 = 1040.808032

Diff = 8.488032 Rs. 8.9


10. A sum of money tripled itself at compound interest in 10 years. In how many years will it become 27 times.

(a) 35 years (b) 31 years (c) 30 years (d) 32 years

Solution : P = 3P

= 3

Suppose after n years it will become 27 times Then,

P = 27 p =

= n = 30.


Practice Set (Simple Interest & Compound Interest)

1. A sum of money placed at compound interest doubles itself in 4 yr. In how many years will it amount to four times itself?

a) 12 yr

b) 13 yr

c) 8 yr

d) 16 yr

2. The difference between the compound interest and simple interest on Rs. 10000 for 2 yr is Rs. 25. The rate of interest per annum is

a) 5%

b) 7%

c) 10%

d) 12%

3. The simple interest on a sum of

money is of the principal and

the number of years is equal to the rate per cent per annum. The rate per cent per annum is equal to

a) 3%

b)

c)

d)

4. Ratio of the principal and the amount after 1 yr is 10 : 12. Then, the rate of interest per annum is

a) 12%

b) 16%

c) 18%

d) 20%

5. At some rate of simple interest, A lent Rs. 6000 to B for 2 yr and Rs. 1500 to C for 4 yr and received Rs. 900 as interest from both of them together. The rate of interest per annum was

a) 5%

b) 6%

c) 8%

d) 10%

6. What annual payment will discharge a debt of Rs. 6450 due in 4 yr at 5% per annum simple interest?

a) Rs. 1400

b) Rs. 1500

c) Rs. 1550

d) Rs. 1600


7. In how many years will a sum of

money double itself at simple

interest per annum?

a) 24 yr

b) 20 yr

c) 16 yr

d) 12 yr

8. In how many years will a sum on Rs. 800 at 10% per annum compound interest, compounded semi-annually becomes Rs. 926.10?

a)

b)

c)

d)

9. A certain sum amount to Rs. 5832 in 2 yr at 8% per annum compound interest, then the sum is

a) Rs. 5000

b) Rs. 5200

c) Rs. 5280

d) Rs. 5400

10. In what time will Rs. 10000 amount to Rs. 13310 at 20% per annum compounded half-yearly?

a)

b)

c)

d)

11. In what time Rs. 8000 will amount to Rs. 9261 at 10% per annum compound interest, when the interest is compounded half-yearly?

a)

b)

c)

d)

12. Simple interest on Rs. 500 for 4 yr at 6.25% per annum is equal to the simple interest on Rs. 400 at 5% per annum for a certain period of time. The period of time is

a) 4 yr

b) 5 yr

c)

d)


13. If the difference between the compound and simple interests on a certain sum of money for 3 yr at 5% per annum is Rs. 15.25, then the sum is

a) Rs. 2000

b) Rs. 1000

c) Rs. 1500

d) Rs. 2500

14. If the simple interest for 6 yr be equal to 30% of the principal, it will be equal to the principal after

a) 20 yr

b) 30 yr

c) 10 yr

d) 22 yr

15. The effective annual rate of interest, corresponding to a nominal rate of 6% per annum payable half-yearly is

a) 6.06%

b) 6.07%

c) 6.08%

d) 6.09%

16. What annual instalment will discharge a debt of Rs. 6450 due in 4 yr at 5% simple interest?

a) Rs. 1500

b) Rs. 1835

c) Rs. 1935

d) Rs. 1950

17. At what rate per cent per annum will the simple interest on a sum of

money be of the amount in 10 yr?

a) 4%

b) 6%

c)

d)

18. The difference between the simple and compound interest on a certain sum of money for 2 yr at 4% per annum is Rs. 4. The sum is

a) Rs. 2500

b) Rs. 2400

c) Rs. 2600

d) Rs. 2000

19. The compound interest on a certain sum of money at a certain rate for 2 yr is Rs. 40.80 and the simple interest on the same sum is Rs. 40 at the same rate and for the same time. The rate of interest per annum is

a) 2%

b) 3%

c) 4%

d) 5%


20. A sum of money becomes eight times of itself in 3 yr at compound interest. The rate of interest per annum is

a) 100%

b) 80%

c) 20%

d) 10%


Simple Interest & Compound Interest Practice Set

(Answers)

1) c

2) a

3) d

4) d

5) a

6) b

7) c

8) a

9) a

10) a

11) b

12) c

13) a

14) a

15) d

16) a

17) d

18) a

19) c

20) a


Chapter: Time and Work

Introduction: -

Pipes and cisterns problems are similar to time and work problem, the only difference is that pipes and cisterns problems have outlets and inlets .

1. Inlet is a pipe connected with a tank which fills it

2. Outlet is a pipe connected with a tank to empty it.

Mathematical use :-

1. If pipe can fill (or empty) a tank in x hours, then the part filled (or emptied) in 1 hour =

.

2. If a pipe P fills a tank in x hours and another pipe Q empties the full tank in y hours and if

both the pipes are opened then the net part filled in 1 hour = ( ) .

Example 1: How much time will it take to fill the tank if a pipe A fill it in 30 hours and another pipe B empties it in 40 hours?

Sol 1: Net part filled in 1 hour =

Hence the time taken to fill the tank = 12 hours.

Direct method:-

The time taken to fill the tank = .

3. If two pipes fills the tank in x and y hours respectively then the net part filled in 1 hours ,

when both the pipes are opened = ( ) .


Example 2: How much time will it take to fill the tank if two pipes A and B fill it in 20 hours and 30 hours respectively and both the pipes are opened?

Sol 2: Net part filled by both pipes A and B together in 1 hour =


Direct method:-

The time taken to fill the tank = .

4. If two pipes fills the tank in x and y hours respectively and a third pipe emptied the full

tank in z hours , then the net part filled in 1 hours , when all the pipes are opened =

( ) .

Example3: How much time will it take to fill the tank if two pipes A and B fill it in 10 hours and 20 hours respectively and a third pipe C empties it in 40 hours?

Sol 3: Net part filled in 1 hour =


Direct method:-

The time taken to fill the tank =

5. If a pipe fills the tank in x hours but due to leakage in bottom it is filled in y hours , then

the time taken by leak to empty the tank if the tank is full = .


Example 4: How much time will the leak take to empty the tank if a pipe A fill it in 10 hours but due to leak in the bottom it is filled in 15 hours ?

Sol 4: Tank empty due to leak in 1 hour = (

Hence the leak will empty the full tank in 30 hours

Direct method:-

The required time =


Solved Examples (Time and Work)

1. A and B together can complete a task in 20 days. Band together can complete the same task in 30 days. A and C together can complete the same task in 40 days. What is the respective ratio of the number of days taken by A when completing the same task alone to the number of days taken by C when completing the same task alone? (IBPS CWE PO MT 2012) (1) 2: 5 (2) 2: 7 (3) 3: 7 (4) 1: 5 (5) 3: 5 Solution: A and B can finish the work in 20 days.

A and B’s one day’s work =

B and C and finish the work in 30 days. B and C’s one day’s work =

A and C can finish the work in 40 days

A and C’s one day's work =

Adding we get 2(A + B + C)’s one day’s work

= + + = =

(A + B + C)’s one day’s work

(A + B + C)’s one day’s work

= =

A’s one day’s work = -

= = =

A alone can finish the work in 48 days.


C’s one day work = - = =

C alone can finish the work in 240 days.

Reqd ratio = = 1 : 5

Ans: (4)

2. Seven girls can do a piece of work in 13 days, six boys can do the same piece of work in 12 days, nine men can do the same piece of work in nine days and six women can do the same piece of work in 14 days. Who are the most efficient? (Corporation Bank PO 2011)

(1) Boys

(2) Girls

(3) Women

(4) Men

(5) Both men and women

Solution: One girl can complete the work in 7 ⤬ 13 = 91 days.

One boy can complete the work in 6 ⤬ 12 = 72 days.

One man can complete the work in 9 ⤬ 9 = 81 days.

One woman can complete the work in 6 ⤬ 14 = 84 days.

Hence, boys are the most efficient.

Ans: (1)

3. A and B together can complete a work in 12 days. A alone can complete in 20 days. If B does the work only half a day daily, then in how many days A and B together will complete the work.

(1) 10 days (2) 20 days (3) 11 days (4) 15 days

Ans: (4)

Solution: (A + B)’s 1 days work =


A’s 1 day work =

B’s 1 day work = -

= =

=

B does the work only half day.

B’s 1 days work = × =

Now, A+B)’s day work = +

= = =

Hence, A and B together will complete the work in 15 days.

4. A can do a piece of work in 70 days and B is 40% more efficient than A. The number of days taken by B to do same work is

(1) 3

(2)

(3) 5

(4)

Ans: (3)

Solution: A can do a piece of work in 70 days while B us 40% more efficient than A.

B do the same work in


= 70 ×

= 70 × = 50 days

5. A and B working separately can do a piece of work in 9 and 12 days, respectively. If they work for a day alternately with A beginning, the work would be completed in

(1) 28 days (2) 14 days

(3) 5 days

(4) days

Ans: (3)

Solution: One day work of A =

One day work of B =

Two days work of (A+B) = + =

=

10 days = 5 × 2 days work of (A + B) = ×5 =

Remaining work = 1- =

Now, A‘s turn.

A, will complete the work in × 9 = day

Hence, the total time = 10 + = day


6. A and B can do a work in 72 days. B and C can do it in 120 days. The number of days needed for A to the work alone is

(1) 20 (2) 22 (3) 33 (4) 44

Ans: (3)

Solution: Work done by (A+B) in 1 day =

Work done by (B+C) in 1 day =

Work done by (C+A) in 1 day =

Work done by 2 (A+B+C) in 1 day =

=

Work done by (A+B+C) in 1 day

= × =

A’s 1 day work = -

=

Hence, A do the work alone in 120 days.

7. A, B and C individually can do a work in 10 days, 12 days and 15 days, respectively. If they start working together, then the number of days required to finish the work is

(1) 16 days (2) 8 days (3) 4 days (4) 2 days


Ans: (3)

Solution: Given, x = 10, y = 12 and z = 15

Required number of days

=

=

= 4 days

8. A cistern has two pipes. One can fill it with water in 8 h and other can empty it in 5 h. In

how many hours will the cistern be emptied, if both the pipes are opened together when of

the cistern is already full of water?

(1) h (2) 10 h (3) 6 h

(4) h

Ans: (b)

Solution: Part of cistern emptied in 1 h

= - = =

part is emptied in 1 h.

part is emptied in × = 10h


Practice Set (Time and Work)

1. Two men A and B are started a job in which A was thrice as good as B and therefore took 60 days less than B to finish the job. How many days will they take to finish the job, if they start working together?

(a) 15 days

(b) 20 days

(c) 22 days

(d) 25 days

2. A and B can separately do a piece of work in 20 and 15 days, respectively. They worked together for 6 days, after which B was replaced by C. The work as finished in next 4 days. The number of days in which C alone could do the work is

(a) 30 days

(b) 45 days

(c) 40 days

(d) 35 days

3. A can do a work in 12 days. When he had worked for 3 days, B joined him. If they complete the work in 3 more days, in how many days can B alone finish the work?

(a) 6 days

(b) 12 days

(c) 4 days

(d) 8 days

4. A and B can complete a piece of work in 8 days, B and C can do it in 12 days, C and A can do it in 8 days. A, Band C together can complete it in

(a) 4 days

(b) 5 days

(c) 6 days

(d) 7 days

5. X is 3 times as fast as Yand is able to complete the work in 40 days less than Y. Then, the time in which they can complete the work together is

(a) 15 days

(b) 10 days

(c) 7 days

(d) 5 days

6. A and B can do a job alone in 12 days and 15 days, respectively. A starts the work and after 6 days B also joins to finish the work together. For how many days B actually worked on the job?

(a) 3


(b) 9

(c) 5

(d) 6

7. Two pipes can fill a cistern separately in 24 min and 40 min, respectively. A waste pipe can drain off 30 L/min. If all the there pipes are opened, the cistern fills in one hour. The capacity of the cistern is

(a) 800 L

(b) 400 L

(c) 600 L

(d) 500 L

8. A cistern is normally filled in 8 h but takes another 2 h longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in

(a) 16 h

(b) 20 h

(c) 25 h

(d) 40 h

9. Two pipes, P and Q can fill a cistern in 12 and 15 min, respectively. Both are opened together but at the end of 3 min, P is turned off. In how many more minutes will Q fill the cistern?

(a) 7min

(b) 7 min

(c) 8 min

(d) 8 min

10. Tapas works twice as fast as Mihir. If both of them together complete a work in 12 days. Tapas alone can complete it in

(a) 15 days

(b) 18 days

(c) 20 days

(d) 24 days

11. One man and one woman together can complete a piece of work in 8 days. A man alone can complete the work in 10 days. In how many days can one woman alone complete the work?

(a)

(b) 30

(c) 40

(d) 42

12. A and B together can do a work in 10 days. Band C together can do the same work in 6 days. A and C together can do the work in 12 days. Then, A, Band C together can do the work in

(a) 28 days

(b) 14 days

(c) 5 days

(d) 8 days


13. A is thrice as good a workman as Band therefore is able to finish a job in 40 days less than B. Working together, they can do it in

(a) 14 days

(b) 13 days

(c) 20 days

(d) 15 days

14. 45 men can complete a work in 16 days. Four days after they started working, 36 more men joined them. How many days will they now take to complete the remaining work?

(a) 6 days

(b) 8 days

(c) 6 day

(d) 7 days

15. A can do a work in 5 days less than the time taken by B to do it. If both of them

together take 11 days, then the time

taken by B alone 9 to do the same work (in days) is

(a) 15

(c) 25

(b) 20

(d) 30

16. A can complete of a work in 5 days

and B, of the work in 10 days. In how

many days both A and B together can complete the work?

(a) 10

(b) 9

(c) 8

(d) 7

17. A boy and girl together fill a cistern with water. The boy pours 4 L of water every 3 min and the girl pours 3 L every 4 min. How much time will it take to fill 100 L of water in the cistern?

(a) 36 min

(b) 42 min

(c) 48 min

(d) 44 min

18. If 28 men complete of a piece of work

in a week, then the number of men, who must be engaged to get the remaining work completed in another week, is

(a) 5

(b) 6

(c) 4

(d) 3


19. While working 7 h a day, A alone can complete a piece of work in 6 days and B alone in 8 days. In what time would they complete it together, working 8 h a day?

(a) 3 days

(b) 4 days

(c) 2.5 days

(d) 3.6 days

20. A and B can complete a piece of work in 12 and 18 days, respectively. A begins to do the work and they work alternatively one at a time for one day each. The whole work will be completed in

(a) 14 days

(b) 15 days

(c) 16 days

(d) 18 days

21. A man and a boy received Rs 800 as wages for 5 days for the work they did together. The man's efficiency in the work was three times that of the boy. What are the daily wages of the boy?

(a) Rs 76

(b) Rs 56

(c) Rs 44

(d) Rs 40

22. A pipe can empty a tank in 40 min. A second pipe with diameter twice much as that of the first is also attached with the tank to empty it. The two pipes together can empty the tank in

(a) 8 min

(b) 13 min

(c) 30 min

(d) 38 min

23. A and B can do a work in 45 days and 40 days, respectively. They began the work together but A left after sometime and B completed the remaining work in 23 days. After how many days of the start of the work did A leave?

(a) 10 days

(b) 9 days

(c) 8 days

(d) 5 days

24. A and B working separately can do a piece of work in 10 days and 15 days, respectively. If they work on alternate days beginning with A, in how many days will the work be completed?

(a) 18 days

(b) 13 days

(c) 12 days

(d) 6 days


25. Two pipes can fill a tank separately in 20 min and 30 min, respectively. If both the pipes are opened simultaneously, then the tank will be filled in

(a) 10 min

(b) 12 min

(c) 15 min

(d) 25 min


Averages Practice Set (Time and Work)

1) c

2) c

3) a

4) c

5) a

6) a

7) c

8) d

9) d

10) b

11) c

12) c

13) d

14) c

15) c

16) b

17) c

18) c

19) a

20) a

21) d

22) b

23) b

24) c

25) b


Chapter: Speed Distance & Time

Speed, Distance and time is one of the most important chapter for the purpose of the maths section in aptitude exams. Time and Distance Formulae relates Time, Distance and Speed. These relationships have many practical applications. The basic concepts of Speed, Distance and time are used in solving questions based on relative motion, circular motion, problem based on trains, problem based on boats, races, etc.

The questions asked in IBPS Bank PO is very diverse in nature, therefore this chapter is very important for Bank PO aspirants. 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= Speed Time, Time=

(ii) S km/hr =(s )m/sec

(iii) S m/sec =(s )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.

Example1. A Chennai Express travelling at of its actual speed and covers 42 km in 1 hr 40

min 48 sec, find the actual speed of the Chennai Express.

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

Then new speed = S

Time taken by Chennai Express with new speed

= 1 hr 40 min 48 sec


= 1hr+40 hr+48 [Because 1hr = 60 minutes,

1 minute = 60 Second,

& 1 minute = hr,

1 Second = minute

= 1hr + hr + hr

= hrs

Now, according to the formula

New Speed × Time taken by Chennai Express with new speed =

Distance covered by Chennai Express

S× = 42

S =

S =55km/hr

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

TRAINS, BOATS AND STREAMS

IMPORTANT POINTS ON TRAINS:

1.) Time taken by a train x metres long to pass a single post or a pole or a standing man =

Time taken by the train to cover x metres.

2.) Time taken by a train x metres long to pass a stationary object of length y metres = Time

taken by the train to cover (x+y) meters.

3.) Suppose two trains or two bodies are moving in the same direction at u kmph and v

kmph such that u > v, then their relative speed = (u-v) kmph.


4.) If 2 trains of lengths x km and y km are moving in the same direction at u kmph and v

kmph , where u > v, then time taken by faster train to cross the slower train = hrs.

5.) Suppose two trains or two bodies are moving in opposite directions at u kmph and v

kmph . Then their relative speed = (u+v) kmph

6.) If two trains of lengths x km and y km are moving in opposite directions at u kmph and v

kmph , then time taken by the trains to cross each other = hrs.

7.) If two trains start at the same time from two points A and B towards each other and

after crossing they take a and b hours in reaching B and A respectively.

Then, A’s speed: B’s speed = (√b: √a).

8.) x kmph = m/sec

9.) y metre/sec = km/hr

EXAMPLES

Example 1: Find the time taken by a train 180 m long, running at 72 kmph, in crossing an electric pole.

Solution: speed of the train = (72 × 5/18) m/sec = 20 m/sec

Distance moved in passing the pole = 180m Required time taken = (180/20)sec = 9 sec.

Example 2: Two trains 137m and 163m in lengths are running towards each other on parallel lines, one at the rate of 42 kmph and another at 48 kmph. In what time will they be clear of each other from the moment they meet?

Solution: Relative speed of the trains = (42+48) kmph = 90 kmph = (90×5/18)m/sec= 25m/sec.

Time taken by the trains to pass each other = Time taken to cover (137+163)m at 25m/sec = (300/25)sec = 12 sec.

IMPORTANT POINT ON BOATS AND STREAMS:

1.) In water, the direction along the stream is called downstream and the direction against

the stream is called upstream.

2.) If speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then

Speed downstream = (u + v) km/hr Speed upstream = (u – v) km/hr

3.) If the speed downstream is a km/hr and the speed upstream is b km/hr, then :


Speed in still water = (a + b)/2 km/hr Rate of stream = (a – b)/2 km/hr

EXAMPLES

Example 1: A man can row upstream at 7 kmph and downstream at 10 kmph. Find man’s rate in still water and the rate of current.

Solution: Rate in still water = (10 + 7)/2 km/hr= 8.5 km/hr

Rate of current = (10 – 7)/2 km/hr = 1.5km/hr

Example 2: A man can row 8 kmph in still water and the river is running at 2 kmph. If the man takes 1 hour to row to a place and back, how far is the place?

Solution: Man’s rate downstream = (8+2) kmph = 10 kmph

Man’s rate upstream = (8-2) kmph = 6 kmph


Solved Examples (Speed Distance & Time)

1. A 476-metre-Iong moving train crosses a pole in 14 seconds. The length of a platform is equal to the distance covered by tile train in 20 seconds. A man crosses the same platform in 7 minutes and 5 seconds. What is the speed of the man in metre/second? (IBPS RRB Group ‘A’ Officers Exam 2012)

(1) 1.8 m/s

(2) 1.4 m/s

(3) 1.6 m/s

(4) 2 m/s

(5) 1.2 m/s

Solution: Speed of train

Length of platform = 34 ⤬ 20 = 680 metre.

(∵ 7 minute 5 second = 7 ⤬ 60 + 5 = 425 second)

Speed of man

Ans: 3

2. Train A crosses a pole in 20 seconds and Train B crosses the same pole in one minute. The length of Train A is half the length of Train B. What is the ratio of the speed of Train A to that of Train B? (Corporation Bank PO 2011)

(1) 3:2

(2) 3:4

(3) 4:3

(4) Cannot be determined

(5) None of these

Solution: Let the length of Train B be 2x and that of Train A be x.

Speed of Train A


Speed of Train B

Ratio:

Ans: 1

3. A 280 metre long train moving with an average speed of 108 km/hr crosses a platform in 12 seconds. A man crosses the same platform in 10 seconds. What is the speed of the man? (Allahabad Bank Probationary Officers Exam 2011)

(1) 5 m/s (2) 8 m/s (3) 12 m/s (4) Cannot be determined (5) None of these

Solution: Speed of the train = 108 km/hr = m/sec

If the length of the platform be x metres, then

= 30

x + 280 = 360

x = (360-280 =) = 80 metres

man’s speed = = 8 m/sec

Ans: 2

4. A student goes to school at the rate of 2 km/h and reaches 6 min late. If he travels at the

speed of 3 km/h, he is 10 min early. The distance (in km) between the school and his house is

(1) 5 (2) 4 (3) 3 (4) 1

Ans: (2)


Solution: Let the distance between school and his house

= x km

According to the given condition.

- =

- =

=

=

x = 4 km

5. The Speeds of two trains are in the ratio 6 : 7. If the second train runs 364 km in 4 h, then the speed of first train is

(1) 60 km/h (2) 72 km/h (3) 78 km/h (4) 84 km/h

Ans: (3)

Solution: Let the speeds of two rain be 6x km/h and 7x km/h, respectively.

By given condition

Speeds of second train = = 91 km/h.

7x = 91

x = = 13 km/h

Speed of first train = 13 × 6 = 78 km/h


6. If a man walks 3 km/h, he is late to his office by 20 min. If he increases his speed to 6 km/h, he reaches the office 30 min early. The distance of his office from the starting place is

(1) 6 km (2) 5 km (3) 5.5 km (4) 4 km

Ans: (2)

Solution: Let the distance of his office from the starting place is x km.

By given condition,

- = +

=

x = = 5 km

7. In covering a distance of 30 km, Abhay takes 2 h more than Sameer. If Abhay doubles his speed, then he would take 1 h less than Sameer. Abhay's speed (in km/h) is

(1) 5 (2) 6 (3) 6.25 (4) 7.5

Ans: (1)

Solution: Let the speeds of Abhay be x km/h and Sameer be y km/h respectively.

Then, = 2 ……………..(i)

And = 1 ……………..(i)

On adding Eqs. (i) and (ii), we get


= 3

30 = 6x

x = 5 km/h

8. A man completed a certain journey by a car. If he covered 30% of the distance at the speed of 20 km/h and the remaining distance at 10 km/h, his average speed for the whole journey was

(1) 25 km/h (2) 28 km/h (3) 30 km/h (4) 33 km/h

Ans: (1)

Solution: Let the total be 100 km.

Average Speed =

= =

=

=

= 25 km/h

9. Two trains started at the time one from A to B and other from B to A. If they arrived at B and A, respectively 4 h and 9 h after they passed each other, the ratio of the speeds of the two trains was

(1) 2 : 1 (2) 3 : 2 (3) 4 : 3 (4) 5 : 4

Ans: (2)

Solution: Required ratio of the speeds of two trains


= = =

10. A man takes 6 h 15 min in walking a distance and ridding to the starting place. He could walk both ways in 7h 45 min. The time taken by him to ride both ways, is

(1) 4 h (2) 4 h 30 min (3) 4 h 45 min (4) 5 h

Ans: (3)

Solution: Time taken in walking both ways

= 7 h 45 min ……………………(i)

Time taken in walking one way and riding back

= 6 h 15 min ……………………(ii)

On multiplying by 2 in Eq. (ii) and then subtracting

Eq . (i) from Eq. (ii), we get

Time taken by the man to ride both ways

= 12 h 30 min – 7h 45 min

= 7 h 45 min


Practice Set (Speed Distance & Time)

1. A person travels 285 km in 6 h. In the first part of the journey, he travels at 40 km/h by bus. In the second part, he travels at 55 km/h by train. The distance travelled by train is

(a) 165 km

(b) 615 km

(c) 561 km

(d) 156 km

2. With average speed of 40 km/h, a train reaches its destination at time. If it goes with an average speed of 35 km/h, it is late by 15 min. The total journey is

(a) 30 km

(b) 40 km

(c) 70 km

(d) 80 km

3. If a train runs at 40 km/h, it reaches its destination late by 11 min but, if it runs at 50 km/h, it is late by 5 min only. The correct time (in min) for the train to complete the journey is

(a) 13

(b) 15

(c) 19

(d) 21

4. Two trains, 80 m and 120 m long, are running at the speed of 25 km/h and 35 km/h, respectively in the same direction on parallel tracks. How many seconds will they take to pass each other?

(a) 48 s

(b) 64 s

(c) 70 s

(d) 72 s

5. A train, 300 m long, passed a man, walking along the line in the same direction at the rate of 3 km/h in 33 s. The speed of the train is

(a) 30 km/h

(b) 32 km/h

(c) 32 km/h

(d) 38 km/h

6. By walking at of his usual speed, a man

reaches his office 20 min later than his usual time. The usual time taken by him to reach his office is

(a) 75 min

(b) 60 min

(c) 40 min


(d) 30 min

7. In a 100 m race, Kamal defeats Bimal by 5 s. If the speed of Kamal is 18 km/h, then the speed of Bimal is

(a) 15.4 km/h

(b) 14.5 km/h

(c) 14.4 km/h

(d) 14 km/h

8. A train, 240 m long, crosses a man walking along the line in opposite direction at the rate of 3 km/h in 10 s. The speed of the train is

(a) 63 km/h

(b) 75 km/h

(c) 83.4 km/h

(d) 86.4 km/h

9. A boy is late by 9 min, if he walks to school at a speed of 4 km/h. If he walks at the rate of 5 km/h he arrives 9 min early. The distance to his school is

(a) 9 km

(b) 5km

(c) 4 km

(d) 6 km

10. Two towns A and B are 500 km apart. A train starts at 8 am from A towards B at a speed of 70 km/h. At 10 am, another train starts from B towards A at a speed of 110 km/h. When will the two trains meet?

(a) 1 pm

(b) 12 noon

(c) 12: 30 pm

(d) 1: 30 pm

11. A train 150 m long passes a pole in 15 s and another train of the same length travelling in the opposite direction in 12 s. The speed of the second train is

(a) 45 km/h

(b) 48 km/h

(c) 52 km/h

(d) 54 km/h

12. A, B and C start together from the same place to walk round a circular path of length 12 km. A walks at the rate of 4

km/h, B 3 km/h and C km/h. They will

meet together at the starting place at the end of

(a) 10 h

(b) 12 h

(c) 15 h

(d) 24 h

13. A train travelling at 48 km/h crosses another train, having half its length and travelling in opposite direction at 42 km/h in 12 s. It also passes a railway platform in 45 s. The length of the railway platform is

(a) 200 m

(b) 300 m

(c) 350 m


(d) 400 m

14. In a race of one kilometer, A gives B a start of 100 m and still wins by 20 s but, if A gives B a start of 25 s, B wins by 50 m. The time taken by A to run one kilo metre is

(a) 17 s

(b) s

(c) s

(d) s

15. A student walks from his house at a

speed of 2 km/h and reaches his school 6

min late. The next day he increases his speed by 1 km/h and reaches 6 min before school time. How far is the school from his house?

(a) km

(b) km

(c) km

(d) km

16. A train passes two persons walking in the same direction at a speed of 3 km/h and 5 km/h respectively, in 10 sand 11 s, respectively. The speed of the train is

(a) 28 km/h

(b) 27 km/h

(c) 25 km/h

(d) 24 km/h

17. A and B run a kilometre and A wins by 25 s. A and C run a kilometre and A wins by 275 m. When Band C run the same distance, B wins by 30 s. The time taken by A to run a kilometre is

(a) 2 min 25 s

(b) 2 min 50 s

(c) 3 min 20 s

(d) 3 min 30 s

18. A train passes a man standing on a platform in 8 s and also crosses the platform which is 264 m long in 20 s. The length of the train is

(a) 188 m

(b) 176 m

(c) 175 m

(d) 96 m

19. A walks at a uniform rate of 4 km/h and 4 h after his start, B bicycles after him at the uniform rate of 10 km/h. How far from the starting point will B catch A?

(a) 16.7 km

(b) 18.6 km

(c) 21.5 km

(d) 26.7 km


20. A constable follows a thief who is 200 m ahead of the constable. If the constable and the thief run at speeds of 8 km/h and 7 km/h respectively, the constable would catch the thief in

(a) 10 min

(b) 12 min

(c) 15 min

(d) 20 min


Speed Distance & Time Practice Set (Answers)

1) a

2) c

3) c

4) d

5) d

6) b

7) c

8) c

9) d

10) b

11) d

12) d

13) d

14) b

15) b

16) c

17) a

18) b

19) d

20) b


Chapter: Mensuration

Mensuration is a branch of Mathematics which deals with lengths of lines, areas of surfaces and volume of solids.

Mensuration may be divided into two parts:

(i) Plane mensuration which deals with the sides, perimeters and areas of plane figures

of different shapes.

(ii) Solid mensuration which deals with the areas and volumes of solid objects.

Perimeter and Area of Triangles:

1.) If a, b, c are three sides of a triangle; then

i) Perimeter = a+b+c

ii) Area = √{s(s-a)(s-b)(s-c)} , where s= semi-perimeter of triangle

i.e., s= (a+b+c)/2

2.) If base and the corresponding altitude of triangle are known, then

Area of triangle = ½ ×base × height

3.) If ‘a’ is side of an equilateral triangle then

i.) Perimeter = 3a

ii.) Area = (√3/4) ×a2

Perimeter and area of rectangles, square:

l

b a a

1.) If ‘l’ and ‘b’ are length and breadth of rectangle,

Then i.) Perimeter = 2(l + b)

ii.) Area = l×b

iii.) Diagonal= √(l2 + b2)


2.) If ‘a’ is side of square , then

i.) Perimeter = 4a

ii.) Area = a2

iii.) Diagonal , d = √a2 + a2 = a√2

iv.) Side = √area

Area of Trapezium

Two sides of trapezium are parallel.

If ‘a’ and ‘b’ are parallel sides and ‘h’ is distance (i.e. height) between them; then

Area of trapezium = ½ (a + b) × h

Parallelogram

Area of parallelogram = b×h where: b = base of the parallelogram h=height of the parallelogram Rhombus

Rhombus has all the sides equal and diagonals of rhombus bisect each other at right angles. If ‘a’ is the side of rhombus and d1 , d2 are diagonals ; then Perimeter of rhombus = 4a Area of rhombus = ½ ×d1 × d2


Circle

If ‘r’ is radius of circle, then diameter= 2r, circumference of circle = 2πr, area of circle = πr2

The Greek letter π (pronounced as Pie) = 22/7 = 3.14

EXAMPLES:

1) Find the area of triangle whose sides are 10cm, 24cm and 26cm.

Solution:

a= 10cm b= 24cm c=26cm

s= (a+ b+ c)/2= (10+ 24+ 26)/2 =30

area of triangle = √{s(s-a)(s-b)(s-c)}

= √{30(30-10)(30-24)(30-26)}

=120cm2

2) If two sides of a triangle are 6cm and 8cm and height of the triangle corresponding to

6cm side is 4cm; find the area of triangle:

Solution: area of triangle= ½ base×height = ½ ×6×4 =12cm2

3.) Find the length and perimeter of the rectangle whose:

Area = 120cm2 and breadth= 8cm Solution: area of rectangle =120cm2 breadth b=8cm Area = l×b l ×8 =120 l = 15cm Perimeter = 2 (l + b) = 2 (15 + 8) = 2×23 = 46cm

4.) The parallel sides of a trapezium are 8.4cm and 12.3cm. If its height is 7.2 cm, find its

area.


Solution: sum of parallel sides = 8.4+ 12.3 = 20.7cm Height = 7.2cm Area of trapezium = ½ (sum of parallel sides) ×height = ½ × 20.7×7.2 = 74.52cm2

3-D FIGURES:

CUBOID:

h lllllllll b l Lateral / curved surface area = 2(l+b)×h Total surface area = 2(lb+bh+lh) Volume = lbh Diagonal of cuboid= √(l2+b2+h2) Where: l= length, b= breadth, h = height

CUBE:

l l

l

Lateral/curved surface area = 4l2

Total surface area = 6l2

Volume = l3

Diagonal of cube = l√3

Where: l = edge of cube


RIGHT CIRCULAR CYLINDER:

Lateral/curved surface area= 2πrh

Total surface area= 2πr(r+h)

Volume = πr2h

Where: r = radius

h = height

RIGHT CIRCULAR CONE:

Lateral/curved surface area = πrl

Total surface area = πr(l+r)

Volume = (1/3)πr2h

Where: r=radius of base

h=height

l=slant height = √(r2+h2)


SPHERE:

Lateral/curved surface area = 4πr2

Total surface area = 4πr2

Volume = (4/3)πr3 ,Where: r= radius of sphere

HEMISPHERE:

Lateral/curved surface area = 2πr2

Total surface area = 3πr2

Volume = (2/3)πr3 where: r= radius of hemisphere

SPHERICAL SHELL:

Total surface area = 4π(R2+r2)

Volume = (4/3)π(R3-r3)

Where: R= external radius

r = internal radius


EXAMPLES:

1.) Find the volume of right circular cylinder which has a height of 21cm, and the base

radius 5cm. Also find the curved surface area of the cylinder.

Solution: Given h=21cm, r=5cm Volume of cylinder=πr2h = (22/7) ×25×21 = 1650cm3

Curved surface area = 2πrh= 2× (22/7) ×5×21= 660cm2

2.) A rectangular sheet of paper 44cm × 18cm is rolled along its length and a cylinder is

formed. Find the volume of cylinder.(Use π=22/7)

Solution: The sheet of paper is rolled along the length. Then, height of the cylinder= 18cm and the circumference of base of the cylinder= 44cm Let the radius of the base of the cylinder. 2 ×(22/7) × r=44 r = 7cm and height = 18cm Volume of the cylinder = πr2h = (22/7)×7×7×18 = 2772cm3

3.) A beam 9m long, 40cm wide and 20cm deep is made up of iron which weighs 50kg per

cubic meters. Find the weight of the beam.

Solution: Volume of the beam = l×b×h = 9× (4/100) × (20/100)m3 = 72/100 m3 Since the beam is made up of iron which weighs 50kg per cubic metre. Weight of the beam = (72/100) × 50 = 36kg

4.) A rectangular reservoir is 120m long and 75m wide. At what speed per hour must water

flow into it through a square pipe of 20cm side so that the water rises by 24m in 18

hours?

Solution: Volume of accumulated in 18 hours = (120×75×24)cu.m. Volume of water accumulated in one hour = (120×75×24)/18 m3 Area of the end of the square pipe = (20/100)2 = (1/25) m2 = 0.4 m2 Speed of the water per hour = (120×75×24)/(18×0.4) = 30,000 m = 30 km Hence water must flow at 30km/hour into the reservoir.


Solved Examples (Mensuration)

1. The sum of the ages of 4 members of a family 5 years ago was 94 years. Today, when the daughter has been married off and replaced by a daughter-in-law, the sum of their ages is 92. Assuming that there has been on other change in the family structure and all the people-are alive, what is the difference in the age of the daughter and the daughter-in-law? (IBPS CWE PO MT 2012)

(1) 22 years (2) 11 years (3) 25 years (4) 19 years (5) 15 years Solution: There are four members in a family. Five years ago the sum of ages of the family members = 94 years Now, sum of present ages of family members = 94 + 5 4 = 114 years

Daughter is replaced by daughter-in-law. Thus, sum of family member's ages becomes 92 years.

Difference = 114 - 92 = 22 years Ans: (1) 2. The area of a square is 1444 square meters. The breadth of a rectangle is 1/4th the side of the square and the length of the rectangle is thrice the breadth. What is the difference between the area of the square and the area of the rectangle? (IBPS CWE PO MT 2012) (1) 1152.38 sq mtr (2) 1169.33 sq mtr (3) 1181.21 sq mtr (4) 1173.25 sq mtr (5) None of these Solution: Area of square = 1444 Let the side of square be a. So a2 = 1444

a= = 38 m Breadth of rectangle = 38 = 9.5 metres

Length = 3 x 9.5 = 28.5 m


Area of rectangle = 28.5 x 9.5 = 270.75 m2 Difference = 1444 - 270.75 = 1173.25 sq m

Ans: (4) 3. The side of a square is half the diameter of a circle. The area of the square is 1225 sq cm. What is the area of the circle? (1) 962.5 sq cm

(2) 3850 sq cm

(3) 15400 sq cm

(4) 15600 sq cm

(5) None of these

Solution: Side of the square

Radius of the circle = 35 cm

Area of the circle

Ans: (2)

4. The adjacent angles of a parallelogram are in the ratio of 5 : 7. The larger angle of the parallelogram is equal to the largest angle of a triangle. The smallest angle of the triangle is 40° less than the smaller angle of the parallelogram. What is the measure of the second largest angle of the triangle? (Corporation Bank PO 2011)

(1) 30°

(2) 40°

(3) 55°

(4) 35°

(5) None of these

Solution: Sum of adjacent angles of a parallelogram = 180˚

Larger angle of the parallelogram

Smaller angle of the parallelogram = 180˚ – 105˚ = 75˚

Second-largest angle of the triangle

= 180˚ – 105˚ – (75˚ – 40˚) = 40˚


Ans: (2)

5. The side of a square is half the diameter of a circle. The area of the square is 1225 sq cm. What is the area of the circle? (Corporation Bank PO 2011)

(1) 962.5 sq cm

(2) 3850 sq cm

(3) 15400 sq cm

(4) 15600 sq cm

(5) None of these

Solution: Side of the square

Radius of the circle = 35 cm

Area of the circle

Ans: (2)

6. The sum of the circumference of a circle and the perimeter a square is equal to 272 cm. The diameter of the circle is 56 cm. What is the sum of the area of the circle and the area of the square? (Allahabad Bank Probationary Officers Exam 2011)

(1) 2464 sq cm (2) 2644 sq cm (3) 3040 sq cm (4) Cannot be determined (5) None of these

Solution: Circumference of the circle = π x diameter

= × 56 = 176 cm

Perimeter of the square = (272 - 176 =) 96 cm

Side of the square = ( ) 24 cm

Area of the square = (24 x 24 =) 576 sq cm


Area of the circle = πr2 = x 28 x 28 = 2464 sq cm

required sum = (576 + 264) sq cm = 3040 sq cm

Ans: (3)

7. The ratio of the three angles of a quadrilateral is 13: 9: 5 respectively. The value of the fourth angle of the quadrilateral is 36°. What is the difference between the largest and the second smallest angles of the quadrilateral? (Allahabad Bank Probationary Officers Exam 2011)

(1) 104° (2) 108° (3) 72° (4) 96° (5) None of these

Solution: Let the three angles of the quadrilateral be13xo, 19xo and 5xo respectively.

Now, according to the question;

13x + 9x + 5x = 360 - 36 = 324

27x = 324

x = = 12

required difference = 13x - 5x = 8x = 8 x 12 = 96°

Ans: (4)


Practice Set (Mensuration)

1. The diameter of garden roller is 1.4

m and it is 2 m long. The area

covered by the roller in 5

revolutions is

a) 4.4 m2

b) 44 m2

c) 16.8 m2

d) 8.8 m2

2. In a cylindrical vessel of diameter

24 cm filled up with sufficient

quantity of water, a solid spherical

ball of radius 6 cm is completely

immersed. Then, the increase in

height of water level is

a) 1.5 cm

b) 2 cm

c) 3 cm

d) 4.2 cm

3. If the length of the diagonal of a

cube is , then its surface

area is

a) 192 cm2

b) 512 cm2

c) 768 cm2

d) 384 cm2

4. A solid sphere of radius 1 cm is

melted to convert into a wire of

length 100 cm. The

radius of the wire (take

) is

a) 0.08 cm

b) 0.09 cm

c) 0.16 cm

d) 0.11 cm

5. A field is in the form of a rectangle

of length 18 m and width 15 m. A

pit, 7.5 m long, 6 m broad and 0.8

m deep, is dug in a corner of the

field and the earth taken out is

evenly spread over the remaining

area of the field. The level of the

field raised is

a) 12 cm

b) 14 cm

c) 16 cm

d) 18 cm

6. A right circular cylinder, a

hemisphere and a right circular

cone stand on the same base and

have the same height. The ratio of

the volumes is

a) 3 : 6 : 1

b) 3 : 4 : 1

c) 3 : 2 : 1

d) 4 : 3 : 1

7. The base of a right pyramid is an

equilateral triangle of side 4 cm.

The height of the pyramid is half of

its slant height. Its volume is

a)

b)


c)

d)

8. A tent is of the shape of a right

circular cylinder upto a height of 3

m and then becomes a right circular

cone with maximum height of 13.5

m above the ground. If the radius of

the base is 14 m, the cost of

painting the inner side of the tent

at the rate Rs. 2 per sq m is

a) Rs. 2050

b) Rs. 2060

c) Rs. 2068

d) Rs. 2080

9. A solid cone of height 9 cm with

diameter of its base 18 cm is cut out

from a wooden solid sphere of

radius 9 cm. The percentage of

wood wasted is

a) 25%

b) 30%

c) 50%

d) 75%

10. The number of spherical bullets

that can be made out of a solid

cube of lead whose edge measures

44 cm, each bullet being of 4 cm

diameter, is

a) 2541

b) 2451

c) 2514

d) 2415

11. From a solid cylinder whose height

is 12 cm and diameter 10 cm, a

conical cavity of same height and

same diameter of the base is

hollowed out. The volume of the

remaining solid is approximately

a) 942.86 cm3

b) 314.29 cm3

c) 628.57 cm3

d) 450.76 cm3

12. The base of a right pyramid is a

square of side 16 cm long. If its

height be 15 cm, then the area of

the lateral surface (in cm 2) is

a) 136

b) 544

c) 800

d) 1280

13. The curved surface area of a

cylindrical pillar is 264 sq m and its

volume is 924 cu m. The ratio of its

diameter to height is

a) 3 : 7

b) 7 : 3

c) 6 : 7

d) 7 : 6

14. The ratio of the volume of a cube

and of a solid sphere is 363 : 49.

The ratio of an edge of the cube

and the radius of the sphere is

a) 7 : 11

b) 22 : 7

c) 11 : 7

d) 7 : 22


15. The base of a solid right prism is a

triangle whose sides are 9 cm, 12

cm and 15 cm. The

height of the prism is 5 cm. Then,

the total surface area of the prism

is

a) 180 cm2

b) 234 cm2

c) 288 cm2

d) 270 cm2

16. Volume of two cones are in the

ratio 1 : 4 and their diameters are in

the ratio 4 : 5. The ratio of their

heights is

a) 1 : 5

b) 5 : 4

c) 5 : 16

d) 25 : 64

17. Water is being pumped out through

a circular pipe whose internal

diameter is 7 cm, if the flow of

water is 12 cm/s then how many

litres of waters being pumped out

in one hour?

a) 1663.2 L

b) 1500 L

c) 1747.6 L

d) 2000 L

18. A right circular cylinder of height 16

cm is covered by a rectangular tin

foil of size

16 cm × 22 cm. The volume of the

cylinder is

a) 352 cm3

b) 308 cm3

c) 176 cm3

d) 616 cm3

19. The volume of a cube (in cm3),

whose diagonal measures cm

is

a) 16

b) 27

c) 64

d) 8

20. Some solid metallic right circular

cones, each with radius of the base

3 cm and height 4 cm, are melted to

form a solid sphere of radius 6 cm.

The number of right circular cones

is

a) 12

b) 24

c) 48

d) 6

21. A cuboidal water tank contains 216

L of water. Its depth is of its

length and breadth is of of the

difference between length and

depth. The length of the tank is

a) 72 dm

b) 18dm

c) 6 dm

d) 2 dm


Mensuration Practice Set (Answers)

1) b

2) b

3) d

4) d

5) c

6) c

7) c

8) c

9) d

10) a

11) c

12) b

13) b

14) b

15) c

16) d

17) a

18) d

19) c

20) b

21) b


Chapter: Permutations, Combinations & Probability

Permutation

The arrangement of a number of things taking some or all of them at a time is called permutation. If there are ‘n’ number of things and we have to select ‘r’ things at a time then

the total number of permutation is denoted by n =

For example if there are 3 candidates A ,B and C for the post of president and vice president of a college , since we have to select only 2 candidates , it can be done in 3! Ways. i.e. (A, B) (B, C) (A, C) (B, A) (C, B) and (C, A). Here order of arrangement matters.

Restricted Permutation:

Sometimes we have to find out the number of permutation keeping few specific objects at specific places. In this case, we find out the number of permutation of filling remaining vacant places by the remaining objects.

If r objects are taken out of n dissimilar objects

(i) A specific object is taken each time: if there are n objects .

Suppose that is taken each time. If takes first place then the remaining (n-1)

objects can be arranged in n-1 ways. Since can take any place so number of

permutation is r n-1 .

(ii) Specific object never taken: then r objects are taken out of (n-1) objects, so number of

permutation is n-1 .

{Note: n = n-1 + r n-1 }

Permutation of things when some are identical:

If we have n things in which p are exactly of one kind , q of second kind , r of third kind and the

rest are different then the number of permutation of n things taken all at a time n =


Example: In how many ways can the letters of the word LEADER be arranged?

Solution: The word LEADER contains total 6 letters namely 1L, 2E, 1A, 1D, 1R

Therefore, the number of ways to arrange the letters of the word LEADER

= = 360.

Repetition of things:

The number of permutation formed by taking r things at a time out of n things in any object

arrangement such that each object can be taken any number of time is .

Circular permutation:

If we fix one of the objects around the circumference of a circle then number of permutation of n different thing taken all at a time is (n-1)! Ways. It will be same as by putting (n-1) objects at (n-1) places.

But if we do not consider the direction i.e. clockwise and anticlockwise then the number of

permutation is .

Combination

From a given group of object each of the number of groups which are formed by taking some objects or all objects at a time without caring about the sequence of the objects is called combination. The number of combination formed by taking r objects at a time out of n object is

denoted by n where C expresses combination.

n =

For example if we have 3 objects A , B and C , 2 objects are taken out at a time then 3 combination are formed AB , BC and CA.

Note:

If r= 0 , then n = = 1


If r= 1 , then n = = n

If r= n , then n = = 1

n = n

Example: Find the value of .

Solution: We have, = = = 100 {Because 1! =1}

Restricted combination:

The combination of r object out of n objects on which p specific objects:

1. Are always included is n-p . We have to keep aside p specific objects and to select

remaining (n-p).

2. Are never included is n-p . Since p specific object are never included we have to form

the combination taking r obects out of (n-p) objects.

The number of ways to select some or all thing out of any number of given thing:

There are 2 ways to select anything i.e. either it will be selected or not. Therefore number of

ways to select n things is 2 2 …………… n times = . In these empty selection is also include.

For non- empty selection is -1.

Note: n + n +…………………………. + n = -1.


Difference between permutation and combination:

Suppose there are 5 objects out of which 2 have to be chosen.

Permutation Combination

Number of required way

=

= = 5 4 = 20

=

= = = 10

So it is clear that in permutation order matters while in combination order does not matter.

Probability

The mathematical measure of the uncertainty is called probability. For example, consider the following questions:

(a) Will it rain today? (b) Which of the three candidates will win? (c) On throwing a dice, the number obtained will be even or odd? (d) On tossing a coin, head will occur or tail will occur?

The answer to all these question is not sure i.e. there is uncertainty .We study the uncertainty of the result of such question in the theory of probability , which may not have one result but more than one result are possible .

Random experiment:

The experiments in which the outcomes cannot be predicted before hand is called random experiments. When these kind of experiment are repeated under identical condition, they do not produce the same outcome every time and there may be many possible outcome which depends upon chance and cannot be predicted. For example, on tossing a coin either the head will come up or the tail will come up, we cannot predict it. This is an example of random event.

Sample Space:

The set of all possible outcomes of experiments is called the sample space and it is denoted by S. And the subset of a sample space is called an event. That is, every subset A of the sample


space S is an event of that random experiment. For example, in an experiment of tossing a coin,

if h is obtained then it is a random event, since here S = {H, T} and {H} S

Now, the probability of any event A can be defined as the ratio between the number of favourable outcomes to the event A and the number of total equiprobable outcomes, that is

P(A) =

Here it should be noted that the probability of a certain or sure event is 1 and that of impossible event is 0.

Now, since the probability of an event to occur is =

So the probability of an event A not to occur is = 1 –

Mutually Exclusive events:

Two events A and B are said to be mutually exclusive if they cannot occur together, that is simultaneously. For example , on throwing a dice , the events A = { 2,4,6 } and B = { 1,3,5 } are

mutually exclusive events , i.e. A B = .

In term of probability if A and B are mutually exclusive events, then

P (A B) = P (A) + P (B) and,

P (S) = P (A) + P (A’) = 1 where A’ is Complement of A.

Example: If a dice is thrown once then the probability of the number appearing on dice is more than 2?

(a) 1/3 (b) 1/2 (c) 2/3 (d) 4.3/4

Solution: As there are 6 faces on a dice, So the total number of possible events are 1, 2 , 3 ……. 6 , that is = 6 Now the number more than are 3, 4 , 5 and 6 So total number of favourable events =


Probability of an event =

Required probability = =

Example : An urn contains 3 green, 6 red, and 4 black balls. 3 balls are drawn. Find the probability that all 3 balls are of same colour?

(a) 3/44 (b) 25/286 (c) 15/286 (d) 5/286

Solution: Total number of balls in an urn is 13.

Number of ways 3 balls can be drawn out of 13 balls = = = 286

Numbers of ways 3 green balls are drawn = = = 1

Numbers of ways 6 red balls are drawn = = = 20

Numbers of ways 4 black balls are drawn = = = 4

Now,

The required probability = + + =


Solved Examples (Permutations, Combinations & Probability)

1. In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together? (IBPS CWE PO MT 2012) (1) 720 (2) 1440 (3) 5040 (4) 3600 (5) 4800 Solution: Total number of letters is 7, and these letters can be arranged in 7! ways . = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 ways There are seven letters in the word THERAPY including 2 vowels. (E, A) and five consonants. Consider two vowels as one letter. We have 6 letters which can be arranged in 6P6 = 6 ways. But vowels can be arranged in 2! ways. Hence, the number of ways, all vowels will come together = 6! x 2! = 1 x 2 x 3 x 4 x 5 x 6 x 2 = 1440 Total number of ways in which vowels will never come together = 5040 - 1440 = 3600 Ans: (4) 2. A bag contains 13 white and 7 black balls. Two balls are drawn at random. What is the probability that they are of the same colour? (IBPS CWE PO MT 2012) (1)

(2)

(3)

(4)


(5)

Solution: Total number of balls = 13 + 7 = 20 Number of sample space = n(S) = 20C2 = 190 Number of events = n(E) = 13C2 + 7C2 = 78 + 21 = 99

P(E) = =

Ans: (4) Directions (1-5): Study the given information carefully to answer the questions that follow. An urn contains 4 green, 5 blue, 2 red and 3 yellow marbles.

3. If two marbles are drawn at random, what is the probability that both are red or at least one is red?

(1)

(2)

(3)

(4)

(5) None of these

Solution: Total number of marbles in the urn

= 4 + 5 + 2 + 3 = 14

Total number of possible outcomes

= Selection of 2 marbles out of 14 marbles

= 14C2= 91

Total number of favourble cases

= 2C2 + 2C1 + 12C1 = 1 + 2 x 12 = 25


required probability =

Ans: (5)

4. If three marbles are drawn at random, what is the probability that at least one is yellow?

(1)

(2)

(3)

(4)

(5) None of these

Solution: Total number of possible outcomes

= 14C3 = = 364

When no marbles is yellow, favourable number of cases

= 14C3 = = 165

Probability that no marble is yellow =

required probability = (Probability that at least one is yellow) = (1 - Probability that no marble is yellow)

1- = =

Ans: (3)


5. If eight marbles are drawn at random, what is the probability that there are equal numbers of marbles of each colour?

(1)

(2)

(3)

(4)

(5) None of these

Solution: Total possible outcomes = 14C8 = 14C6 [ nCr = nCn-r]

= 3003

Total number of favourable cases

= 4C2 x 5C2 x 2C2 X 3C2 = 6 x 10 x 1 x 3 = 180 = 3003

required probability = =

Ans: (3)

6. If three marbles are drawn at random, what is the probability that none is green?

(1)

(2)

(3)

(4)


(5)


= 14C2 = = 364

Now, according tot the question, no marble should be green.

Total number of favourable outcomes

= Selection of 3 marbles out of 5 blue, 2 red and 3 yellow marbles

= 10C3 = = 120

required probability = =

Ans: (5)

7. If three marbles are drawn at random, what is the probability that two are blue and two are red?

(1)

(2)

(3)

(4)

(5) None of these


= 14C4 = = 1001

Total number of favourable cases


= 5C2 x 22C = 10 x 1 = 10

required probability =

Ans: (1)

Directions (Q. 8-10): Study the given information carefully and answer the questions that follow:

A basket contains 4 red, 5 blue and 3 green marbles.

8. If three marbles are picked at random, what is the probability that either all are green or all are red?

(1)

(2)

(3)

(4)

(5) None of these

Solution: P(All Green) + P(All Red) = 3C3 / 12C3 + 4C3 / 12C3 = 1/44 Ans : (4)

9. If two marbles are drawn at random, what is the probability that both are red?

(1)

(2)

(3)

(4)

(5) None of these


Solution: 4C2 / 12C2 = 1/11 Ans: (5)

10. If three marbles are picked at random, what is the probability that at least one is blue?

(1)

(2)

(3)

(4)

(5) None of these

Solution: 1 – P (None Blue)

= 1- (7C3 / 12C3) = 37/44

Ans: (2)


Practice Set-1 (Permutations, Combinations &

Probability)

1. In how many ways can six different

rings be worn on four fingers of one

hand?

a) 10

b) 12

c) 15

d) 16

2. There are three prizes to he

distributed among five students. If

no student gets more than one

prize, then this can be done in

a) 10 ways

b) 30 ways

c) 60 ways

d) 80 ways

3. In a hockey championship, there

were 153 matches played. Every

two teams played one match with

each other. The number of teams

participating in the championship is

a) 18

b) 19

c) 17

d) 16

4. In an examination paper, there are

two groups each containing 4

questions. A candidate is required

to attempt 5 questions but not

more than 3 questions from any

group. In how many ways can 5

questions be selected?

a) 24

b) 48

c) 96

d) None of these

5. After a get together every person

present shakes the hand of every

other person. If there were 105

hands shakes in all, how many

persons were present in the party?

a) 14

b) 13

c) 15

d) 16

6. There are 4 candidates for the post

of a lecturer in Mathematics and

one is to be selected by votes of 5

men. The number of ways in which

the votes can be given is

a) 1048

b) 1072

c) 1024

d) none of these

7. A student is to answer 10 out of 13

questions in an examination such

that he must choose at least 4 from

the first five questions. The number

of choices available to him is

a) 140

b) 280

c) 196

d) 346


8. After a get-together every person present shakes the hand of every other person. If

there were 105 hands shakes in all, how many persons were present in the party?

a) 14

b) 13

c) 15

d) 16

9. Out of eight crew members three particular members can sit only on the left side.

Another two particular members can sit only on the right side. Find the number of

ways in which the crew can be arranged so that four men can sit on each side.

a) 864

b) 865

c) 863

d) 1728


Practice Set-2 (Permutations, Combinations &

Probability)

1. Four different objects 1, 2, 3, 4 are

distributed at random in four places

marked 1, 2, 3, 4. What is the

probability that none of the objects

occupy the place corresponding to

its number?

a)

b)

c)

d)

2. Two dice are tossed. The

probability that the total score is a

prime number is

a)

b)

c)

d)

3. A bag contains 3 white balls and 2

black balls. Another bag contains 2

white balls and 4 black balls. A bag

and a ball are picked at random.

The probability that the ball will be

white is

a)

b)

c)

d)

4. Suppose six coins are flipped. Then

the probability of getting at least

one tail is

a)

b)

c)

d)

5. A bag contains 2 red, 3 green and 2

blue balls. 2 balls are to be drawn

randomly. What is probability that

the balls drawn contain no blue

ball?

a)

b)

c)

d)


6. I forgot the last digit of a 7-digit

telephone number. If I randomly

dial the final 3 digits after correctly

dialling the first four, then what is

the chance of dialling the correct

number?

a)

b)

c)

d)

7. A box contains 6 white balls and 7

black balls. Two balls are drawn at

random. What is the probability

that both are of the same colour?

a)

b)

c)

d)

8. A brother and sister appear for an

interview against two vacant posts

in an office. The probability of the

brother's selection is and that of

the sister's selection is . What is

the probability that one of them is

selected?

a)

b)

c)

d)

9. A room has 3 lamps. From a

collection of 10 light bulbs of which

6 are not good, a person selects 3 at

random and puts them in a socket.

The probability that he will have

light, is

a) 5/6

b) 1/2

c) 1/6

d) none of the above

10. Four boys and three girls stand in

queue for an interview. The

probability that they will stand in

alternate positions is

a) 1/34

b) 1/35

c) 1/17

d) 1/68


Permutations, Combinations & Probability Practice Set-1

(Answers)

1) c

2) a

3) a

4) b

5) c

6) d

7) a

8) c

9) d


Permutations, Combinations & Probability Practice Set-2

(Answers)

1) c

2) b

3) d

4) c

5) b

6) d

7) b

8) b

9) d

10) a


Chapter: Data Interpretation

Introduction:

Analyzing data is the major part of our daily routine. Financial data as in P & L sheet, marketing

and sales data, data on productivity, data on performance appraisal, data on each and every

thing that we can imagine. In some companies entire department generates and manages every

conceivable data that we can imagine.

To collect information from all the amount of data it needs to be presented in a lucid and

concise manner. Therefore we use data representation as it immediately provides the overall

scenario and it is also sufficient to compute any detailed information.

Strategies:

While studying the DI section one should follow few strategies given below:

1. Solve the graph with which you are most comfortable for example some are more

comfortable with line graph while the other may be with pie chart.

2. The problems with numbers with 2 or 3 digits like 82, 114 etc are easier to solve then

the problem with number of 4 or 5 digits like 3457, 52468.

3. In pie charts if sectors are 15% 20% etc then it will be easier calculation then the sectors

like 17.5%, 23.6 % etc.

4. We must also look at the number in answer option, if the answers are 12.5 5 6.33% etc

then it will be easier to arrive at.

5. Wider are the choices in answer, easier will be the elimination process and lesser will be

the calculation.

6. The answer choice “ Cannot be determined “ makes the question much easier as it

becomes data sufficiency question and if the question can be answered then only four

options are left.


7. Pick the alternative which is the middle one and check if your answer is less than, equal

to or greater than this value. In this way only one iteration will give you the correct

answer.

8. Te answer choice “None of these “on the other hand make it more difficult as we have

to calculate the question to the exact value.

9. It is better to solve a line chart with two lines and 5 points than to solve a table with 5

rows and 6 columns. But this may not be the case always looking at the other factors.

Growth rate and Growth:

Growth and Growth rate are two different cases. Growth refers to just increase in the

underlying value, while the Growth rate refers to the percentage increase. For example

following table shows the sales and profit of a company A in Rs. Lakhs

1999 2000 2001 2002 2003

Sales 240 290 320 350 380

Profit 50 55 70 50 40

Here,

Growth during the period from 2000 to 2002 is simple the difference of sales = 350-290 = 60

But the Growth rate during this period is the percentage of growth =

Profit Percentages:

The profit percentage is not but it is . For example in above

table the profit percentage of year 1999 is not = .


But the profit percentage of year 1999 is = as cost price of this year is 190

Also the percentage change in profit percentage is percent of profit percent of two years with

base as previous year profit percent. For example the percentage change in profit percentage in

the year 2002 over that in year 2001 is:

=

=

But in line graph growth rate is related to slope of line. There are few points to understand:

1. The slope of each segment is same across all years of company B but still the growth

rate is not the same in all years. The same slope simply means the sales grow by a

constant amount each year. But the growth rate depends on the base value so in

company B it’s decreasing from 2000 to 2004.

2. The line representing sales of A in 2002-03 is steeper but 2003 is not the year with

highest growth rate of sales of A . The highest growth rate of sale of A is in year 2001.


Chapter: Data Interpretation-Table Chart

In studying problems on statistics, the data collected by the investigator are arranged in systematic form, called the tabular form. In order to avoid some heads again and again, we make tables, consisting of horizontal lines called rows and vertical lines called columns with distinctive heads, known as captions. Units of measurements are given along with the captions.

Example:

The table given below shows the population, literates and illiterates (in thousands) and the percentage of literacy in 3 states, in a year:

State Population Literates Illiterates Percentage of literacy

Madras 49342 6421 …………….. ……………………

Bombay ……………. 4068 16790 ……………………..

Bengal 60314 …………… …………… 16.1

After reading the table, mark a tick (√) against the correct answer in each question given below and hence complete the table.

1) Percentage of literacy in Madras is

(a) 14.9%

(b) 13.01%

(c) 12.61%

(d) 15.04%

2) Percentage of literacy in Bombay is

(a) 19.5%

(b) 16.7%

(c) 18.3%

(d) 14.6%


3) Number of literates in Bengal (in thousands) is:

(a) 50599

(b) 9715

(c) 76865

(d) 9475

Solution:

1) (b) percentage of literacy in Madras = (6421/49342)×100% = 13.01%

2) (a) Population of Bombay = (4068+16790) thousands = 20858 thousands.

Therefore, percentage of literacy in Bombay = (4068/20858)×100% = 19.5% 3) (b) Number of literates in Bengal =(16.1/100)×60314 = 9715 thousands


Solved Examples (Data Interpretation-Table Chart)

Direction (1-5): Study the table carefully to answer the questions that follow :- (IBPS PO Exam 2011)

Number of people visiting six different Super-markets and the percentage of Men, Women and Children visiting those Super-markets

Names of the Total Number of Percentage of

Super- markets

People Men Women Children

A 34560 35 55 10

B 65900 37 43 20

C 45640 35 45 20

0 55500 41 26 33

E 42350 06 70 24

F 59650 24 62 14


1. The Number of men visiting Super-market D forms approximately what percent of the total number of people visiting all the Super-markets together?

(1) 11 (2) 5.5 (3) 13 (4) 9 (5) 7.5

Solution: Number of men visiting supermarket D

= 41% of 55500

= = 41×55500 = 22755

Total number of people visiting all the supermarkets together

= 34560 + 65900 + 45640 + 55500 + 42350 + 59650

= 303600

required probability = × 100 = 7.5% (Aprox)

Ans: (5)

2. The Number of children visiting Super-market C forms what percent of number of children visiting Super- market F? (rounded off to two digits after decimal)

(1) 91.49 (2) 49.85 (3) 121.71 (4) 109.30 (5) None of these

Solution: Number of children visiting supermarkets C

= 20% of 45640 = 20 x 45640 = = 9128

Number of children visiting supermarket F

= 14% of 59650 = = 8351


required percentage = x 100 = 109.30%

Ans: (4)

3. What-is the total number of children visiting Super-markets B and D together?

(1) 18515 (2) 28479 (3) 31495 (4) 22308 (5) None of these

Solution: Total number of children visiting supermarket B and D together

= 20% of 65900 + 33% of 55500

= +

= 13180 + 18315 = 31495

Ans: (3)

4. What is the average number of women visiting all the Super-markets together?

(1) 24823.5 (2) 22388.5 (3) 26432.5 (4) 20988.5 (5) None of these

Solution: Total number of women = 55% of 34560 + 43% of 65900 + 45% of 45640 + 26% of 55500+ 70% of 42350 + 62% of 59650

= 19008 +'28337 + 20538 + 14430 + 29645 + 36983

= 148941

required average = = 24823.5

Ans: (1)


5. What is the ratio of number of women visiting Super-markets A to that visiting Super- market C?

(1) 35: 37 (2) 245: 316 (3) 352: 377 (4) 1041: 1156 (5) None of these

Solution: Required ratio = 19008: 20538 = 1056: 1141

Ans : (3)

(Directions Q.6-10): Study the table carefully to answer the questions that follow :- (IBPS PO Exam 2011)

Percentage of Marks Obtained by Different Students in Different Subject of MBA

SUBJECTS (Maximum Marks)

Students Strategic Brand Compensation Consumer Service Training &

Management Management Management Behaviour Marketing Development

(150) (100) (150) (125) (75) (50)

, Anushka 66 75 88 56 56 90

Archit 82 76 84 96 92 88

- - , - Arpan 76 66 78 88 72 70

Garvita 90 88 96 76 84 86

Gunit 64 70 68 7.2 68 , 74

Pranita 48 56 50 64 64 58

••••

• 1;;' ~ ,'_

6. How many marks did Anushka get in all the Subjects together?

(1) 369 (2) 463 (3) 558 (4) 496 (5) None of these

Solution: Total marks of Anuska

= + 75 + + + + 45


= = 99 + 75 + 132 + 70 + 42 + 45 = 463

Ans: (2)

7. The Marks obtained by Garvita in Brand Management are what percent of marks obtained by Archit in the same Subject? (rounded off to two digits after decimal)

(1) 86.36 (2) 101.71 (3) 115.79 (4) 133.33 (5) None of these

Solution: Marks obtained by Garvita in Brand Management

= 88% of 100 = 88

Marks obtained by Archita in Brand Management

= 76% of 100 = 76

required percentage = x 100 115.79%

Ans: (3)

8. What is the average marks obtained by all students together in Compensation Management?

(1) 116 (2) 120 (3) 123 (4) 131 (5) None of these

Solution: Average marks obtained by all students together in Compensation Management

=


= × 150=116

Ans: (1)

9. Who has scored the highest total marks in all the subjects together?

(1) Archit (2) Gunit (3) Pranita (4) Garvita (5) Arpan

Solution: Total obtained in all the subjects together by

Arapn: 76% of 150 + 66% of 100 + 78% of 150 + 88% of 125 + 72% of 75 + 70% of 50

= + + + + +

= 114 + 66 + 117 + 110 + 54 + 35 = 496

Archit: 82% of 150 + 76% of 100 + 84% of 150 + 96% of 125 + 92% of 75 + 88% of 50

= + + + + +

= 123 + 76 + 126 + 120 + 69 + 44 = 558

Garvita: 90% of 150 + 88% of 100 + 96% of 150 + 76% of 125 + 84% of 75 + 86% of 50 .

= 135 + 88 + 144 + 95 + 63 + 43 = 568

= + + + + +

= 135 + 88 + 144 + 95 + 63 + 43 = 568

Gunit: 64% of 150 + 70% of 100 + 68% of 150 + 72% of 125 + 68% of 75 + 74% of 50

= + + + + +

= 96 + 70 + 102 + 90 + 51 + 37 = 446

Pranita: 48% of 150 + 56% of 100 + 50% of 150 + 64% of 125 + 64% of 75 + 58% of 50


= + + + + +

= 72 + 56 + 75 + 80 + 48 + 29 = 360

Clearly, Garvita scored the highest total marks in all the subjects together.

Ans: (4)

10. How many Students have scored the highest marks in more than one Subject?

(1) three (2) two (3) one (4) none (5) Now of these

Solution: Archit (consumer behaviour and service marketing) and Garvita (strategic management, brand management and compensation management).

Ans: (2)


Practice Set (Data Interpretation-Table Chart)

Directions (Q. 1-5): Study the table carefully to answer the questions that follow

Number of cars (in thousand) of two models (Basic and Premium) produced by five different companies in five different years (IBPS RRB Grade Officer Exam 2012)

Company

→

A

B C D E

Year

↓

Basic

Premium

Basic

Premium

Basic

Premium

Basic

Premium

Basic

Premium

2006 4.4 2.5 5.6 2.4 5.4 6.1 7.6 7.5 2.7 5.1

2007 4.9 7.2 9.4 7.2 7.5 8.3 8.4 4.9 4.2 5.5

2008 13.6 15.5 14.8 9.5 12.8 9.9 9.2 8.2 7.7 11.5

2009 6.6 13.9 11.8 11.4 16.6 18.2 10.6 10.4 7.2 12.8

2010 5.8 14.9 12.2 7.2 19.9 22.3 14.6 12.2 13.2 12.2

1. The number of cars of premium model produced by Company D in the year 2009 was approximately what per cent of the total number of cars (both models) produced by Company C in the year 2007?

(1) 70 (2) 51 (3) 56 (4) 61 (5) 66


2. What was the approximate percentage decrease in the number of cars of basic model

produced by Company B in the year 2009 as compared to the previous year?

(1) 15

(2) 20

(3) 10

(4) 80

(5) 85

3. What was the average number of cars of premium model produced by Company A over all

the years together?

(1) 9000

(2) 8000

(3) 6000

(4) 48000

(5) None of these

4. In which year was the difference between the basic model and the premium model of cars

produced by Company E the second highest?

(1) 2010

(2) 2006

(3) 2007

(4) 2008

(5) 2009

5. In which company did the production of cars of premium model consistently increase from

the year 2006 to the year 2010?

(1) Both C and E

(2) Both C and d

(3) C only

(4) D only

(5) E only


Directions (Q. 6-10): Study the table carefully to answer the questions that follow.

Number of animals in grasslands of four different countries in five different years (RBI Grade’B’ Officer’s Exam 2011)

Year

Country

South Africa China Sri Lanka England

Tiger Lion Bear Tiger Lion Bear Tiger Lion Bear Tiger Lion Bear

1990 145 156 250 320 346 436 280 468 255 423 342 234

1995 134 165 354 445 256 542 354 354 343 368 136 345

2000 120 135 324 583 325 454 433 345 545 354 267 456

2005 110 184 285 466 475 322 343 324 546 562 235 567

2010 160 224 264 411 535 534 535 532 453 349 345 324

6. What is the average of the number of tigers in the grassland of Sri Lanka over all the years together?

(1) 386 (2) 389 (3) 369 (4) 276 (5) None of these

7. What is the difference between the total number of lions and bears in the grassland of England in the year 2005 and the number of tigers in the grassland of South Africa in the year 1995?

(1) 597 (2) 558 (3) 677 (4) 668 (5) None of these


8. The total number of animals together in the grassland of China in the year 1990 is approximately what per cent of the total number of bears in the grassland of Sri Lanka over all the years together?

(1) 44% (2) 56% (3) 41% (4) 47% (5) 51%

9. If 35 per cent of the total number of animals in the grassland of China in the year 2010 died due to an epidemic, how many animals remained in the grassland of China in the year2010?

(1) 976 (2) 952 (3) 986 (4) 962 (5) None of these

10. What is three-fourths of the total number of lions in the grasslands of all the four countries in the year 2000?

(1) 848 (2) 868 (3) 804 (4) 824 (5) None of these


Directions (Q. 11-5): Study the table carefully to answer the questions that follow:

Number of girls and boys (in hundreds) in six different years in five different schools

School→

Years

↓

A B C D E

Boys Girls Boys Girls Boys Girls Boys Girls Boys Girls

2005 3.3 3.6 5.2 3.1 5.5 4.5 2.4 1.4 6.5 6.6

2006 6.6 4.2 4.9 2.2 6.9 3.3 4.4 2.3 5.5 3.6

2007 9.3 6.9 4.7 4.2 5.8 4.9 6.4 3.3 2.7 2.4

2008 5.4 9.6 6.3 5.4 6.6 5.2 5.3 5.4 5.4 5.7

2009 8.4 12.9 7.5 5.9 8.7 6.6 12.1 5.2 6.8 6.5

2010 12.3 14.4 9.8 4.4 11.7 4.2 12.2 9.4 10.8 12.7

11. What is the approximate percentage decrease in the number of boys in School D in the year 2008 as compared to that in the previous year?

(1) 17 (2) 12 (3) 9 (4) 5 (5) 23

12. The number of girls in School B in the year 2009 is approximately what per cent of the total number of students (both boys and girls) in School E in the year 2006?

(1) 46 (2) 52 (3) 70 (4) 58 (5) 65

13. What is the average number of girls in School A in all the years taken together?

(1) 760 (2) 800 (3) 860


(4) 600 (5) None of these

14. What is the ratio of the number of boys in School C in the year 2009 to the number of girls in School A in the year 2009?

(1) 29 : 41 (2) 36 : 11 (3) 29 : 43 (4) 36: 13 (5) None of these

15. In which year is the total number of students (both girls and boys together) the third highest in School E?

(1) 2006 (2) 2007 (3) 2008 (4) 2005 (5) 2010


Directions-(Q. 16-20) Study the table carefully to answer the questions that follow:

Number of Athletes (in hundreds) who participated in a Sports Event from Five Different Countries over the years (Allahabad Bank Probationary Officers Exam 2011)

Countries

Year

A B C D E

Male Fem Male Fem Male Fem Male Fem Male Fem

2005 4.4 3.3 6.3 4.2 4.5 3.1 5.6 4.1 4.7 2.1

2006 6·6 4·2 8·4 6·2 6·9 3·3 8·4 6·3 7·8 5·2

2007 4·6 1·8 7·4 4·8 4·8 2·8 9·3 7·3 8·7 6·5

2008 9.6 4·9 11.4 8·4 6·6 4·2 12·6 9·4 8·9 5·8

2009 11.8 6·4 10·6 5·2 7·9 6·3 14.4 10·2 11·8 9·2

2010 8·2 5·2 6·4 7·2 10·8 6·9 15·6 12·1 13·6 9·8

16. In which of the following years was the total number of participants (athletes) second highest from Country C?

(1) 2005 (2) 2006 (3) 2007 (4) 2008 (5) None of these

17. What was the average number of female athletes who participated from Country B over all the years together?

(1) 1200 (2) 400 (3) 600 (4) 1800 (5) 3600


18. What was the approximate percentage decrease in the number of male athletes who participated from Country C in the year 2007 as compared to the previous year?

(1) 21 (2) 30 (3) 35 (4) 39 (5) 25

19. The Number of female athletes who participated from Country E in the year 2009 was approximately what percentage of the 42 total number of athletes who participated from Country-B in the year 2008?

(1) 40 (2) 46 (3) 50 (4) 56 (5) 60

20. In which of the following country is the difference between the number of male and female participants second highest in the year 2006?

(1) A (2) B (3) C (4) D (5) E


Data Interpretation-Table Chart Practice Set (Answers)

1) 5

2) 2

3) 5

4) 5

5) 3

6) 2

7) 4

8) 5

9) 4

10) 3

11) 1

12) 5

13) 3

14) 3

15) 5

16) 5

17) 3

18) 2

19) 2

20) 5


Chapter: Data Interpretation-Line Graphs

Line graphs of a frequency distribution is obtained from the histogram of the frequency distribution by joining the mid points of respective tops of the rectangles in a histogram.

To complete the line graphs, the mid-points at each end are joined to the immediately lower or higher mid-points (as the case may be) at zero frequency.

Example: Study the following graph and answer the following questions:

1.) The total expenditure of which of the following pairs of years was equal to the income in 1992?

(a) 1987 and 1988 (b) 1987 and 1989 (c) 1988 and 1989 (d) 1988 and 1990 (e) none of these

2.) What was the percentage decrease in expenditure from 1988 and 1989?


(a) 80 (b) 50 (c) 40 (d) 10 (e) none of these

3.) In how many of the given years was the expenditure more than the average expenditure of the given years?


4.) In which of the following years was the percentage of expenditure to income, the highest?


5.) What was the approximate percentage increase in income from 1991 to 1992?

(a) 35 (b) 40 (c) 20 (d) 15 (e) 25

Solution:

1.) (c) : income in 1992= 475 crores

Total expenditure in 1988 and 1989 = Rs.( 250+225) crores

=Rs. 475 crores.

2.) (d) : expenditure in 1988 = Rs. 250 crores

Expenditure in 1989 = Rs. 225 crores.

Decrease % = = 10%

3.) (b): average expenditure =Rs.

=Rs. 287.5

The expenditure is greater than the average expenditure during the years 1987, 1990 and 1992.

Required no. of years =3

4.) (e): the required percentage :

In 1987 is (300×100/450)% = 66.66%

In 1988 is (250×100/400)% = 62.5%

In 1989 is (225×100/350)% = 64.29%


In 1990 is (375×100/425)% = 88.24%

In 1991 is (175×100/375)% = 46.6%

In 1992 is (400 ×100/475)% = 84.21%

Clearly the percentage is highest in 1990.

5.) (e): income in 1991 = 375 crores

Income in 1992 = 475 crores

Therefore increase % = (100×100/375)% = 26.6% = 25% nearly


Solved Examples (Data Interpretation-Line Graphs)

Directions-(Q.1-5): Study the following graph carefully to answer the questions that follow: (Allahabad Bank Probationary Officers Exam 2011)

1. In which state was the total number of trees planted by NGO A and NGO B together second lowest?

(1) Bihar (2) Punjab (3) Haryana (4) Assam (5) Tamil Nadu

Solution: Number of tree planted by NGO-A and NGO-B together in

Bihar: 100 + 60 = 160

Punjab: 120 + 80 = 200

Haryana: 80 + 140 = 220

Assam: 150 + 160 = 310


Tamil Nadu: 140 + 180 = 320

Ans: (2)

2. What was the difference between the trees planted by NGO A in Haryana and the number of trees planted by NGO C in Tamil Nadu?

(1) 90 (2) 60 (3) 120 (4) 100 (5) None of these

Solution: Required difference = 160 – 80 = 80

Ans: (5)

3. What was the average number of trees planted in Haryana by all the NGOs together?

(1) 420 (2) 140 (3) 120 (4) 390 (5) None of these

Solution: Required average = = 129

Ans: (5)

4. The total number of trees planted by NGO A and NGO B together in Bihar was approximately what per cent of the total number of trees planted by NGO-B and NGO-C together in Punjab?

(1) 85 (2) 90 (3) 105 (4) 110 (5) 95

Solution: Required percentage = × 100 = × 100 95%

Ans: (5)


5. What was the ratio of the number of trees planted by NGO B in Tamil Nadu, number of trees planted by NGO C in Assam and the number of trees planted by NGO A in Assam?

(1) 5: 3 : 6 (2) 5: 6 : 3 (3) 6: 4 : 5 (4) 6: 5 : 3 (5) None of these

Solution: Required ratio: 180: 120 + 150 = 6: 4: 5

Ans: (3)

Direction (Q. 6 – 10): Study the graph carefully and answer the questions that follow:

Per cent profit made by two companies over the years

6. If in the year 2006 the expenditures incurred by company P and Q were same, what was

the ratio of the income of company Q to that of company P in that year?

(a) 26 : 27

(b) 27 : 26

(c) 24 : 25

(d) 25 : 24

(e) None of these

Answer: (c)


7. If the amount of profit earned by company Q in the year 2007 was Rs. 2.4 lakhs, what was

its expenditure in that year?

(a) Rs. 13 lakhs

(b) Rs. 15 lakhs

(c) Rs. 24 lakhs

(d) Rs. 16 lakhs

(e) Rs. 20 lakhs

Answer: (d)

8. What is the average per cent profit earned by company P over all the years together?

(a) 30

(b) 25

(c) 40

(d) 33

(e) None of these

Answer: (b)

9. If in the year 2009, the incomes of both the companies P and Q were same, what was the

ratio of the expenditure of company P to the expenditure of company Q in the same year?

(a) 26 : 23

(b) 23 : 26

(c) 24 : 25

(d) 25 : 24

(e) None of these

Answer: (a)

10. What is the ratio of the amount of profit earned by company A to that by company B in

the year 2010?

(a) 27 : 24

(b) 24 : 27

(c) 23 : 24

(d) 24 : 23

(e) None of these Cannot be determined

Answer: (e)


Practice Set (Data Interpretation-Line Graphs)

Directions (1-5): Study the following graph carefully and answer the questions given below: (Allahabad Bank PO Exam: 2010)

Profit earned by Three Companies over the years (Rs. in crores)

1. What was the average profit earned by all the three companies in the year 2008?

(a) Rs. 300 crore

(b) Rs. 400 crore

(c) Rs. 350 crore

(d) Rs. 520 crore

(e) None of these

2. In which of the following years was the difference between the profits earned by company B and company A the minimum?

(a) 2003

(b) 2004


(c) 2005

(d) 2008

(e) None of these

3. In which of the following years was the total profit earned by all three comapnies together the highest?

(a) 2004

(b) 2007

(c) 2008

(d) 2009

(e) None of these

4. What was the approximate percentage increase in the profit earned by Company A from 2006 to 2007?

(a) 36

(b) 24

(c) 40

(d) 20

(e) 54

5. What was the difference between the profit earned by company A in 2004 and the profit earned by company C in 2009?

(a) Rs. 50 crore

(b) Rs. 1 crore

(c) Rs. 100 crore

(d) Rs. 200 crore

(e) None of these


Direction (Q. 6 – 10): Study the given graph carefully and answer the questions that follow:

The line diagram shows the cost of production and profit of six companies for the year 2011-12. (The figures are in 'Lakhs').

Revenue = Cost of Production + Profit.

6. The ratio of profits of company B and D to the profits of A and E is:

(a) 2 : 3 (b) 10 : 9 (c) 3 : 2 (d) 10 : 7 (e) None of these

7. The profit of company C is what percentage of the revenue of company F?

(a) 20% (b) 25% (c) 30% (d) 35% (e) None of these

8. The revenue of company C is how many times of company E's profit?

(a) 5.5 (b) 5.25 (c) 5.75 (d) 5 (e) None of these


9. Which company has the maximum percentage of profit?

(a) C (b) D (c) E (d) F (e) None of these

10. What is the average profit of the last five companies (B, C, D, E and F)?

(a) Rs. 500 (b) Rs. 5,000 (c) Rs. 50,000 (d) Rs. 4,66,667 (e) None of these


Data Interpretation-Line Graphs Practice Set (Answers)

1. (b)

2. (e)

3. (d)

4. (a)

5. (c)

6. (d)

7. (a)

8. (c)

9. (b)

10. (e)


Chapter: Data Interpretation-Bar Graphs

In a bar diagram, information is presented by means of rectangles, whose lengths indicate the quantity of the variable which the bar is representing. The following points are important:

1) All bars are in the form of rectangles and the width of the bars is uniform throughout

the diagram.

2) The height of each bar is proportional to the frequency of the variable.

3) The gap between various bars is uniform.

4) The base line of all the bars is the same.

5) The bars can be either horizontal or vertical depending on the space available.

Example: The expenditure of a company under different heads (in thousands of rupees) is given below:

Head Expenditure(in thousands of rupees)

Salary of employees 400

Travelling allowance (TA) 100

Rent 150

Equipment 200

Miscellaneous 300

Draw a bar chart to depict the above data.


Example: The following bar diagram represents the percentages of total expenditure incurred by a state during the years 1981- 90 for different items. In each bar the blue portion stands for the expenditure during the first five years and the red portion stands for the next five years. Study the graph and answer questions 1-5.

1) Which of the items listed below accounts for the maximum expenditure during the

year 1981 to 1985?


(a) Communication (b) education (c) health (d) housing

2) Which of the items listed below accounts for the maximum expenditure during 1986

to 1990?

(a) Agriculture (b) communication (c) education (d) health

3) The amount of expenditure on Agriculture is approximately what proportion of that

on industry during the year 1986-90?

(a) 1/5 (b) ¼ (c) 1/3 (d) data inadequate

4) If the total expenditure on housing is Rs. 610 crores during 1981-85, the total

expenditure on industry during the same period would (approximately)

(a) Rs 2440 crores (b) Rs 1220 crores (c) Rs 4620 crores (d) none of these

5) Out of every 10,000 rupees spent during 1981-90 approximately, how much was spent

during the years 1981-85 on housing?

(a) Rs 1400 (b) Rs 700 (c) Rs 1000 (d) Rs 2800

Answer:

1) (d) out of the items listed in the question, clearly maximum expenditure during 1981-85

is on housing.

2) (a) out of the items listed in the question, clearly the maximum expenditure during

1986-90 is on agriculture.

3) (a) expenditure on agriculture during 1986-90

a. = (15-10)% of total expenditure = 5x/100 = x/20 b. Expenditure on industry during 1986-90 c. = (52.5-27.5)% of total expenditure = 25x/100 =x/4. d. Required ratio = x/20 : x/4 = 1:5

4) (d) expenditure on housing during 1981-85 = 10% of total expenditure.

a. Let the total expenditure be Rs. x. b. Then , 105 of x = 610 crores or 10x/100 =610 crores c. x=6100 crores d. total expenditure on industry during 1981-85 = 25% of 6100 crores e. = Rs 1525 crores.

5) (c) 20% of total expenditure during 1981-90 was spent on housing.

a. Expenditure on housing during 1981-90 for a total expenditure of Rs 10000 = (20×10000/100) = Rs 2000

b. Ratio of expenditure on housing during 1981-85 and that during 1986-90= 10%/(20-10)% = 1/1

c. Expenditure on housing during 1981-85 = Rs 1000


Solved Examples (Data Interpretation-Bar Graphs)

Directions (Q. 1-5): Study the following graph and answer the questions given below:

1. Out of the total number of students who opted for the given three subjects, in the year 2009, 38% were girls. How many boys opted for Mathematics in the same year?

(1) 1322 (2) 1332 (3) 1312 (4) Cannot be determined (5) None of these

Solution: Number of students who opted for all three subjects in 2009 = (20 + 20 + 5) thousand = 45000

Number of boys = = 27900

Since, we do not know the number of girls in Mathematics, number of boys opted for Mathematics cannot be determined.

Ans : (4)


2. If the total number of students in the University in the year 2007 was 455030, then, the total number of students who opted for the given three subjects were approximately what percent of the total students?

(1) 19 (2) 9 (3) 12 (4) 5 (5) 23

Solution: Required percentage = 100

= 100 9

Ans: (2)

3. What is the total number of students who opted for Hindi and who opted for Mathematics in the years 2006, 2007 and 2009 together?

(1) 97000 (2) 93000 (3) 85000 (4) 96000 (5) None of these

Solution: Required number of students

= (5 + 35 + 15 + 15 + 20 + 5) x 1000

= 95 x 1000 = 95000

Ans: (5)

4. The total number of students who opted for Mathematics in the years 2005 and 2008 together are approximately what percent of the total number of students who opted for all three subjects in the same years?

(1) 38 (2) 28 (3) 42 (4) 32 (5) 48


Solution: Required percentage

= ×100

= × 100 = × 100 32

Ans: (4)

5. What is the respective ratio between the number of students who opted for English in the years 2006 and 2008 together and the number of students who opted for Hindi in the year 2005 and 2009 together?

(1) 11: 5 (2) 12: 7 (3) 11: 7 (4) 12:5 (5) None of these

Solution: Required ratio = (25 + 30): (5+20) = 55:25 = 11:15

Ans: (1)

Directions (Q. 6-10): Study the following graph carefully to answer the questions that follow:

Monthly income (Rs in thousand) of three different persons in six different years (IBPS RRB Grade ‘A’ Officers Exam 2012)


6. What was the difference between the total monthly salary of Arun in all the years together and Suman's monthly income in the year 2007?

(1) Rs. 1.24 Iakh (2) Rs. 1.14 Iakh (3) Rs. 11.4 lakh (4) Rs. 12.4 lakh (5) None of these

Solution. Arun monthly income in all year together

Suman's monthly income in the year 2007

= 15 thousand

∴ Difference = 129 – 15 = 114 = 114 ⤬ 1000 = 114000 lakh

Ans: 2

7. What is the ratio of Arun's monthly income in the year 2006, Suman's monthly income in the year 2007 and Jyoti's monthly income in the year 2005?

(1) 6:3:5

(2) 6:4:5

(3) 5:6:4

(4) 5:4:7

(5) None of these

Solution. Ratio = Arun : Suman : Jyoti

18 : 15 : 9

6 : 5 : 3

Ans: 5


8. In which year was the difference between Jyoti's and Arun's monthly income the second highest?

(1) 2005

(2) 2006

(3) 2007

(4) 2009

(5) 2010

Solution. Difference in 2005 → 14 – 9 = 5

2006 → 18 – 10 = 8

2007 → 23 – 18 = 5

2008 → 27 – 21 = 6

2009 → 27 – 26 = 1

2010 → 35 – 26 = 9

Ans: 2

9. The monthly income of Suman in the year 2009 was approximately what percentage of the monthly income of Jyoti in the year 2010?

(1) 72

(2) 89

(3) 83

(4) 67

(5) 95

Solution. Monthly income of Suman in 2009 = 29000

Monthly income of Jyoti in 2010 = 35000

Ans: 3


10. What was the percentage increase in the monthly income of Jyoti in the year 2008 as compared to the previous year?

(1) 50

(2) 150

(3) 160

(4) 60

(5) None of these

Solution.

Ans: 1


Practice Set (Data Interpretation-Bar Graphs)

Directions (Q. 1-5): Study the following graph carefully to answer the questions that follow: (Corporation Bank PO 2011)

1. What is the percentage increase in the number of runs scored by Team B in Match 4 as compared to that in the previous match (Match 3)?

(1) 40 (2) 30 (3) 20

(4) 25

(5) None of these


2. What is the ratio of the number of runs scored by Team A in Match 2 to the number of runs scored by Team C in Match 6?

(1) 5 : 4

(2) 2 : 5

(3) 2 : 3

(4) 3 : 4

(5) None of these

3. What is the average number of runs scored by Team B in all the matches together?

(1) 250

(2) 275

(3) 200

(4) 300

(5) 225

4. The number of runs scored by all the teams together in Match 3 is approximately what percentage of the total runs scored by Team C in all the matches together?

(1) 37

(2) 57

(3) 52

(4) 47

(5) 42

5. In which match is the total runs scored by all the teams together the second highest?

(1) Match 2 only

(2) Match 6 only

(3) Match 4 only

(4) Both Match 2 and Match 6

(5) Both Match 2 and Match 4


Directions (Q. 6-10): Study the given graph carefully to answer the questions that follow:

6. What is the average number of people using mobile service M for all the years together?

1) 16

2) 14444

3) 16666

4) 14

5) None of these

7. The total number of people using all the three mobile services in the year 2007 is what per cent of the total number of people using all the three mobile services in the year 2008? (rounded off to two digits after decimal)

1) 89.72

2) 93.46

3) 88.18


4) 91.67

5) None of these

8. The number of people using mobile service N in the year 2006 forms a proximately what per cent of the total number of people using all the three mobile services in that year?

1) 18

2) 26

3) 11

4) 23

5) 29

9. What is the ratio of the number of people using mobile service L in the year 2005 to that of those using the same service in the year 2004?

1) 8: 7

2) 3 : 2

3) 19: 13

4) 15: 11

5) None of these

10. What is the total number of people using mobile service M in the years 2008 and 2009 together?

1) 35,000

2) 30,000

3) 45,000

4) 25,000

5) None of these


Data Interpretation-Bar Graphs Practice Set (Answers)

1) 3

2) 3

3) 1

4) 5

5) 4

6) 5

7) 4

8) 1

9) 2

10) 3


Chapter: Data Interpretation-Pie Diagram

In a pie chart, the values of different components of a frequency distribution are represented by the sectors of a circle. These sectors are so constructed that the area of each sector is proportional to the corresponding value of the component.

Since the sum of all the central angles is 360 degrees, we have

Central angle of a component = degrees

Example: The following pie diagram shows the expenditure incurred on the preparation of a book by a publisher, under various heads.

A: Paper 20, B: Printing 25%, C: Binding, Canvassing, Designing etc 30%

D: Miscellaneous 10% E: Royalty 15%

Study the diagram carefully and answer the questions 1-5:

1) What is the angle of pie diagram showing the expenditure incurred on paying the

royalty?

(a) 15 degrees

(b) 24 degrees

(c) 48 degrees

(d) 54 degrees


2) The marked price of a book is 20% more than the C.P. If the marked price of the book be

Rs 30, what is the cost of paper used in a single copy of the book?

(a) Rs. 6

(b) Rs. 5

(c) Rs 4.50

(d) Rs 6.50

3) Which two expenditures together will form an angle of 108 degrees at the centre of the

pie diagram:

(a) A and E

(b) B and E

(c) A and D

(d) D and E

4) If the difference between two expenditures be represented by 18 degrees in the pie-

diagram, these expenditures are :

(a) B and E

(b) A and C

(c) B and D

(d) none of these

Answers:

1) (d) Angle representing royalty D = (15×360/100)degrees = 54 degrees

2) (b) C.P. of a book = Rs ( 100×30/120) = Rs 25

Cost of paper = Rs (20×25/100) = Rs 5 3) (c) angle A = (20×360/100)degrees =72 degrees

Angle B = (25×360/100) degrees = 90 degrees Angle C = (30×360/100)degrees = 108 degrees Angle D = (10×360/100) degrees = 36 degrees Angle E = (15×360/100) degrees = 54 degrees Thus, A and D together will form an angle of 108 degrees. 4) (d) These expenditures are A and B; Band C; D and E; and A and E.


Solved Examples (Data Interpretation-Pie Diagram)

Direction (Q. 1-5): Study the following pie-chart and answer the questions given below:

Preferences of students among six beverages in terms of degree of angle in the pie-chart

Total No. of students = 6800

1. What is the difference between the total number of students who prefer Beverage A and C together and the total number of students who prefer beverage D and F together?

(1) 959 (2) 955 (3) 952 (4) 954 (5) None of these

Solution: Difference of corresponding angles = (122.4 + 21.6)0 - (79.2 + 14.4)0 = 50.40

required difference = x 6800 = 952

Ans: 3


2. What is the ratio of the number of students who prefer beverage F and the number of students who prefer beverage A?

(1) 3: 11 (2) 3: 13 (3) 6: 11 (4) 5: 11 (5) None of these

Solution: Required Ratio = 21.6: 79.2 = 3: 11

Ans: 1

3. The number of student who prefer beverage E and F together is what per cent of the total of student?

(1) 18 (2) 14 (3) 26 (4) 24 (5) None of these

Solution: Required percentage = ( ) × 100 =24%

Ans: 4

4. The number of students who prefer beverage C are approximately what percent of the number of students who prefer Beverage D?

(1) 7 (2) 12 (3) 18 (4) 22 (5) 29

Solution: Required percentage = × 100 = 11.76 12%

Ans: 2

5. How many students prefer beverage B and beverage E together?

(1) 2312 (2) 2313 (3) 2315 (4) 2318


(5) None of these

Solution: Number of students who prefer beverages B and E together

= × 68000 = = 2312

Ans: 1

Directions (Q. 6-10): Study the following pie-chart and answer the following questions. (IBPS RRB Group ‘A’ Officers Exam 2012)

Percentagewise distribution of teachers in six different universities.

Total number of teachers = 6400

6. The number of teachers in University B is approximately what per cent of the total number of teachers in University D and University E together?

(1) 55 (2) 59 (3) 49 (4) 45 (5) 65

Solution: Number of teachers in University B


Number of teachers in University D

Number of teachers in University E

∴ Required percentage

Ans: 3

7. If twenty five per cent of the teachers in University C are females, what is the number of male teachers in University C?

(1) 922 (2) 911 (3) 924 (4) 912 (5) None of these

Solution: Number of teachers in University C

Number of female teachers in University C

Number of male teachers in University C

= 1216 – 304 = 912

Ans: 4

8. The difference between the total number of teachers in University A, University B and University C together and the total number of teachers in University D, University E and University F together is exactly equal to the number of teachers of which University?

(1) University A (2) University B (3) University C (4) University D


(5) University F

Solution: Number of teachers in University A

Number of teachers in University B

Number of teachers in University C

Number of teachers in University D

Number of teachers in University E

Number of teachers in University F

∴ Difference = 3392 – 3008 = 384

Quicker method:

Difference = (D + E + F)% – (A + B + C)%

= (53 – 47) = 6%

6% of 6400 = 384

Hence, University of D is equal to 6%.

Ans: 4

9. If one-thirty sixth of the teachers from University F are professors and the salary of each professor is Rs. 96000, what will be the total salary of all the professors together from University F?

(1) Rs. 307.2 lakh (2) Rs. 32.64 lakh (3) Rs. 3.072 lakh (4) Rs. 3.264 lakh (5) None of these

Solution: Number of teachers in University F


Number of professors in University F

∴ Total Salary of professors in University F

= 32 ⤬ 96000 = 30.72 lakh

Ans: 5

10. What is the average number of teachers in University A, University C, University D and University F together?

(1) 854 (2) 3546 (3) 3456 (4) 874 (5) None of these

Solution: Average

Ans: 5


Practice Set (Data Interpretation-Pie Diagram)

Directions (1-5): Study the following pie-charts carefully and answer the questions given below: Discipline-wise Breakup of the Number of candidates appeared in Interview and Discipline-wise Break up of the Number of candidates selected by an organisation

Discipline-wise Breakup of Number of candidates appeared in Interview

Total Number of Candidates Appeared In the Interview = 25780 Percentage Distribution

Discipline-wise Break-up of Number of candidates selected after Interview by the organization

Total Number of Candidates selected After Interview = 7390

Percentage Distribution


1. What was the ratio of the number of candidates appeared . in interview from other disciplines and the number of candidates selected from Engineering discipline respectively (rounded off to the nearest integer)?

1) 3609: 813

2) 3094: 813

3) 3094: 1035

4) 4125: 1035

5) 3981: 767

2. The total number of candidates appeared in interview from Management and other discipline was what percentage of number of candidates appeared from Engineering discipline?

1) 50

2) 150

3) 200

4) Cannot be determined

5) None of these

3. Approximately what was the difference between the number of candidates selected from Agriculture discipline and number of candidates selected from Engineering discipline?

1) 517

2) 665

3) 346

4) 813

5) 296

4. For which discipline was the difference in number of candidates selected to number of candidates appeared in interview the maximum?

1) Management

2) Engineering

3) Science

4) Agriculture

5) None of these

5. Approximately what was the total number of candidates selected from Commerce and Agricultural discipline together?

1) 1700


2) 1800

3) 2217

4) 1996

5) 1550

Directions (Q.6-10): Study the following pie-chart carefully to answer these questions. (Central Bank of India (PO) 2010)

Percentagewise Distribution of teachers who teach six different subjects Total number of Teachers = 1800

Percentage of teachers

6. If two-ninths of the teachers who teach Physics are female, then the number of male Physics teachers is approximately what percentage of the total number of teachers who teach Chemistry?

1) 57

2) 42

3) 63

4) 69

5) 51

7. What is the total number of teachers teaching Chemistry, English and Biology?

1) 1, 226


2) 1, 116

3) 1, 176

4) 998

5) None of these

8. What is the difference between the total number of teachers who teach English and Physics together and the total number of teachers who teach Mathematics and Biology together?

1) 352

2) 342

3) 643

4) 653

5) None of these

9. What is the ratio of the number of teachers who teach Mathematics to the number of teachers who teach Hindi?

1) 13:7

2) 7:13

3) 7: 26

4) 8: 15

5) None of these

10. If the percentage of Mathematics teachers is increased by 50 per cent and the percentage of Hindi teachers decreased by 25 per cent then what will be the total number of Mathematics and Hindi teachers together?

1) 390

2) 379

3) 459

4) 480

5) None of these


Direction (Q. 11 – 15): Study the given pie-charts carefully and answer the questions that follow:

Discipline-wise breakup of the number of candidates appeared in Interview and Discipline-wise breakup of the candidates selected by and organisation.

Total number of candidates appeared in the interview = 25,600 and total number of candidates selected after interview = 7,500.

11. What was the ratio of the number of candidates appeared in interview from other disciplines and the number of candidates selected from art disciplines?

(a) 256 : 125

(b) 125 : 256

(c) 125 : 216

(d) Cannot be determined

(e) None of these

12. The total number of candidates appeared in interview from Management and Art disciplines was what per cent of the number of candidates from Engineering discipline?

(a) 66.67


(b) 75

(c) 80

(d) 120

(e) 150

13. What was the difference between the number of candidates selected from Science discipline and the number of candidates selected from Commerce discipline?

(a) 1,000

(b) 1,100

(c) 1,200

(d) 1,250

(e) None of these

14. From which discipline was the difference in number of candidates selected to number of candidates appeared in interview the maximum?

(a) Management

(b) Engineering

(c) Commerce

(d) Science

(e) Art

15. What was the total number of candidates selected from Commerce and Art discipline together?

(a) 1,800

(b) 1,950

(c) 2,100

(d) 2,250

(e) 2,400


Data Interpretation- Pie Diagram Practice Set (Answers)

1. 2

2. 2

3. 5

4. 3

5. 1

6. 1

7. 2

8. 2

9. 5

10. 3

11. a

12. e

13. c

14. d

15. b


Chapter: Data Interpretation-Case lets

Direction for questions 1 to 5: Answer these questions on the basis of the following information.

Shekhar bought 10 acres of land for Rs.250000 in 2011. That year he cultivated Sugarcane and Soya bean in the 10 acres with the ratio of area under Sugarcane and Soya bean being 5:4. The profit obtained from Sugarcane and Soya bean was in the ratio 3:2 with the total profit being Rs.58500. This was 15% of the amount he invested in cultivation that year. The next year he again cultivated Sugarcane and Soya bean, with the areas being same as before and reaped a profit of Rs.66000 in total with that from Sugarcane and Soya bean being in the ratio 8:7 but his return on his investment that year was only 14%.

1. What is the amount invested by Shekhar for cultivation in 2011?

(a) Rs.356000

(b) Rs.374800

(c) Rs.380000

(d) Rs. 390000

Solution: (d) Shekhar had a profit of Rs.58500 and this profit was 15% of the money he invested, his investment was

2. What is the profit obtained by Shekhar by cultivating Sugarcane in 2011?

(a) Rs.43800

(b) Rs.35100

(c) Rs.36200

(d) None of these

Solution: (b)

The profit obtained by Shekhar by cultivating Sugarcane in 2012 =


3. What is the profit obtained by cultivating Soya bean in 2012?

(a) Rs.30800

(b) Rs.36100

(c) Rs.24200

(d) None of these

Solution: (a)

The profit obtained by cultivating Soya bean in 2012 =

4. What is the ratio of the profit obtained from Sugarcane and Soya bean in the two years together?

(a) 89 : 79

(b) 167 : 211

(c) 703 : 542

(d) None of these

Solution: (c)

The profit obtained in 2011 from Sugarcane = Rs.35100

The profit obtained in 2012 from Sugarcane = 66000 - 30800 = Rs.35200

Total profit from Sugarcane = Rs.70300

Total profit in the two years = Rs.58500 + Rs.66000 = Rs.124500

Therefore, Profit from Soya bean Rs.54200.

Now, the required ratio = 70300: 54200 = 703 : 542.


5. What is the approximate amount invested by Shekhar for cultivation in 2012?

(a) Rs.428500

(b) Rs.471400

(c) Rs.495300

(d) Rs.518650

Solution: (b) Here the profit of 66,000 is 14% of the amount invested, therefore, the invested =


Solved Examples (Data Interpretation-Caselets)

Directions: (1-5) Study the following information and answer the questions that follow: (IBPS CWE PO MT 2012) The premises of a bank are to be renovated. The renovation is in terms of flooring. Certain areas are to be floored either with marble or wood. All rooms/halls and pantry are rectangular. The area to be renovated comprises of a hall for customer transaction measuring 23 m by 29 m, branch manager’s room measuring 13 m by 17 m, a pantry measuring 14 m by 13 m, a record keeping cum server room measuring 21 m by 13 m and locker area measuring 29 m by 21 m. The total area of the bank is 2000 square meters. The cost of wooden flooring is f 170/- per square meter and the cost of marble flooring is Rs. 190/- per square meter. The locker area, record keeping cum server room and pantry are to be floored with marble. The branch manager's room and the hall for customer transaction are to be floored with wood. No other area is to be renovated in terms of flooring. 1. What is the respective ratio of the total cost of wood en flooring to the total cost of marble flooring? (1) 1879: 2527 (2) 1887: 2386 (3) 1887: 2527 (4) 1829: 2527 (5) 1887: 2351 Solution: Total flooring area with marble = locker area + record keeping + pantry = 182+273 +609 = 1064 sqm Cost of flooring = 1064 190 Total flooring area with wooden = Branch Manager room + Hall = 221 + 667 = 888 sqm Cost of flooring = 888 170 Ratio= 888 170: 1064 190 = 888 17: 1064 19 = 15096 : 20216 = 1887: 2527 Ans: 3


2. If the four walls and ceiling of the branch manager’s room (The height of the room is 12 meters) are to be painted at the cost off 190/- per square meter, how much will be the total cost of renovation of the branch manager's room including the cost of flooring? (1) Rs. 1, 36,800/- (2) Rs. 2, 16,660/- (3) Rs. 1, 78,790/- (4) Rs. 2, 11,940/- (5) None of these Solution: Cost of flooring of branch manager room =221 170= Rs. 37570 Cost of painting = [2(17 12+ 13 12)+ 13 x 17] 190 = [2( 204 + 156) + 221] 190= (2 360 + 221) 190 = (720 + 221) 190 = 941 190 = Rs. 178790 Total cost = 178790 + 31570 = Rs.216360 Ans: (5) 3. If the remaining area of the bank is to be carpeted at the rate of Rs. 110/- per square meter, how much will be the increment in the total cost of renovation of bank premises? (1) Rs. 5,820/- (2) Rs. 4,848/- (3) Rs. 3,689/- (4) Rs. 6,890/- (5) None of these Solution: Total area of bank = 2000 sqm Total flooring area = 1952 sqm Remaining area = 2000 - 1952 = 48 sqm

Cost of carpeting = 48 110 = Rs.5280 Ans: (5) 4. What is the percentage area of the bank that is not to be renovated? (1) 2.2 % (2) 2.4 % (3) 4.2 % (4) 4.4 %


(5) None of these Solution: Area not to be renovated = 48 sq m

Reqd % = l00 = 2.4%

Ans: 2 5. What is the total cost of renovation of the hall for customer transaction and the locker area? (1) Rs. 2, 29,100 (2) Rs. 2, 30,206 (3) Rs. 2, 16,920 (4) Rs. 2, 42,440 (5) None of these Solution: Cost of renovation of hall + locker area = 667 170 + 609 190 = 113390 + 115710= Rs. 229100 Ans: (1)

Directions (Q. 6 – 10): Study the given information carefully to answer the questions that

follow:

An organization consists of 3500 employees working in different departments, viz HR,

Marketing, IT, Production and Accounts. The ratio of male to female employees in the

organisation is 3 : 2. 8% of the males work in the HR department. 22% of the female work in the

account department. The ratio of males to females working in the HR department is 3 : 5. One-

seventh of the females work in the IT department. 46% of the males work in the Production

department. The number of females is one-sixth of the males working in the same. The

remaining females work in the Marketing department. The total number of employees working

in the IT department is 375. 22% of the males work in the Marketing department and remaining

work in the Account department.

6. The number of males working in the Account department forms approximately what per

cent of the total number of males in the organisation?

(a) 6

(b) 8

(c) 10


(d) 11

(e) 12

Answer: (a)

7. How many females work in Production department?

(a) 140

(b) 200

(c) 180

(d) 160

(e) None of these

Answer: (e)

8. The total number of employees working in the Account department forms approximately

what per cent of the total number of female employees in the organisation?

(a) 28

(b) 32

(c) 29

(d) 31

(e) 30

Answer: (d)

9. The ratio of the numbers of females working in IT department to the numbers of males

working in the same department is

(a) 15 : 8

(b) 1 : 2

(c) 8 : 15

(d) 2 : 1

(e) 7 : 11

Answer: (c)

10. What is the total number of employees working in the Marketing and Production

departments together?

(a) 1900

(b) 2040

(c) 2020

(d) 2031

(e) 2042

Answer:(b)


PracticeSet-(Data Interpretation- Caselets)

Directions (Q. 1-5): Study the information carefully to answer the questions that follow.

A company produced five different products, viz mobile phone, pen drive, calculator, television and washing machine. The total number of all the five products is 1650.24% of the total number of products is mobile phones. One-sixth of the total number of products is pen drives. 14% of the total number of products is calculators. Remaining products are either television or washing machine. The number of washing machines is 50 more than the number of televisions produced. (IBPS RRB Grade Officer Exam 2012)

1. What is the ratio of the number of washing machines to the number of calculators produced by the company?

(1) 17:11

(2) 19:11

(3) 11:17

(4) 19: 13

(5) None of these

2. If 24 per cent of the pen drives are

defective, what is the number of pen

drives which are not defective?

(1) 209

(2) 215

(3) 219

(4) 225

(5) None of these

3. The number of televisions produced is

approximately what per cent of the total

number of calculators and washing

machines produced together?

(1) 63

(2) 55

(3) 59

(4) 51

(5) 67

4. What is the difference between the total

number of televisions and mobile phones

together and the number of calculators

produced?

(1) 534

(2) 524

(3) 511

(4) 523

(5) None of these

5. What is the total number of pen drives,

calculators and washing machines produced

by the company?

(1) 907

(2) 917

(3) 925

(4) 905

(5) None of these


Directions (Q. 6-10): Study the following information carefully to answer the questions that follow: (RBI Grade’B’ Officer’s Exam 2011)

There are two trains, Train A and Train B. Both trains have four different types of coaches, viz General, Sleeper, First Class and AC. In Train A, there are total 700 passengers. Train B has thirty per cent more passengers than Train A. Twenty per cent of the passengers of Train A are in General Coach. One-fourth of the total number of passengers of Train A are in AC coach. Twenty three per cent of the passengers of Train A are in Sleeper Coach. Remaining passengers of Train A are in First Class Coach. The total number of passengers in AC Coach in both the trains together is 480. Thirty per cent of the number of passengers of Train B are in Sleeper Coach. Ten per cent of the total passengers of Train B are in First Class Coach. The remaining passengers of Train B are in General Coach.

6. What is the ratio of the number of passengers in First Class Coach of Train A to the number of passengers in Sleeper Coach of Train B?

(1) 13 : 7 (2) 7 : 13 (3) 32 : 39 (4) Data Inadequate (5) None of these

7. What is the total number of passengers in the General Coach of Train A and the AC Coach of Train B together?

(1) 449 (2) 459

(3) 435 (4) 445 (5) None of these

8. What is the difference between the number of passengers in the AC Coach of Train A and the total number of passengers in Sleeper and First Class Coach together of Train B?

(1) 199 (2) 178 (3) 187 (4) 179 (5) None of these

9. The total number of passengers in General Coaches of both the trains together is approximately what percentage of the total number of passengers in Train B?

(1) 35 (2) 42 (3) 46 (4) 38 (5) 31

10. If the cost per ticket of First Class coach is Rs.450, what will be the total amount generated from First Class Coach of Train A?

(1) Rs.1, 00, 080 (2) Rs.1, 08, 000 (3) Rs.1, 00, 800 (4) Rs.10, 800 (5) None of these


Directions (Q. 11-15): Study the information carefully to answer the questions that follow:

A company produces five different products, viz Television, Refrigerator, Mobile Phone, Oven and Water Heater. The total number of all the five products manufactured is 1200. 15 per cent of the total number of products are Televisions. Three-tenths of the total number of products are Refrigerators. The number of Mobile Phones manufactured is 40 more than the number of Televisions. 22 per cent of the total number of products is Oven and the remaining number of products is Water Heaters. (Corporation Bank PO 2011)

11. What is the total number of Refrigerators and Water Heaters?

(1) 436

(2) 476

(3) 576

(4) 536

(5) None of these

12. The number of Mobile Phones is

approximately what per cent of the

number of Televisions?

(1) 115

(2) 140

(3) 135

(4) 130

(5) 120

13. If the cost of one Oven is Rs. 8,200, what is the cost of all the Ovens manufactured by the company?

(1) Rs. 2,16,480

(2) Rs. 21,68,400

(3) Rs. 2,16,48,000

(4) Rs. 2,16,840

(5) None of these

14. What is the difference between the number of Refrigerators and the number of Mobile Phones manufactured?

(1) 160

(2) 140

(3) 120

(4) 130

(5) None of these

15. If 25 per cent of the number of Ovens

are defective, what is the number of non-

defective Ovens?

(1) 176

(2) 188

(3) 198

(4) 186

(5) None of these


Directions-(Q. 16-20): Study the information carefully to answer the questions that follows:

In a ship there are 1200 passengers. 18 per cent of the total number of passengers is from Britain. Two- fifth of the total number of passengers is from South Africa 6 per cent of the total number of passengers is from Madagascar. Remaining number of passengers is from India. 25 per cent of the number of passengers from Britain is females. Half the numbers of passengers from South Africa are male. There is no female passenger from Madagascar. Two- third of the number of passengers from India are females. (Allahabad Bank Probationary Officers Exam 2011)

16. What is the ratio of the number of passengers from Madagascar, number of female passengers from South Africa and the total number of passengers from India?

(1) 2: 5 : 18 (2) 3: 10 : 18 (3) 3: 11 : 18 (4) 2: 18 : 5 (5) None of these

17. The number of male passengers from South Africa is approximately what percentage of the total number of passengers from Britain?

(1) 111 (2) 115 (3) 120 (4) 125 (5) 131

18. What is the average number of male passengers from all the four countries?

(1) 154.5 (2) 164.5 (3) 145 (4) 164 (5) None of these

19. What is the difference between the number of male passengers from Madagascar and the number of male passengers from India?

(1) 64 (2) 82 (3) 74 (4) 72 (5) None of these

20. What is the total number of male from Britain passengers’ female passengers from India together?

(1) 340 (2) 420 (3) 350 (4) 460 (5) None of these


Data Interpretation-Caselets Practice Set (Answers)

1) 2

2) 1

3) 2

4) 5

5) 4

6) 3

7) 4

8) 5

9) 2

10) 3

11) 4

12) 5

13) 5

14) 2

15) 3

16) 2

17) 1

18) 1

19) 4

20) 5


Solved Examples (Data Interpretation-Miscellaneous)

Directions: (1-5) Study the following graph and table carefully and answer the questions given below:

Time Taken To Travel (In Hours) By Six Vehicles On Two Different Days

Distance covered (in kilometers) by six vehicles on each day

(IBPS CWE PO MT 2012)

1. Which of the following vehicles travelled at the same speed on both the days? (1) Vehicle A (2) Vehicle C (3) Vehicle F


(4) Vehicle B (5) None of these Solution: The speed of Vehicle B on both the days is 43 km/hr. Ans: (4) 2. What was the difference between the speed of vehicle A on day 1 and the speed of vehicle C on the same day? (1) 7 km/hr (2) 12 km/hr (3) 11 km/hr (4) 8 km/hr (5) None of these Solution: Speed of A on 1st day = 52 km/hr Speed of C on 1st day = 63 km/hr

Difference = 65 - 52 = 11 km/hr Ans: 2 3. What was the speed of vehicle C on day 2 in terms of meters per second? (1) 15.3 (2) 12.8 (3) 11.5 (4) 13.8 (5) None of these Solution: Speed of Vehicle C on 2nd day = 45 km/hr

= 45 × = 2.5 × 5 = 12.5 m/sec

Ans: (5) 4. The distance travelled by vehicle F on day 2 was approximately what percent of the distance travelled by it on day 1? (1) 80 (2) 65 (3) 85


(4) 95 (5) 90

Solution: Reqd % = 100 = 90.46 90%

Ans: (5) 5. What is the respective ratio between the speeds of vehicle 0 and vehicle E on day 2? (1) 15: 13 (2) 17: 13 (3) 13: 11 (4) 17: 14 (5) None of these

Solution: Reqd Ratio =

= = = 17:13

Ans: (2)


Directions (6-10) Study the following pie-chart and table carefully and answer the questions given below: Percentagewise distribution of the number of mobile phones sold a shopkeeper during six months Total number of mobile phones sold = 45000

The respective ratio between the numbers of mobile phones sold of company A and company B during six months

6. What is the respective ratio between the number of mobile phones sold of company B during July and those sold during December of the same company? (1) 119: 145 (2) 116: 135 (3) 119: 135 (4) 119: 130 (5) None of these Solution: Total number of mobiles sold in the month of July


= 45000 = 7650

Mobile phones sold by Company B in the month of

July = 7650 = 3570

Total number of mobile phones sold in the month of

December = 45000 = 7200

Mobile phones sold by Company B in the month of

December = 7200 = 4050

Reqd ratio = = = = 119: 135

Ans: (3) 7. If 35% of the mobile phones sold by company A during November were sold at a discount, how many mobile phones of company A during that month were sold without a discount? (1) 882 (2) 1635 (3) 1638 (4) 885 (5) None of these Solution: Number of mobile phones sold in the month of

November = 45000 = 5400

Number of mobile phones sold by Company A in

the month of November = 5400 = 2520

Number of mobile phones without discount in the month of November by Company A

= 2520 = 2520 0.65 = 1638

Ans: 3 8. If the shopkeeper earned a profit of Rs. 433/- on each mobile phone sold of company B during October, what was his total profit earned on the mobile phones of that company during the same month? (1) Rs. 6,49,900 (2) Rs. 6,45,900 (3) Rs. 6,49,400 (4) Rs 6,49,500 (5) None of these


Solution: Number of mobile phones sold in the month of

October = 45000 = 3600

Number of mobile phones sold by Company B in

the month of October = 3600 = 1500

Total profit earned by Company B in the month of October = 1500 433 = 649500 Ans: (4) 9. The number of mobile phones sold of company A during July is approximately what percent of the number of mobile phones sold of company A during December? (1) 110 (2) 140 (3) 150 (4) 105 (5) 130 Solution: Number of mobile phones sold in the month of July

= 45000 = 7650


the month of July = 7650 = 4080

Number of mobile phones sold in the month of

December = 45000 = 7200


the month of December = 7200 = 3150

Reqd % = = 129.52 130

Ans: (5) 10. What is the total number of mobile phones sold of company B during August and September together? (1) 10000 (2) 15000 (3) 10500 (4) 9500 (5) None of these


Solution: Number of mobile phones sold in the month of

August = 45000=9900

Number of mobile phones sold in the month of

September = 45000 = 45000 = 11250

Number of mobile phones sold by Company B in the month of August

= 9900 = 5500

Number of mobile phones sold by Company B in

September= 11250 = 4500

Total number of mobile phones sold in August and September by Company B = 5500+4500= 10000 Quicker Method: Total number of mobile phones sold by Company B in August and September .

= 10000

Ans: (1)


Directions: (11-15) Study the following information and answer the questions that follow: The graph given below represents the production (in tonnes) and sales (in tonnes) of company a from 2006-2011.

The table given below represents the respective ratio of the production (in tonnes) of Company A to the production (in tonnes) of Company B, and the respective ratio of the sales (in tonnes) of Company A to the sales (in tonnes) of Company B.

11. What is the approximate percentage increase in the production of Company A (in tonnes) from the year 2009 to the production of Company A (in tonnes) in the year 2010? (1) 18% (2) 38% (3) 23% (4) 27% (5) 32% Solution: Production of Company A in year 2009 = 550 Production of Company A in year 2010 = 700


Reqd % = 100 = 100

= = 27.27 27%

Ans: (4) 12. The sale of Company A in the year 2009 was approximately what percent of the production of Company A in the same year? (1) 65% (2) 73% (3) 79% (4) 83% (5) 69% Solution: Sales of Company A in year 2009 = 400 Production of Company A in year 2009 = 550

Reqd % = 100 = = 72.72 73%

Ans: (2) 13. What is the average production of Company B (in tonnes) from the year 2006 to the year 2011? (1) 574 (2) 649 (3) 675 (4) 593 (5) 618 Solution: Average production of Company B

=

= = 675

Ans: (3)


14. What is the respective ratio of the total ‘production (in tonnes) of Company A to the total sales (in tonnes) of Company A? (1) 81: 64 (2) 64: 55 (3) 71: 81 (4) 71: 55 (5) 81: 55 Solution:

= = = 81: 55

Ans: (5) 15. What is the respective ratio of production of Company B (in tonnes) in the year 2006 to production of Company B (in tonnes) in the year 2008? (1) 2: 5 (2) 4: 5 (3) 3: 4 (4) 3: 5 (5) 1: 4 Solution: Production of Company B in the year 2006 = 150 = 600 Production of Company B in the year 2008 = 200 Ratio = = 3: 4

Ans: (3)


Practice Set (Data Interpretation-Miscellaneous)

Directions (Q. 1-5): Study the radar-graph carefully and answer the questions that follow.

Monthly salary (in thousands) of five different persons in three different years (RBI Grade’B’ Officer’s Exam 2011)

1. What is the average of the monthly salary of Sumit in the year 2008, Anil in the year 2009 and Jyoti in the year 2010?

(1) Rs.20, 000 (2) Rs.5, 000 (3) Rs.45, 000 (4) Rs.15, 000 (5) None of these

2. The total monthly salary of Arvind in all years together is what per cent of the total monthly salary of all the five persons together in the year 2008?

(1) 55% (2) 60% (3) 75% (4) 70% (5) None of these


3. Among the five persons, whose earning per month over all the years together is the second lowest?

(2) Sumit (3) Anil (4) Jyoti (5) Arvind (6) Poonam

4. What is the per cent decrease in the monthly salary of Poonam in the year 2009 as compared to her monthly salary in the previous year?

(1) 40% (2) 10% (3) 20% (4) 80% (5) None of these

5. If Jyoti's monthly salary in the year 2010 was increased by 30 per cent what would her monthly salary be in that year?

(1) Rs.36, 000 (2) Rs.39, 000 (3) Rs.45, 000 (4) Rs.42, 000 (5) None of these


Directions (Q. 6-10): Study the following pie-chart and bar-graph and answer the following questions. (Corporation Bank PO 2011)

Percentage wise distribution of teachers in six different districts

Total number of Teachers = 4500

Percentage of Teachers


6. What is the total number of male teachers in District F, female teachers in District C and female teachers in District B together?

(1) 1080 (2) 1120 (3) 1180 (4) 1020 (5) None of these

7. The number of female teachers in District D is approximately what per cent of the total number of teachers (both male and female) in District A?

(1) 70 (2) 75 (3) 80 (4) 95 (5) 90

8. In which district is the number of male teachers more than the number of female teachers?

(1) B only (2) D only (3) Both B and E (4) Both E and F (5) None

9. What is the difference between the number of female teachers in District F and the total number of teachers (both male and female) in District E?

(1) 625 (2) 775 (3) 675 (4) 725 (5) None of these

10. What is the ratio of the number of male teachers in District C to the number of female teachers in District B?

(1) 11 : 15 (2) 15 : 11 (3) 15 : 8 (4) 30 : 13 (5) None of these


Directions-(Q. 11-15) Study the following pie-chart and bar chart and answer the following questions. (Allahabad Bank Probationary Officers Exam 2011)

Percentagewise Distribution of students in six different Schools

Total number of Students = 6000

Percentage of Students

Number of Boys out of the 6000 Students in each School Separately.


11. What is the sum of the number of girls in School C. Number of girls in School-E and the number of boys in school-D together?

(1) 1700 (2) 1900 (3) 1600 (4) 1800 (5) None of these

12. What is the respective ratio of the number of boys in School-C, number of girls in School B and total number of students in School E?

(1) 45: 7: 97 (2) 43: 9: 97 (3) 45: 7: 87 (4) 43: 9: 87 (5) None of these

13. What is the difference between the total number of students in School-P and the number of boys in School E?

(1) 820 (2) 860 (3) 880 (4) 900 (5) None of these

14. In which school the total numbers of students (both boys and girls) together are equal to the number of girls in School E?

(1) A (2) B (3) C (4) D (5) F 15. The Number of girls in School A is approximately what per cent of total number of students in School B? (1) 55 (2) 50 (3) 35 (4) 45 (5) 4


Data Interpretation-Miscellaneous Practice Set

(Answers)

1 (1)

2 (2)

3 (4)

4 (4)

5 (2)

6 (1)

7 (5)

8 (3)

9 (2)

10 (3)

11 (4)

12 (3)

13 (5)

14 (2)

15 (5)


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