__Quantitative Aptitude Practice Questions | IBPS 2017__**Dear Aspirants,**

Welcome to

**Mentor for Bank Exams**Quantitative Aptitude Quiz section. The following quiz covers Number Series (5 Questions), Inequalities (5 Questions), Data Interpretation (5 Questions), Word Problems (5 Questions). All the best for upcoming IBPS Exams 2017.**Directions (1 – 5): What should come in place of question mark (?) in the following number series?**

**1. 8, 7, 11, 12, 14, 17, 17, 22, ?**

(a) 27

(b) 20

(c) 22

(d) 24

(e) 25

**2. 11, 29, 83, 245, 731, ?**

(a) 2193

(b) 2189

(c) 2139

(d) 2389

(e) 2219

**3. 3, 8, 20, 46, 100, 210, ?**

(a) 436

(b) 438

(c) 416

(d) 432

(e) 430

**4. 5, 7, 17, 55, 225, 1131, ?**

(a) 6973

(b) 6379

(c) 7639

(d) 7369

(e) 6793

**5. 1, 5, 14, 30, 55, 91, ?**

(a) 128

(b) 140

(c) 135

(d) 138

(e) 142

**Directions (6 – 10): Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly. Give answer,**

(a) Quantity I > Quantity II

(b) Quantity I ≥ Quantity II

(c) Quantity II > Quantity I

(d) Quantity II ≥ Quantity I

(e) Quantity I = Quantity II or Relation cannot be established

**6. If the perimeter of square and circle is same, then**

**Quantity I:**

**the area of square**

**Quantity II:**

**the area of circle**

**7. The principal amount is Rs 12,000**

**Quantity I:**

**Simple Interest on the sum at the rate of 10% per annum for 3 years**

**Quantity II:**

**Compound Interest on the sum at rate of 10% compounded semi-annually for one and a half year.**

**8. If the cost price of an article is increased by 40% to make it as the market price of article,**

**Quantity I:**

**Selling price of article after allowing a discount of 20%, if the cost price is Rs 1,75,000**

**Quantity II:**

**The cost price of article if after selling article at 20% profit is Rs 2,52,000**

**9. Four years ago, the ratio of ages of A and B is 5 : 6 and five years after the ratio will become 6 : 7, then**

**Quantity I:**

**present age of A**

**Quantity II:**

**age of B 4 years ago**

**10. Quantity I:**

**The largest number if average of 5 consecutive odd numbers is 39.**

**Quantity II:**

**The smallest number if average of 5 consecutive of odd numbers is 47.**

**Directions (Q. 11-15): The following information is about the performance of Akhilesh in SBI PO mains exam. Read the information carefully and answer the following question.**

**The exam consists of 200 marks, with 5 sections i.e. Reasoning, quant, English, G.A., Computers. Akhilesh attempted 22 questions in Reasoning with an accuracy of 850/11%. Each question of reasoning consists of 2 marks with a negative marking of 25%. (if right question is of 2 mark, then 0.5 mark will be deducted for each wrong answer).**

**Each section of the exam have the 25% of negative marking for each wrong question. The total number of questions in reasoning is 30. Each question of computer consists of 1/2 marks and maximum marks in computer are 10. Total 16 questions are attempted by Akhilesh in computer with the ratio of right questions to wrong questions 3 : 1.**

**The number of questions in English is equal to maximum marks of English. Akhilesh attempted 26 questions with 50% accuracy. The number of questions attempted in English is 65% of the total number of questions in English.**

**GA section consists of 40 questions with each question 0.75 marks. Akhilesh attempted 23 questions out of which 8 are wrong. Quant section contains 40 questions out of which Akhilesh attempted 35 questions and got 52.5 marks.**

**11. Another student Arunoday attempted 70% questions in the same exam, then find the number of questions left by Arunoday.**

(a) 119

(b) 68

(c) 51

(d) 65

(e) None of these

**12. Find the marks obtained by Akhilesh in GA.**

(a) 8.75

(b) 9.25

(c) 9.75

(d) 10.75

(e) None of these

**13. The number of correct questions in reasoning is how much more than the number of incorrect questions in the same subject?**

(a) 12

(b) 7

(c) 18

(d) 9

(e) None of these

**14. Find the total marks obtained by Akhilesh in the exam .**

(a) 101

(b) 105

(c) 109

(d) 102

(e) None of these

**15. Find the total number of incorrect questions attempted by Akhilesh in the exam.**

(a) 27

(b) 15

(c) 28

(d) 18

(e) None of these

**16. The average marks of Sameer decreased by 1, when he replaced the subject in which he has just scored 40 marks by the other two subjects in which he has just scored 23 and 25 marks respectively. Later he has also included 57 marks of Computer Science, then the average marks increased by 2. How many subjects were there initially?**

(a) 6

(b) 12

(c) 15

(d) Can’t be determined

(e) None of these

**17. 14 men can do a work in 18 days ,15 women can do a work in 24 days. If 14 men work for first 3 days and 10 women, work after that for 3 days, find the part of work left after that?**

(a) 3/4

(b) 1/4

(c) 1/2

(d) 1/6

(e) 1/5

**18. Two trains crosses each other in 14 seconds when they are moving in opposite direction, and when they are moving in same direction, they crosses each other in 3 minute 2 seconds. Find the speed of the faster train by what percent more than the speed of the slower train?**

(a) 16.67%

(b) 17.33%

(c) 16.33%

(d) 17.67%

(e) 18.33%

**19. A, B and C started a business with their investments in the ratio 1 : 2 : 4. After 6 month A invested the half amount more as before and B invested same the amount as before while C withdrew 1/4th of his investment. Find the ratio of their profits at the end of the year.**

(a) 5 : 12 : 13

(b) 5 : 11 :14

(c) 5: 12 : 14

(d) 5 : 12 : 10

(e) None of the above

**20. A’s age is twice C’ age. Ratio of age of B 2 years hence to age of C 2 years ago is 5 : 2. C is 14 years younger than D. Difference in ages of D and A is 4 years. Find the average of their ages.**

(a) 36

(b) 25

(c) 27

(d) 13

(e) 18

**Solutions:**

**1. B)**There are two series in the given pattern i.e (8, 11, 14, 17, 20) and (7, 12, 17, 22) increasing be 3 and 5

**2. B)**The series is *3 – 4;

So, ? = 731 * 3 – 4 = 2189

**3. D)**Pattern of the series is *2 + 2, *2 + 4, *2 + 6, *2 + 8, *2 + 10, *2 + 12

So, ? = 210 * 2 + 12 = 432

**4. E)**Pattern of the series is *1 + 2, *2 + 3, *3 + 4, *4 + 5, *5 + 6, *6 + 7

So, ? = 1131 * 6 + 7 = 6793

**5. B)**Pattern of the series is +2^2, +3^2, +4^2, +5^2, …..

So, ? = 91 + 7^2 = 140

**6. C)**When perimeter of square, circle and rectangle are equal then area in descending order is

area of circle > area of square > area of
rectangle

Hence II > I

**7. A)**I: SI = 3*10% = 30% of 12000 = Rs 3600

II: CI:

Rate at half = 10/2 = 5%, time doubles so t = 1
½ = 3 hours

So CI is Rs 1891.5

Hence I > II

**8. C)**40% = 2/5 [CP = 5, MP = 7] ………………..(1)

I: discount = 20% = 1/5 [MP = 5, SP = 4] ………………(2)

make MP same by multiplying (1) by 5 and (2) by 7

CP………..SP…………MP is

25………..28…………35

So SP = 196000

II: profit = 20% = 1/5 [CP = 5, SP = 6] ………………(3)

CP is same, so from (3) and (1):

CP………..SP…………MP is

5………….6…………..7

So CP = 210000

Hence II > I

**9. C)**

A………..B

4 years ago
==== 5…………6

5 years hence ==
6…………7

1 == 9 and 6 ==54

5 == 45

I: A’s present age = 45+4 = 49

II: B’s age 4 years ago = 54

Hence II > I

**10. E)**I: 35, 37, 39, 41,43

II: 43, 45, 47, 49, 51

So CSA = 2πrh = 528cm<sup>2</sup>

Hence I = II

**11. C)**No. of questions in Reasoning = 30; No. of questions in Computer = 20; No. of questions in English = 40; Number of questions in Quant = 40; Number of questions in GA = 40

Total no. of questions = 170

Arunoday attempt = 70/100 * 170 = 119

Number of questions left = 170 – 119 = 51

**12. C)**Total attempts in GA = 23 questions

Wrong attempts = 8; Right attempts = 23 – 8 = 15

Marks obtained for correct attempts = 15 * 0.75 =
11.25

Marks deducted for wrong attempts = 8 * 0.75/4 = 1.5

Therefore marks obtained in GA = 11.25 – 1.5 = 9.75

**13. A)**No. of questions attempted in Reasoning = 22

Accuracy = 850/11% = 77.27%

100% => 22

77.27% => ? (Number of correct attempts)

? = 22 * 77.27/100 = 17

Therefore no. of wrong attempts = 22 – 17 = 5

Therefore the required difference = 17 – 5 = 12

**14. C)**Marks obtained in GA = 9.75

Marks obtained in Reasoning = 17 * 2 – 5 * 0.5 = 34
– 2.5 = 31.5

Marks obtained in English = 13 * 1 – 13 * 0.25 = 13
– 3.25 = 9.75

Marks obtained in Computer = 12 * 0.5 – 4 * 0.5/4 =
6 – 0.5 = 5.5

Marks obtained in Quant = 52.5

Therefore total marks obtained = 9.75 + 31.5 + 9.75
+ 5.5 + 52.5 = 109

**15. E)**No. of incorrect questions attempted in GA = 8

No. of incorrect questions attempted in English = 13

No. of incorrect questions attempted in Computer = 4

No. of incorrect questions attempted in Reasoning =
5

No. of incorrect questions attempted in Quant = 0

Therefore total no. of question attempted incorrect
= 8 + 13 + 4 + 5 + 0 = 30

**16. C)**Let the no. of subjects be ‘n’ and the average marks be ‘x’.

Hence the total marks = nx

ATQ, (n + 1)(x – 1) = (nx – 40) + (23 + 25) => x
– n = 9 ……… (i)

And (n + 2)(x + 1) = (nx – 40) + (23 + 25) + 57
=> 2x + n = 63 ……… (ii)

Solving eq (i) and eq (ii) we get n = 15

**17. A)**14 Men’s 1 day work = 1/18

14 Men’s 3 days work = 3/18 = 1/ 6

Now, 15 women’s 1 day work = 1 / 24

1 women 1 day work = 1 / (15 x 24)

10 women’s 1 day work = 10 / (15 x 24) = 1/36

10 women’s 3 days work = 3 x (1 / 36) = 1/12

Total work done = 1/6 + 1/12 = (2 + 1) / 12 = 3/12 =
1/4

Now, remaining work = 1 —1/4 = 3 / 4

**18. A)**Let, distance = D and speed of fast train = S

_{1}

and speed of slow train = S

_{2}
D =14(S

_{1 }+ S_{2}) …..(i)
(when both trains are in opposite direction)

and D =182(S

_{1}– S_{2}) ……(ii)
(when both trains are in same direction)

Now, from Eqs. (i) and (ii), we have

14(S

_{1}+ S_{2}) = 182(S_{1}– S_{2})
(S1 + S2) = 13(S

_{1}— S_{2})
13S

_{1 }– 13S_{2}= S_{1}+ S_{2}
12S

_{1}= 14S_{2}
S

_{1}/ S_{2 }= 14 / 12 = 7 / 6
S

_{1}= 7/6 S_{2}
Required percentage = [(7/6) – 1] × 100%

= (100 / 6) % = 16.666% = 16.67%

**19. C)**Let, money invested by A, B and C are respectively Rs. x, Rs. 2x and Rs. 4x. Then, ratio of profit of A, B and C at the end of the year = {x x 6 + ( +x/2) × 6)

= {2x x 6 + (2x + 2x) x 6} : { 4x × 6 + (4x – (1/4)
x 4x ) × 6}

= (6x + (3x / 2) x 6) : (12x + 24x) : (24x + 18x)

=15x : 36x : 42x = 5 : 12 : 14

**20. E)**A = 2C

(B+2)/(C-2) = 5/2

C = D – 14

D – A = 4

Solve, A = 20, B = 18, C = 10, D = 24