__Quantitative Aptitude Quiz for IBPS Exams 2017__**Dear Aspirants,**

Welcome to

**Mentor for Bank Exams**Quantitative Aptitude Quiz section. The following quiz covers Number Series (5 questions), Simplifications (5 Questions), Data Interpretation (5 Questions), Word Problems (5 Questions). All the best for upcoming**IBPS Exams 2017**.**Directions (1 – 5): Find the Number missed in the following questions:**

**1. 4, 5, 6, 14, ?, 100.5**

A) 32.5

B) 47.5

C) 67.5

D) 37.5

E) 27.5

**2. 2, 2, 7, ?, 87, 342**

A) 21

B) 26

C) 23

D) 24

E) 22

**3. 4, ?, 14, 40, 88, 170**

A) 9

B) 5

C) 6

D) 7

E) 2

**4. 6, 6, 7, ?, 91, 463**

A) 33

B) 43

C) 38

D) 25

E) 44

**5. 2, 9, 39, 161, ?, 2613**

A) 675

B) 670

C) 665

D) 651

E) 655

**Directions (6 – 10): What will come in place of question-mark (?) in the following question?**

**6. (7171 + 3854 + 1195) ÷ (892 + 214 + 543) = ?**

A) 7

B) 11

C) 17

D) 9

E) None of these

**7. 1164 * 128 ÷ 8.008 + 969.007 = ?**

A) 13684

B) 17694

C) 12694

D) 19594

E) None of these

**8. 69.008% of 699.998 + 32.99% of 399.999 = ?**

A) 645

B) 615

C) 715

D) 725

E) 675

**9. 4374562 * 64 = ? * 7777**

A) 36040

B) 36560

C) 36750

D) 38700

E) 39700

**10. 43931.03 ÷ 2011.02 x 401.04 = ?**

A) 8850

B) 8300

C) 7500

D) 7550

E) 8800

**Directions (11 – 15): Following Line Graph represents the uniform speed of Trains (A, B, C, D and E) on two different days in meter per second and the Bar Chart represents the length of each train in meter.**

**11. Approximately what is the average time taken by train D on day 1 and day 2 to cross the platform of length 500 meter?**

A) 6 Sec

B) 9 Sec

C) 12 Sec

D) Data Insufficient

E) None of these

**12. At what speed must train C travel on Day 2 such that it crosses the platform of 500 meter in half of the expected time?**

A) 139 m/s

B) 149 m/s

C) 159 m/s

D) Data insufficient

E) None of these

**13. What is the average speed of all the trains on Day 1 in Km/Hr?**

A) 449

B) 459

C) 429

D) Data insufficient

E) None of these

**14. What will be the total time travelled by train B, C & D on Day 1?**

A) 554 minutes

B) 224 minutes

C) 334 minutes

D) Data insufficient

E) None of these

**15. What is the ratio of Percentage increase or decrease in speed of Train A from day 1 to day 2 to Train D from day 1 to day 2?**

A) 4:1

B) 3:1

C) 2:1

D) Data insufficient

E) None of these

**16. A man rows to a place 210 km away and comes back to the starting point. If the speed of the stream is 3 kmph and the speed of the boat in still water is 18 kmph, then what is the total time taken by him?**

a) 24 hours

b) 25 hours

c) 26 hours

d) 22 hours

e) None of these

**17. A circular road runs around a circular ground. If the radius of the ground is 3.5m and the difference between the circumference of the outer circle and that of the innercircle is 88m, then the area of the road is**

a) 920m

^{2}
b) 918m

^{2}
c) 924m

^{2}
d) 926m

^{2}
e) None of these

**18. Pipe A and B can fill a tank in 20 minutes and 25 minutes respectively. A third pipe C can empty it in 10 minutes. The first two pipes are kept open for 8 minutes in the beginning and then the third pipe is opened. In what time will the tank become empty?**

a) 72 minutes

b) 64 minutes

c) 56 minutes

d) 48 minutes

e) None of these

**19. Abhinav was conducting an experiment in which the average of 11 observations came to 82, while the average of the first five observations was 77, and that of the last five was 78. What was the measure of the 6th observation?**

a) 127

b) 125

c) 123

d) 120

e) None of these

**20. The simple interest accrued on an amount of Rs. 18500 at the end of three years is Rs. 6150. What would be the compound interest accrued on the same amount at the same rate of interest for the same period?**

a) Rs. 6801.17

b) Rs. 6900

c) Rs. 7000

d) Rs. 7801.23

e) None of these

**Solutions:**

**1. A)**4 * 1 + 1 = 5

5 * 1.5 – 1.5= 6

6 * 2 + 2 = 14

14 * 2.5 – 2.5 = 32.5

32.5 * 3 + 3 = 100.5

**2. E)**2 + 1² – 1 = 2

2 + 2² + 1= 7

7 + 4² – 1= 22

**3. C)**4 + 1² + 1= 6

6 + 3² – 1= 14

14 + 5² + 1= 40……..

**4. D)**6*1 – 1 + 1 = 6

6*2 – 2 – 3 = 7

7*3 – 1 + 5 = 25

**5. D)**2 * 4 + 1 = 9

9 * 4 + 3 = 39

39 * 4 + 5 = 161

161 * 4 + 7 = 651

651 * 4 + 9 = 2613

**6. A)**? = (7171 + 3854 + 1195) ÷ (892 + 214 + 543)

? = 12220 / 1649 ≈ 7

**7. D)**? = 1164 * 128 ÷ 8.008 + 969.007 = 1164 * 16 + 970 = 19594

**8. B)**? = (69 * 700 + 33 * 400) / 100 = 61500 / 100 = 615

**9. E)**? = = 4374562 * 64 / 7777 = 36000

**10. E)**? = (44000 ÷ 2000) x 400 = 8800

**11. B)**Time= Distance/Speed

Time taken on day 1, t1= (train length + Platform
Length)/speed = (225+500)/200 = 3.6 sec

Time taken on day 2, t2= (train length + Platform
Length)/speed = (225+500)/50 = 14.5 sec

Average time = (t1+t2)/2 = 9 seconds

**12. A)**Expected time = (125+500)/ 70 = 9 seconds

Speed needed to cross the platform in 4.5 sec =
(125+ 500)/4.5 = 139 m/s

**13. E)**Average speed on day 1 = total speed/5 = (60+75+105+200+170)/5 = 610/5 = 122 m/s

Average speed on day 1 in km/hr = 122*(18/5) =
439 km/hr

**14. D)**Without knowing the distance travelled time cannot be calculated, so data is not sufficient.

**15. C)**% increase in speed of Train A= [(150-60)/60]*100 = 150 %

% decrease in speed of Train D=
[(200-50)/200]*100 = 75 %

Ratio = 150: 75 = 2: 1

**16. A)**Total time = 210/(18+3) + 210/(18-3)

= 210/21 + 210/15 = 10 + 14 = 24 hours

**17. C)**Let the radius of the outer circle be R m.

And the radius of the inner circle be r m.

Then, according to the question,

2ПR — 2Пr = 88

Or, R — r = (88 × 7)/(2 × 22) = 14

or, R = 14 + r = 14 + 3.5 = 17.5 m

Now, area of the road = П (17.5

^{2}— 3.5^{2}) = 22/7 × 21 × 14 = 924m^{2}**18. A)**LCM of 20, 25 and 10 = 100 units

A can fill (100/120 = ) 5 units in one hour.

A can fill (100/25 = ) 4 units in one hour

Third pipe empties (100/10 =) 10 units in one hour.

In the beginning the tank filled by (A + B) in 8
minutes = 8 x (5 + 4) = 72 units

Third pipe empties 10 units every hours

Third pipe empties the tank in 72 minutes.

**19. A)**The sum of 11 observations =11 x 82 =902

The sum of first five observations = 5 x 77 = 385

The sum of last five observations = 5 x 78 = 390

6th observation = 902 — (385 + 390) =127

**20. A)**given that P = Rs. 18500

T = 3 years

SI – Rs.6150

R = (6150 × 100) / (18500 × 3) = 11%

Now, CI = 18500 [1 + (11/100)]

^{3}– 18500
= 18500 × (111/100) × (111/100) × (111/100) – 18500

= 25301.17 – 18500 = Rs. 6801.17