# Quantitative Aptitude Quiz for IBPS Clerk Prelims: 3rd November 2018

**Dear Students,**

**Numerical Ability or Quantitative Aptitude**Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the

**maximum marks in the examination.**Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.

**Directions (1-5): Study the following table carefully and answer the questions: The table represents the different types of shirts sold by company M in 6 different days.**

**Q1. Find the total number of casual shirts and Denim shirts together sold by company M?**

5000

4450

5250

5005

5025

Solution:

Required total

= (500 + 480 + 175 + 350 + 360 + 645) + (425 + 125 + 620 + 550 + 275 + 500)

= 5,005

= (500 + 480 + 175 + 350 + 360 + 645) + (425 + 125 + 620 + 550 + 275 + 500)

= 5,005

**Q2. Number of formal shirts sold on Tuesday and Wednesday together is what approximate percent of the number of flannel shirts sold on Monday and Thursday together ?**

100%

105%

125%

75%

120%

**Q3. Find the difference between the total number of shirts sold on Monday and Tuesday together and the number of shirts sold on Thursday ?**

1380

1050

1280

1400

1475

Solution:

Total number of shirts sold on Monday and Tuesday

= (500 + 350 + 75 + 125 + 425) + (480 + 50 + 250 + 500 + 125)

= 2880

Total number of shirts sold on Thursday

= (350 + 200 + 190 + 210 + 550)

= 1500

Required difference

= 1380

= (500 + 350 + 75 + 125 + 425) + (480 + 50 + 250 + 500 + 125)

= 2880

Total number of shirts sold on Thursday

= (350 + 200 + 190 + 210 + 550)

= 1500

Required difference

= 1380

**Q4. Find the ratio of total number of shirts sold on Thursday to the total number of shirts sold on Saturday ?**

19 : 15

15 : 38

15 : 19

25 : 19

5 : 7

Solution:

Total number of shirts sold on Thursday

= (350 + 200 + 190 + 210 + 550)

= 1500

Total number of shirts sold on Saturday

= (645 + 321 + 179 + 255 + 500)

= 1900

Required ratio

= 1500 : 1900

= 15 : 19

= (350 + 200 + 190 + 210 + 550)

= 1500

Total number of shirts sold on Saturday

= (645 + 321 + 179 + 255 + 500)

= 1900

Required ratio

= 1500 : 1900

= 15 : 19

**Q5. Find the difference between the total number of Denim shirts and total number of sleeve shirts sold ?**

230

245

280

375

290

Solution:

Total number of Denim shirts

= (425 + 125 + 620 + 550 + 275 + 500)

= 2495

Total number of sleeve shirts

= (125 + 500 + 475 + 210 + 640 + 255)

= 2205

Required difference

= 2495 – 2205

= 290

= (425 + 125 + 620 + 550 + 275 + 500)

= 2495

Total number of sleeve shirts

= (125 + 500 + 475 + 210 + 640 + 255)

= 2205

Required difference

= 2495 – 2205

= 290

**Q6. A can do a piece of work in 45 days. He works only 10 days and left and then B join the job. He finished the remaining work in 42 days. In how many days B alone can finish the whole work ?**

45 days

54 days

63 days

36 days

48 days

**Q7. A man can cover a distance in 3 hours with running at the speed of 9 kmph. If he goes by bike at the speed of 27 kmph, the man cover the same distance in ? (Answer will be in minute)**

40 min

25 min

48 min

60 min

65 min

Solution:

Distance covered by man = 3 × 9 = 27 km

Time (bike) = 27/27 = 1 hour = 60 min

Time (bike) = 27/27 = 1 hour = 60 min

**Q8. Rajan can do a piece of work in 12 days and Mohan can do the same work in 18 days. Rajan and Mohan undertake to do it for Rs. 4800. If they completed the same work in 6 days with the help of Sohan. How much is to be paid to Sohan ?**

Rs. 800

Rs. 900

Rs. 650

Rs. 600

Rs. 525

**Q9. What is time taken by a boy to run in a rectangular field around its perimeter of length 16 meters and width 24 meters, if he runs at the speed of 8 km/hr ?**

36 sec

40 sec

45 sec

27 sec

30 sec

**Q10. Some workers can do a work in 80 days. If 20 workers join the work, the same work finished in 60 days. Find the initial number of workers ?**

72

60

45

56

63

Solution:

Let the initial number of workers is x

80 × x = (x + 20) 60

8x = 6x + 120

2x = 120

x = 60 workers

80 × x = (x + 20) 60

8x = 6x + 120

2x = 120

x = 60 workers

**Directions (11-15): For the two given equations I and II. Give answer-**

**Q11. I. p²+5p+6=0**

**II. q²+3q+2=0**

if p is greater than q.

if p is smaller than q.

if p is equal to q.

if p is either equal to or greater than q.

if p is either equal to or smaller than q.

Solution:

I. p² + 5p + 6 = 0

⇒ (p + 2) (p + 3) = 0

⇒ p = –2, –3

II. q² + 3q + 2= 0

⇒ (q+ 1) (q+ 2) = 0

⇒ q= –1, –2

⇒ p≤ q

⇒ (p + 2) (p + 3) = 0

⇒ p = –2, –3

II. q² + 3q + 2= 0

⇒ (q+ 1) (q+ 2) = 0

⇒ q= –1, –2

⇒ p≤ q

**Q12. I. p²=4**

**II. q²+4q=-4**

if p is greater than q.

if p is smaller than q.

if p is equal to q.

if p is either equal to or greater than q.

if p is either equal to or smaller than q.

Solution:

I. p² = 4

⇒ p = 2, –2

II. q² + 4q +4= 0

⇒ (q+ 2) (q+ 2) = 0

⇒ q = –2, –2

⇒ p ≥ q

⇒ p = 2, –2

II. q² + 4q +4= 0

⇒ (q+ 2) (q+ 2) = 0

⇒ q = –2, –2

⇒ p ≥ q

**Q13. I. p²+p=56**

**II. q²-17q+72=0**

if p is greater than q.

if p is smaller than q.

if p is equal to q.

if p is either equal to or greater than q.

if p is either equal to or smaller than q.

Solution:

I. p² + p -56 = 0

⇒ (p + 8) (p – 7) = 0

⇒ p = –8, 7

II. q² –17q + 72 = 0

⇒ (q – 9) (q – 8) = 0

⇒ q = 8, 9

⇒ p < q

⇒ (p + 8) (p – 7) = 0

⇒ p = –8, 7

II. q² –17q + 72 = 0

⇒ (q – 9) (q – 8) = 0

⇒ q = 8, 9

⇒ p < q

**Q14. I. 3p+2q-58=0**

**II. q+p=23**

if p is greater than q.

if p is smaller than q.

if p is equal to q.

if p is either equal to or greater than q.

if p is either equal to or smaller than q.

Solution:

I. 3p + 2q =58

&

II. p + q = 23

Solving I & II we get

P = 12, q = 11

⇒ p > q

&

II. p + q = 23

Solving I & II we get

P = 12, q = 11

⇒ p > q

**Q15. I. 3p²+17p+10=0**

**II. 10q²+9q+2=0**

if p is greater than q.

if p is smaller than q.

if p is equal to q.

if p is either equal to or greater than q.

if p is either equal to or smaller than q.

Solution:

I. 3p² + 17p + 10 = 0

⇒ 3p² + 15p + 2p + 10 = 0

⇒ (p + 5) (3p + 2) = 0

⇒ p = –5, -2/3

II. 10q² + 9q+ 2= 0

⇒ 10q² + 5q + 4q+ 2 = 0

⇒ (2q + 1) (5q + 2) = 0

⇒ q=-1/2, -2/5

⇒ p < q

⇒ 3p² + 15p + 2p + 10 = 0

⇒ (p + 5) (3p + 2) = 0

⇒ p = –5, -2/3

II. 10q² + 9q+ 2= 0

⇒ 10q² + 5q + 4q+ 2 = 0

⇒ (2q + 1) (5q + 2) = 0

⇒ q=-1/2, -2/5

⇒ p < q

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