# Quantitative Aptitude For IBPS PO/Clerk Prelims: 28th January 2019

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**Quantitative Aptitude Quiz For **IBPS PO/Clerk Prelims

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.

**Q1. The average of first nine integral multiples of ‘3’ is:**

17

15

18

20

12

**Q2. A number x when divided by 289 leaves 18 as remainder. The same number when divided by 17 leaves y as a remainder. The value of y is:**

1

2

3

5

0

Solution:

Here, the first divisor (289) is a multiple of second divisor (17).

∴ Required remainder = 18/17 = 1

∴ Required remainder = 18/17 = 1

**Q3. If (x + 7954 × 7956) be a square number, then what is the value of x?**

16

9

4

1

25

**Q4. Ronald and Alan are working on an assignment. Ronald take 6 hours to type 32 pages on a computer, while Alan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?**

7 hr. 15 min.

4 hr. 15 min.

8 hr. 15 min.

6 hr. 15 min.

8 hr. 45 min.

**Q5. In four consecutive prime numbers that are in ascending order, the product of the first three is 385 and that of the last three is 1001. The largest given prime number is:**

11

13

17

19

23

**Q6. The average of 11 results is 50. If the average of first 6 results is 49 and that of the last six is 52, then the sixth result is:**

64

48

46

56

65

Solution:

Direct: Sixth result=49×6+52×6–50×11=56

**Q7. Which of the largest of the following fractions?**

3/5

2/3

8/11

11/17

None of these

Solution:

2/3 =.66, 3/5 = .6,

8/11 = .72,

11/17 = .64,

8/11 = .72,

11/17 = .64,

**Q8. When 75 is added to 75% of a number, the answer is the number. Find 40% of that number.**

140

120

110

100

220

**Q9. If the arithmetic mean of 3a and 4b is greater than 50, and a is twice of b, then the smallest possible integer value of a is:**

19

25

23

21

15

**Q10. Find the greatest number, which will divide 215, 167 and 135, so as to leave the same remainder in each case.**

10

20

18

16

12

Solution:

Greatest number which will divide 215, 167 and 135, so as to leave the same reminder in each case will be the HCF of (215 – 167), (167 – 135) and (215 – 135) i.e., the HCF of 48, 32 and 80. Since, HCF of 48, 32 and 80 is 16.

So, the greatest number which divide 215, 167 and 135, so as to leave the same remainder in each case is 16.

So, the greatest number which divide 215, 167 and 135, so as to leave the same remainder in each case is 16.

**Q11. Find the least number which when divided by 10, 11, 15 and 22 leaves 3, 4, 8 and 15 as remainders, respectively.**

332

323

233

325

245

Solution:

Least number which when divided by 10, 11, 15 and 22 so as to leave 3, 4, 8 and 15 as remainders, respectively will be equal to [LCM (10, 11, 15, 22) –7]

Since, LCM of 10, 11, 15 and 22 is 330

∴ The required number = 330 – 7 = 323

Since, LCM of 10, 11, 15 and 22 is 330

∴ The required number = 330 – 7 = 323

**Q12. The L.C.M. of two numbers is 48. The numbers are in the ratio of 2: 3. Find sum of numbers.**

50

60

40

45

30

Solution:

Let the numbers be 2x and 3x.

So, their LCM = 6x

However, the LCM is given as 48

∴ 6x=48 ⇒ x=8

So, the numbers are 16 and 24.

Hence, their sum is 40.

So, their LCM = 6x

However, the LCM is given as 48

∴ 6x=48 ⇒ x=8

So, the numbers are 16 and 24.

Hence, their sum is 40.

**Q13. Find the greatest possible length which can be used to measure 4 m 3 cm, 4 m 34 cm and 4 m 65 cm exactly.**

31 cm

41 cm

35 cm

21 cm

51 cm

Solution:

Required measure

= HCF of 403, 434 and 465=31 cm.

= HCF of 403, 434 and 465=31 cm.

**Q14. Six bells commence tolling together and toll at the intervals of 5, 10, 15, 20, 25 and 30 seconds respectively, in 60 min. How many times do they toll together?**

17

15

13

19

11

**Q15. The LCM and HCF of two numbers are 240 and 16 respectively. If two numbers are in the ratio 3: 5, then the bigger number is:**

90

85

70

80

75

Solution:

Suppose two numbers are 3x and 5x Then, 3x×5x=HCF×240

⇒ 15x²=16×240

⇒ x²=256

⇒ x=16

So, the numbers are

3x = 3×16 = 48 and 5x × 16 = 80

⇒ 15x²=16×240

⇒ x²=256

⇒ x=16

So, the numbers are

3x = 3×16 = 48 and 5x × 16 = 80

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