**IBPS RRB Quantitative Aptitude Quiz**

Here, we are providing

**IBPS RRB**Study plan as there is left not enough time to deal in details. The questions asked in the quantitative aptitude 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): What approximate value should come in place of question mark (?) in the following questions?**

**Q1. 24.002 × 14.005 – 7.895 × 5.964 =?**

388

288

188

244

222

Solution:

? ≈ 24 × 14 – 8 × 6

≈ 336 – 48 = 288

6597

6527

6777

5677

6577

**Q3. 40.01² – 23.98² - ? = 31.97²**

4

12

32

36

0

Solution:

≈ 40² – 24² – ? = 32²

⇒ 1600 – 576 – ? = 1024

⇒ 1024 – ? = 1024

⇒ ? = 0

9219

9391

9319

9129

9643

217

107

227

237

307

**Q6. The ages of Raghav and Priti are 40 years and 60 years, respectively. How many years before the ratio of their ages was 3 : 5?**

15 years

20 years

37 years

10 years

16 years

Solution:

**Q7. Eighteen years ago, a father was three times as old as his son. Now the father is only twice as old as his son. Then the sum of the present ages of the son and the father is:**

54

72

105

108

112

Solution:

Let present age of son = x year

∴ Present age of father = 2x years

ATQ,

2x – 18 = 3 (x – 18)

⇒ x = 36 years = son’s age

∴ Father’s present age = 72 years

∴ Required sum = 72 + 36 = 108 years

**Q8. Sushil got married 6 years ago. His age is 7/6 times of his age at the time of his marriage. Three years ago, his son was 3 years old. The ratio of their (Sushil’s son & Sushil) present age is:**

1 : 6

1 : 7

2 : 7

6 : 1

8 : 1

Solution:

**Q9. 1 year ago, a mother was 4 times older to her son. After 6 years, her age becomes more than double her son’s age by 5 years. The present ratio of mother and son age will be:**

13 : 12

3 : 1

11 : 3

25 : 7

7 : 25

Solution:

Let present age of mother and son be x and y years respectively.

Then, x - 1 = 4(y - 1)

⇒ x = 4y – 3 ...(i)

And, x + 6 = 2 (y + 6) + 5 ...(ii)

⇒ 4y – 3 = 2y + 11

⇒ y = 14/2 = 7 years

And, x = 25 years

And, Required ratio = 25 : 7

**Q10. The sum of present age of P and Q is 54 years. After 4 years, ratio of their ages will be 2 : 3. Find the present age of P.**

24.6 years

24.6 years

21.8 years

20.8 years

22.6 years

**Directions (11 - 15): In the following Questions, two equations I and II are given. You have to solve both the equations and Give answer**

**Q11. I. x² – 7x + 6 = 0**

**II. 2y² – 8y + 6 = 0**

If x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or the relationship between x and y cannot be established.

Solution:

I. x² – 7x + 6 = 0

x² – 6x– x + 6 = 0

(x – 6) (x – 1) = 0

x= 1, 6

II. 2y² – 8y + 6 = 0

⇒ y² – 4y + 3 = 0

⇒ y² – 3y – y + 3 = 0

⇒ (y – 1) (y – 3) = 0

⇒ y = 1, 3

No relation

**Q12. I. 3x² + 13x – 16 = 0**

**II. y² – 5y + 4 = 0**

If x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or the relationship between x and y cannot be established.

Solution:

I. 3x² + 13x – 16 = 0

⇒ 3x² + 16x – 3x – 16 = 0

⇒ (3x + 16) (x – 1) = 0

⇒ x = 1, -16/3

II. y² – 5y + 4 = 0

⇒ y² – 4y – y + 4 = 0

⇒ (y – 4) (y – 1) = 0

⇒ y = 4, 1

y ≥ x

**Q13. I. y² + 17y + 72= 0**

**II. x² + 11x + 30 = 0**

If x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or the relationship between x and y cannot be established.

Solution:

I. y² + 17y+ 72 = 0

⇒ y² + 8y + 9y + 72 = 0

⇒ (y + 8) (y + 9) = 0

⇒ y = –8, –9

II. x² + 11x + 30 = 0

⇒ x² + 5x + 6x + 30 = 0

⇒ (x + 5) (x + 6) = 0

⇒ x = –5, –6

x > y

**Q14. I. x + 3y = 8**

**II. 2x +y = 6**

If x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or the relationship between x and y cannot be established.

Solution:

I. x + 3y = 8

II. 2x + y = 6

Multiplying equation (ii) by 2 and then substracting (ii) from (i) we get

x = 2, y = 2

**Q15. I. 2x² – 9x + 10 = 0**

**II. 3y² – 14y + 16 = 0**

If x > y

if x ≥ y

if x < y

if x ≤ y

if x = y or the relationship between x and y cannot be established.

Solution:

I. 2x² – 9x + 10 = 0

⇒ 2x² – 4x – 5x + 10 = 0

⇒ 2x (x – 2) – 5 (x – 2) = 0

⇒ (x – 2) (2x – 5) = 0

x = 2, 5/2

II. 3y² – 14y + 16 = 0

⇒ 3y² – 6y – 8y + 10 = 0

⇒ 3y (y – 2) – 8 (y – 2) = 0

⇒ (y – 2) (3y – 8) = 0

⇒ y = 2, 8/3

No relation