# IBPS RRB 2019 Prelims Quantitative Aptitude: PO/Clerk | 7th July

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
Q2.
6597
6527
6777
5677
6577
Solution:
Q3. 40.01² – 23.98² - ? = 31.97²
4
12
32
36
0
Solution:
≈ 40² – 24² – ? = 32²
⇒ 1600 – 576 – ? = 1024
⇒ 1024 – ? = 1024
⇒ ? = 0
Q4.
9219
9391
9319
9129
9643
Solution:
Q5.
217
107
227
237
307
Solution:
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:
Let a year ago, the ratio of Raghav and Priti’s ages was 3: 5.
⇒ 200 – 5a = 180 – 3a
⇒ a = 10 years
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:
Let Sushil’s present age = x years
∴ His age during marriage = (x – 6) years
ATQ,
⇒ x = 42 years
∴ present age of son = 3 + 3 = 6 years
∴ Required ratio = 42: 6
= 7: 1 or 1: 7
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
Solution:
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

You May also like to Read: