# Quant Test for IBPS Clerk 2018 Prelim Exam Set – 19

Directions(1-5): What will come in place of question mark “?” in the following questions.

1. (1575/63*24+48)*3 = ?^2*6^3
4
5
6
8
3
Option E
(1575/63*24+48)*3 = ?^2*6^3
=> (25*24+48)*3 = ?^2*216
=>?^2 = (648*3)/216 = 9
=> ? = 3

2. 1610*(17.64)^1/2/49 = ?^2 – (961)^1/2
18
13
10
16
14
Option B
1610*(17.64)^1/2/49 = ?^2 – (961)^1/2
=> 1610*4.2/49 = ?^2 – 31
=> ?^2 = 138 – 31
=> ?^2 = 169
=> ? = 13

3. (4095.8)^1/3 + 106.97% of 1599.8 = ?^3
8
15
10
12
13
Option D
(4095.8)^1/3 + 106.97% of 1599.8 = ?^3
=> (4095)^1/3 + 107% of 1600 = ?^3
=> 16+1.07*1600 = ?^3
=> ?^3 = 1728
=> ? = 12

4. 81^3 /27^5 * 9^7 = 3^?
9
13
11
14
17
Option C
81^3 /27^5 * 9^7 = 3^?
=> 3^12 / 3^15*3^14 = 3^?
=> 3^11 = 3^?
=> ? = 11

5. 32*45+16*35 + ? = 48*55+14*65
1440
1550
1300
1250
1050
Option B
32*45+16*35 + ? = 48*55+14*65
=> 1440+560+? = 2640+910
=> ? = 1550

6. A boat goes 90 km upstream and comes back downstream to its original place in overall 16 hours. The speed of boat in still water is x km/hr. Find the value of x if the speed of stream is 3 km/hr.
14
10
15
12
16
Option D
Let speed of boat in still water be x km/hr. 90/[x+3]+ 90/[x-3] = 16 => 4x^2 -45x – 36 = 0
=> 4x(x-12)+3(x-12) = 0
=> x = 12,-3/4
Hence, the value of x = 12 km/hr.

7. A cricket player has an average of x runs after 10 matches. He scored 90 runs in his 11th match due to which the average of his runs becomes (x+7) in the 11 matches. Find his average runs in the 11 matches.
18
20
15
22
24
Option B
His average runs after 10 matches be x.
10x+90 = 11(x+7)
=> x= 13
Average of his runs after 11th match = 13+7 = 20

8. Akash bought a laptop at Rs. x and marked it 40% above the cost price and sold it to Rajan after giving 15% discount thus made a profit of Rs. 9880. What would have been his amount of profit or loss if he had sold it at 30% discount.
1040
1250
1440
1200
1000
Option A
Let CP be x. MP = 1.40*x = Rs. 1.40x
SP = 0.85*1.40x= Rs. 1.19x
Profit = 1.19x – 1.40x = 9880
=> x = Rs. 52000
New SP = 0.70*1.40x = Rs. 0.98x
Loss = x – 0.98x = 0.02x = 0.02*52000 = Rs. 1040

9. The present ages of Gaurav and Ashwani are in the ratio 11:13 resp. And the ratio of their age 12 years ago was 4:5 resp. If after Y years the ratio of their age would be 13:15 resp. The find the value of Y.
9
10
6
8
5
Option D
Now , [11x-12]/[13x-12] = 4/5
=> x = 4
Gaurav = 44 years and Ahswani = 52 years
After Y years, 44+Y/52+Y = 13/15
=>Y = 8 years

10. A bag contains 30 oranges out of which 8 are rotten. 3 oranges are drawn randomly and the probability that at most one orange is rotten is “p” . Find the value of “1-p”.
24/141
23/145
22/143
20/141
24/145
Option E
At most one rotten orange means either no rotten orange or one rotten orange.
Probability = [22C3+22C2*8C1]/30C3 = 121/145
1-P = 1 – 121/145 = 24/145