# Quant Test for NIACL AO Prelims Exam Set – 2

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

1. 23.97% of 7624.89 + 9.988% of ? – 14.78% of 1779.89 = 64.78% of 2959.9
1005
1100
1215
1000
1805
Option E
23.97% of 7624.89 + 9.988% of ? – 14.78% of 1779.89 = 64.78% of 2959.9 =>24% of 7625 + 10% of ? – 15% of 1780 = 65% of 2960
=> 1830 + 0.20* ? – 267 = 1924
=>? = 1805

2. (4912.7)^1/3 – (5329.10)^1/2 + ?^2 = (7920.99)^1/2 + (13823.78)^1/3
5
10
8
13
6
Option D
(4912.7)^1/3 – (5329.10)^1/2 + ?^2 = (7920.99)^1/2 + (13823.78)^1/3
=> (4912)^1/3 – (5329)^1/2 + ?^2 = (7921)^1/2 + (13824)^1/3
=> 17 – 73 + ?^2 = 89 + 24
=> ?^2 = 89 + 24 + 73 – 17
=> ? = 13

3. (24.55)^5 / (8.42)^6 * (2.9)^3 = (2.9)^(? – 2)
5
4
8
7
9
Option C
(24.55)^5 / (8.42)^6 * (2.9)^3 = (2.9)^(? – 2)
=> (2.9^3)^5 / (2.9^2)^6 * (2.9)^3 = (2.9)^(?-2)
=> ? = 8

4. 17.98*7(1/7)% of 363.98 + 39.98% of 519.98 = ?^2
26
20
25
12
19
Option A
17.98*7(1/7)% of 363.98 + 39.98% of 519.98 = ?^2
=> ?^2 = 18*1/14*364 + 0.4*520
=> ?^2 = 18*26 + 208
=> ? = 26

5. 14(3/5) + 8(2/3) – 5(4/5) = (?)^1/2 + 6(7/15)
92
100
121
80
111
Option C
14(3/5) + 8(2/3) – 5(4/5) = (?)^1/2 + 6(7/15)
=> (?)^1/2 = 73/5 + 26/3 – 29/5 – 97/15
=> (?)^1/2 = 44/5 + 26/3 – 97/15
=> (?)^1/2 = [132 + 130 – 97]/15
=> ? = 121

6. A boat can cover a “r” km upstream in 7 hours. It can cover “r+16” km downstream in 5 hours. Find the time taken by boat to cover 66 km upstream and 70 km downstream if the speed of the boat in still water and speed of stream are in the ratio 4:1 resp.
8 hours
9 hours
7 hours
6 hours
5 hours
Option B
Let the speed of the stream be x km/hr and speed of the still water be 4x km/hr.
Upstream =4x – x = 3x kmhr.
Downstream = 5x km/hr.
r = 7*3x = 21x
And r + 16 = 5*5x
=>21x + 16 = 25x
=> x = 4
Downstream speed = 20 km/hr.
Upstream speed = 12 km/hr.
Time taken to cover 66 km upstream and 70 km downstream speed
= 66/12+ 70/20 = 9 hours

7. Radius of the circular filed is “r” metre. Find the area of the circular field if the cost of fencing the field along its circumference at the rate of Rs. 18.5 per meter is Rs. 4652.
44567 m^2
41400 m^2
40450 m^2
49896 m^2
44050 m^2
Option D
Circumference of the circular field = 2*22/7*r = 14652 /18.5
=> r = 126
Area of the circular field = 22/7*126*126 = 49896 m^2

8. A.B and C together can complete work in “x” days. B alone can complete the work in “3x” days. A is 20% efficient than B. C alone can complete the work in 30 days. In how many days A and C together can complete the work ?
7 days
9 days
12 days
10 days
8 days
Option C
Time taken by A alone to complete the work = 3x*(100/120) = 5x/2 days
Now, 1/3x + 2/5x + 1/30 = 1/x
=> [15-5-6]/15x = 1/30
=> x = 8
A alone can complete the work = 5*8/2 = 20 days
Time taken by A and C together to complete the work = 1/[1/20+1/30] = 12 days

9. Present ages of Radha and Geeta are in the ratio 3:4 resp. Mohan is 6 years older than Geeta. 8 years ago the ratio of the ages of Mohan and Sohan was 11:24 resp. If the present age of Sohan is two more than thrice teh age of Radha then find the present average age of Radha,Geeta, Mohan and Sohan.
28 years
33 years
32 years
25 years
30 years
Option C
Let the present ages of Radha and Geeta be 3x and 4x resp. and the present ages of Mohan and Sohan be 11y and 24y resp.
Present age of Mohan and Sohan = 11y+8 years and 24y+8 years resp.
Now, 4x+6 = 11y+8
=> x= [11y+2]/4
Also, 24y + 8 = 3*3x+2
=> y = 2
=> x = 6
Present ages of Radha, Geeta, Mohan and Sohan are 18 years, 24 years, 30 years and 56 years resp.
Required average age = (18+24+30+56)/4 = 32 years

10. A bag contains 10 red, 11 blue and “x” pink balls. Two balls are randomly drawn from the bag and the probability that both the balls are pink is 1/8. What is the probability that one ball is red and other ball is blue?
7/20
3/22
5/24
6/23
5/21
Option C
Probability of both balls are pink = xC2/(10+11+x)C2 = 1/8
=>xC2/(21+x)C2 = 1/8
=> x(x-1)/(21+x)(20+x) = 1/8
=> 7x^2 – 49x – 420 = 0
=> x = 12, -5
Bag contains pink balls = 12
Probability of one red and one blue ball = 10C1*11C1/33C2 = 5/24

## One Thought to “Quant Test for NIACL AO Prelims Exam Set – 2”

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