# Quant Test for NIACL AO Prelims Exam Set – 12

Directions(1-5): Find the missing term ‘?’ of the following series.

1. 8,13,26,51,92,?
133
153
140
132
150
Option B
+(1^2+2^2)
+(2^2+3^2)
+(3^2+4^2)
+(4^2+5^2)
+(5^2+6^2)
? = 153

2. 630,210,84,42,?,28,56
15
20
28
16
13
Option C
/3
/2.5
/2
/1.5
/1
/0.5
? = 28

3. 463,453,433,403,?,313
333
363
350
344
362
Option B
-10.-20,-30,-40,-50
? = 363

4. 24,16,28,74,?,1187.5
222
255
241
263
272
Option D
*0.5+4
*1.5+4
*2.5+4
*3.5+4
*4.5+4
? = 263

5. 23,40,61,86,?,148
95
100
88
115
90
Option D
+17,+21,+25,+29,+33
? = 115

6. At what rate of simple interest must Rs.15900 be deposited for 4 years to receive the same interest that is obtained by investing Rs.15000 at 12% per annum compound interest compounded for 2 years ?
8
6
7
5
4
Option B
Compound interest earned = 15000*{(1+0.12)^2 – 1} = Rs.3816
Let rate be r%.
Simple interest earned = 15900*4*r/100 = 3816
=> r = 6

7. If the numerator of a fraction is increased by 40% and denominator in increased by 25% then the resultant fraction will become 7/5. What is the original fraction?
3/5
11/9
7/2
4/3
5/4
Option E
Let the original fraction be x/y. (x*1.4)/(y*1.25) = 7/5
=> x/y = 5/4

8. Find the area of the circular field having radius equal to the breadth of a room with area of four walls 840 m^2 and height 10 m if it is given that ratio of length and breadth of the room is 4:3 resp.(Take pi = 3)
900 m^2
818 m^2
919 m^2
972 m^2
700 m^2
Option D
2(l+b)*h = 840
=> 2(4x+3x)*10 = 840
=> x = 6
Breadth = 3*6 = 18 m
Area of the circular field = pi*(18)^2 = 972 m^2

9. Present ages of A and B are in the ratio of 2:3 resp. Present ages of C and D are in the ratio of 8:7 resp. Present average age of A ,B and C is 46 years. Present average age of B,C and D is 48 years. Find the difference between the present ages of B and C.
4 years
2 years
5 years
8 years
6 years
Option E
Let the present ages of A,B,C and D are 2x,3x,8y and 7y resp.
Now, 2x+3x+8y = 3*46 5x+8y = 138 —-(1)
Also, 3x+8y+7y = 3*48 —–(2)
On solving (1) and (2), we get
x = 18
y = 6
Difference between present ages of B and C = 54 â€“ 48 = 6 years

10. A cricket team of 11 players is to be selected from a group of 8 batsmen, 5 bowlers and 3 wicketkeepers. In how many ways a team a having at least 4 bowlers and exactly 1 wicketkeeper is selected?
500
555
588
610
585
Option C
Required number of ways = 3C1*5C4*8C6 + 3C1*5C5*8C5
= 420+168 = 588