# Mixed Quantitative Aptitude Questions Set 109

Directions(1-5): Find the wrong number from the following series.

1. 22,46,21,43,20,42
42
20
46
21
43
Option E
*2+2
/2-2
*2+2
/2-2
*2+2
44 should be in place of 43.

2. 4,9,27,125,726
726
27
9
125
4
Option B
2!+2
3!+3
4!+4
5!+5
6!+6
28 should be in place of 27.

3. 12,21,39,65,102,147
21
12
39
65
147
Option D
+(9*1)
+(9*2)
+(9*3)
+(9*4)
+(9*5)
66 should be in place of 65.

4. 39,88,152,233,345,454
152
88
345
39
233
Option C
+7^2
+8^2
+9^2
+10^2
+11^2
333 should be in place of 345.

5. 120,80,60,90,180,450
120
80
60
90
450
Option B
*0.5
*1
*1.5
*2
*2.5
60 should be replaced in place of 80.

6. P and Q can clean a room in 18 hours and 15 hours resp. If P works for 3 hours and Q works for the remaining time only, in how many hours would Q complete the remaining work?
10(1/6) hours
12(1/2) hours
8(1/4) hours
13(1/3) hours
10(1/5) hours
Option B
Work done in first 3 hours = 3*1/18 = 1/6
Remaining work = 1-1/6 = 5/6
Work can be completed by Q = 5/6*15 = 12(1/2) hours

7. A train running at a speed of 54 km/hr crosses a man standing on the platform in 28 seconds. What would be the length of the platform which is 33.33% more than the length of the train?
550 m
444 m
440 m
560 m
500 m
Option D
Speed of train = 54 km/hr. = 15 m/s.
Length of train = 15*28 = 420 m
Length of platform = 420 + 33.33% of 420 = 420+140 = 560 m

8. The average runs scored by the team India in the first 20 overs of a one-day match (50 overs) is 5.45 and in the next 10 overs is 6.1. If the total runs scored by team India in 50 overs is 320, find the average runs scored in the last 20 overs.
7.5
6.6
7.7
5.2
8.8
Option A
Runs scored by team India in the first 30 overs
= 5.45*20+6.1*10 = 170
Runs scored in last 20 overs = 320 – 170 = 150
Required average runs scored in the last 20 overs = 150/20 = 7.5

9. A cylinder of radius 27 cm and height 20 cm is melted down and cast into small cones of radius 9 cm and height 10 cm. Find the number of cones formed.
62
40
50
45
54
Option E
Number of cones formed = Volume of cylinder/Volume of cone
= (pi*r*r*h)/(1/3*pi*r1*r1*h) = 54

10. In how many ways can 2 women and 4 men stand in a queue such that none of the women stand on either ends of the queue?
257 ways
288 ways
250 ways
222 ways
280 ways
Option B
Number of ways for the women to stand = 4C2*2!
= 12
Number of ways for the remaining 4 men to stand = 4! = 24
Required number of possible arrangements = 12*24 = 288 ways