# Mixed Quantitative Aptitude Questions Set 87

Directions(1-5): In each of these questions a number series is given. Below the series one
number is given followed by (a), (b), (c), (d) and (e) You have to complete this series following the
same logic as in the original series and answer the question that follows.

1. 5, 9, 25, 91, 414, 2282.5
3, (a), (b), (c), (d), (e)
What will come in place of (c) ?

55.42
43.25
54.12
60.15
64.75
Option E
(a) 3 × 1.5 + 1.5 = 4.5 + 1.5 = 6
(b) 6 × 2.5 + 2.5 = 15 + 2.5 = 17.5
(c) 17.5 × 3.5 + 3.5 = 61.25 + 3.5 = 64.75

2. 7, 6, 10, 27, 104, 515
9, (a), (b), (c), (d), (e)
What will come in place of (d) ?

152
131
111
117
105
Option A
(a) 9 × 1 – 1 = 8
(b) 8 × 2 – 2 = 14
(c) 14 × 3 – 3 = 39
(d) 39 × 4 – 4 = 152

3. 6, 16, 57, 244, 1245, 7506
4, (a), (b), (c), (d), (e)
What will come in place of (d) ?

1151
1005
1453
1231
1147
Option B
(a) 4 × 2 + 22 = 8 + 4 = 12
(b) 12 × 3 + 32 = 36 + 9 = 45
(c) 45 × 4 + 42 = 180 + 16 = 196
(d) 196 × 5 + 52 = 980 + 25 = 1005

4. 15, 9, 8, 12, 36, 170
19, (a), (b), (c), (d), (e)
What will come in place of (b) ?

16
17
13
10
21
Option A
(a) 19 × 1 – 1 × 6 = 19 – 6 = 13
(b) 13 × 2 – 2 × 5 = 26 – 10 = 16

5. 8, 9, 20, 63, 256, 1285
5, (a), (b), (c),(d), (e)
What will come in place of (e) ?

845
919
925
909
898
Option C
(a) 5 × 1 + 1 = 6
(b) 6 × 2 + 2 = 14
(c) 14 × 3 + 3 = 45
(d) 45 × 4 + 4 = 184
(e) 184 × 5 + 5 = 925

6. A, B, C are partners. A receives 2/7 of the profit and B & C share the remaining profit equally. A’s income is increased by Rs. 240 when the profit rises from 10% to 15%. Find the capital invested by A, B and C respectively?
Rs. 2100, Rs. 6000, Rs. 8000
Rs. 3800, Rs. 4000, Rs. 6000
Rs. 3100, Rs. 6000, Rs. 5000
Rs. 4800, Rs. 6000, Rs. 6000
Rs. 1500, Rs. 9000, Rs. 6000
Option D
A’s share = 2/7
B’s share = 5/14 = c′𝑠 share A’s share in total profit
when profit is 10% = 2/7*10 = 20/7%
A’s share in total profit when profit is 15% = 30/7%
Difference = 30/7% – 20/7% = 240
profit = 24*7*100 = 16,800
Capital investment by A = 16,800 *2/7 = 4,800
Capital investment by B = (16,800 – 4,800)/2 = 6,000
Capital invested by A, B & C = Rs. 4800, Rs. 6000, Rs. 6000

7. Profit on selling 10 candles equals selling price of 3 bulbs. While loss on selling 10 bulbs equals selling price of 4 candles. Also profit percent-age equals to the loss percentage and cost of a candle is half of the cost of a bulb. What is the ratio of selling price of candle to the sellings price of a bulb?
3:2
5:3
4:1
5:2
3:1
Option A
Let CP of one candle = 𝑥
C.P. of one bulb = 2𝑥
Let SP of one candle = c & SP of one Bulb = b
According to question
3b/10𝑥 × 100 = 4c/(10 × 2𝑥)
=> c/b = 3/2

8. Three men A, B and C working together 8 hours per day can print 960 pages in 20 days. In a day B prints as many pages more than A as C prints as many pages more than B. The number of pages printed by A in 4 hours equal to the number of pages printed by C in 1 hours. How many pages C prints in each hour?
4
2
5
3
6
Option B
(A + B + C)per hour = 960/20×8 = 6 … (i)
Let C prints 4𝑥 pages per hour.
All print 𝑥 pages per hour.
According to question,
C – B = B – A
=> 2B = A + C
=> B = 5𝑥/2
From (i) 𝑥 + 5𝑥/2 + 4𝑥 = 6
=> 𝑥 = 4/5
No. of pages print by B per hour = 5/2 × 4/5 = 2

9. A square floor of the dimensions 72 cm × 72 cm has to be laid with rectangular tiles whose length and breadth are in the ratio 3 : 2. What is the difference between the maximum number of tiles and minimum numbers of tiles, given that the length and the breath are integers ?
858
890
845
800
987
Option A
Let length and breadth of tiles are 3𝑥 and 2𝑥.
No. of tiles = 72×72/6𝑥^2 = 864/𝑥^2
Maximum no. of tiles is when 𝑥 = 1
And no. of tiles = 864
Minimum no. of tiles is when 𝑥^2 = 144
=> 𝑥 = 12 then no. of tiles = 864/144 = 6
Required difference = 864 − 6 = 858

10. A box contains 6 bottles of variety1 drink, 3 bottles of variety 2 drink and 4 bottles of variety 3 drink. Three bottles of them are drawn at random, what is the probability that the three are not of the same variety?
837/872
871/892
842/877
800/793
833/858
Option E
Total number of drink bottles = 6 + 3 + 4 = 13.
Let S be the sample space.
Then, n(S) = number of ways of taking 3 drink bottles out of 13.
Therefore, n(S) = 13C3
= (13 x 12 x 11)/(1 x 2 x 3)
= 66 x 13 = 858.
Let E be the event of taking 3 bottles of the same variety.
Then, E = event of taking (3 bottles out of 6) or (3 bottles out of 3) or (3 bottles out of 4)
n(E) = 6C3 + 3C3 + 4C3
= 6 x 5 x 4 / 1 x 2 x 3 + 1 + 4 x 3 x 2 / 1 x 2 x 3
= 20 + 1 + 4 = 25.
The probability of taking 3 bottles of the same variety = n(E)/n(S) = 25/858.
Then, the probability of taking 3 bottles are not of the same variety
= 1 – 25/858 = 833/858.