# Mixed Quantitative Aptitude Questions Set 132

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

1. 8, 4, 4, 8, 32, ?
300
220
245
256
233
Option D
8 × 0.5 = 4
4 × 1 = 4
4 × 2= 8
8 × 4 =32
32 × 8 = 256

2. 129, 128, 124, 115, ?, 74
78
80
99
76
89
Option C
Series is -1², -2², -3², -4²…..
? = 115 – 16 = 99

3. 5, 3.5, 5, ?, 21.5, 56.75
8.8
7.2
9.5
9.7
7.7
Option C
Series is ×0.5+ 1, ×1 + 1.5, ×1.5+2, ×2+2.5
? = 5 × 1.5 + 2 = 9.5

4. 0.5, 1.5, 5, 18, 76, ?
333
385
300
345
359
Option B
*1+1
*2+2
*3+3
*4+4
*5+5
? = 385

5. 255, 230, 250, 235, 245, ?
220
230
240
250
260
Option C
Alternate difference of 5
? = 240

6. Cost price of two articles is same, trade man got profit of 40% on first article, selling price of second article is 25% less than first article, then find over all profit percent
19(1/3)%
21(1/2)%
20(1/4)%
22(1/2)%
22(1/4)%
Option D
Let cost price of both article = 100x
Profit on sell of 1st article = 40% of 100x = 40x
Selling price of 1st article = 140x
Selling price of 2nd Article = 140x – 140×25/100 = 105x
Profit of on 2nd Article = 105x – 100x = 5x
Total profit percent = 45x /200x ×100 = 22(1/2)%

7. Average age of A and B, 2 years ago was 26. If age of A 5 years hence is 40 yrs, and B is 5 year younger to C, then find difference between age of A and C?
6
5
8
7
9
Option E
Average age of A and B 2 year ago = 26
Present Average age (A + B)/ 2 = 28
Present age A + B = 56 A’s age after 5 year = 40
Now A’s age = 40 – 5 = 35
B’s age = 56 – 35 = 21
C’s age = 21 + 5 = 26
Required difference = 35 – 26 = 9

8. Two different amounts are invested in two schemes. In scheme A, amount X is invested at 8 % per annum and in scheme B amount (X+1400) is invested at 12% per annum.After 2 years difference between both interests is 880, then find value of X?
5600
6000
6800
4700
4000
Option C
In scheme A Interest = X×8×2/ 100
In scheme B Interest (X + 1400)×12×2/100
ATQ, (X + 1400)×12×2 /100 – X×8×2 /100 = 880
X = 6800

9. Ratio of speed of boat in down stream and speed of stream is 9:1, if speed of current is 3 km per hr, then find distance travelled(in km) upstream in 5 hours.
100
90
105
95
88
Option C
speed of current = y = 3
Down stream speed = 9 × 3 = 27 km/h
Speed of boat = x
x + y = 27
x = 24
y= 3
Distance travelled upstream in 5 hr = speed × Time
= (x – y) ×5 = (24 – 3) × 5
= 21 × 5= 105 km

10. Sum of 4 consecutive even numbers are greater than three consecutive odd numbers by 81. If sum of least odd and even number is 59, then find the sum of largest odd and even numbers.
56
69
70
74
80
Option B
Sum of least even and odd number = 59
x + y = 59
Now 4th consecutive even number is x + 6 and 3rd consecutive odd number is y + 4.
Required value
x + 6 + y + 4
= x + y + 10
= 59 + 10 = 69