# Mixed Quantitative Aptitude Questions Set 139

1. Rs. 1800 is given at 20% per annum SI while Rs (1800 – P) is given at 30% per annum CI. If the difference between both interests at the end of two years is Rs 315. Find P.
200
400
300
100
500
Option C
Total S.I. = 1800×2×20/ 100 = Rs. 720
Total C.I. = (1800 − 𝑝) [(1 + 30 100 ) 2 − 1] = (1800 − 𝑃) × 69/ 100
ATQ, (1800 − 𝑃) × 69/ 100 − 720 = 315
(1800 − 𝑃) = 100/ 69 × 1035
1800 − 𝑃 = 1500
⇒ 𝑃 = 300

2. Three partners A , B and C invested their amounts in ratio 7:5:3. At the end of six months, A withdraws his amount such that his total investment will be equal to C’s initial investment. If C’ share in annual Profit is Rs. 3600. A’s annual profit will be ?
5000
4000
3000
6000
2000
Option D
Let their initial investments be 7𝑥, 5𝑥 and 3𝑥 respectively.
Ratio of their profit = 7𝑥 × 6 + 3𝑥 × 6 ∶ 5𝑥 × 12 ∶ 3𝑥 × 12
= 60𝑥 ∶ 60𝑥 ∶ 36𝑥 = 15 ∶ 15 ∶ 9
A’s profit = 3600 × 39/ 9 × 15/ 39
= Rs. 6000

3. B’s age is 10 years older than A. If the ratio of B’s age 11 years hence and C’s present age is 3:2. At present C’s age is thrice A’s age. What will be the age of C after 7 years?
22
27
25
28
21
Option C
Let, age of A be ‘𝑥’ years.
Then age of B = ‘𝑥 + 10’ years
And age of C = 3𝑥 years
ATQ, (𝑥+10+11)/ 3𝑥 = 3/ 2
(𝑥 + 21)2 = 9𝑥
7𝑥 = 42
⇒ 𝑥 = 6
Age of C after 7 years = 3𝑥 + 7
= 18 + 7 = 25 years

4. If the ratio of curved surface area to the volume of cylinder is 4:7 while the ratio of diameter to the height of cylinder is 14:5. Find the total surface area of cylinder.?
118
120
125
110
132
Option E
2𝜋𝑟ℎ/ 𝜋𝑟^ 2ℎ = 4/ 7
2 𝑟 = 4/ 7
𝑟 = 7/ 2
2𝑟/ ℎ = 14/ 5
⇒ ℎ = 5 /2
Total surface area = 2 × 22/ 7 × 7 /2 ( 7 /2 + 5/ 2 )
= 22 × 12 /2 = 132

5. A and B sold two articles at 25% profit and 40% profit respectively. If total profit is Rs. 178 and the cost price of A is Rs 120 less than B. find the CP of B.
280
200
320
220
300
Option C
Let, CP of B be 𝑥 + 120
And that of A be 𝑥
Then, 25/100 × 𝑥 + 40/100 (𝑥 + 120) = 178
65𝑥 /100 + 48 = 178
𝑥 = 200
C.P. of B = 𝑥 + 120 = Rs. 320

6. Directions(6-10): What approximate value will come in place of question mark ‘?’ in the following questions.

7. 24.001 × 14.999 × 9.998 = ?
3230
3600
3125
3000
3250
Option B
24.001 × 14.999 × 9.998 = ?
24 × 15 × 10 ≈ 3600

8. 14.003√?+ 53.0345√? = 67/26.999 × (? )
729
772
670
789
770
Option A
14.003√?+ 53.0345√? = 67/26.999 × (? )
14√?+ 53√? = 67/27 ×?
67√?= 67/ 27 ×?
√? = 27
? = 729

9. 10.11×36.93+√48.875 × 19.99 = 17.231 × √?
700
900
600
800
1000
Option B
10.11×36.93+√48.875 × 19.99 = 17.231 × √?
10 × 37 + 7 × 20 = 17 × √?
√? = 30
? = 900

10. 1898.88 ÷ 189.921 + 9.99 + (? )^2 = 83.89
6
7
9
8
4
Option D
1898.88 ÷ 189.921 + 9.99 + (? )^2 = 83.89
≈ 1900 ÷ 190 + 10 +?^2 = 84
?^ 2 = 84 − 10 − 10
? = 8

11. 39.7% 𝑜𝑓 801 − 250.17 = ? −63% 𝑜𝑓 801
567
647
663
574
600
Option D
39.7% 𝑜𝑓 801 − 250.17 = ? −63% 𝑜𝑓 801
≈ 40 × 8 – 250
=? −63 × 8
?= 574