# Aptitude Questions: Ratio & Proportion Set 6

Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample in Ratio & Proportion , which is common for all the competitive  exams. We have included Some questions that are repeatedly asked in exams !!

1. Three cars travel same distance with speeds in the ratio 2 : 4 : 7. What is the ratio of the times taken by them to cover the distance?
A) 12 : 6 : 7
B) 14 : 7 : 4
C) 10 : 5 : 9
D) 7 : 4 : 14
E) 14 : 10 : 7
B) 14 : 7 : 4
Explanation:

s = d/t
Since distance is same, so ratio of times:
1/2 : 1/4 : 1/7 = 14 : 7 : 4

2. Section A and section B of 7th class in a school contains total 285 students. Which of the following can be a ratio of the ratio of the number of boys and number of girls in the class?
A) 6 : 5
B) 10 : 9
C) 11 : 9
D) 13 : 12
E) Cannot be determined
B) 10 : 9
Explanation:

The number of boys and girls cannot be in decimal values, so the denominator should completely divide number of students (285).
Check each option:
6+5 = 11, and 11 does not divide 285 completely.
10+9 = 19, and only 19 divides 285 completely among all.

3. 180 sweets are divided among friends A, B, C and D in which B and C are brothers also such that sweets divided between A and B are in the ratio 2 : 3, between B and C in the ratio 2 : 5 and between C and D in ratio 3 : 4. What is the number of sweets received by the brothers together?
A) 78
B) 84
C) 92
D) 102
E) 88
B) 84
Explanation:

A/B = N1/D1 B/C = N2/D2 C/D = N3/D3
A : B : C : D = N1*N2*N3 : D1*N2*N3 : D1*D2*N3 : D1*D2*D3
A/B = 2/3 B/C = 2/5 C/D = 3/4
A : B : C : D
2*2*3 : 3*2*3 : 3*5*3 : 3*5*4
4 : 6 : 15 : 20
B and C together = [(6+15)/(4+6+15+20)] * 180

4. Number of students in 4th and 5th class is in the ratio 6 : 11. 40% in class 4 are girls and 48% in class 5 are girls. What percentage of students in both the classes are boys?
A) 62.5%
B) 54.8%
C) 52.6%
D) 55.8%
E) 53.5%
B) 54.8%
Explanation:

Total students in both = 6x+11x = 17x
Boys in class 4 = (60/100)*6x = 360x/100
Boys in class 5 = (52/100)*11x = 572x/100
So total boys = 360x/100 + 572x/100 = 932x/100 = 9.32x
% of boys = [9.32x/17x] * 100

5. Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?
A) 8 : 5
B) 7 : 5
C) 5 : 11
D) 4 : 9
E) 5 : 7
E) 5 : 7
Explanation:

Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg
Total brass = 30+40 = 70 kg
So copper in mixture is (50+70) – 70 = 50 kg
So copper to brass = 50 : 70

6. Ratio of A and B is in the ratio 5 : 8. After 6 years, the ratio of ages of A and B will be in the ratio 17 : 26. Find the present age of B.
A) 72
B) 65
C) 77
D) 60
E) None of these
A) 72
Explanation:

A/B = 5/8 , A+6/B+6 = 17/26
Solve both, B = 72

7. A bag contains 25p, 50p and 1Re coins in the ratio of 2 : 4 : 5 respectively. If the total money in the bag is Rs 75, find the number of 50p coins in the bag.
A) 45
B) 50
C) 25
D) 40
E) None of these
D) 40
Explanation:

2x, 4x, 5x
(25/100)*2x + (50/100)*4x + 1*5x = 75
x = 10, so 50 p coins = 4x = 40

8. A is directly proportional to B and also directly proportional to C. When B = 6 and C = 2, A = 24. Find the value of A when B = 8 and C = 3.
A) 42
B) 40
C) 58
D) 48
E) None of these
D) 48
Explanation:

A directly proportional B, A directly proportional to C:
A = kB, A = kC
Or A = kBC
When B = 6 and C = 2, A = 24:
24 = k*6*2
k = 2
Now when B = 8 and C = 3:
A = 2*8*3

9. A is directly proportional to B and also inversely proportional to the square of C. When B = 16 and C = 2, A = 36. Find the value of A when B = 32 and C = 4.
A) 25
B) 20
C) 18
D) 32
E) None of these
C) 18
Explanation:

A = kB, A = k/C2
Or A = kB/ C2
When B = 16 and C = 2, A = 36:
36 = k*16/ 22
k = 9
Now when B = 32 and C = 4:
A = 9*32/ 42

10. A is directly proportional to the inverse of B and also inversely proportional to C. When B = 36 and C = 9, A = 42. Find the value of A when B = 64 and C = 21.
A) 24
B) 40
C) 32
D) 48
E) None of these
A) 24
Explanation:

A = k√B, A = k/C
Or A = k√B/C
When B = 36 and C = 9, A = 42:
42 = k√36/9
k = 63
Now when B = 64 and C = 21:
A = 63*√64/21

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