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 !!

**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**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.**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**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**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**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**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**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**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/C^{2}

Or A = kB/ C^{2}

When B = 16 and C = 2, A = 36:

36 = k*16/ 2^{2}

k = 9

Now when B = 32 and C = 4:

A = 9*32/ 4^{2}**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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