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QUESTION: 1

Divide Rs.2340 into three parts, such that first part be double that of second part and second part be 1/3 of the third part.Find the Third part amount?

Solution:

Answer – B.Rs.1170 Explanation : First: Second: Third = 2:1:3 Third part = 3*2340/6 = 1170

QUESTION: 2

The ratio of income of A and B is 2:3. The sum of their expenditure is Rs.8000 and the amount of savings of A is equal to the amount of expenditure of B.What is the their ratio of sum of income to their sum of savings?

Solution:

Answer -A.5:3 Explanation : 2I-E + E = 8000

I = 4000

Sum of their Income = 5*I = 5*4000 = 20,000 Sum of their Savings = 20000-8000 = 12,000 20000:12000 = 5:3

QUESTION: 3

There are 2 containers of equal capacity. The ratio of milk to water in the first container is 4:5 and in the second container is 3:7.If they are mixed up then the ratio of milk to water in the mixture will be

Solution:

Answer – D.67:113 Explanation : 4+5 = 9=> 40:50

3+7 = 10=> 27:63

40+27 : 50:63 = 67:113

QUESTION: 4

There are two numbers. When 25% of the first number is added to the second number, the resultant number is 1.5times th first number.What is the ratio of 1^{st} number to the 2^{nd} number ?

Solution:

Answer – C.4:5 Explanation : A+25/100 + B = 1.5A

A/4 + B = 15A/10

10A+40B/40 =60A/40

10A+40B = 60A

50A = 40B

A/B = 4/5

QUESTION: 5

A bag contains 10p,25p and Rs50p coins in the ratio of 5:2:1 respectively. If the total money in the bag is Rs.120.Find the number of 25p coins in that bag?

Solution:

Answer – A.160 Explanation : 10*5 : 25*2 : 50*1 = 50:50:50 = 1:1:1

120/3 = Rs.40 Rs. 1 = 4 Rs.40 = 4*40 = 160 coins

QUESTION: 6

The ratio of Ganesh’s age and his mother’s age is 5:12.The difference of their ages is 21.The ratio of their ages after 4 years will be

Solution:

Answer – D.19:40 Explanation : 12x – 5x = 21 7x = 21 X = 3

5:12 = 15:36

After 4 years = 19:40

QUESTION: 7

The ratio of students of three classes is 2:3:4. If 15 students are increased in each classes then their ratio turns into 13:18:23. What was the total number of students in all the three classes originally ?

Solution:

Answer – C.225 Explanation : 50:75:100

15 students increased 65:90:115 => 13:18 :23

Total no of students = 50+75+100 = 225

QUESTION: 8

Ravi and Govind have money in the ratio 5 : 12 and Govind and Kiran also have money in the same ratio 5 : 12. If Ravi has Rs. 500, Kiran has

Solution:

Answer – B.Rs.2880 Explanation : Ravi : Kiran = 5/12* 5/12 = 25/144 Kiran = 144*500/25 = 2880

QUESTION: 9

A town with a population of 1000 has provision for 30days, after 10 days 600 more men added, how long will the food last at the same rate ?

Solution:

Answer – C.12 ½ days Explanation : 1000*20/1600 = 12 1/2 days

QUESTION: 10

A man spends Rs.2480 to buy lunch box Rs.120 each and bottles at Rs.80 each,What will be the ratio of maximum number of bottles to lunch box are bought ?

Solution:

Answer – A.13:12 Explanation : Check the ans using option 13*80+ 12*120 = 1040+1440 = 2480

QUESTION: 11

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?

Solution:

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

QUESTION: 12

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?

Solution:

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.

QUESTION: 13

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?

Solution:

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

QUESTION: 14

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?

Solution:

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

QUESTION: 15

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?

Solution:

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

QUESTION: 16

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.

Solution:

A) 72

Explanation: A/B = 5/8 , A+6/B+6 = 17/26

Solve both, B = 72

QUESTION: 17

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.

Solution:

D) 40

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

QUESTION: 18

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.

Solution:

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

QUESTION: 19

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.

Solution:

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}

QUESTION: 20

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.

Solution:

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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