Ratio and Proportion Practice Test Questions – Aptitude -- Free Practice

Question 1

The students in three classes are in the ratio 2 : 3 : 5. If 40 students are increased in each class the ratio changes to 4 : 5 : 7. Originally the total number of students was:

A

180✓✗

B

400✓✗

C

100✓✗

D

200✓✗

Explanation

Let the original numbers of students be 2x, 3x and 5x.
After the increase: (2x+40) : (3x+40) : (5x+40) = 4 : 5 : 7
From the first two terms: 5(2x+40) = 4(3x+40)
10x + 200 = 12x + 160
2x = 40, so x = 20
Original total = 2x + 3x + 5x = 10x = 10 × 20 = 200

Question 2

A bag contains coins of denominations 1 rupee, 2 rupee and 5 rupees. Their numbers are in the ratio 4:3:2. If the bag has a total of Rs. 1800, then find the number of 2 rupee coins.

A

270✓✗

B

230✓✗

C

180✓✗

D

210✓✗

Explanation

Let the number of 1-rupee, 2-rupee and 5-rupee coins be 4k, 3k and 2k respectively.
Total value = 1(4k) + 2(3k) + 5(2k) = 4k + 6k + 10k = 20k
20k = 1800, so k = 90
Number of 2-rupee coins = 3k = 3 × 90 = 270

Question 3

A bag contains 25 paise, 10 paise and 5 paise coins in the ratio 3:2:1. If the total value is Rs. 40, the number of 5 paise coins is:

A

45✓✗

B

48✓✗

C

40✓✗

D

20✓✗

Explanation

Let the number of 25p, 10p, 5p coins be 3k, 2k, k respectively.
Total value (in paise) = 25(3k) + 10(2k) + 5(k) = 75k + 20k + 5k = 100k paise = k rupees
k = 40 (since total value = Rs. 40)
Number of 5 paise coins = k = 40

Question 4

What must be added to each term of the ratio 49:68 so that it becomes 3:4?

A

3✓✗

B

5✓✗

C

8✓✗

D

9✓✗

Explanation

Let x be added to each term.
49 + x68 + x = 34
4(49 + x) = 3(68 + x)
196 + 4x = 204 + 3x
x = 8

Question 5

The expenditure and savings of a person are in the ratio 4:1. If his savings are increased by 25% of his income, then what is the new ratio of his expenditure and savings?

A

11:9✓✗

B

8:5✓✗

C

7:5✓✗

D

7:4✓✗

Explanation

Let income = I. Expenditure = 4I5, Savings = I5
New savings = I5 + 0.25I = 4I20 + 5I20 = 9I20
New expenditure = I - 9I20 = 11I20
New ratio = 11I20 : 9I20 = 11 : 9

Question 6

P, Q and R are three cities. The ratio of average temperature between P and Q is 11:12 and that between P and R is 9:8. The ratio between the average temperature of Q and R is:

A

22 : 27✓✗

B

27 : 22✓✗

C

32 : 33✓✗

D

None of these✓✗

Explanation

P:Q = 11:12 ⇒ P = 11k, Q = 12k
P:R = 9:8 ⇒ P = 9m, R = 8m
Equating P: 11k = 9m ⇒ k:m = 9:11. Take k=9, m=11:
P = 99, Q = 12(9) = 108, R = 8(11) = 88
Q:R = 108:88 = 27:22

Question 7

The third proportional between (a²-b²) and (a+b)² is:

A

a + ba - b✓✗

B

a - ba + b✓✗

C

(a - b)²a + b✓✗

D

(a + b)³a - b✓✗

Explanation

If x is the third proportional: (a²-b²) : (a+b)² = (a+b)² : x
x = (a+b)⁴a²-b² = (a+b)⁴(a-b)(a+b) = (a+b)³a-b

Question 8

The ratio of the number of boys and number of girls in a school is found to be 15:32. How many boys and an equal number of girls should be added to bring the ratio to 2:3?

A

19✓✗

B

20✓✗

C

23✓✗

D

27✓✗

Explanation

Let boys = 15, girls = 32 (matching the given ratio). Let x be added to each.
15 + x32 + x = 23
3(15 + x) = 2(32 + x)
45 + 3x = 64 + 2x
x = 19

Question 9

The ratio of the prices of two Fans was 16 : 23. Two years later, when the price of the first has increased by 10% and that of the second by Rs. 477, the ratio of the prices becomes 11 : 20. Find the original prices of the two Fans.

A

Rs. 848, Rs. 1,219✓✗

B

Rs. 838, Rs. 1,119✓✗

C

Rs. 828, Rs. 1,219✓✗

D

Rs. 848, Rs. 1,229✓✗

Explanation

Let the original prices = 16x and 23x.
New prices: 16x × 1.1 = 17.6x and (23x + 477)
17.6x / (23x + 477) = 1120
352x = 253x + 5247
99x = 5247, so x = 53
Original prices = 16 × 53 = Rs. 848 and 23 × 53 = Rs. 1,219

Question 10

If a : b = 3 : 4, the value of (2a + 3b) : (3a + 4b) is

A

54 : 25✓✗

B

8 : 25✓✗

C

17 : 24✓✗

D

18 : 25✓✗

Explanation

Let a = 3k, b = 4k.
2a + 3b = 6k + 12k = 18k
3a + 4b = 9k + 16k = 25k
Required ratio = 18k : 25k = 18 : 25

Question 11

The third proportional to 49 and 21 is

A

6✓✗

B

9✓✗

C

12✓✗

D

28✓✗

Explanation

Third proportional to a and b = b²/a
= 21²/49 = 44149 = 9

Question 12

The ratio of income of A and B is 5 : 4 and their expenditure is 3 : 2. If at the end of the year each saves Rs. 1,600, then the income of A is:

A

Rs. 3,400✓✗

B

Rs. 3,600✓✗

C

Rs. 4,000✓✗

D

Rs. 4,400✓✗

Explanation

Let income of A = 5x, income of B = 4x, expenditure of A = 3y, expenditure of B = 2y.
Savings: 5x - 3y = 1600 ... (i)
4x - 2y = 1600 ⇒ 2x - y = 800 ⇒ y = 2x - 800 ... (ii)
Substituting (ii) in (i): 5x - 3(2x - 800) = 1600 ⇒ 5x - 6x + 2400 = 1600 ⇒ x = 800
Income of A = 5 × 800 = Rs. 4,000

Question 13

The mean proportional between 12x² and 27y² is:

A

18xy✓✗

B

81xy✓✗

C

8xy✓✗

D

19.5xy✓✗

Explanation

Mean proportional = √12x² × 27y² = √324x²y² = 18xy

Question 14

The ratio of two numbers is 3 : 4. The difference of their squares is 28. The greater number is:

A

8✓✗

B

12✓✗

C

24✓✗

D

64✓✗

Explanation

Let the numbers be 3x and 4x.
(4x)² - (3x)² = 28
16x² - 9x² = 7x² = 28
x² = 4, so x = 2
Greater number = 4x = 8

Question 15

The prices of a scooter and a moped are in the ratio 7 : 9. The price of the moped is Rs. 1,600 more than that of the scooter. Then the price of the moped is:

A

Rs. 7,200✓✗

B

Rs. 5,600✓✗

C

Rs. 800✓✗

D

Rs. 700✓✗

Explanation

Let scooter price = 7x, moped price = 9x.
9x - 7x = 1600
2x = 1600, so x = 800
Moped price = 9 × 800 = Rs. 7,200

Question 16

Four numbers are in the ratio 2:3:4:5 and add up to give a sum of 1344. Find the biggest number.

A

240✓✗

B

540✓✗

C

480✓✗

D

680✓✗

Explanation

Total parts = 2 + 3 + 4 + 5 = 14. Each part = 134414 = 96. Biggest number = 5 × 96 = 480.

Question 17

A 140 cm long tape is to be cut into two pieces such that one piece will be 25 as long as the other. Calculate the length of the shorter piece.

A

30 cm✓✗

B

20 cm✓✗

C

40 cm✓✗

D

35 cm✓✗

Explanation

Let the longer piece be L, so the shorter piece = 2L5. Then, L + 2L5 = 140, i.e. 7L5 = 140, giving L = 100 cm. Shorter piece = 2 × 1005 = 40 cm.

Question 18

x varies inversely as the square of y. Given that y = 2 for x = 1, find the ratio of the original value of x to its value when y = 6.

A

1:7✓✗

B

1:8✓✗

C

8:1✓✗

D

9:1✓✗

Explanation

Since x varies inversely as y², x = ky². When y = 2, x = 1: 1 = k4, so k = 4. When y = 6: x = 436 = 19. So the original x (1) to the new x (19) is in the ratio 9:1.

Question 19

The compounded ratio of (2:3), (6:11) and (11:2) is:

A

1:2✓✗

B

2:1✓✗

C

3:1✓✗

D

4:1✓✗

Explanation

Multiply numerators and denominators: (2 × 6 × 11):(3 × 11 × 2) = 132:66 = 2:1.

Question 20

The ratio of the third proportional to 12 and 30, and the mean proportional between 9 and 25, is:

A

2:3✓✗

B

3:1✓✗

C

4:5✓✗

D

5:1✓✗

Explanation

Third proportional to 12,30 = 30²/12 = 75. Mean proportional between 9,25 = √9 × 25 = 15. Ratio = 75:15 = 5:1.

Question 21

If m:n=o:p, then the value of (m²+n²)/(o²+p²) is

A

mn/op✓✗

B

mnop✓✗

C

0✓✗

D

1✓✗

Explanation

Let m:n=o:p=r. Writing m=ar, n=a, o=br, p=b for some scale factors a,b: (m²+n²)/(o²+p²)=a²(r²+1)/[b²(r²+1)]=a²/b², and mn/op=(ar.a)/(br.b)=a²/b² also. So they are equal.

Question 22

A mixture contains alcohol and water in the ratio 3:2. If 4 litres of water is added to the mixture, the ratio becomes 3:4. Find the quantity of alcohol in the given mixture.

A

4 L✓✗

B

3 L✓✗

C

6 L✓✗

D

5 L✓✗

Explanation

Let alcohol=3k, water=2k. After adding 4L water: 3k/(2k+4)=34, giving 12k=6k+12, so k=2. Alcohol=3 × 2=6 L.

Question 23

The third proportional to m²-n² and m-n is

A

(m+n)/(m-n)✓✗

B

(m+n)/n✓✗

C

(m+n)/m✓✗

D

(m-n)/(m+n)✓✗

Explanation

Third proportional c = b²/a where a=m²-n²=(m-n)(m+n), b=m-n. c=(m-n)²/[(m-n)(m+n)]=(m-n)/(m+n).

Question 24

A rope is 70 cm long. It is cut into two parts. One part is 25 as long as the other part. What will be the length of the longer part?

A

10 cm✓✗

B

14 cm✓✗

C

20 cm✓✗

D

50 cm✓✗

Explanation

Let shorter=x, longer=y, with x=(25)y. x+y=70 gives (75)y=70, so y=50 cm.

Question 25

A carton containing a dozen glasses is dropped. Which of the following cannot be the ratio of broken glasses to unbroken glasses?

A

2:1✓✗

B

3:1✓✗

C

3:2✓✗

D

7:5✓✗

Explanation

With 12 total glasses, the ratio parts must sum to a divisor of 12. 2:1(sum3), 3:1(sum4), and 7:5(sum12) all divide 12 evenly, but 3:2(sum5) does not (125 is not an integer).