Practice MCQ Questions and Answer on Ratio

31.

The current ages of Sonali and Monali are in the ratio 5 : 3. Five years from now, their ages will be in the ratio 10 : 7. Then Monali's current age is = ?

(A) 3 years
(B) 5 years
(C) 9 years
(D) 15 years

Solution: Let, ⇒ Sonali's age = 5x ⇒ Monali's age = 3x According to the question, $\begin{aligned} \Rightarrow \frac{5 x + 5}{3 x + 5} = \frac{10}{7} \\ \Rightarrow \frac{x + 1}{3 x + 5} = \frac{2}{7} \\ \Rightarrow 7 x + 7 = 6 x + 10 \\ \Rightarrow x = 3 \end{aligned}$ ⇒ So, Monali's present age = 3x = 3 × 3 = 9 years

32.

A person bought some rice and wheat for Rs. 380. The ratio of weight of rice and wheat is 4 : 3 and the price equal amount of rice and wheat is in the ratio 5 : 6. The rice was bought of worth = ?

(A) Rs. 380
(B) Rs. 300
(C) Rs. 200
(D) Rs. 180

33.

Two numbers are such that the ratio between them is 4 : 7. If each is increase by 4, the ratio become 3 : 5. The larger number is = ?

(A) 36
(B) 48
(C) 56
(D) 64

Solution: A : B 4x : 7x Now 4 is added to each number $\begin{aligned} \frac{4 x + 4}{7 x + 4} = \frac{3}{5} \\ \Rightarrow 20 x + 20 = 21 x + 12 \\ \Rightarrow x = 8 \end{aligned}$ ∴ Smaller number is 4 × 8 = 32 Larger number is 7 × 8 = 56

34.

Incomes of x and y are in the ratio 4 : 3. Their expenditure are in the ratio 12 : 7. Both save Rs. 3200 at the end of the month, then the income of x is = ?

(A) Rs. 4000
(B) Rs. 8000
(C) Rs. 6000
(D) Rs. 2000

Solution: Let their income be 4x and 3x Their savings = Rs. 3200 each According to the question, $\begin{aligned} \Rightarrow \frac{4 x - 3200}{3 x - 3200} = \frac{12}{7} \\ \Rightarrow \frac{x - 800}{3 x - 3200} = \frac{3}{7} \\ \Rightarrow 7 x - 5600 = 9 x - 9600 \\ \Rightarrow 2 x = 4000 \\ \Rightarrow x = 2000 \\ \Rightarrow A = 2000 \\ \Rightarrow Income of A = 4 x \\ \Rightarrow 4 \times 2000 = Rs . 8000 \end{aligned}$

35.

The ratio between the third proportional of 12 and 30 and mean proportional of 9 and 25 is -

(A) 2 : 1
(B) 5 : 1
(C) 7 : 15
(D) 9 : 14

Solution: Let the third proportional to 12 and 30 be x. Then, $\begin{aligned} = 12 : 30 :: 30 : x \\ \Rightarrow 12 x = 30 \times 30 \\ \Rightarrow x = \frac{(30 \times 30)}{12} = 75 \end{aligned}$ ∴ Third proportional to 12 and 30 = 75 Mean proportional between 9 and 25 $\begin{aligned} = \sqrt{9 \times 25} = 15 \\ ∴ Required ratio = 75 : 15 \\ = 5 : 1 \end{aligned}$

36.

How many sides does a regular polygon have whose interior and exterior angle are in the ratio 2 : 1 ?

(A) 3
(B) 5
(C) 6
(D) 12

Solution: Each exterior angle of n sided polygon is $= (\frac{360}{n})$ And each interior angle of n sided polygon $\begin{aligned} = \frac{(n - 2) \times 180}{n} \\ ∴ \frac{\frac{(n - 2) \times 180}{n}}{\frac{360}{n}} = \frac{2}{1} \\ \Rightarrow \frac{(n - 2)}{2} = 2 \\ \Rightarrow n - 2 = 4 \\ \Rightarrow n = 6 \end{aligned}$

37.

The ratio of milk to water in 80 litres of a mixture is 7 : 3. The water (in litres) to be added to it to make the ratio 2 : 1 is -

(A) 4
(B) 5
(C) 6
(D) 8

Solution: $\begin{aligned} Quantity of milk \\ = (80 \times \frac{7}{10}) litres \\ = 56 litres . \\ Quantity of water \\ = (80 - 56) litres \\ = 24 litres . \end{aligned}$ Let the quantity of water to be added be x litres Then, $\begin{aligned} = \frac{56}{24 + x} = \frac{2}{1} \\ \Rightarrow 2 x = 8 \\ \Rightarrow x = 4 \end{aligned}$

38.

What is the ratio in Rs. 2.80 and 40 paise?

(A) 1 : 7
(B) 2 : 7
(C) 7 : 1
(D) 1 : 14

39.

8 litres are drawn from a cask filled with wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the total solution is 16 : 81. How much wine did the cask hold originally?

(A) 24 litres
(B) 45 litres
(C) 49 litres
(D) 44 litres

Solution: Let the quantity of the wine in the cask originally be x litres. Using formula: Final Amount of solute that is not replaced = Initial Amount × ${(\frac{Vol . after removal}{Vol . after replacing})}^{N}$ Where N = No. of operation done. Then ratio of wine to total solution in cask after 4 operations, $\begin{aligned} 1 \times {(\frac{x - 8}{x})}^{4} = \frac{16}{81} \\ \Rightarrow 1 \times {\frac{x - 8}{x}}^{4} = {(\frac{2}{3})}^{4} \\ \Rightarrow \frac{x - 8}{x} = \frac{2}{3} \\ \Rightarrow 3 x - 24 = 2 x \\ \Rightarrow x = 24 litres \end{aligned}$

40.

A bucket contains a mixture of two liquids A and B in the proportion 7 : 5. If 9 litres of mixture is replaced by 9 liters of liquid B, then the ratio of the two liquids becomes 7 : 9. How much of the liquid A was there in the bucket ?

(A) 21 liters
(B) 23 liters
(C) 25 liters
(D) 27 liters

Solution: Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively. When 9 litres of mixture are drawn off, quantity of A in mixture left: $\begin{aligned} = [7 x - \frac{7}{12} \times 9] litres \\ = [7 x - \frac{21}{4}] litres \\ Similarly quantity of B in mixture left, \\ = [5 x - \frac{5}{12} \times 9] litres \\ = [5 x - \frac{15}{4}] litres \\ ∴ ratio becomes, \\ \frac{7 x - \frac{21}{4}}{(5 x - \frac{15}{4}) + 9} = \frac{7}{9} \\ \Rightarrow \frac{28 x - 21}{20 x + 21} = \frac{7}{9} \\ \Rightarrow 252 x - 189 = 140 x + 147 \\ \Rightarrow 112 x = 336 \\ \Rightarrow x = 3 \\ So the can contained, \\ = 7 \times x \\ = 7 \times 3 \\ = 21 litres of A initially . \end{aligned}$

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Practice MCQ Questions and Answer on Ratio

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