Practice MCQ Questions and Answer on Ratio

11.

A mixture contains wine and water in the ratio 3 : 2 and another mixture contains them in the ratio 4 : 5. How many litres of the latter must be mixed with 3 litres of the former so that the resulting mixture may contain equal quantities of wine and water = ?

(A) $5 \frac{2}{5} litres$
(B) $5 \frac{2}{3} litres$
(C) $4 \frac{1}{2} litres$
(D) $3 \frac{3}{4} litres$

Solution: The first mixture contains wine and water in the ratio of 3 : 2 Wine in 3 liters of mixture = $\frac{3}{5} \times 3 = \frac{9}{5}$ Water in 3 liters of mixture = $\frac{2}{5} \times 3 = \frac{6}{5}$ Let us consider that the first mixture is mixed with the second mixture that has quantity as 9x liters (4x liters of wine and 5x liters of water). After mixing, The total quantity of wine = Total quantity of water $\begin{aligned} \Rightarrow \frac{9}{5} + 4 x = \frac{6}{5} + 5 x \\ \Rightarrow x = \frac{9}{5} - \frac{6}{5} \\ \Rightarrow x = \frac{3}{5} \end{aligned}$ Second Mixture required = 9x = $9 \times \frac{3}{5}$ = $5 \frac{2}{5} litres$

12.

If $\frac{x}{y} = \frac{3}{4},$ ÃÂÃÂ the ratio of (2x + 3y) and (3y - 2x) is = ?

(A) 2 : 1
(B) 3 : 2
(C) 3 : 1
(D) 1 : 1

Solution: $\begin{aligned} Given \\ \frac{x}{y} = \frac{3}{4} \\ \Rightarrow \frac{2 x + 3 y}{3 y - 2 x} \\ = \frac{2 \times 3 + 3 \times 4}{3 \times 4 - 2 \times 3} \\ = \frac{6 + 12}{12 - 6} \\ = \frac{18}{6} \\ = 3 : 1 \end{aligned}$ This is the required ratio 3 : 1

13.

If (x³ - y³) : (x² + xy + y²) = 5 : 1 and (x² - y²) : (x - y) = 7 : 1, then the ratio 2x : 3y equals = ?

(A) 3 : 2
(B) 2 : 3
(C) 4 : 3
(D) 4 : 1

Solution: According to question, $\begin{aligned} \frac{x^{3} - y^{3}}{x^{2} + x y + y^{2}} = \frac{5}{1} \\ \frac{(x - y) (x^{2} + x y + y^{2})}{x^{2} + x y + y^{2}} = \frac{5}{1} \\ x - y = 5. . . . . (i) \\ \frac{x^{2} - y^{2}}{x - y} = \frac{7}{1} \\ \frac{(x + y) (x - y)}{x - y} = \frac{7}{1} \\ x + y = 7. . . . . (ii) \\ Solve equations (i) and (ii) \\ x = 6 \\ y = 1 \\ ∴ \frac{2 x}{3 y} = \frac{2 \times 6}{3 \times 1} \\ = \frac{12}{3} \\ = \frac{4}{1} \\ = 4 : 1 \end{aligned}$

14.

If a : b = 5 : 7 and c : d = 2a : 3b then ac : bd is = ?

(A) 20 : 38
(B) 50 : 147
(C) 10 : 21
(D) 50 : 151

Solution: $\begin{aligned} a : b c : d \\ 5 : 7 2 a : 3 b \\ \frac{a}{b} = \frac{5}{7}, \frac{c}{d} = \frac{2 a}{3 b} \\ = \frac{2}{3} \times \frac{5}{7} \\ = \frac{10}{21} \\ ∴ a c : b d \\ = \frac{ac}{bd} \\ = \frac{5}{7} \times \frac{10}{21} \\ = \frac{50}{147} \\ = 50 : 147 \end{aligned}$

15.

Which of the following is the lowest ratio ?

(A) 7 : 15
(B) 15 : 23
(C) 17 : 25
(D) 21 : 39

Solution: $\begin{aligned} \Rightarrow 7 : 15 = \frac{7}{15} = 0.466 \\ \Rightarrow 15 : 23 = \frac{15}{23} = 0.652 \\ \Rightarrow 17 : 25 = \frac{17}{25} = 0.68 \\ \Rightarrow 21 : 39 = \frac{21}{39} = 0.538 \end{aligned}$ Clearly, 7 : 15 is the lowest

16.

If $\frac{2}{3}$ of A=75% of B = 0.6 of C, then A : B : C is

(A) 2 : 3 : 3
(B) 3 : 4 : 5
(C) 4 : 5 : 6
(D) 9 : 8 : 10

Solution: $\begin{aligned} According to thequestion, \\ \frac{2 A}{3} = \frac{75 B}{100} = \frac{C \times 6}{10} \\ Above relation gives; \\ \frac{A \times 2}{3} = \frac{B \times 3}{4} \\ \to \frac{A}{B} = \frac{9}{8} \\ And, \\ \frac{B \times 3}{4} = \frac{C \times 3}{5} \\ \to \frac{B}{C} = 4 : 5 \\ \to \frac{B}{C} = 8 : 10 (multiple by 2) \\ Thus, \\ A : B : C = 9 : 8 : 10 \end{aligned}$

17.

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

18.

A sum of money is divided among A, B, C and D in the ratio of 3 : 4 : 9 : 10 respectively. If the share of C is Rs. 2580 more than the share of B, then what is the total amount A and D together ?

(A) Rs. 5676
(B) Rs. 6192
(C) Rs. 6708
(D) Rs. 7224

19.

If x varies as y then x² + y² varies as

(A) x + y
(B) x - y
(C) x2 - y2
(D) x3 - y3

20.

Two numbers are in the ratio $1 \frac{1}{2}$ : $2 \frac{2}{3}$ When each of these is increased by 15, they become in the ratio $1 \frac{2}{3}$ : $2 \frac{1}{2}$ . The greater of the numbers is = ?

(A) 27
(B) 36
(C) 48
(D) 64

Solution: $\begin{aligned} A : B \\ \frac{3}{2} : \frac{8}{3} \end{aligned}$ take L.C.M. of denominator and multiply) $\Rightarrow \frac{3}{2} \times 6 : \frac{8}{3} \times 6 = 9 x : 16 x$ After adding 15 in each we get $\begin{aligned} ∴ \frac{9 x + 15}{16 x + 15} = \frac{5}{3} \times \frac{2}{5} \\ \Rightarrow \frac{9 x + 15}{16 x + 15} = \frac{2}{3} (cross multiply) \\ \Rightarrow 27 x + 45 = 32 x + 30 \\ \Rightarrow 5 x = 15 \\ \Rightarrow x = 3 \\ ∴ Smaller number \\ = 9 \times 3 = 27 \\ Greater number \\ = 16 \times 3 = 48 \end{aligned}$

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Practice MCQ Questions and Answer on Ratio

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