Practice MCQ Questions and Answer on Ratio

91.

x, y, z, u are real numbers such that x : y = y : z = z : u and x : u = 64 : 27. the value of x : z is -

(A) 64 : 27
(B) 16 : 9
(C) 4 : 3
(D) 3 : 4

Solution: $\begin{aligned} Let \frac{x}{y} = \frac{y}{z} = \frac{z}{u} = k . \\ Now, \frac{x}{u} = \frac{64}{27} \\ \Rightarrow \frac{x}{y} \times \frac{y}{z} \times \frac{z}{u} = \frac{64}{27} \\ \Rightarrow k^{3} = {(\frac{4}{3})}^{3} \\ \Rightarrow k = \frac{4}{3} . \\ So, x : y = y : z = z : u = 4 : 3. \\ ∴ \frac{x}{z} = \frac{x}{y} \times \frac{y}{z} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9.} \end{aligned}$

92.

A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 : 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, then the ratio of gold and copper in alloy C will be.

(A) 5 : 7
(B) 5 : 9
(C) 7 : 5
(D) 9 : 5

Solution: $\begin{aligned} Gold in C \\ = (\frac{7}{9} + \frac{7}{18}) units \\ = \frac{7}{6} units \\ Copper in C \\ = (\frac{2}{9} + \frac{11}{18}) units \\ = \frac{5}{6} units \\ ∴ Gold : Copper \\ = \frac{7}{6} : \frac{5}{6} \\ = 7 : 5 \end{aligned}$

93.

The ratio of water and milk in a 30 liter mixture is 7 : 3. Find the quantity of water to be added to the mixture in order to make this ratio 6 : 1.

(A) 30
(B) 32
(C) 33
(D) 35

94.

The salaries of A, B and C are in the ratio 1 : 3 : 4. If the salaries are increased 5%, 10% and 15% respectively. Then the increased salaries will be in the ratio = ?

(A) 20 : 66 : 95
(B) 21 : 66 : 95
(C) 21 : 66 : 92
(D) 19 : 66 : 92

95.

If a : b = c : d = e : f = 1 : 2, then (3a + 5c + 7e) : (3b + 5d + 7f) is = ?

(A) 8 : 7
(B) 2 : 1
(C) 1 : 4
(D) 1 : 2

Solution: a : b = 1 : 2 c : d = 1 : 2 e : f = 1 : 2 ∴ a = x, b = 2x, c = x, d = 2x, e = x, f = 2x $\begin{aligned} ∴ \frac{3 \times x + 5 \times 7 + 7 \times x}{3 \times 2 x + 5 \times 2 x + 7 \times 2 x} \times 1 \\ \Rightarrow \frac{15 x}{30 x} \\ \Rightarrow \frac{1}{2} \end{aligned}$

96.

The mean proportional between $(3 + \sqrt{2})$ ÃÂÃÂ and $(12 - \sqrt{32})$ ÃÂÃÂ is-

(A) $\sqrt{7}$
(B) $2 \sqrt{7}$
(C) $6 \sqrt{7}$
(D) $\frac{15 - 3 \sqrt{2}}{2}$

Solution: Required mean proportional $\begin{aligned} = \sqrt{(3 + \sqrt{2}) (12 - \sqrt{32})} \\ = \sqrt{(3 + \sqrt{2}) (12 - 4 \sqrt{2})} \\ = \sqrt{36 - 8} \\ = \sqrt{28} \\ = 2 \sqrt{7} \end{aligned}$

97.

If p : q = 3 : 4, r : s = 8 : 5 and x : y = 10 : 6, then psx : qry is equal to.

(A) $\frac{16}{25}$
(B) $\frac{25}{32}$
(C) $2$
(D) $\frac{75}{32}$

Solution: $\begin{aligned} = \frac{p}{q} = \frac{3}{4}, \frac{r}{s} = \frac{8}{5}, \frac{x}{y} = \frac{10}{6} \\ \Rightarrow p = \frac{3 q}{4}, r = \frac{8 s}{5}, x = \frac{5 y}{3} \\ ∴ \frac{p s x}{q r y} = \frac{\frac{3 q}{4} \times s \times \frac{5 y}{3}}{q \times \frac{8 s}{5} \times y} \\ = \frac{\frac{5}{4} q s y}{\frac{8}{5} q s y} \\ = \frac{5}{4} \times \frac{5}{8} \\ = \frac{25}{32} \end{aligned}$

98.

If $\frac{a}{b} = \frac{c}{d} = \frac{e}{f},$ ÃÂÃÂ ÃÂÃÂ then each of them is equal to = ?

(A) $\frac{a + 3 c + 5 e}{b + d - f}$
(B) $\frac{3 a + 3 c - 5 e}{b + 3 d - f}$
(C) $\frac{a + c - e}{b + d - 5 f}$
(D) $\frac{a + 3 c - 5 e}{b + 3 d - 5 f}$

Solution: $\begin{aligned} Let, \\ \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{1}{2} \end{aligned}$ Now check from option to save your valuable time $\begin{aligned} Option D \\ \frac{a + 3 c - 5 e}{b + 3 d - 5 f} \Rightarrow \frac{c}{d} [\frac{\frac{a}{c} + 3 - \frac{5 e}{c}}{\frac{b}{d} + 3 - \frac{5 f}{d}}] \\ \Rightarrow \frac{c}{d} = \frac{1}{2} (Satisfy) \end{aligned}$

99.

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, $\begin{aligned} 1 \times {(\frac{x - 8}{x})}^{4} = \frac{16}{81} \\ \Rightarrow 1 \times {\frac{x - 8}{x}}^{4} = {(\frac{2}{3})}^{4} \\ \Rightarrow \frac{x - 8}{x} = \frac{2}{3} \\ \Rightarrow 3 x - 24 = 2 x \\ \Rightarrow x = 24 litres \end{aligned}$

100.

What must be added to each term of the ratio 7 : 11, So as to make it equal to 3 : 4?

(A) 8
(B) 7.5
(C) 6.5
(D) 5

Solution: $\begin{aligned} Let x be added to each term . \\ According to question, \\ \frac{7 + x}{11 + x} = \frac{3}{4} \\ O r, 33 + 3 x = 28 + 4 x \\ O r, x = 5 \end{aligned}$

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Practice MCQ Questions and Answer on Ratio

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