Practice MCQ Questions and Answer on Ratio

81.

A person divided Rs. 10800 among his three sons in the ratio 3 : 4 : 5. Second son kept Rs. 1000 for himself, gave Rs. 600 to his wife and divided the remaining money among his two daughters in the ratio 11 : 9. Then one of his daughters received.

(A) Rs. 1000
(B) Rs. 1050
(C) Rs. 1100
(D) Rs. 1150

Solution: $\begin{aligned} Second son's share \\ = Rs . (10800 \times \frac{4}{12}) \\ = Rs . 3600 \end{aligned}$ Money distributed between the two daughters = Rs. [3600 - (1000 + 600)] = Rs. 2000 $\begin{aligned} First daughter's share \\ = Rs . (2000 \times \frac{11}{20}) \\ = Rs .1100 . \\ Second daughter's share \\ = Rs . (2000 \times \frac{9}{20}) \\ = Rs . 9 00 \end{aligned}$

82.

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

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

Solution: $\begin{aligned} (3 + \sqrt{2}) : x : (12 - \sqrt{32}) \\ a : b : c \\ mean proportion \\ \Rightarrow b^{2} = a \times c \\ \Rightarrow x^{2} = (3 + \sqrt{2}) \times (12 - \sqrt{32}) \\ \Rightarrow x^{2} = (3 + \sqrt{2}) \times (12 - 4 \sqrt{2}) \\ \Rightarrow x^{2} = (3 + \sqrt{2}) \times 4 (3 - \sqrt{2}) \\ \Rightarrow x^{2} = 4 \times {{(3)}^{2} - {(\sqrt{2})}^{2}} \\ \Rightarrow x^{2} = 28 \\ \Rightarrow x = 2 \sqrt{7} \end{aligned}$

83.

If p : q : r = 1 : 2 : 4, then $\sqrt{5 p^{2} + q^{2} + r^{2}}$ ÃÂÃÂ ÃÂÃÂ is equal to =

(A) 5
(B) 2q
(C) 5p
(D) 4r

Solution: $\begin{aligned} p : q : r \\ 1 : 2 : 4 \\ x : 2 x : 4 x \\ ∴ \sqrt{5 p^{2} + q^{2} + r^{2}} \\ = \sqrt{5 x^{2} + 4 x^{2} + 16 x^{2}} \\ = \sqrt{25 x^{2}} \\ = 5 x \\ = 5 p \end{aligned}$

84.

Six coins of gold and silver of equal weights are melted and new coins are cast. The ratio of gold and silver in one of the coins is 2 : 1, in another two coins 3 : 5 and 7 : 5 in the remaining coins. What will be the ratio between gold and silver respectively in the new coins ?

(A) 1 : 1
(B) 12 : 11
(C) 42 : 25
(D) 19 : 17

Solution: $\begin{aligned} Gold in new coins \\ = (\frac{2}{3} + 2 \times \frac{3}{8} + 3 \times \frac{7}{12}) units \\ = (\frac{2}{3} + \frac{3}{4} + \frac{7}{4}) units \\ = \frac{19}{6} units \\ Silver in new coins \\ = (\frac{1}{3} + 2 \times \frac{5}{8} + 3 \times \frac{5}{12}) units \\ = (\frac{1}{3} + \frac{5}{4} + \frac{5}{4}) units \\ = \frac{17}{6} units . \\ ∴ Required ratio \\ = \frac{19}{6} : \frac{17}{6} \\ = 19 : 17 \end{aligned}$

85.

Two alloys both are made up of copper and tin. The ratio of the copper and tin in the first alloy is 1 : 3 and in the second alloy is 2 : 5. In what ratio should the two alloys be mixed to obtain a new alloy in which the ratio of tin and copper be 8 : 3 ?

(A) 3 : 5
(B) 4 : 7
(C) 3 : 8
(D) 5 : 11

Solution: Given: ⇒ Let the alloys be mixed in the ratio of $x$ : $y$ (Assumption) ⇒ In 1st alloy, Copper = $\frac{x}{4}$ ⇒ In 1st alloy, Tin = $\frac{3 x}{4}$ ⇒ In 2nd alloy, Copper = $\frac{2 y}{7}$ ⇒ In 2nd alloy, Tin = $\frac{5 y}{7}$ To find: ⇒ The ratio in which 2 alloys must be mixed to get a new alloy with a ratio of copper and tin be 3 : 8 =? Now we have, $\begin{aligned} (\frac{x}{4} + \frac{2 y}{7}) : (\frac{3 x}{4} + \frac{5 y}{7}) = 3 : 8 \\ \Rightarrow \frac{(\frac{x}{4} + \frac{2 y}{7})}{(\frac{3 x}{4} + \frac{5 y}{7})} = \frac{3}{8} \\ \Rightarrow \frac{\frac{7 x + 14 y}{28}}{\frac{21 x + 20 y}{28}} = \frac{3}{8} \\ \Rightarrow \frac{7 x + 14 y}{21 x + 20 y} = \frac{3}{8} \\ \Rightarrow 56 x + 64 y = 63 x + 60 y \\ \Rightarrow 64 y - 60 y = 63 x - 56 x \\ \Rightarrow 4 y = 7 x \\ ∴ \frac{x}{y} = \frac{4}{7} \end{aligned}$ ⇒ Ratio in which 2 alloys must be mixed to get a new alloy with a ratio of copper and tin be 3 : 8 = 4 : 7

86.

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 $\begin{aligned} = \sqrt{9 \times 25} = 15 \\ ∴ Required ratio = 75 : 15 \\ = 5 : 1 \end{aligned}$

87.

If W₁ : W₂ = 2 : 3 and W₁ : W₃ = 1 : 2, then W₂ : W₃ is -

(A) 3 : 4
(B) 4 : 3
(C) 2 : 3
(D) 4 : 5

Solution: $\begin{aligned} = \frac{W_{2}}{W_{1}} = \frac{3}{2} and \frac{W_{1}}{W_{3}} = \frac{1}{2} \\ \Rightarrow \frac{W_{2}}{W_{3}} = (\frac{W_{2}}{W_{1}} \times \frac{W_{1}}{W_{3}}) \\ = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} \\ \Rightarrow W_{2} : W_{3} = 3 : 4 \end{aligned}$

88.

In a college the students in Arts and commerce faculties were in the ratio of 4 : 5 respectively. When 65 more students joined commerce faculty the ratio becomes 8 : 11 respectively. How many students are there in Arts faculty?

(A) 520
(B) 650
(C) 715
(D) Cannot be determined

Solution: Let the number of students in Arts and Commerce faculties be 4x and 5x respectively. Then, $\begin{aligned} = \frac{4 x}{5 x + 65} = \frac{8}{11} \\ \Rightarrow 44 x = 8 (5 x + 65) \\ \Rightarrow 4 x = 520 \\ \Rightarrow x = 130 \end{aligned}$ ∴ Number of students in Arts faculty = (4 × 130) = 520

89.

The students of three classes are in the ratio 4 : 6 : 9. If 12 students are increased in each class the ratio changes to 7 : 9 : 12. Then the total number of students in the three classes before the increase is = ?

(A) 95
(B) 76
(C) 100
(D) 114

90.

A dealer buys dry fruit at the rate of Rs. 100, Rs. 80 Rs. 60 per kg. He bought them in the ratio 12 : 15 : 20 by weight. He in total gets 20% profit by selling the first two and at last he finds he has no gain no loss in selling the whole quantity which he had. What was the percentage loss he suffered for the third quantity ?

(A) 20%
(B) 30%
(C) 40%
(D) 50%

Solution: Let the weights of the three varieties be 12x, 15x and 20x kg respectively Then, Total C.P. = Rs. (100 × 12x + 80 × 15x + 60 × 20x) = Rs. (3600x) S.P. of the first two verities $\begin{aligned} = Rs . (\frac{120}{100} \times 2400 x) \\ = Rs . (2880 x) \\ S .P . of the third variety \\ = Rs . (3600 x - 2880 x) \\ = Rs . 720 x \\ Loss on third variety \\ = Rs . (1200 x - 720 x) \\ = Rs . 480 \\ loss % \\ = (\frac{480 x}{1200 x} \times 100) % \\ = 40 % \end{aligned}$

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Practice MCQ Questions and Answer on Ratio

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