Practice MCQ Questions and Answer on Ratio

71.

If the ratio of the ages of Maya and Chhaya is 6 : 5 at present, and fifteen years from now, the ratio will get changed to 9 : 8, then find Maya's present age.

(A) 24 years
(B) 30 years
(C) 18 years
(D) 33 years

Solution: Let Maya's and Chhaya's present age is 6x and 5x respectively And, $\frac{6 x + 15}{5 x + 15} = \frac{9}{8}$ Or, 48x + 120 = 45x = 135 Or, 3x = 15 Or, x = 5 Present age of Maya = 6x = 30

72.

If A : B : C = 2 : 3 : 4, then ratio $\frac{A}{B}$ : $\frac{B}{C}$ : $\frac{C}{A}$ is equal to?

(A) 8 : 9 : 16
(B) 8 : 9 : 12
(C) 8 : 9 : 24
(D) 4 : 9 : 12

Solution: $\begin{aligned} A : B : C = 2 : 3 : 4 \\ Let \\ A = 2 x, B = 3 x, C = 4 x \\ ∴ \frac{A}{B} : \frac{B}{C} : \frac{C}{A} = \frac{2 x}{3 x} : \frac{3 x}{4 x} : \frac{4 x}{2 x} \\ \Rightarrow \frac{2}{3} : \frac{3}{4} : \frac{2}{1} \end{aligned}$ Multiply by the L.C.M of denominator to remove fraction $\begin{aligned} So, \\ L .C .M of (3, 4, 1) = 12 \\ ∴ \frac{A}{B} : \frac{B}{C} : \frac{C}{A} \\ \frac{2}{3} \times 12 : \frac{3}{4} \times 12 : \frac{2}{1} \times 12 \\ 8 : 9 : 24 \end{aligned}$

73.

There are Rs. 225 consisting of one rupee, 50 paise and 25 paise coins. The ratio of their numbers in that order is 8 : 5 : 3. The number of one - rupee coins is = ?

(A) 80
(B) 112
(C) 160
(D) 172

Solution: Rs. 1 : 50 paise : 25 paise Number of coins 8x : 5x : 3x $\begin{aligned} Value of coins \\ = \frac{8 x}{1} : \frac{5 x}{2} : \frac{3 x}{4} \\ ∴ 8 x + \frac{5 x}{2} + \frac{3 x}{4} = Rs . 225, \\ Given \frac{32 x + 10 x + 3 x}{4} = 225 \\ \Rightarrow 45 x = 225 \times 4 \\ \Rightarrow x = \frac{225 \times 4}{45} \\ \Rightarrow x = 5 \times 4 = 20 \end{aligned}$ ∴ Number of one rupees coins = 20 × 8 = 160

74.

If x : y = 3 : 4 and a : b = 1 : 2, then the value of $\frac{2 x a + y b}{3 y b - 4 x a}$ ÃÂÃÂ is

(A) $\frac{5}{6}$
(B) $\frac{6}{5}$
(C) $\frac{6}{7}$
(D) $\frac{7}{6}$

Solution: $\begin{aligned} = \frac{x}{y} = \frac{3}{4} and \frac{a}{b} = \frac{1}{2} \\ \Rightarrow \frac{x a}{y b} = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} \\ \frac{2 x a + y b}{3 y b - 4 x a} \\ = \frac{2 (\frac{x a}{y b}) + 1}{3 - 4 (\frac{x a}{y b})} \\ = \frac{2 \times \frac{3}{8} + 1}{3 - 4 \times \frac{3}{8}} \\ = \frac{7}{4} \times \frac{2}{3} \\ = \frac{7}{6} \end{aligned}$

75.

The ratios of copper to zinc in alloys A and B are 3 : 4 and 5 : 9, respectively. A and B are taken in the ratio 2 : 3 and melted to form a new alloy C. What is the ratio of copper to zinc in C?

(A) 3 : 5
(B) 8 : 15
(C) 9 : 11
(D) 27 : 43

Solution: $\begin{aligned} A \to 3 : 4 = 7_{\times 2 \times 2} \\ B \to 5 : 9 = 14_{\times 3} \\ Copper : Zinc \\ 12 : 16 \\ 15 : 27 \\ \overset{―}{27 : 43} \end{aligned}$

76.

Two numbers are in the ratio 17 : 45. One third of the smaller is less than $\frac{1}{5}$ of the bigger by 15. The smaller number is -

(A) $25 \frac{1}{2}$
(B) $67 \frac{1}{2}$
(C) $76 \frac{1}{2}$
(D) $86 \frac{1}{2}$

Solution: Let the numbers be 17x and 45x Then, $\begin{aligned} = \frac{1}{5} of 45 x - \frac{1}{3} of 17 x = 15 \\ \Leftrightarrow 9 x - \frac{17}{3} x = 15 \\ \Leftrightarrow \frac{10 x}{3} = 15 \\ \Leftrightarrow x = 15 \times \frac{3}{10} = \frac{9}{2} \\ ∴ Smaller number \\ = 17 x = (17 x \times \frac{9}{2}) \\ = \frac{153}{2} \\ = 76 \frac{1}{2} \end{aligned}$

77.

The ratio of the length and the perimeter of a rectangle is 2 : 7. What is the ratio of the length and breadth of the rectangle?

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

Solution: $\begin{aligned} \frac{l}{2 (l + b)} = \frac{2}{7} \\ \Rightarrow 7 l = 4 l + 4 b \\ \Rightarrow 3 l = 4 b \\ \Rightarrow \frac{l}{b} = \frac{4}{3} = 4 : 3 \end{aligned}$

78.

The ratio of boys and girls in a school is 27 : 23. If the difference between the number of boys and girls is 200, then find the number of boys.

(A) 1350
(B) 1250
(C) 1300
(D) 1200

79.

The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?

(A) 8 : 9
(B) 17 : 18
(C) 21 : 22
(D) Cannot be determined

Solution: Originally, let the number of boys and girls in the college be 7x and 8x respectively Their increased number is (120% of 7x) and (110% of 8x) $\begin{aligned} \Rightarrow (\frac{120}{100} \times 7 x) and (\frac{110}{100} \times 8 x) \\ \Rightarrow \frac{42 x}{5} and \frac{44 x}{5} \\ ∴ The required ration \\ = \frac{42 x}{5} : \frac{44 x}{5} \\ = 21 : 22 \end{aligned}$

80.

The ratio of the incomes of Ram and Shyam is 7 : 17 and that of Shyam and Sohan is 7 : 17. If the income of Ram is Rs. 490, what is the income of Sohan ?

(A) Rs. 490
(B) Rs. 1190
(C) Rs. 2790
(D) Rs. 2890

Solution: $\begin{aligned} Ram : Shyam = 7 : 17 \\ Shyam : Sohan = 7 : 17 \\ = (7 \times \frac{17}{7}) : (17 \times \frac{17}{7}) \\ = 17 : \frac{289}{7} \\ Ram : Shyam : Sohan \\ = 7 : 17 : \frac{289}{7} \\ = 49 : 119 : 289 \end{aligned}$ Let Sohan's income be Rs. x Then, = 49 : 289 :: 490 : x or x = 2890

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Practice MCQ Questions and Answer on Ratio

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