Practice MCQ Questions and Answer on Ratio

71.

Mrs. Richi Rich inherits 3224 gold coins and divides them amongst her 3 daughters Lalita, Palita and Salita in a certain ratio. Out of the total coins each of them received, Lalita sells her 50 coins, Palita donates 85 of her coins and Salita makes jewellery out of her 39 coins. Now the ratio of gold coins with them is 24 : 21 : 16 respectively. How many coins did Lalita receive from her mother ?

(A) 1050
(B) 1135
(C) 1200
(D) 1250

72.

25% of A's income is equal to 35% of B's income. The ratio of the incomes of A and B is -

(A) 5 : 7
(B) 7 : 5
(C) 13 : 15
(D) 15 : 13

Solution: $\begin{aligned} = 25 % of A = 35 % of B \\ \Rightarrow \frac{25}{100} A = \frac{35}{100} B \\ \Rightarrow \frac{A}{4} = \frac{7 B}{20} \\ \Rightarrow \frac{A}{B} = \frac{7}{20} \times 4 = \frac{7}{5} \\ \Rightarrow A : B = 7 : 5 \end{aligned}$

73.

A and B are two alloys in which ratios of gold and copper are 5 : 3 and 5 : 11 respectively. If these equally amount of two alloys are melted and made alloy C. What will be the ratio of gold and copper in alloy C?

(A) 25 : 23
(B) 33 : 25
(C) 15 : 17
(D) 17 : 15

Solution: Ratio of Gold and Copper in Alloy A = 5 : 3 Ratio of Gold and Copper in Alloy B = 5 : 11 Amount of Gold in Alloy A = $\frac{5}{8}$ Amount of Gold In Alloy B = $\frac{5}{16}$ Amount of Copper in A = $\frac{3}{8}$ Amount of Copper in B = $\frac{11}{16}$ Amount of Gold In C, = (Amount of gold in A + Amount of gold in B) = $\frac{5}{8}$ + $\frac{5}{16}$ = $\frac{10 + 5}{16}$ = $\frac{15}{16}$ Amount of Copper in C, = Amount of Copper in A + Amount of Copper in B = $\frac{3}{8}$ + $\frac{11}{16}$ = $\frac{17}{16}$ So, Ratio of Gold and Copper in C, $= \frac{15}{16} : \frac{17}{16} = 15 : 17$

74.

Let $\frac{a}{b} : - \frac{b}{a} = x : y .$ ÃÂÃÂ ÃÂÃÂ If $(x - y) =$ ÃÂÃÂ ${\frac{a}{b} + \frac{b}{a}},$ ÃÂÃÂ then x is equal to -

(A) $\frac{a - b}{a}$
(B) $\frac{a + b}{a}$
(C) $\frac{a + b}{b}$
(D) None of these

Solution: $\begin{aligned} \Rightarrow \frac{x}{y} = \frac{(\frac{a}{b})}{(- \frac{b}{a})} = - \frac{a^{2}}{b^{2}} \\ \Rightarrow y = (- \frac{b^{2}}{a^{2}}) x \\ ∴ x - y = \frac{a}{b} + \frac{b}{a} \\ \Rightarrow x + \frac{b^{2}}{a^{2}} x = \frac{a^{2} + b^{2}}{a b} \\ \Rightarrow x (\frac{a^{2} + b^{2}}{a^{2}}) = \frac{a^{2} + b^{2}}{a b} \\ \Rightarrow x = \frac{a^{2}}{a b} = \frac{a}{b} \end{aligned}$

75.

Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit's salary?

(A) Rs. 17,000
(B) Rs. 20,000
(C) Rs. 25,500
(D) Rs. 38,000

Solution: Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively $Then, \frac{2 x + 4000}{3 x + 4000} = \frac{40}{57}$ ⇒ 57(2x + 4000) = 40(3x + 4000) ⇒ 6x = 68,000 ⇒ 3x = 34,000 Sumit's present salary = (3x + 4000) = Rs. (34000 + 4000) = Rs. 38,000

76.

The ratio of the length of a school ground to its width is 5 : 2. If the width is 40m, then the length is = ?

(A) 200m
(B) 100m
(C) 50m
(D) 80m

77.

The numerator of a fraction is 3 more than the denominator. When 5 is added to the numerator and 2 is subtracted from the denominator, the fraction becomes $\frac{8}{3} .$ When the original fraction is divided by $5 \frac{1}{2},$ ÃÂÃÂ the fraction so obtained is:

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

Solution: $\begin{aligned} Let denominator = x \\ Numerator = x + 3 \\ Fraction become = \frac{x + 3}{x} \\ According to the question, \\ \frac{x + 3 + 5}{x - 2} = \frac{8}{3} \\ 3 x + 24 = 8 x - 16 \\ 5 x = 40 \\ x = 8 \\ ∴ Denominator = 8 \\ Numerator = 8 + 3 = 11 \\ Fraction = \frac{11}{8} \\ Then, \frac{11}{8} \div \frac{11}{2} = \frac{11}{8} \times \frac{2}{11} = \frac{1}{4} \end{aligned}$

78.

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 $\begin{aligned} ∴ \frac{9 x + 15}{16 x + 15} = \frac{5}{3} \times \frac{2}{5} \\ \Rightarrow \frac{9 x + 15}{16 x + 15} = \frac{2}{3} (cross multiply) \\ \Rightarrow 27 x + 45 = 32 x + 30 \\ \Rightarrow 5 x = 15 \\ \Rightarrow x = 3 \\ ∴ Smaller number \\ = 9 \times 3 = 27 \\ Greater number \\ = 16 \times 3 = 48 \end{aligned}$

79.

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

80.

The present ratio of ages of A and B is 4 : 5. 18 years ago, this ratio was 11 : 16. Find the sum total of their present ages.

(A) 90 years
(B) 105 years
(C) 110 years
(D) 80 years

Solution: Let present age of A and B be 4x and 5x 18 years ago their ages $\frac{4 x - 18}{5 x - 18} = \frac{11}{16}$ Or, 64x - 288 = 55x - 198 Or, 64x - 55x = -198 + 288 Or, 9x = 90 Or, x = $\frac{90}{9}$ = 10 Sum of the present ages = 40 + 50 = 90 years

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Practice MCQ Questions and Answer on Ratio

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