Practice MCQ Questions and Answer on Ratio

21.

A container contains two liquids A and B in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with B, the ratio of A and B becomes 1 : 1. How many litres of liquid A was in the container initially = ?

(A) 26
(B) $16 \frac{1}{2}$
(C) $36 \frac{3}{4}$
(D) $26 \frac{3}{4}$

Solution: Quantity of A in mixture 7x and B in Mixture in 5x Quantity of A in Mixture left after 9 liter Drawn = $(7 x - \frac{7}{12} \times 9) = (7 x - \frac{21}{4})$ liters Quantity of B in Mixture left after 9 liter Drawn = $(7 x - \frac{5}{12} \times 9) = (7 x - \frac{15}{4})$ liters $\begin{aligned} ∴ \frac{(7 x - \frac{21}{4})}{(5 x - \frac{15}{4}) + 9} = \frac{1}{1} \\ \Rightarrow \frac{28 x - 21}{20 x + 21} = 1 \\ \Rightarrow 28 x - 21 = 20 x + 21 \\ \Rightarrow x = \frac{42}{8} = \frac{21}{4} \end{aligned}$ ∴Quantity of A in mixture = $7 x$ $= 7 \times \frac{21}{4} = \frac{147}{4} = 36 \frac{3}{4}$ liter.

22.

Rs. 600 are divided among A, B, C so that Rs. 40 more than $\frac{2}{5}$ of A's share, Rs. 20 more than $\frac{2}{7}$ of B's share and Rs. 10 more than $\frac{9}{17}$ of C's share may all be equal. What is A's share ?

(A) Rs. 150
(B) Rs. 170
(C) Rs. 200
(D) Rs. 280

Solution: $\begin{aligned} = \frac{2}{5} A + 40 = \frac{2}{7} B + 20 \\ \Rightarrow \frac{2}{7} B = \frac{2}{5} A + 20 \\ \Rightarrow B = \frac{7}{2} (\frac{2}{5} A + 20) \\ = (\frac{7}{5} A + 70) \\ And, \\ = \frac{2}{5} A + 40 = \frac{9}{17} C + 10 \\ \Rightarrow \frac{9}{17} C = \frac{2}{5} A + 30 \\ \Rightarrow C = \frac{17}{9} (\frac{2}{5} A + 30) \\ = \frac{34}{45} A + \frac{170}{3} \\ = A + B + C = 600 \\ \Rightarrow A + (\frac{7}{5} A + 70) + (\frac{34}{45} A + \frac{170}{3}) = 600 \\ \Rightarrow \frac{142 A}{45} = 600 - \frac{380}{3} \\ = \frac{1420}{3} \\ \Rightarrow A = \frac{1420}{3} \times \frac{45}{142} \\ = 150 \end{aligned}$

23.

A barrel contains a mixture of wine and water in the ratio 3 : 1. How much fraction of the mixture must be drawn off and substituted by water so that the ratio of wine and water in the resultant mixture becomes 1 : 1 = ?

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

Solution: Original ratio of the mixture 3:1 Taking out the mixture actually means just taking out the wine because the water anyways is going to be added back. if we remove 1 part of wine, it makes the ratio 2 : 2 ( 1 part water is added to keep the volume constant ) so, what we have actually done is remove 1 part wine from 3 part Wine. i.e. $\frac{1}{3}$ So, $\frac{1}{3}$ mixture drawn.

24.

Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:

(A) 27
(B) 33
(C) 49
(D) 55

Solution: $\begin{aligned} Let the number be 3 x and 5 x \\ Then, \frac{3 x - 9}{5 x - 9} = \frac{12}{23} \\ \Rightarrow 23 (3 x - 9) = 12 (5 x - 9) \\ \Rightarrow 9 x = 99 \\ \Rightarrow x = 11 \\ ∴ The smaller number \\ = 3 \times 11 \\ = 33 \end{aligned}$

25.

The sides of a triangle are in the ratio $\frac{1}{2}$ : $\frac{1}{3}$ : $\frac{1}{4}$ and its perimeter is 104 cm. The length of the longest side (in cm) is -

(A) 26
(B) 32
(C) 48
(D) 52

Solution: $\begin{aligned} Ratio of sides \\ = \frac{1}{2} : \frac{1}{3} : \frac{1}{4} \\ = 6 : 4 : 3 \end{aligned}$ Let the sides be 6x, 4x and 3x Then, ⇒ 6x + 4x + 3x = 104 or 13x = 104 or x = 8 ∴ Longest side = 6x = (6 × 8)m = 48m

26.

Two natural numbers are in the ratio 3 : 5 and their product is 2160. The smaller of the numbers is -

(A) 12
(B) 18
(C) 24
(D) 36

27.

Railway fares of 1st, 2nd and 3rd classes between two stations were in the ratio of 8 : 6 : 3. The fares of 1st and 2nd class were subsequently reduced by $\frac{1}{6}$ and $\frac{1}{12}$ respectively. If during a year the ratio between the passengers of 1st, 2nd and 3rd classes was 9 : 12 : 26 and the total amount collected by the sale of tickets was Rs. 1088, then find the collection from the passengers of 1st class.

(A) Rs. 260
(B) Rs. 280
(C) Rs. 300
(D) Rs. 320

Solution: Let the initial fares of 1st, 2nd and 3rd class be Rs. 8x, Rs. 6x, and Rs. 3x respectively $\begin{aligned} Revised fare of 1st class \\ = \frac{5}{6} of Rs .8 x \\ = Rs . (\frac{20 x}{3}) \\ Revised fare of 2 nd class \\ = \frac{11}{12} of Rs .6 x \\ = Rs . (\frac{11 x}{2}) \end{aligned}$ Let the number of passengers of 1st, 2nd and 3rd class be 9y, 12y and 26y respectively Then, $\begin{aligned} = \frac{20 x}{3} \times 9 y + \frac{11 x}{2} \times 12 y + 3 x \times 26 y = 1088 \\ \Rightarrow 60 x y + 66 x y + 78 x y = 1088 \\ \Rightarrow 204 x y = 1088 \\ \Rightarrow x y = \frac{1088}{204} = \frac{16}{3} \end{aligned}$ ∴ Collection from passengers of 1st class = 60xy $\begin{aligned} = Rs . (60 \times \frac{16}{3}) \\ = Rs .320 \end{aligned}$

28.

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

29.

In two alloys A and B the ratio of Zinc and Tin is 5 : 2 and 3 : 4 respectively. 7 kg of the alloy A and 21 kg of the alloy B are mixed together to form a new alloy. What will be the ratio of Zinc and Tin tin the new alloy ?

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

Solution: Zinc : Tin A 5x : 2x = 7x B 3y : 4y = 7y ⇒ A ⇒ 7x = 7 kg x = 1 kg ∴ Zinc in alloy A ⇒ 5kg Tin in alloy A ⇒ 2 kg ⇒ B ⇒ 7y = 21 kg y = 3 kg Zinc in alloy B ⇒ 3 × 3 = 9 kg Tin in alloy B ⇒ 3 × 4 = 12 kg ∴ After mix - up the ratio of Zinc and Tin in new alloy $\begin{aligned} Zinc : Tin \\ A 5 : 2 \\ B 9 : 12 \\ \overset{―}{A + B : 14 14} \\ 1 : 1 \end{aligned}$

30.

A box contains 280 coins of one rupee, 50 paise and 25 paise. The value of each kind of the coins are in the ratio of 8 : 4 : 3. Then the number of 50 paise coins is = ?

(A) 70
(B) 60
(C) 80
(D) 90

Solution: Rs. 1 : 50 P : 25 p Values of coins 8x : 11x : 3x Number of coins 8x×1 : 4x×2 : 3x×4 8x : 8x : 12x $\begin{aligned} ∴ Total coins \\ \Rightarrow 8 x + 8 x + 12 x = 28 x \\ \Rightarrow 28 x = 280 (given) \\ \Rightarrow x = \frac{280}{28} \\ \Rightarrow x = 10 \end{aligned}$ ∴ Number of 50 P coins are 8x = 8 × 10 = 80

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Practice MCQ Questions and Answer on Ratio

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