Practice MCQ Questions and Answer on Alligation

51.

Three vessels whose capacities are in the ratio of 3 : 2 : 1 are completely filled with milk mixed with water. The ratio of milk and water in the mixture of vessels are 5 : 2, 4 : 1 and 4 : 1 respectively. Taking $\frac{1}{3}$ of first, $\frac{1}{2}$ of second and $\frac{1}{7}$ of third mixture, a new mixture kept in a new vessel is prepared. The percentage of water in the new mixture is -

(A) 32%
(B) 28%
(C) 30%
(D) 24%

Solution: Capacities of vessels $3 : 2 : 1$ $\begin{aligned} M : W T Mixture \\ V - 1 \to (5 : 2 = 7)_{\times 5} \\ V - 2 \to (4 : 1 = 5)_{\times 7} \\ V - 3 \to (4 : 1 = 5)_{\times 7} \end{aligned}$ Equate the mixture $\begin{aligned} M : W T Mixture \\ (V - 1) \to 25 : 10 = 35 \\ (V - 2) \to 28 : 7 = 35 \\ (V - 3) \to 28 : 7 = 35 \end{aligned}$ $\begin{aligned} Capacities M : W = Total Mix . \\ (V - 1) \times 3 \to 75 : 30 = 105 \\ (V - 2) \times 2 \to 56 : 14 = 70 \\ (V - 1) \times 1 \to 28 : 7 = 35 \end{aligned}$ Water taken out $\Rightarrow \frac{1}{3} of water in (V - 1)$ $+ \frac{1}{2} of water in (V - 2)$ $+ \frac{1}{7} of water in (V - 3)$ $\begin{aligned} \Rightarrow \frac{1}{3} \times 30 + \frac{1}{2} \times 14 + \frac{1}{7} \times 7 \\ \Rightarrow 10 + 7 + 1 \\ \Rightarrow 18 \end{aligned}$ Similarly mixture will be $\Rightarrow \frac{1}{3} \times 105 + \frac{1}{2} \times 70 + \frac{1}{7}$ × 35 ⇒ 75 $\begin{aligned} ∴ % of water = \frac{18}{75} \times 100 \\ = 24 % \end{aligned}$

52.

A vessel contains 60 litre of milk. 12 litres of milk taken out from it and replaced with water. Then again from mixture, 12 litres is taken out and replaced with water. The ratio of milk and water in the resultant mixture is ?

(A) 16 : 9
(B) 15 : 10
(C) 16 : 10
(D) 9 : 5

Solution: Total milk = 60 litresDrawn off = 12 litres $\frac{Final quantity}{Initial quantity}$ $= {(1 - \frac{x}{c})}^{t}$ X = Replaced quantity C = Capacity T = Number of process $\begin{aligned} \frac{Final quantity}{Initial quantity} = {(1 - \frac{12}{60})}^{2} \\ = {(\frac{4}{5})}^{2} \\ = \frac{16}{25} \end{aligned}$ Ratio of milk and water in the resultant mixture : = 16 : 9

53.

40 litres of 60% concentration of acid solution is added to 35 litres of 80% concentration of acid solution. What is the concentration of acid in the new solution?

(A) 66%
(B) $66 \frac{2}{3} %$
(C) $69 \frac{1}{3} %$
(D) 69%

Solution: $\begin{aligned} I \to \frac{40 \times 60}{100} = 24 \\ II \to \frac{35 \times 80}{100} = 28 \\ Acid % = \frac{52}{75} \times 100 = 69 \frac{1}{3} % \end{aligned}$

54.

An alloy contains zinc, copper and tin in the ratio 2 : 3 : 1 and another contains copper, tin and lead in the ratio 5 : 4 : 3. If equal weights of both alloys are melted together to form a third alloy, then the weight of lead per kg in new alloy will be:

(A) $\frac{1}{2}$ kg
(B) $\frac{1}{8}$ kg
(C) $\frac{3}{14}$ kg
(D) $\frac{7}{9}$ kg

Solution: Ratio of Zinc, Copper and Tin is given as, Z : C : T = 2 : 3 : 1. Now, let the first alloy be 12 kg (taken as 4 kg Zinc, 6 kg Copper and 2 Kg Lead). Weight of second alloy = 12 kg as, C : T : L = 5 : 4 : 3. (taken as 5 kg Copper, 4 kg Tin and 3 Kg Lead.) Alloys are mixed together to form third alloy. Then the ratio of content in it, Z : C : T : L = 4 : (6 + 5) : (2 + 4) : 3 Weight of third alloy = 12 + 12 = 24 Kg. So, weight of the Lead = $\frac{3}{24}$ = $\frac{1}{8}$ kg.

55.

In what ratio must water be mixed with milk costing Rs. 12 per litre to obtain a mixture worth of Rs. 8 per litre?

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

56.

The milk and water in a mixture are in the ratio 7 : 5. When 15 litres of water are added to it. The ratio of milk and water in the new mixture becomes 7 : 8. The total quantity of water in the new mixture is -

(A) 35 litres
(B) 40 litres
(C) 60 litres
(D) 96 litres

Solution: According to the question, $Milk : Water$ $\begin{matrix} 7 : 5 \\ 7 : 8 \end{matrix}) 3 unit$ ∴ Remember water is added and not milk, so make milk equal but here milk is already equal $\begin{aligned} 3 units = 15 litres \\ 1 unit = 5 litres \\ 8 units = 40 litres \end{aligned}$ Total quantity of water in the new mixture = 40 litres

57.

A vessel contains a 32 litre solution of acid and water in which the ratio of acid and water is 5 : 3. If 12 litres of the solution are taken out and $7 \frac{1}{2}$ ÃÂÃÂ litres of water are added to it, then what is the ratio of acid and water in the resulting solution?

(A) 5 : 6
(B) 4 : 7
(C) 4 : 9
(D) 8 : 11

Solution: $\begin{aligned} After removed 12L \\ Acid : Water \\ \Rightarrow 12 \times \frac{5}{8}, 12 \times \frac{3}{8} \\ \Rightarrow - 7.5, - 4.5 \\ Acid Water \\ 20L 12L \\ - 7.5 - 4.5 \\ + 7.5 \\ \overset{―}{12.5 : 15} \\ 5 : 6 \end{aligned}$

58.

20 litres of a mixture contains milk and water in the ratio 3 : 1. Then the amount of milk to be added to the mixture so as to have milk and water in ratio 4 : 1 is-

(A) 7 litres
(B) 4 litres
(C) 5 litres
(D) 6 litres

Solution: In 20 litres of mixture Quantity of milk $\Rightarrow \frac{3}{4} \times 20 = 15 litres$ Quantity of water $\Rightarrow \frac{1}{4} \times 20 = 5 litres$ Let the quantity of milk added be x litres. According to the question, $\begin{aligned} \Rightarrow \frac{15 + x}{5} = \frac{4}{1} \\ \Rightarrow 15 + x = 4 \times 5 \\ \Rightarrow x = 20 - 15 \\ \Rightarrow x = 5 litres \end{aligned}$

59.

A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

(A) 26.34 litres
(B) 27.36 litres
(C) 28 litres
(D) 29.16 litres

Solution: $\begin{aligned} Amount of milk left after 3 operations \\ = [40 {(1 - \frac{4}{40})}^{3}] litres \\ = (40 \times \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10}) litres \\ = 29.16 litres \end{aligned}$

60.

A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is:

(A) 400 kg
(B) 560 kg
(C) 600 kg
(D) 640 kg

Solution: By the rule of alligation, we have: Ratio of 1st and 2nd parts = 4 : 6 = 2 : 3 ∴ Quantity of 2nd kind $\begin{aligned} = (\frac{3}{5} \times 1000) kg \\ = 600 kg \end{aligned}$

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Practice MCQ Questions and Answer on Alligation

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