Practice MCQ Questions and Answer on Alligation

11.

Lala has lent some money to Arun at 5% p.a. and Bhatia at 8% p.a. At the end of the year, he has gained an overall interest at the rate 6%. In what ratio has he lent the money to Arun and Bhatia ?

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

12.

7 kg of tea costing Rs. 280 per kg is mixed with 9 kg of tea costing Rs. 240 per kg. The average price per kg of the mixed tea is ?

(A) Rs. 255.80
(B) Rs. 257.50
(C) Rs. 267.20
(D) Rs. 267.50

Solution: According to the question,Average price of mixed tea $\begin{aligned} = \frac{280 \times 7 + 240 \times 9}{16} \\ = \frac{1960 + 2160}{16} \\ = \frac{4120}{16} \\ = Rs . 257 .50 \end{aligned}$

13.

Alcohol and water in two vessels A and B are in the ratio 5 : 3 and 5 : 4 respectively. In what ratio, the liquid of both the vessels be mixed to obtain a new mixture in vessel C in the ratio 7 : 5 ?

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

Solution: According to the question, $\begin{aligned} Alcohol Water \\ Vessel A 5 : 3 \\ Vessel B 5 : 4 \end{aligned}$ Now using alligation,

14.

In what ratio must water be mixed with milk to gain $16 \frac{2}{3} %$ ÃÂÃÂ on selling the mixture at cost price?

(A) 1 : 6
(B) 6 : 1
(C) 2 : 3
(D) 4 : 3

Solution: Let C.P. of 1 litre milk be Rs. 1 S.P. of 1 litre of mixture = Rs. 1, Gain = $\frac{50}{3}$ % ∴ C.P. of 1 litre of mixture = $100 \times \frac{3}{350} \times 1$ = $\frac{6}{7}$ BY the rule of alligation, we have: ∴ Ration of water and milk = $\frac{1}{7}$ : $\frac{6}{7}$ = 1 : 6

15.

A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

(A) 10
(B) 20
(C) 21
(D) 25

Solution: $\begin{aligned} Suppose the can initially contains \\ 7 x and 5 x of mixtures A and B respectively . \\ Quantity of A in mixture left \\ = (7 x - \frac{7}{12} \times 9) litres \\ = (7 x - \frac{21}{4}) litres \\ Quantity of B in mixture left \\ = (5 x - \frac{5}{12} \times 9) litres \\ = (5 x - \frac{15}{4}) litres \\ ∴ \frac{(7 x - \frac{21}{4})}{(5 x - \frac{15}{4}) + 9} = \frac{7}{9} \\ \Rightarrow \frac{28 x - 21}{20 x + 21} = \frac{7}{9} \\ \Rightarrow 252 x - 189 = 140 x + 147 \\ \Rightarrow 112 x = 336 \\ \Rightarrow x = 3 \end{aligned}$ So, the can contained 21 litres of A

16.

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

17.

From a container, full of pure milk, 20% is replaced by water and this process is repeated three times. At the end of third operation, the quantity of pure milk reduces to:

(A) 40.0%
(B) 50.0%
(C) 51.2%
(D) 58.8%

Solution: Let pure milk was 100L. So, Water is replaced 20% in per process = 20% of 100 = 20L. Now, we use short-cut formula for it. Quantity of Milk reduced to, $\begin{aligned} = X \times {[1 - \frac{Y}{X}]}^{n} \\ = 100 \times {[1 - \frac{20}{100}]}^{3} \\ = \frac{100 \times 64}{125} \\ = 51.2 % \end{aligned}$ Here, X = Initial quantity of milk. Y = Replaced water in per process. n = No. of process repeated. Note: The formula used in above problem is quite similar to depreciation formula or Compound interest formula. Alternatively, Let pure milk be 100 litres initially. After third operation, milk will be 100 == 20%↓(- 20L) ⇒ 80 == 20%↓(- 16L) ⇒ 64 == 20%↓(-12.8L) ⇒ 51.2 %

18.

A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:

(A) 4%
(B) 16%
(C) 20%
(D) 25%

Solution: Let C.P. of 1 litre milk be Rs. 1 Then, S.P. of 1 litre of mixture = Rs. 1, Gain = 25% C.P. of 1 litre mixture = Rs. $(\frac{100}{125} \times 1)$ = $\frac{4}{5}$ By the rule of alligation, we have: ∴ Ratio of milk to water = $\frac{4}{5}$ : $\frac{1}{5}$ = 4 : 1 Hence, percentage of water in the mixture = $(\frac{1}{5} \times 100)$ % = 20%

19.

In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg?

(A) 3 : 7
(B) 5 : 7
(C) 7 : 3
(D) 7 : 5

20.

A vessel contains milk and water in the ratio 3 : 2. The volume of the contents is increased by 50% by adding water to it. From this resultant solution 30 L is withdrawn and then replaced with water. The resultant ratio of milk water in the final solution is 3 : 7. Find the original volume of the solution.

(A) 80 L
(B) 65 L
(C) 75 L
(D) 82 L

Solution: Let the original volume be x. then, quantity of milk and water, = $\frac{3 x}{5}$ and $\frac{2 x}{5}$ respectively. After adding water to it, the volume becomes 150%, the quantity of milk and water, = $\frac{3 x}{5}$ and $\frac{9 x}{10}$ $\frac{\frac{3 x}{5} - 12}{\frac{9 x}{10} + 12} = \frac{3}{7}$ 14(3x - 60) = 3(9x + 120) Or, x = 80 L

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Practice MCQ Questions and Answer on Alligation

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