Question 1

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:

Accepted Answer

By the rule of alligation, we have: Profit on 1st part Profit on 2nd part 8% Mean Profit 14% 18% 4 6 Ration of 1st and 2nd parts = 4 : 6 = 2 : 3 ∴ Quantity of 2nd kind = ❨ 3 x 1000 ❩kg = 600 kg. 5

Question 2

The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs. 20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is:

Accepted Answer

Let the price of the mixed variety be Rs. x per kg. By rule of alligation, we have: Cost of 1 kg of Type 1 rice Cost of 1 kg of Type 2 rice Rs. 15 Mean Price Rs. x Rs. 20 (20 - x) (x - 15) ∴ (20 - x) = 2 (x - 15) 3 ⟹ 60 - 3x = 2x - 30 ⟹ 5x = 90 ⟹ x = 18.

Question 3

A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

Accepted Answer

Let the cost of 1 litre milk be Re. 1 Milk in 1 litre mix. in 1st can = 3 litre, C.P. of 1 litre mix. in 1st can Re. 3 4 4 Milk in 1 litre mix. in 2nd can = 1 litre, C.P. of 1 litre mix. in 2nd can Re. 1 2 2 Milk in 1 litre of final mix. = 5 litre, Mean price = Re. 5 8 8 By the rule of alligation, we have: C.P. of 1 litre mixture in 1st can C.P. of 1 litre mixture in 2nd can 3 4 Mean Price 5 8 1 2 1 8 1 8 ∴ Ratio of two mixtures = 1 : 1 = 1 : 1. 8 8 So, quantity of mixture taken from each can = ❨ 1 x 12 ❩ = 6 litres. 2

Question 4

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?

Accepted Answer

By the rule of alligation: Cost of 1 kg pulses of 1st kind Cost of 1 kg pulses of 2nd kind Rs. 15 Mean Price Rs. 16.50 Rs. 20 3.50 1.50 ∴ Required rate = 3.50 : 1.50 = 7 : 3.

Question 5

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:

Accepted Answer

Since first and second varieties are mixed in equal proportions. So, their average price = Rs. ❨ 126 + 135 ❩ = Rs. 130.50 2 So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x. By the rule of alligation, we have: Cost of 1 kg of 1st kind Cost of 1 kg tea of 2nd kind Rs. 130.50 Mean Price Rs. 153 Rs. x (x - 153) 22.50 ∴ x - 153 = 1 22.50 ⟹ x - 153 = 22.50 ⟹ x = 175.50

Question 6

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?

Accepted Answer

Amount of milk left after 3 operations = [ 40 ❨ 1 - 4 ❩ 3 ] litres 40 = ❨ 40 x 9 x 9 x 9 ❩ = 29.16 litres. 10 10 10

Question 7

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:

Accepted Answer

Let C.P. of 1 litre milk be Re. 1 Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%. C.P. of 1 litre mixture = Re. ❨ 100 x 1 ❩ = 4 125 5 By the rule of alligation, we have: C.P. of 1 litre of milk C.P. of 1 litre of water Re. 1 Mean Price Re. 4 5 0 4 5 1 5 ∴ Ratio of milk to water = 4 : 1 = 4 : 1. 5 5 Hence, percentage of water in the mixture = ❨ 1 x 100 ❩% = 20%. 5

Question 8

Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.

Accepted Answer

By the rule of alligation: Cost of 1 kg of 1st kind Cost of 1 kg of 2nd kind 720 p Mean Price 630 p 570 p 60 90 ∴ Required ratio = 60 : 90 = 2 : 3.

Question 9

How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg?

Accepted Answer

S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%. ∴ C.P. of 1 kg of mixture = Rs. ❨ 100 x 9.24 ❩ = Rs. 8.40 110 By the rule of allilation, we have: C.P. of 1 kg sugar of 1st kind Cost of 1 kg sugar of 2nd kind Rs. 9 Mean Price Rs. 8.40 Rs. 7 1.40 0.60 ∴ Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3. Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind. Then, 7 : 3 = x : 27 ⟹ x = ❨ 7 x 27 ❩ = 63 kg. 3

Question 10

In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?

Accepted Answer

S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%. C.P. of 1 kg of the mixture = Rs. ❨ 100 x 68.20 ❩ = Rs. 62. 110 By the rule of alligation, we have: Cost of 1 kg tea of 1st kind. Cost of 1 kg tea of 2nd kind. Rs. 60 Mean Price Rs. 62 Rs. 65 3 2 ∴ Required ratio = 3 : 2.

Question 11

A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is:

Accepted Answer

By the rule of alligation, we have: Strength of first jar Strength of 2nd jar 40% Mean Strength 26% 19% 7 14 So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2 ∴ Required quantity replaced = 2 3

Question 12

A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

Accepted Answer

Suppose the vessel initially contains 8 litres of liquid. Let x litres of this liquid be replaced with water. Quantity of water in new mixture = ❨ 3 - 3x + x ❩ litres 8 Quantity of syrup in new mixture = ❨ 5 - 5x ❩ litres 8 ∴ ❨ 3 - 3x + x ❩ = ❨ 5 - 5x ❩ 8 8 ⟹ 5x + 24 = 40 - 5x ⟹ 10x = 16 ⟹ x = 8 . 5 So, part of the mixture replaced = ❨ 8 x 1 ❩ = 1 . 5 8 5

Question 13

In what ratio must water be mixed with milk to gain 16⅔% on selling the mixture at cost price?

Accepted Answer

Let C.P. of 1 litre milk be Re. 1. S.P. of 1 litre of mixture = Re.1, Gain = 50 %. 3 ∴ C.P. of 1 litre of mixture = ❨ 100 x 3 x 1 ❩ = 6 350 7 By the rule of alligation, we have: C.P. of 1 litre of water C.P. of 1 litre of milk 0 Mean Price Re. 6 7 Re. 1 1 7 6 7 ∴ Ratio of water and milk = 1 : 6 = 1 : 6. 7 7

Question 14

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?

Accepted Answer

Suppose the can initially contains 7x and 5x of mixtures A and B respectively. Quantity of A in mixture left = ❨ 7x - 7 x 9 ❩ litres = ❨ 7x - 21 ❩ litres. 12 4 Quantity of B in mixture left = ❨ 5x - 5 x 9 ❩ litres = ❨ 5x - 15 ❩ litres. 12 4 ∴ ❨ 7x - 21 ❩ 4 = 7 ❨ 5x - 15 ❩ + 9 4 9 ⟹ 28x - 21 = 7 20x + 21 9 ⟹ 252x - 189 = 140x + 147 ⟹ 112x = 336 ⟹ x = 3. So, the can contained 21 litres of A.

Question 15

8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65. How much wine did the cask hold originally?

Accepted Answer

Let the quantity of the wine in the cask originally be x litres. Then, quantity of wine left in cask after 4 operations = [ x ❨ 1 - 8 ❩ 4 ] litres. x ∴ ❨ x(1 - (8/x))4 ❩ = 16 x 81 ⟹ ❨ 1 - 8 ❩ 4 = ❨ 2 ❩ 4 x 3 ⟹ ❨ x - 8 ❩ = 2 x 3 ⟹ 3x - 24 = 2x ⟹ x = 24.

Question 16

In a mixture of 45 liters, the ratio of milk and water is 4:1. How much water must be added to make the mixture ratio 3:2?

Accepted Answer

Let us assume the ratio factor is x. According to question, Quantity of milk = 4x liters and Quantity of Water = x liters. According to question, 4x + x = 45 ⇒ 5x = 45 ⇒ x = 9 Quantity of milk = 4x = 9 x 4 = 36 liters Quantity of water = x = 9 liters Let y liters of water be added to make the ratio 3:2 Then, 36/(9 + y) = 3/2 ⇒ 72 = 27 + 3y ⇒ y = 15 liters If you add 15 liters of water, the ratio will become 3:2.

Question 17

In a class of 30 students , the average weight of boys is 20 kg and the average weight of the girls is 25 kg . The fraction of boys out of the total students of the class is

Accepted Answer

Let us assume the number of boys = B and number of girls = G. According to question, B + G = 30 Lets us assume total weight of boys = W1 and total weight of girls = W2 average weight of boys = total weight of boys/number of boys total weight of boys/number of boys = 20 W1/B = 20 W1 = 20B average weight of girls = total weight of girls/number of girls 25 = W2/G W2 = 25G Data is not sufficient to solve the equation. since we do not know either the average weight of the whole class or the ratio of no. of boys to girls.

Question 18

In what ratio should milk and water be mixed so that after selling the mixture at the cost price a profit of 50/3 % is made?

Accepted Answer

Method 1 to solve the equation. Let us assume Cost Price (C.P) of 1 liter milk be Y rupees. According to question, Selling Price (S.P) of 1 liter of mixture Y rupees. (Selling price should be same as Cost price) Profit = 50/3 % Let us assume Cost price of 1 Liter of Mixture of Water and Milk = C Cost price of 1 Liter of Mixture of Water and Milk = Selling price of Mixture - Profit C = Y - C x 50/3% ⇒ C = Y - C x 50/3x100 ⇒ C = Y - C x 1/3x2 ⇒ C = Y - C/6 ⇒ Y = C + C/6 ⇒ Y = (6C + C)/6 ⇒ Y = (6C + C)/6 ⇒ Y = 7C/6 ⇒ C = 6Y/7 Since in Y rupees we will get 1 liter milk. Hence in 1 rupees we will get 1/Y liter milk. Hence in 6Y/7 rupees we will get 1 x 6Y/7 x Y liter milk. Hence in 6Y/7 rupees we will get 6/7 liter milk. We need 1 liter mixture of milk and water for sold on the same price, we need to mix the water. So water quantity in mixture = 1 - 6/7 = 1/7 Ratio of Milk and water in mixture = quantity of Milk in Mixture/quantity of Water in Mixture Ratio of Milk and water in mixture = 6/7 / 1/7 Ratio of Milk and water in mixture = 6 x 7 / 7 Ratio of Milk and water in mixture = 6 / 1 Ratio of Milk and water in mixture = 6 : 1 Method 2 to solve the question. Let the original amount of milk be 1 liter and the cost price is 1 rupees per liter. Cost price of milk = 1 rupees. Selling price of Mixture = 1 rupees. When the milk is mix with x liters of water milk remaining is 1- x. Let us assume the cost price of milk in mixture is Y and given that the Sold Price is 1. we should use the Profit formula in algebra, Y + Y x 50/3 % = 1 Y + Y x 50/3 x 100 = 1 Y + Y x /3 x 2 = 1 7Y/6 = 1 Y = 6/7 Y = 6/7 is the cost price of Mixture for 1 liter. Since water is free so there is not cost of water in mixture. So price of milk is 6/7 rs in mixture. Since as per question, In 1 rupees we will get 1 liter of milk. hence in 6/7 rupees we will get 1 x 6/7 = 6/7 liter of milk. Now we have to make 1 liter of mixture with water and milk. quantity of Water + Quantity of Milk = 1 liter quantity of Water + 6/7 = 1 quantity of Water = 1 - 6/7 = (7 - 6)/7 quantity of Water = 1/7 Ratio of milk and Water = Quantity of Milk / quantity of Water Ratio of milk and Water = (6/7) / (1/7) Ratio of milk and Water = (6/7) x (7/1) Ratio of milk and Water = 6 x 7/ 7 x 1 Ratio of milk and Water = 6 :1

Question 19

Two solution of 90% and 97% purity and mixed resulting in 21 liters of mixture of 94% purity. How much is the quantity of the first solution in the resulting mixture?

Accepted Answer

Method 1 to solve the equation. Let us assume the number of liters of the 90% purity solution = A and the number of liters of the 97% purity solution = B. According to question, Since there are 21 liters of the solution, A + B = 21 ...................... (1) Since after mixing the two solutions the new mixture has 94% purity, Concentrate of A + Concentrate of B = Concentrate of (A + B) A x 90% + B x 94% = (A+ B) x 97% ⇒ A x 90/100 + B x 97/100 = (A + B) x 94/100 ⇒ 90A + 97B = (A + B) x 94 ⇒ 90A + 97B = 94A + 94B ⇒ 94A + 94B - 90A- 97B = 0 ⇒ 4A - 3B = 0 ........................(2) Multiply the 3 with Equation (1) and add with Equation (2), 3A + 3B + 4A - 3B = 63 + 0 ⇒ 7A = 63 ⇒ A = 63/7 = 9 Put the value of A in Equation (1) , we will get 9 + B = 21 B = 21 - 9 B = 12 The first solution would be A = 9 liters. Method 2 to solve the equation. Hit and trail method. 94% is closer to 97% but barely meaning the mixtures will not be equal parts but will be slightly more of the higher purity. Quickly eliminate A and B. Out of the others 9 is the easy choice. If the other choices were closer to half this wouldn't work.

Question 20

In what proportion must tea at Rs. 62 per kg be mixed with tea at Rs. 72 per kg in order to obtain the mixture worth Rs. 65 per kg?

Accepted Answer

Using Alligation rule, (Quantity of cheaper tea) / (Quantity of dearer tea) = (d - m) / (m - c) = 7/3 Therefore, they must be mixed in the ratio of 7 : 3.

Alligation or Mixture Questions