Alligation or Mixture problems
In a 40 litre mixture of alcohol & water, the ratio of alcohol and water is 5 : 3. If 20% of this mixture is taken out and the same amount of water is added then what will be the ratio of alcohol and water in final mixture?
Quantity of alohol in the mixture = 40 x 5/8 = 25 lit Quantity of water = 40 - 25 = 15 lit According to question, Required ratio = 20 - 40 x 20 100 x 5 8 15 - 40 x 20 100 x 3 8 + 40 x 20 100 = 20 15 - 3 + 8 = 1 : 1
Ratio of milk and water = 2 : 1
Quantity of milk = 60 X 2/3 = 40 litre
Quantity of water = 20 litre
To make ratio, 1: 2, we have to double the water that of milk
So, water should be 80 litre.
That means 80 ? 20 = 60 litre water to be added.
Here first two varieties of tea are mixed in equal ratio.So their average price = (126 + 135) /2 = Rs. 130.50 Let price of the third variety per kg be Rs. A; then now mixture is formed by two varieties one at Rs. 130.50 per kg and other at Rs. A per kg in the same ratio 2 : 2 i.e, 1 : 1 By the rule of alligation,(A - 153) / (22.50) = 1 ? A ? 153 = 22.50 ? A = Rs. 175.50
Let C.P. of 1 liter milk be Re. 1, Gain = 16 2/3 % = 50/3 %and S.P. of 1 liter mixture = Re. 1 then C.P. of 1 liter mixture = (1 x (100 x 3) / 350) = Re. (6 / 7) By the rule of alligation,Hence, required ratio = (1/ 7) : (6 / 7) = 1 : 6
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.
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