Alligation or Mixture problems
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?
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.
Let cost price of spirit be Re. 1 per liter. Then SP of mixture = Re. 1 per liter Gain = 25% So, CP of mixture = 1 × (100 / 125) = Re. 4 / 5We assume that CP of water is zero. Using allegation rule on cost price, Water should be mixed to spirit in the ratio (1 / 5) : (4 / 5) or 1 : 4
Ratio of milk and water = 2 : 1Quantity of milk = 60 X 2/3 = 40 litreQuantity of water = 20 litreTo make ratio, 1: 2, we have to double the water that of milkSo, water should be 80 litre.That means 80 ? 20 = 60 litre water to be added.
Milk in 1-litre mixture of A = 4/7 litre. Milk in 1-litre mixture of B = 2/5 litre. Milk in 1-litre mixture of C = 1/2 litre. By rule of alligation we have required ratio X:Y X : Y 4/7 2/5 \ / (Mean ratio) (1/2) / \ (1/2 ? 2/5) : (4/7 ? 1/2) 1/10 1/1 4 So Required ratio = X : Y = 1/10 : 1/14 = 7:5
EXAMIANS