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.
As per the given question , We are concerned with solid part of the fruit (pure portion). Assume Q kg of dry fruit is obtained.? Solid part in fresh fruit = Solid part in dry fruit? 0·28 × 100 = 0·8 × Q? Q = 35 kg? 35 kg of dry fruit can be obtained from 100 kg fresh fruit.
The given solution has 75% milk.
Milk to be added has 100% milk.
Milk should be added to the given mixture in the ratio 15 : 10 or 3 : 2
Quantity of milk to be added = (3 / 2) × 6 = 9 liters.
Suppose the two liquids A and B are 7x litres and 5x litres respectivel Now, when 9 litres of mixture are taken out, Now, when 9 liters of liquid B are added
Let the capacity of the pot be 'P' litres.Quantity of milk in the mixture before adding milk = 4/9 (P - 8)After adding milk, quantity of milk in the mixture = 6/11 P.6P/11 - 8 = 4/9(P - 8)10P = 792 - 352 => P = 44. The capacity of the pot is 44 liters.
Using Alligation rule, (Quantity of cheaper tea) / (Quantity of dearer tea) = (d - m) / (m - c) = 7/3Therefore, they must be mixed in the ratio of 7 : 3.
EXAMIANS