Alligation or Mixture problems
The ratio of petrol and kerosene in the container is 3:2 when 10 liters of the mixture is taken out and is replaced by the kerosene, the ratio become 2:3. Then total quantity of the mixture in the container is:
pool : kerosene 3 : 2(initially) 2 : 3(after replacement) R e m a i n i n g Q u a n t i t y I n i t i a l Q u a n t i t y = 1 - R e p l a c e d Q u a n t i t y T o t a l Q u a n t i t y (for petrol) 2 3 = 1 - 10 k => K = 30 Therefore the total quantity of the mixture in the container is 30 liters.
Let he mixes the oils in the ratio = x : y Then, the cost price of the oils = 60x + 65y Given selling price = Rs. 68.20 => Selling price = 68.20(x+y) Given profit = 10% = SP - CP => 10/100 (60x + 65y) = 68.20(x+y)-(60x + 65y) => 6x + 6.5y = 8.20x + 3.20y =>2.2x = 3.3y => x : y = 3 : 2
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.
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.
Number of liters of water in 150 liters of the mixture = 20% of 150 = 20/100 x 150 = 30 liters.P liters of water added to the mixture to make water 25% of the new mixture.Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).(30 + P) = 25/100 x (150 + P)120 + 4P = 150 + P => P = 10 liters.
EXAMIANS