Alligation or Mixture problems
A container is full of milk. One-third of milk is taken out of it and replaced by same quantity of water. Then again one-third of the mixture is taken out of it and replaced by the same quantity of water. The process is repeated 4 times. If 16 litres of milk is left in the container at the end of 4th operation, find the capacity of the container.
Ratio of milk andwater = 4:1 Quantity of water = 35/5 = 7 litres Quantity of milk = 35 X 4/5 = 28 litres If 7 litre of water is added, new quantity of water = 14 litre New ratio of milk and water = 28:14 = 2:1
Quantity of milk @ Rs.10 per liter / Quantity of milk @ Rs. 16 per liter = 1 / 2 So, quantity of milk @ Rs. 10 per liter = 26 / 2 = 13 liter.2nd Method Let us assume shopkeeper buy P liter milk of price @ Rs. 10 per liter.Buy price of 26 liter of milk @ Rs. 16 per liter = 26 x 16Buy price of P liter of milk @ Rs. 10 per liter = P x 10Sell price of total milk ( P + 26 ) @ Rs. 14 per liter = 14 x ( P + 26 ) According to question there is no loss or no profit.Then Buy Price = Sell Price 26 x 16 + P x 10 = 14 x ( P + 26 ) ? 26 x 16 + 10P = 14P + 14 x 26 ? 26 x 16 - 26 x 14 = 14P - 10P? 2 x 26 = 4P? 4P = 2 x 26? P = 2 x 26 / 4 = 13So, quantity of milk @ Rs. 10 per liter = 13 liter.
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.
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.