Alligation or Mixture problems
From the 50 liters of milk, 5 liters of milk is taken out and after it 5 liters of water is added to the rest amount of milk. Again 5 liters of milk and water is drawn out and it was replaced by 5 liters of water. If this process is continued similarly for the third time, the amount of milk left after the third replacement:
General Formula: Final or reduced concentration = initial concentration x 1 - a m o u n t b e i n g r e p l a c e d i n e a c h o p e r a t i o n t o t a l a m o u n t n where n is the number of times the same operation is being repeated. The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture. Here amount being replaced denotes the quantity which is to be withdrawn in each time. Therefore, 50 × 1 - 5 50 3 = 36.45 L
Wine Water 8L 32L 1 : 4 20 % 80% (original ratio) 30 % 70% (required ratio) In ths case, the percentage of water being reduced when the mixture is being replaced with wine. so the ratio of left quantity to the initial quantity is 7:8 Therefore , 7 8 = 1 - K 40 => K = 5 Lit
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, quantity of wine left after third operation = [1 - (5 / 25)]3 x 25 = (4 / 5)3 x 25 = (64 / 125) x 25 = (64 / 5) = 12 4/5 liters. Final ratio of wine to water = (64 / 125) / (1- 64 /125) = (64 / 125) /(61 / 125) Wine : Water = (64 / 61)
EXAMIANS