Alligation or Mixture problems
In a container, there is 960 ltr of pure milk from which 48 ltr of milk is replaced with 48 ltr of water, again 48 ltr milk is replaced by same amount of water, as this process is done once more. Now, what is the amount of pure milk?
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 : 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 total parts of milk and water in the solution is 6+2 = 8 parts. 1part = 640/8 = 80 old mixture contains 6parts of milk and 2 parts of water(6:2). To get new mixture containing half milk and half water, add 4parts of water to the old mixture then 6:(2+4) to make the ratio same. i.e, 4 x 80 = 320 ml.
Ratio of Milk and water in a vessel A is 4 : 1 Ratio of Milk and water in a vessel B is 3 : 2 Ratio of only milk in vessel A = 4 : 5 Ratio of only milk in vessel B = 3 : 5 Let 'x' be the quantity of milk in vessel C Now as equal quantities are taken out from both vessels A & B => 4/5 : 3/5 x 3/5-x x - 4/5 => 3 5 - x x - 4 5 = 1 1 (equal quantities) => x = 7/10 Therefore, quantity of milk in vessel C = 7 => Water quantity = 10 - 7 = 3 Hence the ratio of milk & water in vessel 3 is 7 : 3
EXAMIANS