Mixture 'A' contains 75% fruit juice and rest water.20% of the mixture is taken out in container 'P' and 9 litres of water is added in container ' P' such that ratio of fruit juice to water in container 'P' becomes 3:2 .Find the initial quantity of water in mixture 'A'.

quantity of water in the initial mixture is 45L
because only 20% of the initial mixture was taken out in container p.

45l

Let the initial amount of juice in A be 3x and the initial amount of water in A be x. (Given that juice to water ratio in mixture A is 3:1). The total quantity of solution will be 4x.
20% of mixture in A is transferred to P i.e 20% of 4x = 0.8x L. Therefore initial amount of juice in P = (3/4)*0.8x = 0.6x. Similarly, initial amount of water in P = (1/4)*0.8x = 0.2x.
Given that 9L water is added to the solution in P and the resulting ratio of juice to water becomes 3:2;
(0.6x)/(0.2x + 9) = 3/2
On solving this we get x = 45L, i.e the initial amount of water in container A was 45L.

9L

x 9
75% 0%
60%

reslt ratio 4:1

which implies initial quantity 36L
1/4 of 36 =9L

9 liters