two vessels a and b contain equal quantity of a solution of milk and water. the concentration of milk is same in both A & B. If 8 ltrs of solution of vessel A replaced with pure milk, then the concentration of milk in vessel A becomes twice of what it was initially. If 16 ltrs of solution of vessel B replaced with pure milk, then find the ratio of the final concentration of milk in vessel B to the initial concentration of milk in B?
Answers
Given : two vessels a and b contain equal quantity of a solution of milk and water. the concentration of milk is same in both A & B . If 8 ltrs of solution of vessel A replaced with pure milk, then the concentration of milk in vessel A becomes twice of what it was initially
To find : If 16 ltrs of solution of vessel B replaced with pure milk, then find the ratio of the final concentration of milk in vessel B to the initial concentration of milk in B
Solution:
Let say Milk in Vessel A & B = M
& Total Solution = T
as Quantity & concentration are same
Concentration of Milk = M/T
8 ltrs of solution of vessel A replaced with pure milk,
=> Milk now = M - 8M /T + 8
concentration of milk becomes twice of what it was initially
=> Concentration of Milk = 2M/T
=> Milk = 2M
=> M - 8M /T + 8 = 2M
=> 8 - 8M/T = M
=> 8T - 8M = MT
=> T(8 - M) = 8M
=> T = 8M / (8 - M)
Now 16 litre Solution in replaced with pure milk
Hence Milk = M - (16M/T) + 16
= M - (16M/(8M/(8 - M))) + 16
= M - 2(8 - M) + 16
= M - 16 + 2M + 16
= 3M
ratio of the final concentration of milk in vessel B to the initial concentration of milk in B
= 3M : M
= 3 : 1
Ratio of the final concentration of milk in vessel B to the initial concentration of milk in B 3 : 1
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