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:
Vessels = 2 = a and b
Quantity one = 8L
Quantity two = 16L
To Find:
Ratio of final concentration of milk in vessel b to the initial concentration of milk in b
Solution:
Let first quantity of milk be = p
Let the total quantity of solution be = q
Therefore,
First concentration of milk in both vessels = p/q
Now,
8L of the solution from a is replaced with milk.
New quantity of milk in a = p - 8p/q + 8
New concentration of milk in a = (p - 8p/q + 8)/q
Since, the milk concentration becomes double in a, hence -
= (p - 8p/q + 8)/q = 2p/q
= p - 8p/q + 8 = 2p
= p/q = (8-p)/8 ----- Eq 1
Similarly,
16L of the solution from b is replaced with milk.
New quantity of milk in b = p- 16p/q + 16
New concentration of milk in b = (p- 16p/q + 16)/q
= [p - 16(8-p)/8 + 16]/q ( Substituting from eq 1)
= [p - 16 + 2p + 16]/q
= 3p/q
Ratio = new concentration in b / old concentration in b
= (3p/q) / (p/q)
= 3
Answer: The required ratio is 3:1
Answer:
3.5
Step-by-step explanation: