Alligation & Mixture

Question

A vessel is filled with a solution of which 3 parts soda and 5 parts rum. How much of the solution must be taken out and replaced with soda so that the solution contains an equal amount of soda and rum?
(A) 8/5
(B) 2/5
(C) 1/5
(D) 5/8

Answer 1

Vessel contains 3 parts of soda and 5 parts of rum.

∴ Total solution in the vessel = (3+5) = 8 parts

Assuming,

initially in the vessel, there are 8 litres of solution in which 3 litres are soda and 5 litres is rum.

Now, some solution (assuming S litre) has been taken out and replaced with soda (taken as same amount as S litre) so that the solution contains an equal amount of soda and rum (1:1).

As, S litre of solution has been taken out-

remaining amount of soda in the (8-S) litre solution = 3-3/8S  {As, S litre solution also has 3 parts of soda and 5 parts of rum  }

remaining amount of rum in the (8-S) litre solution = 5- 5/8S

Now, S litre soda also been added in the vessel.

Total amount of soda in the vessel = 3-3/8S+S

Now, as per the criteria, (3-3/8S) & $ (5- 5/8s) needs to be same.

∴ 3-(3/8)S+S = 5- (5/8)S

Or, (24-3S+8S)/8 = (40-5S)/8

Or, 24+5S = 40-5S

Or, 10S = 40-24

Or, S = 16/10 = 8/5

∴ 8/5 Lt. solution has been removed initially, which is equal to1/5 of 8 Lt. OR 1/5 of initial solution has been taken out and replaced with soda so that the solution contains equal amount of soda and rum.

Answer 2

We want 1/2 of our solution to be soda, but right now it is 3/8 soda. The way we are going to do that is to remove and replace X amount of the solution.

1/2 = X + (3/8)*(1-X)

1/2 = X - X*(3/8) + 3/8

1/2 - 3/8 = X - X*(3/8)

1/8 = X*(5/8)

1/8 * 8/5 = 1/5 = X

We want to remove and replace 1/5 of the contents, or 1.6 parts with soda