Question

Vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively. A certain volume is taken out from A and poured into B. After stirring, the same volume is taken out from B and poured into A. If the resultant ratio of alcohol and water in A is 15 : 4, then the volume, in litres, initially taken out from A is

Hint:

In transfer-and-mix problems, track the \emph{fraction} of each component after mixing, then multiply by the transferred volume. Ratios often simplify nicely when common denominators cancel.

Answer 1

Correct Answer: 16

Step 1: Initial state. 
Vessel A: \(60\) L alcohol, \(0\) L water.
Vessel B: \(0\) L alcohol, \(60\) L water.
Step 2: First transfer (from A to B).
Let \(x\) litres be taken from A and poured into B.
Vessel A now has \(60 - x\) L alcohol, \(0\) L water.
Vessel B now has \(x\) L alcohol and \(60\) L water.
Total volume in B is \(60 + x\) L, with ratio of alcohol to water \(= x : 60\).
Step 3: Second transfer (from B to A).
Again, \(x\) L is taken from B and poured into A.
Fraction of alcohol in B: \[ \frac{x}{60 + x}, \quad \text{and fraction of water in B: } \frac{60}{60 + x}. \] So, in the \(x\) L taken from B: \[ \text{Alcohol taken} = x \cdot \frac{x}{60 + x} = \frac{x^2}{60 + x}, \quad \text{Water taken} = x \cdot \frac{60}{60 + x} = \frac{60x}{60 + x}. \] Step 4: Final contents of vessel A.
Alcohol in A: \[ \text{Alc}_A = (60 - x) + \frac{x^2}{60 + x} = \frac{(60 - x)(60 + x) + x^2}{60 + x} = \frac{3600 - x^2 + x^2}{60 + x} = \frac{3600}{60 + x}. \] Water in A: \[ \text{Wat}_A = \frac{60x}{60 + x}. \] Step 5: Use the given ratio in vessel A.
Given final ratio of alcohol to water in A is \(15 : 4\): \[ \frac{\text{Alc}_A}{\text{Wat}_A} = \frac{15}{4}. \] Substitute: \[ \frac{\frac{3600}{60 + x}}{\frac{60x}{60 + x}} = \frac{15}{4} \Rightarrow \frac{3600}{60x} = \frac{15}{4}. \] Simplify: \[ \frac{60}{x} = \frac{15}{4} \Rightarrow 15x = 60 \times 4 = 240 \Rightarrow x = \frac{240}{15} = 16. \] So, the volume initially taken out from A is \(\boxed{16}\) litres.