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

Entered answer:

✅ Correct Answer: 16

Vessel A has 60 litres of pure alcohol and Vessel B has 60 litres of pure water.

Let the volume taken out each time be $x$ litres.

Since A has only alcohol, the $x$ litres taken from A is pure alcohol.

After pouring into B:

Vessel A has $(60 - x)$ litres of alcohol.

Vessel B has $x$ litres of alcohol and $60$ litres of water, totalling $(60 + x)$ litres.

After stirring B, the mixture is uniform. In every litre of B:

Fraction of alcohol $= \dfrac{x}{60 + x}$

Fraction of water $= \dfrac{60}{60 + x}$

So the $x$ litres taken from B contains:

Alcohol $= \dfrac{x^2}{60+x}$

Water $= \dfrac{60x}{60+x}$

After pouring this back into A:

Alcohol in A $= (60 - x) + \dfrac{x^2}{60+x}$

$= \dfrac{(60-x)(60+x) + x^2}{60+x}$

$= \dfrac{3600 - x^2 + x^2}{60+x}$

$= \dfrac{3600}{60+x}$

Water in A $= \dfrac{60x}{60+x}$

The ratio of alcohol to water in A is given as $15 : 4$ .

$\dfrac{3600/(60+x)}{60x/(60+x)} = \dfrac{15}{4}$

The $(60 + x)$ cancels from numerator and denominator:

$\dfrac{3600}{60x} = \dfrac{15}{4}$

$\dfrac{60}{x} = \dfrac{15}{4}$

$x = \dfrac{60 \times 4}{15}$

$x = 16$

The volume initially taken out from A is $16$ litres.

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