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+2 votes

From a container of wine, a thief has stolen 15 litres of wine and replaced it with same quantity of water. He repeated the same process and thus in three attempts the ratio of wine and water became 343:169. The initial amount of wine in the container was ____ litres.

+4 votes

Let the initial amount of wine be $x$.

$x - 15 - 15 - 15 : 15+ 15 + 15 = 343: 169$

$x - 45 : 45 = 343:169$

$x = \frac{343 \times 45}{169} + 45$

$=45 \times \left[ \frac{343}{169} + 1\right]$

$=45 \times \frac{512}{169}$

$=136.33$

The above approach is wrong because after the first replacement, we are no longer having pure wine but a mixture of wine and water.

Amount of wine remaining $= \frac{343x}{343+169} = \frac{343x}{512}.$

Amount of wine after first add/remove $= x - 15 = x\left(1-\frac{15}{x}\right)$, after second add/remove $=x\left(1-\frac{15}{x}\right)^2$ and similarly after the third add/remove $x\left(1-\frac{15}{x}\right)^3$.

Thus we get $\frac{343x}{512} = x\left(1-\frac{15}{x}\right)^3$

$\implies \frac{7}{8} = 1 - \frac{15}{x}$

$\implies \frac{15}{x} = \frac{1}{8}$

$\implies x = 15 \times 8 = 120.$

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