A vessel is filled with liquid, $3$ parts of which are water and $5$ parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

\frac{1}{5}

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\frac{2}{5}

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\frac{3}{8}

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\frac{1}{8}

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Solution

The correct option is A $\frac{1}{5}$
Given that $\frac{quantity of water}{quantity of syrup} = \frac{3}{5}$
Let the vessel initially contains $8 k$ litres of liquid where $k \in N$
Let $x$ litres of this liquid be replaced with water.
Quantity of water in new mixture $= (3 k - \frac{3}{8} x + x)$ litres
Quantity of syrup in new mixture $= (5 k - \frac{5}{8} x)$ litres
Now, as per the question $3 k - \frac{3}{8} x + x = 5 k - \frac{5}{8} x$
$\Rightarrow 5 x + 24 k = 40 k - 5 x$
$\Rightarrow 10 x = 16 k$
$\Rightarrow x = \frac{8}{5} k$
So, part of the mixture replaced is $\frac{(8 k) / 5}{8 k} = \frac{1}{5}$

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