A vessel contains a mixture of milk and water in the ratio of $5 : 3$ respectively. How much of the mixture must be siphoned off and replaced with water, so that the mixture may be half milk and half water?

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Solution

The correct option is A $1 / 5$
Let the vessel contains $8$ litres of liquid.
Let $x$ litres of this liquid be replaced with water
when $x$ litre of mixture is taken out it means $\frac{5 x}{8}$ litre milk and $\frac{3 x}{8}$ litre water taken out
Quantity of water in new mixture $= (3 - \frac{3 x}{8} + x) l i t r e s$
Quantity of milk in new mixture $= (5 - \frac{5 x}{8}) l i t r e s$
$∴ (3 - \frac{3 x}{8} + x) = (5 - \frac{5 x}{8})$
$\Rightarrow 5 x + 24 = 40 - 5 x$
$\Rightarrow 10 x = 16$
$∴ x = \frac{8}{5}$
So, part of the mixture replaced $= (\frac{8}{5} \times \frac{1}{8}) = \frac{1}{5}$ .

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A vessel contains a mixture of milk and water in the ratio of 5:3 respectively. How much of the mixture must be siphoned off and replaced with water, so that the mixture may be half milk and half water?

A vessel contains a mixture of milk and water in the ratio of $5 : 3$ respectively. How much of the mixture must be siphoned off and replaced with water, so that the mixture may be half milk and half water?