Question : 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?
Option 1: $\frac{1}{3}$
Option 2: $\frac{1}{4}$
Option 3: $\frac{1}{5}$
Option 4: $\frac{1}{7}$
Correct Answer: $\frac{1}{5}$
Solution :
Given: Suppose the vessel initially contains 8 litres of liquid.
In this liquid, water is 3 litres and syrup is 5 litres.
So the ratio is 3 : 5.
Let $x$ litres of this liquid be replaced with water.
Quantity of water in $x$ litre of liquid $=\frac{3x}{8}$
Quantity of syrup in $x$ litre of liquid $=\frac{5x}{8}$
As per given condition,
The quantity of water in the new mixture is $=3-\frac{3x}{8}+x$
The quantity of syrup in the new mixture is $=5-\frac{5x}{8}$
After replacement, the quantity should be the same.
$3-\frac{3x}{8}+x=5-\frac{5x}{8}$
⇒ $\frac{5x}{8}-\frac{3x}{8}+x=2$
⇒ $\frac{10x}{8}=2$
$\therefore x=\frac{8}{5}$
So, part of the mixture replaced $=\frac{\frac{8}{5}}{8}=\frac{8}{5}×\frac{1}{8}=\frac{1}{5}$
Hence, the correct answer is $\frac{1}{5}$.
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