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?

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{5}$
• D. $\frac{1}{7}$

Answer 1

The correct option is C

$\frac{1}{5}$

Suppose the vessel initially contains 8 litre of liquid.
Initially, water: syrup= 3:5
i.e., water is 3 litres and syrup is 5 litres.
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 = $\frac{5x}{8}$
Now, Quantity of water in new mixture= $(3-\frac{3x}{8}+x)$ litres
Quantity of syrup in new mixture= $(5-\frac{5x}{8})$ litres
According to question,
Quantity of water= Quantity of syrup
$\therefore 3-\frac{3x}{8}+x=5-\frac{5x}{8}$
5x+24=40-5x
10x=16
$x=\frac{8}{5}$
Now, $\frac{8}{5}$l of liquid is replaced by water in 8 litre of liquid.
So, part of the mixture replaced $=\frac{\frac{8}{5}}{8}=\frac{1}{5}$
Hence, correct answer is (c).