Question

A large watermelon weighs 20 kg with 96% of its weight being water. It is allowed to stand in the sun and due to water evaporation, only 95%, of its weight is water. Its reduced weight of the watermelon will be :

• A. 18 kg
• B. 17 kg
• C. 16.5 kg
• D. 16 kg

Answer 1

The correct option is D 16 kg

Weight of water in 20 kg watermelon $=\frac{96}{100}\times 20=19.2kg$

Let ‘x’ kg of water be evaporated, then
$\Rightarrow (19.2-x)=\frac{95\times (20-x)}{100}$
$\Rightarrow 1920-100x=1900-95x$
$\Rightarrow 5x=20$
$\Rightarrow x=4kg$
$\therefore$ Reduced weight = 20 - 4 =16 kg