Question

A watermelon is kept under sun for 3 days.Due to which its water content reduces from85% to 80%. If 3 kg of water got evaporated,then what was the initial weight (in kg) of thewatermelon?​

Answer 1

Step-by-step explanation:

Reduction in water content = 85% - 80% = 5%

Reduction in weight = 3 kg

Let initial weight of the watermelon was x kg.

Therefore,

5% of x = 3

$\frac{5}{100} \times x = 3 \\ \implies \: x = \frac{3 \times 100}{5} = 3 \times 20 = 60 \\ thus \: initial \: weight \: of \: \\ watermelon \: was \: 60 \: kg.\\ \\Alternative \: method\colon\\ 5\% = 3\: kg\\ \therefore \: 100 \% = \frac{3 \times 100}{5}= 60 \: kg$