cfc198 wrote:
A large watermelon weighs 20 kg with 96% of its weight being water. It is allowed to stand in the sun for a while and some of the water evaporates so that now only 95% of it is water. Its
reduced weight will be?
1. 16 kg
2. 19.2 kg
3. 18 kg
4. 19.8 kg
5. 19 kg
Since 96% of the 20 kg watermelon is water, 4% of the 20 kg is non-water:
\((0.04)(20) = 0.8\) kg
Since 95% of the post-evaporation watermelon is water, the remaining 5% must be composed of the 0.8 kg of non-water:
\(0.05x = 0.8\)
\(x = \frac{0.8}{0.05} = \frac{80}{5} = 16\) kg
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