Don't be fooled

99% was water, which means 1% of the watermelon is the content that cannot evaporate. This content, in Kg, is (1 / 100) x 10 = 0.1 Kg. Let the weight of the complete watermelon after evaporation be x Kg. Then, we know that 95% of x is water and so, 5% of x is the content that can't evaporate. Thus, 5% of x = 0.1 , meaning:

(5 / 100) * x = 0.1

so, x = (0.1 * 100) / 5 = 2

therefore, weight after evaporation = 2 Kg

If our fractions represent $\frac{water mass}{watermelonmass}$ , then we have a ratio of $\frac{9.9 kg}{10 kg}$ ( $9.9$ being $99\%$ of $10$ ). Now, since the water and watermelon mass decrease at the same rate (since the watermelon mass includes the water mass), and the water is now $95\%$ of the total mass, then $\frac{9.9 kg - x kg}{10 kg - x kg} = 0.95$ , where $x$ is the mass of evaporated water.

$9.9 kg-x kg = 9.5 kg - 0.95x kg$

$0.4 kg = 0.05x kg$

$8 kg= x kg$

Plugging it into our watermelon mass... $10 kg - x kg = 10 kg - (8) kg = \boxed{2 kg}$ .

Though I don't understand why this is a logic question. Seems more Algebra to me.

Even though trying to work the problem it becomes obvious, the question probably should clearly state that the percentages are "by weight." If it were "by volume" one would need to know the relative density between water and watermelon.

Otherwise, nice problem and nice solutions prior to this comment.

I did it this way: If 99 percent is water, then the ratio before the melon dried out would be: Water to melon meat=99:1=10kg. After it dried out, the ratio would be: Water to melon meat=95:5=unknown weight. Let's compare them: 99:1=10kg 95:5=unknown The melon meat never changed, so we can change the previous ratio to: 99x5:1x5=495:5 So now the ratio comparison will be: 495:5=10kg 95:5=unknown Let's find the unit value. 495+5=500, so the unit value will be 10 divided by 500, which equals to 0.02kg. Now it is simple. 95+5=100,100x0.02=2kg. So now the melons weighs 2kg.