Weight of watermelons after percentage of water is evaporated.

Question

A stock of watermelons of the initial weight of $500 \space\text{kg}$ has been put in a store for a week.

Initially the percentage of water in the watermelons makes up $99 \% $ of the weight,at the end of the week,due to evaporation, such percentage dropped to $98 \%$.

How much do the watermelons weight at the end of the week ?

$A)250 \space\text{kg}$,$B) 400 \text{kg} $,$C)480 \text{kg} $,$D)495 \text{kg}$

My effort

I have that water makes up $99 \%$ of the weight of the watermelons,so it weights $495 \space\text{kg}$ .

After the week I have that $2 \% $ of the water evaporated so we loose $\cfrac{2}{100}\cdot 495 =9,9 \text{kg}$ so the watermelons should weight $490,1 \text{kg}$ which is none of the answers given ...

The solution is given as option $A)250 \text{kg}$ ,but I don't understand why.

Answer 1

As you said, initially water weights $495 \text{kg}$ and dry mass weights $5\text{kg}$. After the week, dry mass weights still $5\text{kg}$ and makes up $2$% of the watermelon. Then watermelons should weight $5\cdot 50 = 250(\text{kg})$.

Answer 2

Your fault is that you assumed that the percentage change in water is 1%. That is not so. Rather, try solving using simultaneous equations.

Let $x$ be the water lost and $y$ be the new mass of the watermelon.

1. $500 - x = y$

2.$\dfrac{495-x}{y}\times 100$ = $98$

Then we get the simultaneous equations:

1. $100(495 - x) = 98y$

2. $500 - x = y$

The solution of $y$ here is your answer.