Weight of watermelons after percentage of water is evaporated. 
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.
 A: 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 $5\cdot 50 = 250(\text{kg})$.
A: 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. 


*

*$500 - x = y$


2.$\dfrac{495-x}{y}\times 100$ = $98$
Then we get the simultaneous equations:


*

*$100(495 - x) = 98y$

*$500 - x = y$
The solution of $y$ here is your answer.
A: At the beginning you count with 1% equivalent to 5 kg of solid mass, this mass shouldn't change for obvious reasons (it can't evaporate), nevertheless the problem is suggesting a drop of 1%, 99% to 98%.
=> Initial equivalence:
.01 => 5 kg
From the water compound now represent 98% of the total mass, where is the other 2%? well it stills being 5 kg nevertheless now represent a 2%.
to know what is the total weight, the 100% we use the following operation.
$\frac{5 kg}{0.02} = 250$.
