The drying water melon puzzle I couldn't find an explanation to this problem that I could understand.  
A watermelon consist of 99% water and that water measures 2 litre. After a day in the sun the water melon dries up and now consist of 98% water. How much water is left in the water melon?  
I know the answer is ~1 litre, but why is that? I've read a couple of answers but I guess I'm a bit slow because I don't understand why.  
EDIT
I'd like you to assume that I know no maths. Explain it like you would explain it to a 10 year old.

You start with 2 litres of water and $x$ litres (say we measure by volume) of "non water". The percentage of water is
$$ \frac{2}{2 + x} = 99\% = \frac{99}{99+1}$$
You solve this to get that $x : 1 = 2 : 99$ or that $x = 2/99$. 
After drying, you have $y$ litres of water and $x$ litres of "non water". Since the non-water bits don't dry, the $x$ is same as before: that is $x = 2/99$. The percentage of water is
$$ \frac{y}{y+x} = \frac{y}{y+ 2/99} = 98\% = \frac{98}{98 + 2} $$
So solving this you get that $y : 98 = 2/99 : 2 = 1 : 99$. Or, in other words, $y = 98 /99 \approx 1$. That's how much water you have left. 

To intuitively understand the problem, it is more helpful to think of the proportion of "non-water". The non water started out at 1%. It ended up in 2%. Since the amount of "non water" didn't change, to have its proportion go from 1% to 2% means that the total volume must have decreased by half. 
$$ \frac{\text{non water}}{\text{total starting volume}} = 1\% \longrightarrow \frac{\text{non water}}{\text{total final volume}} = 2\% $$
Since the watermelon started out almost all water, for the total volume to decrease by half you must lose at least (and almost exactly) half of the water. 
