# Potato water percentage

There is a $100\,\rm g$ potato whose $99\%$ is made up of water. After leaving it, the water percentage was lowered to $98\%$. How much would the potato weight?

Initial amount of non-water substance is $1-(100$g$\cdot99\%)=1$g

Now if the total weight of resultant potato be $100x$ g, water will be $(100x-1)$ g

But, the $98\%$ of water in $100x$ g potato means $100x\cdot98\%$g$=98x$ g

So, $100x-1=98x$, $x=\frac{1}{2}$, result is 50g.

• $50$g...? Obviously wrong
– user126540
Commented May 8, 2014 at 19:44
• @user126540: No, it is correct. Many find it surprising, but with 1g non-water, $49$ grams of water gives a water content of $\frac {49}{50}=98\%$ Commented May 8, 2014 at 19:59
• @RossMillikan Damn... you are right. I guess 3 years at the university didn't count for much.
– user126540
Commented May 8, 2014 at 20:10
• Is it really valid to write, e.g., $5\cdot 80\%=4$? If I wrote that equality in a research paper, would it be perfectly correct? I have my doubts here... I'd say you can only write it using either one of the two following ways: $5\cdot \frac{80\%}{100\%}=4$ or $5\cdot 0.8=4$. I'm interested, though, is your notation agreed upon? Commented May 9, 2014 at 9:51

Let us look at it differently. Say there are $100$ students in a classroom and $99\%$ are girls. That means there are $99$ girls and $1$ boy. Now in order for the girls to be $98\%$, $50$ girls must leave the classroom, leaving $49$ girls and $1$ boy, then the percentage of the girls in the classroom becomes $98\%$. Similarly, $50$ g of water must evaporate in this question to make it $98\%$. The weight of the non-water component of potato still remains $1$ g.