Potato paradox

One explanation begins by saying that initially the non-water weight is 1 kg, which is 1% of 100 kg. Then one asks: 1 kg is 2% of how many kg? In order for that percentage to be twice as big, the total weight must be half as big.

A visualisation where blue boxes represent kg of water and the orange boxes represent kg of solid potato matter. Left, prior to dehydration: 1 kg matter, 99 kg water (99% water). Middle: 1 kg matter, 49 kg water (98% water).

In the beginning (left figure), there is 1 part of non-water and 99 parts of water. This is 99% water, or a non-water to water ratio of 1:99. To double the ratio of non-water to water to 1:49, while keeping the one part of non-water, the amount of water must be reduced to 49 parts (middle figure). This is equivalent to 2 part of non-water to 98 parts water (98% water) (right figure).

In 100 kg of potatoes, 99% water (by weight) means that there is 99 kg of water, and 1 kg of solids. This is a 1:99 ratio.

If the percentage decreases to 98%, then the solids must now account for 2% of the weight: a ratio of 2:98, or 1:49. Since the solids still weigh 1 kg, the water must weigh 49 kg to produce a total of 50 kg.

After the evaporating of the water, the remaining total quantity, ${\displaystyle x}$, contains 1 kg pure potatoes and (98/100)x water. The equation becomes:

${\displaystyle {\begin{aligned}1+{\frac {98}{100}}x&=x\\\Longrightarrow 1&={\frac {1}{50}}x\end{aligned}}}$

resulting in ${\displaystyle x}$ = 50 kg.

The weight of water in the fresh potatoes is ${\displaystyle 0.99\cdot 100}$.

If ${\displaystyle x}$ is the weight of water lost from the potatoes when they dehydrate then ${\displaystyle 0.98(100-x)}$ is the weight of water in the dehydrated potatoes. Therefore:

${\displaystyle 0.99\cdot 100-0.98(100-x)=x}$

Expanding brackets and simplifying

${\displaystyle {\begin{aligned}99-(98-0.98x)&=x\\99-98+0.98x&=x\\1+0.98x&=x\end{aligned}}}$

Subtracting the smaller ${\displaystyle x}$ term from each side

${\displaystyle {\begin{aligned}1+0.98x-0.98x&=x-0.98x\\1&=0.02x\end{aligned}}}$

Which gives the lost water as:

${\displaystyle 50=x}$

And the dehydrated weight of the potatoes as:

${\displaystyle 100-x=100-50=50}$

After the potatoes are dehydrated, the potatoes are 98% water.

This implies that the percentage of non-water weight of the potatoes is ${\displaystyle (1-.98)}$.

If x is the weight of the potatoes after dehydration, then:

${\displaystyle {\begin{aligned}(1-.98)x&=1\\.02x&=1\\x&={\frac {1}{.02}}\\x&=50\end{aligned}}}$