The potato paradox is a mathematical calculation that has a counter-intuitive result. The Universal Book of Mathematics states the problem as such:

Fred brings home 100 kg of potatoes, which (being purely mathematical potatoes) consist of 99% water. He then leaves them outside overnight so that they consist of 98% water. What is their new weight? The surprising answer is 50 kg.

## Simple explanations

### Method 1

Initially, the non-water weight is 1 kg, which is 1% of 100 kg. If after leaving them overnight, the water weight shrinks to 98%, then 1 kg is 2% of how many kg? For non-water percentage to be twice as big, the total weight must be half as big. The non-water weight cannot grow or shrink in this question, therefore it will always be 1 kg, and the water weight must change around it.

Another way to word it starting from the beginning:

1% of 100 kg is 1 kg of solid potato. This amount does not change. So we have 1 kg of solid potato comprising 1% of the total weight.

When water evaporates so that the water is 98% of the total weight, meaning that 2% of the total weight is the UNCHANGED 1 kg of solid.

Simple algebra: 1 kg is 2% of what? The answer is 50 kg.

### Method 2 A visualization 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 non-water and 99 parts 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 parts 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 non-water. This is a 1:99 ratio.

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

## Explanations using algebra

### Method 1

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

{\begin{aligned}1+{\frac {98}{100}}x&=x\\\Longrightarrow 1&={\frac {1}{50}}x\end{aligned}} resulting in $x$ = 50 kg.

### Method 2

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

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

$0.99\cdot 100-0.98(100-x)=x$ Expanding brackets and simplifying

{\begin{aligned}99-(98-0.98x)&=x\\99-98+0.98x&=x\\1+0.98x&=x\end{aligned}} Subtracting the smaller $x$ term from each side

{\begin{aligned}1+0.98x-0.98x&=x-0.98x\\1&=0.02x\end{aligned}} Which gives the lost water as:

$50=x$ And the dehydrated weight of the potatoes as:

$100-x=100-50=50$ ### Method 3

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

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

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

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

The answer is the same as long as the concentration of the non-water part is doubled. For example, if the potatoes were originally 99.999% water, reducing the percentage to 99.998% still requires halving the weight.