Potato paradox

By Wikipedia Contributors

Visualisation of the potato paradox: Blue boxes represent 99 water and orange represents 1 non-water parts (left figure). To double the ratio of non-water to water to 1:49, the amount of water is reduced to 49 to retain same amount of non-water part (middle figure). This is equivalent to doubling the concentration of the non-water part (right figure).
White potatoes are actually around 79% water;[1] agar is 99% water.[2]

The potato paradox is a mathematical calculation that has a counter-intuitive result. The so-called paradox involves dehydrating potatoes by a seemingly minuscule amount, and then calculating a change in mass which is larger than expected.


The paradox has been stated as:

You have 100 lb of potatoes, which are 99 percent water by weight. You let them dehydrate until they're 98 percent water. How much do they weigh now?

The Universal Book of Mathematics states the problem as follows:

Fred brings home 100 pounds of potatoes, which (being purely mathematical potatoes) consist of 99 percent water. He then leaves them outside overnight so that they consist of 98 percent water. What is their new weight? The surprising answer is 50 pounds.[3]

In Quine's classification of paradoxes, the potato paradox is a veridical paradox.

Simple explanations

Method 1

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

Method 2

100 lb of potatoes, 99% water (by weight), means that there's 99 lb of water, and 1 lb of solids. It's a 1:99 ratio.

If the water decreases to 98%, then the solids account for 2% of the weight. The 2:98 ratio reduces to 1:49. Since the solids still weigh 1 lb, the water must weigh 49 lb for a total of 50 lbs for the answer.

Explanations using algebra

Method 1

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

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

resulting in x {\displaystyle x} x = 50 lb.

Method 2

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

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

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

Expanding brackets and simplifying

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

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

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

And solving:

1 0.02 = 0.02 x 0.02 {\displaystyle {\frac {1}{0.02}}={\frac {0.02x}{0.02}}} {\displaystyle {\frac {1}{0.02}}={\frac {0.02x}{0.02}}}

Which gives the lost water as:

50 = x {\displaystyle 50=x} 50=x

And the dehydrated weight of the potatoes as:

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


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.


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