Classic Potato Problem

[BMAD]

Fred has 100 lbs of potatoes which consist of 99 percent water.  Leaving them outside overnight, Fred finds his potatoes only consist of 98 percent water.  What is their new weight?  What was the weight of the 1% of water lost?  What is the weight of the 98% water remaining?

[Pickett]

Problems like this always SEEM so straightforward, but the answers are always a little shocking to see:

New weight: 50lbs
Weight of the 1% water lost: 50lbs
Weight of the 98% water remaining: 49lbs

 

First let's do the obligitory/obvious initial analysis of the original amounts:

0.99 = water weight / total weight

0.99 = water weight / 100

water weight = 99lbs

 

So, now we need to find the new weights:

0.98 = new water weight / new total weight

 

we know that the new water weight is: 99 - water weight lost

and we know the new total weight is: 100 - water weight lost

 

Let x = water weight lost

This gives us the equation: 

0.98 = (99 - x) / (100 - x)

98 - .98x = 99 - x

0.02x = 1

x = 50

 

This means the water weight lost was 50lbs...so using that, we can find the answers above. We can certainly validate this answer by showing that if we have 50lbs of potatoes, and 49lbs of it are water, we indeed have 98% water weight, and both the total weight and water weight decreased by the same amount (50lbs)...or another way of seeing that is that the original weight of the non-water parts of the potatoes were 1 pound (100 - water weight)...since that does not change after the water evaporates overnight, the end result should still have 1 pound of non-water weight...which we do have (50 - 49).

Edited November 6, 2014 by Pickett