## Saturday, November 12, 2011

### embarrassed to say it took all day

Fairfax County Public Schools Math League challenge problem (worded differently from the original because I didn't memorize the wording):

A 100-pound watermelon is 95% water. If you dehydrate the watermelon to the point where it is 90% water, how much will the dehydrated watermelon weigh?

Yesterday, a student handed me a list of six FCPS Math League problems, all of which had to be done in 36 minutes. I'm embarrassed to say that the above problem, first on the list, took me all day to solve. Sure, part of the reason was that I was teaching, so I didn't have the luxury of staring at the problem for several minutes at a time.* All the same, it would have been nice had the flash of insight on how to solve the problem come five hours sooner than it did.

*Not that that should matter much: in principle, a capable Math Leaguer is able to solve each such problem in six minutes.

_

Kevin Kim said...

The solution to this problem is embarrassingly obvious once you figure out how to set the problem up.

What do we know?

Let:

t = 100 = total poundage of original watermelon
w = 95 = original water weight
s = 5 = solid weight of watermelon
x = weight of water removed to get watermelon to its new fighting weight

Keep in mind that, to get the "95% water" figure for the original watermelon, the expression is:

w/t = .95

Given the above, let:

(t - x) = the watermelon's new fighting weight

(w - x) = the new amount of water in the dehydrated melon

The equation describing the leaner, meaner watermelon is therefore:

(w - x)/(t - x) = .9

Multiply both sides by (t - x), and this becomes

w - x = .9t - .9x

Since w and t represent the watermelon's original stats, we can just plug those numbers in:

95 - x = .9(100) - .9x

or

95 - x = 90 - .9x

Subtract 90 from both sides:

5 - x = -.9x

5 = .1x

Multiply by 10 on both sides:

x = 50

So x is the amount of water removed from the watermelon to get it from being 95% water to being 90% water. Since the original watermelon weighed 100 pounds, and we now know that 50 pounds of water was removed, we know the watermelon's new fighting weight is-- ta-dah-- 50 pounds!

QED.

Kevin Kim said...

You can check your answer this way by asking yourself: is the 50-pound watermelon in fact 90% water? This brings us back to the variable we didn't use: s, the weight of the solid matter in the watermelon, which is 5 pounds. The value of s remains the same both before and after the dehydration process, so in our new, leaner watermelon:

w + 5 = 50 (where w = the weight of water in the leaner watermelon)

w = 45

and

45/50 = 90%

QED.

ttuface said...

That doesn't seem right. It's 19-1 ratio and then we lose an overall relationship of 5%. I'm puzzled.

Kevin Kim said...

Yeah, before I did the math, I thought the loss in weight would be a lot smaller, too, but it works out: the 50-pound watermelon is 45 pounds of water and 5 pounds of solid material. 45 out of 50 is 90% water. The original watermelon was 100 pounds, with 95 pounds of that being water (and the same 5 pounds of solid matter), so sure enough, that's 95% water. I agree: it doesn't seem right, but that's the math. (And for what it's worth, 50 pounds turns out to be the official correct answer.)

ttuface said...

oh, that's tricky! 5 lbs was 5%, now 5 lbs is 10%.

5 pounds = 10%
45 pounds = 90%

50 lbs