Math question - how much of the watermelon is water? | Tech Off

Matthew van Eerde
AKA Maurits

My son got the following math problem on a homework assignment. We had a bit of fun discussing it, drawing diagrams, coming up with sanity checks, etc.

But then his teacher marked it wrong -- she pointed out that the teacher's guide had a different answer. I asked her which answer she thought was right, and she sided with the teacher's guide.

I'm trying to convince myself that my understanding of the problem is wrong -- or that the problem is ambiguous -- but I don't see it.

What do you think? Here's the problem statement:

18. A farmer raised a watermelon that weighed 20 lb. From his experience with raising watermelons, he estimated that 95% of the watermelon's weight is water.

a. How much of the watermelon is water?

b. How much of the watermelon is not water?

c. The watermelon was shipped off to market. There it sat, until it had dehydrated (lost water). If the watermelon is still 90% water, what percent of it is not water?

d. The solid part of the watermelon still weighs the same. What was the weight of the watermelon at this point?

To avoid poisoning the well, I will hold off on revealing my son's answer and the teacher's guide answer for now.

Minh
WOOH! WOOH!

a.

95% of 20 lbs = water weight
19 lbs = water weight

b.

1 lb = non-water weight

c.

Trick question? 10%

d.

if the watermelon is now 90% water, then 1 lb (non-water) is 10%, therefore, water is 9x of non-water, therefore water melon is now 10 lbs.

Unless you read c. as "the watermelon only has 90% of its original water left" -- then it's an entirely different question. (which I think maybe what the teacher's hinting at?)

Then...

c2.

90% of 19 lbs = 17.1 lbs

1 lb / (18.1 lbs [entire watermelon]) = 55.2%

d2.

18.1 lbs

(hey, this is more fun than real work!)

CannotResolveSymbol
{insert caption here}

I get the same answer Minh did. I don't think the problem's asking for the weight if you only have 90% of the water left.

Antitorgo

a) and b) are somewhat ambiguous too. It isn't specifically asking for the weight in those questions, just asking "how much". I could conceivably answer.

a) 95%
b) 5%
c) 10%
d) 10 lbs

Was the question really worded as "how much"? If so I'd say that teacher needs some English classes.

d2 could be proven to be the wrong answer since 10% of 18.1 is definitely not 1lb. (which was shown in step b)

Antitorgo

Found this:
http://www.nctm.org/middle/asolutions.asp?ID=445

Matthew van Eerde
AKA Maurits

Antitorgo wrote:
Found this:
http://www.nctm.org/middle/asolutions.asp?ID=445

That's very interesting.

The problem he brought home was worded exactly as I worded it in my post, even down to the italicized "not."

JohnAskew
9 girl in pink sweater

Matthew van Eerde wrote:

Antitorgo wrote: Found this:
http://www.nctm.org/middle/asolutions.asp?ID=445

That's very interesting.

The problem he brought home was worded exactly as I worded it in my post, even down to the italicized "not."

That is very interesting.

It points us to think of the problem as a 'logarithmic curve' vs. straight division or subtraction.

ianlaurin

Mihn is right! I read and solved it with the same answers (sidebar on vista before seeing others had answered it). Remember when doing school math there is usually a trick question like (c) and answers are 99.9% of the time pretty close to whole numbers for someone has to grade it.
thanks

Antitorgo

thumbtacks wrote:

What took up the space? Air?

Who says the volume remained the same? I would assume that the watermelon shrank due to the evaporation and enough shrinkage would cause shriveling of the outer "skin". (note to self, never use shrinkage and shriveling together in a sentance ever again).

phreaks

So what were your answers and the teachers answers?

I'm curious.

Cornelius Ellsonpeter

I'm curious what the volume of displacement would be if a watermelon was hit with a sledgehammer. Would it be equal to the volume of the hammer head?

Matthew van Eerde
AKA Maurits

Generalizing from the two problem statements, the problem "in the large" is this:

A watermelon weighs w lbs and is (100 - p) % water (0 < p << 100)
It dehydrates until it's (100 - 2p) % water.
How much does it weigh?
Answer: w/2 lbs.

The answer that my son turned in to the original problem:

a. 19 pounds, 95%
b. 1 pound, 5%
c. 10%
d. 10 pounds

The answer the teacher gave:

a. 19 pounds
b. 1 pound
c. 10%
d. 19 pounds

She gave an explanation for d. but I don't think it would be fair to quote it here.

BitFlipper

I got this exact same question in one of my interviews. Luckily I was able to solve the problem. The only difference was that the percentages I got asked was 99% and 98%, instead of 95% and 90%. But the final answer is still the same: 10lb

BitFlipper

BitFlipper

Matthew van Eerde wrote:
Generalizing from the two problem statements, the problem "in the large" is this:

A watermelon weighs w lbs and is (100 - p) % water (0 < p << 100)
It dehydrates until it's (100 - 2p) % water.
How much does it weigh?
Answer: w/2 lbs.

The answer that my son turned in to the original problem:

a. 19 pounds, 95%
b. 1 pound, 5%
c. 10%
d. 10 pounds

The answer the teacher gave:

a. 19 pounds
b. 1 pound
c. 10%
d. 19 pounds

She gave an explanation for d. but I don't think it would be fair to quote it here.

The teacher's answer for d is wrong! The correct answer is 10lb

Antitorgo

Matthew van Eerde wrote:
Generalizing from the two problem statements, the problem "in the large" is this:

d. 19 pounds

She gave an explanation for d. but I don't think it would be fair to quote it here.

Can we guess?

I'm going to guess that she said the watermelon lost 5% of its weight (95% - 90% = 5%) and thus the watermelon weighed 5% less.

Could you have her explain how if her answers for b) and c) are correct, that the answer for d) is now incorrect? If the solid is now 10% and the watermelon weights 19 lbs, shouldn't the answer for b) be 1.9 lbs now?

Please, tell me that you pointed this out to her when you spoke with her. I really would have liked to know what the her excuse could have been for the "new" math she had devised... Hopefully the link I provided you helps you out.

I'm betting she is one of those teachers who doesn't like it when parents point out that they are fallable and will go to any lengths to deny it. The best teachers I had were the ones who acknowledged that they were wrong and pointed out to the class the mistake they made.

JeremyJ
The pioneers would be appalled!

19 pounds is correct.

For each 5% of water is 1 lb.
So 90% water is 18 lbs. + 1 lb for the solid parts is 19 lbs.

Or another way to look at it is that if you include the solid stuff it is 95% of its original weight because it lost 5% water.

100% total watermelon.
95% water + 5% solid = 100%
After dehydration:
90% water + 5% solid = 95%

95% of 20 lbs is 19 lbs.

Antitorgo

JeremyJ wrote:
19 pounds is correct.

Is that sarcasm or are you being serious?

The answers from the previous questions were definitely showing that the watermelon after dehydration was 90% water / 10% solid (not 90%/5%) which means that the ratio changed and on a logarithmic scale. Since the solid was unchanged and was 1 lb from another previous answer, it is clear that the new weight was 10 lbs. (10% solid = 1 lb -> 90% water = 9 lbs -> total 10 lbs)

What is annoying to me is that the questions for a, b & c are leading you to solution for d, but then apparently it was tossed out the window.

JeremyJ
The pioneers would be appalled!

To go from 20lbs to 10lbs it would have to lose 50% of its weight... not sure why you think that losing 5% water means that it loses 10lbs. I think you are trying to make it harder than it is.

Antitorgo

Wait, are you the school teacher in question? Perplexed

The rest of the people in this thread and the ~~smart~~ stupid folks at the National Council of Mathematics Teachers are complete idiots "overcomplicating" a simple problem.

Heck, the National Council of Mathematics Teachers were even worse, since their example the watermelon lost 1/2 its wieght by losing 1% water.

JeremyJ
The pioneers would be appalled!

Yeah I guess I don't understand this "new math". If you take 50lbs of water and let it set out until 1% evaporates it will not weigh 50% less.

Sven Groot
My name has 9 letters. Coincidence? I think not...

The thing is, the problem isn't saying 5% of the water evaporated. It's saying that after evaporation, 90% of the melon is water.

It's easier to understand if you come at this reversed: you have a melon that consists of 19lbs water, and 1lbs solid. So the melon is 19/20th water, i.e. 95%. Now remove 10lbs of water. This leaves a melon that is 9lbs water, and still 1lbs solid. Now 9/10th of the melon is water, i.e. 90%.

So despite the fact that the relative amount ofwater content of the melon dropped only 5%, the absolute weight of the water dropped by more than 50%.

JeremyJ
The pioneers would be appalled!

I need to retake a reading comprehension class. Perplexed I was thinking about the question in the wrong way. I was ignoring the non-water parts and just taking the water by itself.

I was basically thinking of it as if you put the watermelon through a wringer and sqeezed out all the water then measured it. Let 5% evaporate then measure again.

Thanks Sven for helping clear that up.

Ping
Je pense, donc je suis!

Interesting question, I'm too late to join this discussion..sigh.. Big Smile

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