# It’s the Friday Puzzle!

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Please do NOT post your answer, but do say if you think you have solved the puzzle and how long it took. Solution on Monday.

I have a 100 pound watermelon laying out in the sun. 99% of the watermelon’s weight is water. After a few hours 98% of the watermelon’s weight is water. How much water evaporated?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.

## 61 comments on “It’s the Friday Puzzle!”

1. Simon Scott says:

Pretty quick today.

• ian says:

A surprising result (like all good puzzles).

Also – that’s one big-ass watermelon.

• Richard M says:

Ian, I agree – I had to double check my workings … 1% is such a small change. I had a calculator out at first in preparation for decimal places.

• Marsmallow says:

This took me longer than I care to admit. After initially reading it, I decided that I needed to get breakfast and coffee before I wanted to deal with it. Then I looked at it again and it looked unpleasant so I took a walk. Then I came back and realized that it was really easy and from there it took me about thirty seconds. So about two hours.

• Arneb says:

Wow. I couldn’t figure it out quickly and so did the formal calculation only to find how far off the mark I had been. Of course, it’s all so easy to see in retrospect. Nice one for leading me astray.

2. RobJ says:

Took me a bit of playing in a spreadsheet to get there, so a minute or so.

• Eddie says:

What. You mean 10 minutes?

3. mgm75 says:

Dismissed the obvious answer then after about a minute or two the penny dropped.

4. John Green says:

What is it laying? Eggs? Seeds?

• Greg says:

About 30 mins for the answer I have. Took a little while to determine how to work it out. Then a few guesses on the calculator and I had my answer.

5. ChrisR says:

about 2 minutes – including pondering how to approach it, then jotting down some algebra, pondering the answer and considering if it was sensible, checking my working and deciding I’d got a valid answer. I could still be wrong though.

6. Barry Goddard says:

This is very clever but we are not given enough information to answer the puzzle.

After a “few hours” the weight of a watermelon may have changed in many ways. Perhaps it has been sprayed by insecticide. It will have grown some more seeds. Perhaps it has ingested a slug.

Even if we research the growth patterns of a watermelon as perhaps Richard wants us to do we still do not have enough information because we do not know the energy budget. IE how hot the is shining or what chemical nutriments are available in the soil.

It is not uncommon for prize vegetable growers to use growth hormones and such the like on their prize fruits. That adds another dimension especially if they have cheated by injecting gycerole or other substances.

Puzzles should be challenging not impossible.

• Garry Boddard says:

Life must be quite a challenge for you, you poor old dear.

• MathMiles says:

Careful – your parodies of Barry are getting so good they are beginning to sound like the real Mr Goddard!

• Barry Goddard says:

Thank you @mathmiles for your support.

It is completely obvious that Richard intends that it is a watermelon to be part of the puzzle. A watermelon is a living breathing vegetable – a fauna in technical botany terms.

If Richard had said “I have a bucket that wieghs one pound with 99 pound water in it. What percentage of water remains if I empty 50 pounds out?” That would be the puzzle that most people here appear to be solving.

Only those of us who respect the question and read it properly will be directed to the deeper answers.

My experience in producing star charts where precise readings are melded with intuitive insights helps me go further like this.

• MathMiles says:

My mistake – you must be the real BG, as you are spouting nonsense! Very entertaining.

7. Anne Elk says:

I’ve got an answer but I doubt it’s right. I did it on paper with a bit of Heath Robinson algebra. Couple of minutes.

8. daverickey says:

That is an extremely hot sun. Must be a dry heat. Took me about ten seconds.

Hint: One pound of the watermelon is not water.

–Dave

• One Eyed Jack says:

The glass is always full and the sound of one hand clapping is the sound of one hand clapping.

9. Steve Jones says:

I can only think that the watermelon must have been left on the sunny side of Mercury.

• Terrence Williams says:

This would explain the rapid rate of evaporation of water!

10. Luke Meyers says:

44.45182

Luke Meyers Email: [lmeyers4@icloud.com] or [luke.meyers2@education.nsw.gov.au]

>

11. Miss Chili says:

a. 100-pound watermelon?! That is enormous, unless…
b. 100-pound watermelon?! That is expensive!
c. Okay, who cut into the watermelon to just have a bite?!

• Lazy T says:

b. and if it’s like sun-dried tomatoes it’s 5000% more expensive than the fresh.

12. mittfh says:

It took me a few seconds to realise that as the water evaporated, the overall weight of the watermelon also reduced. One quick fill down in a spreadsheet later, and I had my answer (although given what the answer is, there’s probably a far quicker algebraic method).

That is, of course, assuming water is the only component of the melon’s composition which reduces in weight over time…

• Anonymussel says:

Good spoiler, mittfh. Well done, wee man.

13. noname says:

This should help people figuring it out, initially 1% is not water and this 1% equals 1 pound. Later 1 pound is still not water but now this 1 pound is 2%.

14. jaycee319 says:

20 seconds of head calculations and then two minutes to read the comments to find confirmation —- but this time it was wasted time since there is no answer — yet. Whats going on here,
😀

15. ChrisR says:

From Wikipedia…
A watermelon contains about 6% sugar and 91% water by weight….
The more than 1200 varieties of watermelon range in weight from less than one to more than 200 pounds; the flesh can be red, orange, yellow or white…

I’d never have thought to expand my knowledge about watermelons if this puzzle had never been posed. Thank you Richard.

16. Michael says:

2 minutes to be sure that the (somewhat surprising) answer was correct.

17. Stan says:

I agree it’s a great puzzle. I had to use a pen to write down an equation.
Then it made sense. Took maybe 4 minutes partly to scratch my head when I got the answer.

18. Duncan says:

I got it – before reading the comments. No one’s given the answer yet but hints are just as annoying.

19. One Eyed Jack says:

2 seconds. The answer is intuitive. No calculation required, and no, it’s not 1%.

20. NonTrivialZero says:

Despite my three years of Sixth grade, I am floundering again between what was, what is…and what may ever be.

21. Walter says:

You cannot tell how much water evaporated, you only can tell how many percent of water evaporated!

• Alex says:

Walter,
How did you come to this conclusion?

• One Eyed Jack says:

Yes you can. You are given the initial mass of he watermelon.

22. Alex says:

How could algebraic equations be posted here?

• Alex says:

Testing jqMath:

$$y-y_0=m(x-x_0)$$

• Alex says:

Testing for LaTex:

$y-y_0=m(x-x_0)$

23. Lee Smith says:

about 4-5 minutes scribbling the calculation on a bit of scrap paper with a pencil. Then deriving a proper algebraic expression to be certain it was correct.

24. James says:

About 5 minutes to get the wrong answer, find my math error, and end up with an even more surprising (and correct) result.

One of the best “puzzles” you’ve had in a long while!

25. Eddie says:

Is this a first? No one’s posted the answer.

26. John Cartwright says:

It took me about 10 seconds to get the obvious and correct answer, then several minutes getting confused by reading the thread and seeing the comments. Lots of people seem to be giving lots of strange cryptic clues which suggest that they find it more confusing than it actually is.

27. Gene says:

I encountered (and eventually solved) this one in a coding interview.

28. Anonymous says:

30 seconds

29. Alex says:

It’s easy to say it in math.\\

$Given\\ m_1=100 lb\\ r_1=99 \%\\ r_2=98 \%\\ Find out\\ m_e_-_w=?\\ \\ \\ $\begin{cases} m_w_1 = m_1 * r_1 (1)\\ m_w_2 = m_2 * r_2 (2)\\ m_2 = m_1 - m_e_-_w (3)\\ m_e_-_w = m_w_1 - m_w_2 (4)\\ \end{cases}$\\ \\ \\ m_w_2 = (m_1 - m_e_-_w) * r_2 (5) (from 2, 3) => \\ m_e_-_w = m_1*r_1 - (m_1 - m_e_-_w)*r_2 (from 5, 1, 4) => \\ m_e_-_w = m_1*r_1 - m_1*r_2 + m_e_-_w*r_2 => \\ m_e_-_w * (1-r_2)= m_1 * (r_1 - r_2) => \\ Final result: m_e_-_w = m_1 * \frac{r_1 - r_2}{1 - r_2} => \\ Add numbers: m_e_-_w = 100 * \frac{0.99 - 0.98}{1 - 0.98} => \\ m_e_-_w = 100 * \frac{0.01}{0.02} => \\ m_e_-_w = 100 * \frac{1}{2} => \\ m_e_-_w = 50 lb => \\ Any questions?$

• Alex says:
• Alex says:

Given
$m_1=100 lb\\ r_1=99 \%\\ r_2=98 \%$
Find out
$latex m_e_-_w=?\\ • Alex says: Find out $m_e_-_w=?$ • Alex says:$ m_e_-_w=?\$

• Alex says:

Find out
$m_e_-_w=?$

30. Alex says:

$$\begin{cases} m_w_1 = m_1 * r_1 (1)\\ m_w_2 = m_2 * r_2 (2)\\ m_2 = m_1 - m_e_-_w (3)\\ m_e_-_w = m_w_1 - m_w_2 (4)\\ \end{cases}$$

31. Alex says:

System of equations:
$\begin{cases} m_w_1 = m_1 * r_1 (1)\\ m_w_2 = m_2 * r_2 (2)\\ m_2 = m_1 - m_e_-_w (3)\\ m_e_-_w = m_w_1 - m_w_2 (4)\\ \end{cases}$

Final result:
$m_e_-_w = m_1 * \frac{r_1 - r_2}{1 - r_2}$
$m_e_-_w = 100 * \frac{0.99 - 0.98}{1 - 0.98} = 100 * \frac{0.01}{0.02} = 100 * \frac{1}{2} = 50 lb$

32. Alex says:

I give up on wordpress LaTex engine. Answer 50 lb

• MathMiles says:

• Terrence Williams says:

You weren’t supposed to give an answer, merely the time that the calculation took!

• pineapple01 says:

It’s easy to say it in math.

Given
$m_1=100lb\\ r_1=99\%\\ r_2=98\%\\$
Find out
$m_{e-w}=?\\$

System of equations is translated from given conditions.
$\begin{cases} m_{w1} = m_1 * r_1 (1)\\ m_{w2} = m_2 * r_2 (2)\\ m_2 = m_1 - m_{e-w} (3)\\ m_{e-w} = m_{w1} - m_{w2} (4)\\ \end{cases}$
Solution for system of equations is:
$m_{w2} = (m_1 - m_{e-w}) * r_2 (5) (from 2, 3) => \\ m_{e-w}= m_1*r_1 - (m_1 - m_{e-w})*r_2 (from 5, 1, 4) => \\ m_{e-w} = m_1*r_1 - m_1*r_2 + m_{e-w}*r_2 => \\ m_{e-w} * (1-r_2)= m_1 * (r_1 - r_2) \\$

Final result:
$m_{e-w} = m_1 * \frac{r_1 - r_2}{1 - r_2}$
$m_{e-w} = 100 * \frac{0.99 - 0.98}{1 - 0.98} = 100 * \frac{0.01}{0.02} = 100 * \frac{1}{2} = 50 lb$