Thursday, April 29, 2010

How many gumballs fit in the machine?

Inspired by my daughter's wishing for a hot pink gumball and my son's wish for anything but pink, and all the puzzles that followed, I decided to bring a gumball machine to the Cambridge Science Carnival. What could be better than gumballs to demonstrate a fascinating side of math to kids (and many adults)?


I bough a 10" diameter gumball machine and a giant box of 1" gumballs. How many gumballs, approximately, do you think we need to use to fill the machine to the top?

Submit your answer on our Family Puzzle Marathon Be first to solve three puzzles and get a prize! One puzzle per week, please.

11 comments:

Kim said...

If there were no wasted space, then you could calculate the number of balls that would fit by dividing the volume of the gumball machine by the volume of a single gumball.

That would be:
4/3 * (10)^3
------------
4/3 * 1 ^3

= 10^3/1 = 1000

So 1000 is a maximum upper limit. There will be some (relatively minimal) wasted space, let's say 5%... (it's probably possibly to calculate the wasted space, but since this is not in the "hard" category, I'm going to go out on the limb and assume my waste estimate is good enough)

With 5% wasted space, we'd fit 950 gumballs.

Anonymous said...

radius cubed is 125.
times pi is almost 393
times 4/3 is just over 523 cubic inches

each individual gumball has a volume of maybe 1 cubic inch (1x1x1?)?

there's some considerable wasted space, but there's also some settling into that wasted space. tom is guessing 500 to 510. tom

Kim said...

Akkk! I missed the diameter and radius thing, so my math is still right, but I should have used 5^3 and .5^3 or 125/(1/8) = 1000

I'm sticking with my not that much wasted space, but it could easily be more than the 5% I made up.

I think Tom's estimate of the volume of the gumball machine looks good, but then using 1" cubes rather than balls with a .5" radius isn't consistent. (Should be 4/3 * pi * (1/8) = .523)

Kim said...

BTW, didn't mean to omit the pi from my original post -- I had intended originally to show that the 4/3 * pi cancelled from top and bottom, but forgot... sorry for a convoluted set of answers!

Katrina said...

V = 4/3 Pi * r^3

Gumball Machine volume (d = 10) = 4/3 pi * 5^3 = 523.5983333

Gumball volume (d=1) + 4/3 pi * 1^3 = 4.188786

Machine volume/gumball volume = 125 gumballs assuming no wasted space.

Maria said...

Ok, let's trace what happened here:

Kim was first, did all the calculations perfectly but forgot that radius rather than diameter should be cubed.

Tom came second. I believe he reasonably assumed that gumball volume could be approximated as a cube. There is some space in-between the gumballs after all.

Kim stopped by again correcting the radius-diameter confusion and pointing out that for gumballs, sphere volume would be a better approximation that a cube.

Katrina came third, but she also used diameter instead of radius.

Looking closely at the gumball packing, I believe the truth is somewhere in the middle between Tom's and Kim's solution, between 500 and 1000, in the 700th. There is some space in-between the gumballs but not as much as if each would be wrapped in a cube.

I had to check it in practice and I did. Oooo, the smell of these gumballs when you pour them into the machine! And I don't even like gum. It looks like we need around 720 to fill the machine up.

Everyone who participated here deserves a puzzle point. But Tom already got one today for the tiles puzzle. So, it's just girls - Kim & Katrina.

Maria said...

Katrina - this is your third puzzle point, and as everyone who solves three puzzles, you earned a prize. Please email me your snail mail address.

Monica said...

Volume of machine is approximately 523.60cubic inches; volume of gumball is approximately 4.19cubic inches. 523.60/4.19 is approximately 125. Therefore, 125 gumballs with room to give.

Monica

Tom said...

Seems like I (Tom) remember solving that gumball problem in high school, long time ago, and only approximately. But i can't remember it! It was a geometry class, and not calculus which I never got to take in the 50s.

Instinct makes us/me think, at first, the "wasted space" will decline with the size of the interior balls. But wait. If there was only one big gumball inside, there'd be no air volume. If there's two fairly big gumballs, there's quite a lot. If 3, still quite a lot, and if 4, quite a lot.

Heck with it, on to other fun problems. Tom

Monica said...

How do we find out what the actual answer is?

Maria said...

720 gumballs as practice showed

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