# Brain teaser revisited: Maybe I didn't ask it correctly. . . .

Let’s say I have a straw, a tube from a roll of “Bounty” paper towels, and two balls the size of basketballs made of clay.

I poke the straw thru one ball, and the paper towel tube thru the other–both perfectly thru the center.

I now have two perfect spheres with cores of different sizes cut out of them.

And the volume of the remaining clay is the same.

Huh?

That was the way I read this riddle when it appeared in the paper, that the remaining volume would be the same regardless of the diameter of the core cut out.

What am I missing?

Volume, as I am sure you have already taken into consideration, is not weight or mass. Since the problem clearly said two spheres, they seem to mean the general shape, as a sphere with a tube missing is not a sphere, it is …I don’t know, I guess a wheel of some sort. So the spheres themselves, regardless of the missing mass are still the same size.

Volume, as I am sure you have already taken into consideration, is not weight or mass. Since the problem clearly said two spheres, they seem to mean the general shape, as a sphere with a tube missing is not a sphere, it is …I don’t know, I guess a wheel of some sort. So the spheres themselves, regardless of the missing mass are still the same size.

*Mjollnir: Let’s say I have a straw, a tube from a roll of “Bounty” paper towels, and two balls the size of basketballs made of clay.
I poke the straw thru one ball, and the paper towel tube thru the other–both perfectly thru the center.

I now have two perfect spheres with cores of different sizes cut out of them.

And the volume of the remaining clay is the same.

Huh?

That was the way I read this riddle when it appeared in the paper, that the remaining volume would be the same regardless of the diameter of the core cut out.

What am I missing?*

They’re creating confusion, wrongly making you think of the volume of just the clay and the shaped clay as the same. (Tough to explain, example at end.)

Immersing either sphere in water before and after its coring and noting the displacements would show that the volume of clay is reduced in both cases, more so in the paper-towel-tube-cored sphere.

Example: Take a letter size sheet of paper. It’s area is 8.5" x 11" = 93.5 in[sup]2[/sup]. Now cut a 1" x 1" square piece out of the middle. The sheet as a whole still encompasses 93.5 in[sup]2[/sup]. But the amount of paper is 92.5 in[sup]2[/sup].

If you have a link to this question (it was in a newspaper?), post it and let’s see exactly what they’re talking about.

Nope, I don’t have a link.

It was in an “Ask Marilyn” in the past two years.

Oh, an “Ask Marilyn” question. No wonder the problem got screwed up!

The woman who thinks square feet and feet squared are the same.

She must have a web site. I’ll look.

Oh, an “Ask Marilyn” question. No wonder the problem got screwed up!

The woman who thinks square feet and feet squared are the same.

She must have a web site. I’ll look.

The way I understand it…the volume would definatly change by definition. The radius would stay the same, but what does that have to do with anything?

Is there any way for the volume to be equal?

-Frankie

I’m not a shrimp, I’m a King Prawn.
-Pepe the Prawn

Like AWB said, it’s a question about creating confusion. Forget basketballs, the clay itself keeps the same volume. the core sampling exercise is just a red herring. The question does not say that the remaining basketball has the same volume, only that the clay does.

To clarify:

The remaining clay is the clay remaining after the coring: the punctured sphere and the cylinder.

I cannot possibly see how. Lets take the problem to extremes.

Lets take two 12" spheres of clay. I core one with a 1" diameter straw. I core the other one with a 11.9" thing. It is impossible the the volume of clay from the barely cored sphere would be remotely equal to the volume of the thin clay sphere.

Not only that but it is obvious that volumes are additive. If I take a 1 cubic foot cube and add to perfectly to another 1 cubic foot cube I have a 2 cubic foot box (2x1x1). Remove a 1 cubic foot cube and I would be back to a 1 cubic foot cube.

In the sphere riddle the volume of sphere A = sphere A after coring + core A. Sphere B = sphere B after coring + core B. core A <> core B (they are almost cylinders and volumn of a cylinder = height x (pi x d), since d is very different the volumes must be very different). Therefore, sphere A after coring <> sphere B after coring.

Is is possible she might mean surface area, anybody know if that would be true? I can at least imagine that being true. Hmmm… have to ponder.

“Glitch … Anything.” - Bob the Guardian

I have some clay in my possession, It’s one ball. I break it in two. I still have the same volume of clay in my possession, it’s just in two pieces now. That’s the gist of my “answer.”

To clarify further, Mjollnir didn’t ask the question correctly. The length of the resulting holes in the spheres must be the same. If you start with two spheres of identical diameter, and stick different size tubes through them, the lengths of the holes will be different.

Since the Bounty tube is wider in diameter, the sphere this is stuck through had to be bigger to begin with, for the length of the resulting holes to be the same.

In this case, the ramaining volume is the same as the the remaining volume of the other sphere.

It is too clear, and so it is hard to see.

Here’s a link to the Marilyn’s Wrong explanation of the sphere problem.

The volume she’s talking about is the volume of the solid of revolution. She’s basically revolving the sphere remnant about an axis perpendicular to the tube-cut hole. Sure, that’s the same. So’s the volume of the sphere if a cylinder of matter 1 millimeter short of its diameter is removed.

I can’t find her original problem verbatim (“Marilyn’s Wrong” can’t quote it legally), so this is the best explanation of her explanation.

I’m thinking that on a math exam, if they give you the radius of a sphere and ask for the volume, it doesn’t matter if the sphere is hollow or filled with water or Jell-O or sawdust or whatever. The math is the same.

So if the spheres were made of swiss cheese the math would be the same even with all the holes.

That’s just my thinking.

ZenBeam got the right answer. If you have a sphere with a hole of length X drilled through its center, the remainder of the sphere will always have a constant volume regardless of the radius of the hole. If you tried to duplicate this by the method Mjollnir described in the OP, you would fail because you are starting with two identical spheres and drilling two non-identical holes through them.