Here is what the car guys said about Part A.

RAY: You could write equations for all this, or you could do it the easy way. The easy way is to say you've got 12 quarts, so four of them are antifreeze and eight of them are water.

So, if you have a one third mixture that means if you took three quarts of that mixture out, two of them would be water and one of them would be antifreeze. You're going to wind up with three quarts of antifreeze and six quarts of water, are you not?

TOM: Right.

RAY: If you now add three quarts of pure antifreeze back you'll have six quarts of antifreeze and --

TOM: And six quarts of water!

RAY: You drain out three quarts of the mixture. That's the answer to part A.

That is a cogent argument that "three" is the correct answer.

Ray's easy explanation unfortunately suggests that finding the answer was also easy. Ray does not tell us how.

Take a look at these slight changes to the problem.

`Suppose you add a mixture that is 2/3 antifreeze instead of pure antifreeze, and we want to end up with a 50/50 mixture as before?``Suppose you add pure antifreeze as before, but we want to end up with a mixture that is 2/3 antifreeze instead of a 50/50 mixture?`

Ray told you "the easy way" so can you figure out the answer to these problems?

Before we tackle these problems, let's look at their answer to Part B.

RAY: Now, for part two: how much of this mixture do you have to drain out if you want to get to 50% water and 50% coolant, if you're adding a 50/50 mixture of water and antifreeze?

TOM: You can't.

RAY: Right. You have to drain out all 12 quarts in order for it to work.

Clever and cute. Are you ready to tackle the two problem variations now?

Tom and Ray are entertainers, not educators. They are extraordinarily good at entertaining. They are also good at stimulating interest in puzzles, for which I am grateful. I recognize that if they went into this much detail on their puzzles, they would have nobody left to listen.

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© 2006 |

Peter C. Scott |