> What with discussing quantum theory in general terms, and also seeing today's poll on picking numbers, why not? I would like to know the "correct" way to solve the math problem that plagued Pettermint Patty in Wednesday's "Peanuts" strip. Schulz's question was:
>
> / / / / "A man has twenty coins consisting of dimes and quarters. If the dimes were quarters and the quarters were dimes, he would have ninety cents more than he has now... How many dimes and quarters does he have?"
>
> The way I solved it (read: the stoopid way) was to realize that here the only way to make up 90 cents is by having 4 dimes and 2 quarters. This is a rough ratio of 2:1 for the two types of coins, so, plugging in this result, I found that the man at the beginning would have started with $3.05 in change. Now my question is: What's the REAL way to solve this problem?? I don't recall getting anything like this in high school! Arrgh.
>
> Wolf "As Pat says: HELP!!!" spirit

Algebra!

We'll say that d is the number of coins the man has, and q is the number of quarters.

We know that the man has 20 coins, so q + d = 20. That's our first equation.

Since quarters are worth 25 cents and dimes are worth 10 cents, we know that man has 25q + 10d cents.

If we change the dimes to quarters and the quarters to dimes, he would have 10q + 25d cents.

And, that would be 90 cents more than he has to start, so we add 90 to the first one, and now they're equal. 25q + 10d + 90 = 10q + 25d

Now, let's solve for one of the variables. How about q:

25q - 10q = 25d - 10d - 90

15q = 15d - 90

q = d - 6

Plug that value for q into the first equation:

d - 6 + d = 20

Now solve for d:

2d = 20 + 6

2d = 26

d = 13

So, d = 13, and q = 7. So, the man has 13 dimes and 7 quarters, for a total of $3.05.