> > > 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 $3.05 in change. Now my question is: What's the REAL way to solve this problem?? I don't recall getting anything like this in high school! Arrgh.
> > >
> > > Wolf "As Pat says: HELP!!!" spirit
> >
> > Let's see if I can get it w/o looking at the answer Jimmy gave:
> >
> > a = # of dimes he has
> > b = # of quarters he has
> >
> > .1a + .25b + .9 = .1b + .25a
> > .1a - .1a + .25b + .9 = .1b + .25a - .1a
> > .25b + .9 = .1b + .15a
> > .25b - .1b + .9 = .1b - .1b + .15a
> > .15b + .9 = .15a
> > (.15b + .9) / .15 = .15a / .15
> > b + 6 = a
> >
> > a + b = 20
> > b + b + 6 = 20
> > 2b + 6 = 20
> > 2b + 6 - 6 = 20 - 6
> > 2b = 14
> > b = 7
> >
> > 20 - 7 = 13
> >
> > 7 dimes($.70) + 13 quarters($3.25) = $3.95
> > 13 dimes($1.30) + 7 quarters($1.75) =$3.05
>
>
> uhhhhh, isn't 13 quarters $3.25 and 7 quarters $1.75?
>
>
> > The man has 13 dimes and 7 quarters.
> >
> > Nyperold

Ohhh.. right.

Ah well, still works out.

Nyper"10q"old