At 02:47 AM 10/3/02 -0400, [EMAIL PROTECTED] wrote:
>How do I transcribe this into an algebraic equation and solution?
>
>dimes and nickels, 63 coins in all, totaling $5.05 - how many nickels? how
>many dimes?
There is a very simple way and since nobody has contributed it yet, here goes.
$5.05 is 101 nickels. Consider each dime to be a nickel with a nickel
"bonus" added in. Each coin contributes one nickel to the total, no matter
what its denomination, and each dime contributes an extra nickel to the
total. Here, then, is an equation for the number of dimes, d:
# nickels contributed by all coins + # extra nickels contributed
by dimes = total value = 101 nickels;
63 + d = 101.
Now, consider each nickel to be a dime that is short one nickel. From this
point of view, each coin contributes two nickels to the total, no matter
what its denomination, but each nickel causes the total to be a nickel
short. An equation for the number of nickels, n, is:
# nickels contributed by all coins - # nickels = total value = 101
nickels;
2 * 63 - n = 101.
Those are the algebraic equations you requested, whose solution is easily d
= 101 - 63 = 38 and n = 2*63 - 101 = 25.
--Bill Huber
Quantitative Decisions
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