total value of exactly 10 coins is 27¢.
many of those coins are nickels?
1: Solve a simpler problem.
Since all coins other than pennies must end
in a value of 0¢ or 5¢, let's determine
how many of the coins must be pennies. The number
of pennies must end in 2 or 7.
There can't be 2 pennies because the value of 8
other coins is more than 25¢ and there can't
be 12 or more pennies since that's already more
than 10 coins. So there are 7 pennies.
The other three coins, worth 20¢, must be 1
dime and 2 nickels.
METHOD 2: Use algebra.
1 quarter is already 25¢ and there must
be 9 other coins, so there are no quarters. Represent
the number of pennies by p, the number of nickels
by n, and the number of dimes by d.
We know there are 10 coins, so p + n + d = 10. The
value of the coins is 27¢, so p + 5n +10d =
Since p appears in both equations, subtract the
first equation from the second to eliminate p. This
gives 4n + 9d = 17.
Since 4n can't equal the odd number 17, d must be
at least 1. Since 10¢ x 2 is greater than 17¢,
d must be smaller than 2. Thus d = 1, leaving 4n
+ 9(1) = 17.
This simplifies to 4n = 8, and n = 2. There are
METHOD 3: Make a table.
Construct a table with each row equaling a combination
of coins totaling 27¢.