cheri,kelsey,ramik,taylor had16 coins consisting 4 pennies 4 dimes 8 nickels they had the same # of coins
*Ramik had no pennies, but had the most money
*Cheri and kelsey are the only ones who had at least one of each coin
*Taylor had the least amount of money, but just 1 cent less than kelsey
4 * 1 + 4 * 10 + 8 * 5 = 84 cents
In order for Taylor to have 1 cent less than Kelsey, Kelsey has to
have only one penney. We are told that Ramik has no pennies, so Cheri
has to have three pennies. We are also told that Kelsey has at least one
of each coin, which means at least one nickel and one dime. That means
that Taylor must also have one nickel and one dime. Cheri has to have one
dime, too. Ramik has to have the fourth dime if he is to have the most money.
At this point, we know that Kelsey is locked in at one penney, one nickel and one dime.
We also know that Taylor is locked in at one nickel and one dime.
We know that Cheri has three pennies and one dime, plus at least one nickel.
Ramik has the last dime, and some number of nickels.
If Ramik has the five remaining nickels, the conditions of the problem have
been met. But what if Ramik has four nickels and Cheri has two nickels? The
conditions of the problem are still satisfied by this combination. So, there
are two solutions to this problem.
Pennies Nickels Dimes Amount
Ramik 0 5 1 35 cents
Cheri 3 1 1 18 cents
Kelsey 1 1 1 16 cents
Taylor 0 1 1 15 cents
total 4 8 4 84 cents
Ramik 0 4 1 30 cents
Cheri 3 2 1 23 cents
Kelsey 1 1 1 16 cents
Taylor 0 1 1 15 cents
total 4 8 4 84 cents
Taking one more nickel away from Ramik and giving it to Cheri
would give Cheri 28 cents and Ramik 25 cents, which violates
one of the conditions of the problem.
