# Janice was given a piggy bank on her seventh birthday, and she put it to use immediately. Each time she puts one or more coins into the piggy bank, she keeps track of the number of coins she has collected to date and the accumulated value of her collection. Janice collects only nickels, dimes, and quarters. Six months after her seventh birthday, Janice looked at her record and ascertained that she had collected 240 coins, which were worth \$36. (a) How many combinations of coins are possible in Janice's collection? (b) Janice counted 100 quarters in her savings. How many nickels and dimes are in her collection?

There are 41 possible combinations of coins. If there are 100 quarters, then there are 60 nickels and 80 dimes.

We are told that Janice has \$36 dollars worth of coins which are either nickels, dimes, or quarters. The total number of coins is 240.

(a) How many combinations of coins can there be?

Let x be the number of nickels, y the number of dimes, and z the number of quarters. Then, we know the following:
(1) x, y, and z are all integers greater than or equal to zero.
(2) .05x+.1y+.25z=36, as this is the total amount of money.
(3) x+y+z=240, as this is the total number of coins.

So, we have a system of two equations and three unknowns. This system is consistent (there are solutions) and dependent (there are multiple solutions). One approach is to set up an equation matrix (or augmented matrix) and put the matrix in reduced row-echelon form.

`([.05,.1,.25,36],[1,1,1,240])` reduces to `([1,0,-3,-240],[0,1,4,480])` .

So, we see that x-3z=-240 and y+4z=480. We can write all three variables in terms of z.

(1) x-3z=-240 => x=3z-240
(2) y+4z=480 => y=-4z+480.

So, the general solution set to the equations is (3z-240,-4z+480,z). But there are restrictions on z for our situation, since all of the variables are nonnegative.

`3z-240>=0 => z>=80, -4z+480>=0 => z<=120`,

so `80<=z<=120` . Thus, there are 41 different combinations of coins that work.

(b) If there are 100 quarters (z=100), we are asked to find the number of nickels and dimes.

x=number of nickels=3z-240=60
y=number of dimes=-4z+480=80

Note that 100+60+80=240, so there are 240 coins, and .05(60)+.1(80)+.25(100)=36, as required.