Question

A man has 32 coins in his pocket, all of which are dimes and quarters. If the total value of his change is 545 cents, how many dimes does he have?

Answer 1

A dime is worth 10 cents and a quarter is worth 25 cents.

Let D be the number of dimes and Q be the number of quarters.

Since the total amount of coins in the pocket is 32, then:

$D+Q=32$

On the other hand, the total value of D dimes is 10D, while the total value of Q dimes is 25Q. Then, the total value of D dimes and Q cuarters is 10D+25Q, which must be equal to 545. Then:

$10D+25Q=545$

Notice that we have found a 2x2 system of equations:

$\begin{gathered} D+Q=32 \\ 10D+25Q=545 \end{gathered}$

Solve the system using the substitution method. To do so, isolate D from the first equation and replace the expression for D into the second equation to obtain a single equation in terms of Q:

$\begin{gathered} D+Q=32 \\ \Rightarrow D=32-Q \\ \\ 10D+25Q=545 \\ \Rightarrow10(32-Q)+25Q=545 \\ \Rightarrow320-10Q+25Q=545 \\ \Rightarrow25Q-10Q=545-320 \\ \Rightarrow15Q=225 \\ \Rightarrow Q=\frac{225}{15} \\ \\ \therefore Q=15 \end{gathered}$

Replace back Q=15 into the expression for D to find the amount of dimes:

$\begin{gathered} D=32-Q \\ =32-15 \\ =17 \end{gathered}$

Therefore, the amount of dimes that the man has is 17.