Carmina has a total of $\$ 3.60$ in nickels and dimes. If her dimes were nickels and her nickels were dimes, then she would have $\$ 5.40$. How many nickels and dimes does Carmina have?
(A) 56
(B) 57
(C) 58
(D) 60
(E) 61
Solution 1

When Carmina trades her nickels for dimes and her dimes for nickels, she gains $\$ 1.80$. Since trading a dime for a nickel results in a loss of 5 cents, and trading a nickel for a dime results in a gain of 5 cents, then by doing her trade she gains 5 cents in $\frac{180}{5}=36$ more cases than she loses 5 cents. Thus, she must have 36 more nickels than dimes.

These extra 36 nickels account for $\$ 1.80$. So her initial coins are worth $\$ 1.80$ and she has an equal number of nickels and dimes. A nickel and dime together are worth 15 cents, so she must have $\frac{180}{15}=12$ sets of a nickel and dime, or 12 nickels and 12 dimes.

So in total, Carmina has 48 nickels and 12 dimes, or 60 coins.

## Solution 2

Suppose that Carmina has $n$ nickels and $d$ dimes.

Then looking at the total number of cents Carmina has, $5 n+10 d=360$.

If we reverse the nickels and dimes and again look at the total number of cents that Carmina has, we see that $10 n+5 d=540$.

If we add these two equations together, we get

$$
\begin{aligned}
5 n+10 d+10 n+5 d & =360+540 \\
15 n+15 d & =900 \\
n+d & =60
\end{aligned}
$$

So the total number of coins is 60. (Notice that we didn't in fact have to calculate the number of nickels or the number of dimes!)

## Solution 3

Suppose that Carmina has $n$ nickels and $d$ dimes.

Then looking at the total number of cents Carmina has, $5 n+10 d=360$ or $n+2 d=72$ or

$n=72-2 d$.

If we reverse the number of nickels and the number of dimes and look at the number of cents,

$$
\begin{aligned}
5 d+10 n & =540 \\
5 d+10(72-2 d) & =540 \\
5 d+720-20 d & =540 \\
180 & =15 d \\
d & =12
\end{aligned}
$$

Since $d=12$, then $n=72-2 d=48$.

Therefore, the total number of nickels and dimes that Carmina has is 60 .

ANSwer: (D)