2004 AMC 10B Problem 15

Let $n$ be the number of nickels and $d$ the number of dimes, so $n + d = 20$.

Consider what the swap does coin by coin: each nickel becoming a dime adds $5$ cents, and each dime becoming a nickel removes $5$ cents. The net change is $5n - 5d$ cents, and we are told it equals $+70$:
$$5(n - d) = 70 \quad\Longrightarrow\quad n - d = 14.$$

Adding this to $n + d = 20$ gives $2n = 34$, so $n = 17$ and $d = 3$.

Her coins are worth $17 \cdot 5 + 3 \cdot 10 = 85 + 30 = 115$ cents.

The answer is $\boxed{\textbf{(A)}\ \$1.15}$.

Instead of computing both totals in full, track the change caused by the swap; it depends only on $n - d$. Together with the count $n + d$, a sum and a difference pin down both unknowns instantly. "Would have $70$ cents more" is a statement about a difference, so model the difference directly.

Let $V$ be the current value and $V'$ the swapped value, in cents. Every coin is worth $5$ in one configuration and $10$ in the other, so $V + V' = 15 \cdot 20 = 300$. With $V' - V = 70$, we get $2V = 230$ and $V = 115$, without ever finding $n$ and $d$.

• Setting up the swap with the sign reversed (dimes outnumbering nickels), which gives 3 nickels and 17 dimes worth $1.85, not among the choices. - Reporting the swapped value ($185$cents) or mixing dollars and cents when writing$70$ into the equation.