2004 AMC 10B Problem 15

1. Look at what the swap does to one coin at a time. A nickel turned into a dime gains $5$ cents. A dime turned into a nickel loses $5$ cents.
2. So each nickel and dime cancel out in pairs. The overall gain of $70$ cents comes from the extra nickels: $70 \div 5 = 14$. She has $14$ more nickels than dimes.
3. She has $20$ coins in all. Two numbers that add to $20$ and differ by $14$: the bigger is $(20 + 14) \div 2 = 17$ and the smaller is $(20 - 14) \div 2 = 3$. So $17$ nickels and $3$ dimes.
4. Check the story: now she has $17 \times 5 + 3 \times 10 = 85 + 30 = 115$ cents. After swapping she would have $17 \times 10 + 3 \times 5 = 170 + 15 = 185$ cents. The difference is $70$ cents. It works.
5. Her coins are worth $115$ cents.

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

Instead of working out both totals, track the change the swap causes. Each coin changes by exactly $5$ cents, so the change only depends on how many more nickels than dimes there are. A sum ($20$ coins) plus a difference ($14$ more nickels) tells you both numbers.

Each coin is worth $5$ cents in one version and $10$ cents in the other, so the two totals add up to $20 \times 15 = 300$ cents. The swapped total is $70$ more than the real one. Two numbers that add to $300$ and differ by $70$: the smaller is $(300 - 70) \div 2 = 115$. So she has $\$1.15$, and you never need to count the coins.

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 $\boxed{\textbf{(A)}\ \$1.15}$.