2014 AMC 10B Problem 1

1. Think about what "one more nickel would make them equal" tells you. Adding a nickel closes the gap, so right now the pennies are ahead by exactly one coin.
2. So we need two numbers that add to $13$ and differ by $1$. Half of $13$ is $6.5$, so the two piles are $7$ and $6$.
3. The pennies are the bigger pile: $7$ pennies and $6$ nickels. Check: with one more nickel there would be $7$ pennies and $7$ nickels, which are equal.
4. The question asks for the value, not the number of coins. The nickels are worth $6 \times 5 = 30$ cents and the pennies are worth $7 \times 1 = 7$ cents.
5. Together that is $30 + 7 = 37$ cents.

The answer is $\boxed{\textbf{(C)}\ 37}$.

The sentence "one more nickel would make them equal" is really a statement about the difference between the two piles: pennies minus nickels is $1$. When you know a total and a difference, split the total almost in half. Always reread the question at the end, because here it asks for cents, not coins.

Use letters. Let $n$ be the number of nickels and $p$ the number of pennies. Then $n + p = 13$ and $p = n + 1$. Putting the second into the first gives $2n + 1 = 13$, so $n = 6$ and $p = 7$. The value is $5 \cdot 6 + 7 = 37$ cents. Notice that the swapped mix, $6$ pennies and $7$ nickels, is worth $41$ cents, which is choice (E), waiting for anyone who reads the sentence backwards.

Let $n$ be the number of nickels and $p$ the number of pennies, so $n + p = 13$. One extra nickel would balance the counts, so $p = n + 1$.

Substituting, $n + (n + 1) = 13$, which gives $n = 6$ and $p = 7$.

The value is $6 \cdot 5 + 7 \cdot 1 = 30 + 7 = 37$ cents.

The answer is $\boxed{\textbf{(C)}\ 37}$.