2014 AMC 10B Problem 1

P1-10~1 minPrint

Linear Equations & Word Problems Set up the equation/formula and compute; no special trick needed

Leah has $13$ coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies as nickels. In cents, how much are Leah's coins worth?

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Key insight

One more nickel would make the two piles equal, so right now there is one more penny than nickel: 7 pennies and 6 nickels.

Solution

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.
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 $6$ .
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.
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.
Together that is $30 + 7 = 37$ cents.

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

Why this works

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.

Alternative approach

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(C)}\ 37}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)