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2014 AMC 10B Problem 1

2014 AMC 10B Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 1 min · position P1-10

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?

A)
$33$
B)
$35$
C)
$37$
D)
$39$
E)
$41$

Answer

Key insight

Adding one nickel would make the counts equal, so there are 7 pennies and 6 nickels among the 13 coins.

Solution

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

Why this works

"One more nickel would make them equal" translates directly to $p = n + 1$ ; combining it with the total count gives a one-variable equation. Always finish by answering the question actually asked (value in cents), not the intermediate count.

Alternative approach

Thirteen coins with a one-coin gap must split as $7$ and $6$ . Since pennies are the larger group, $7$ pennies and $6$ nickels gives $37$ cents. Note that $6$ pennies and $7$ nickels would give $41$ , choice (E), the distractor for the swapped reading.

The trap

Swapping the counts (7 nickels, 6 pennies) and reporting 41 cents, or answering with the coin count instead of the value.

Common mistakes

Swapping the counts (7 nickels, 6 pennies) and reporting 41 cents, or answering with the coin count instead of the value.
Adding the hypothetical extra nickel into the value, giving $42$ or counting $14$ coins.

Techniques

Set up the equation/formula and compute; no special trick needed