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2011 AMC 10A Problem 12

2011 AMC 10A Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 2 of 5 · about 2 min · position P11-15

The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?

A)
$13$
B)
$14$
C)
$15$
D)
$16$
E)
$17$

Answer

Key insight

Let y be the number of two-pointers; then three-pointers also contribute 2y points and free throws contribute y + 1, so 5y + 1 = 61.

Solution

Let $y$ be the number of two-point shots. They scored $2y$ points from these, and the three-point shots produced the same number of points, $2y$ . The free throws numbered $y + 1$ , worth $y + 1$ points.

Total score:
$2y + 2y + (y + 1) = 5y + 1 = 61,$
so $y = 12$ . The number of free throws is $y + 1 = 13$ .

(Check: $12$ two-pointers give $24$ points, $8$ three-pointers give $24$ points, $13$ free throws give $13$ ; total $61$ .)

The answer is $\boxed{\textbf{(A)}\ 13}$ .

Why this works

Everything is expressed in terms of one unknown because each condition ties another quantity to the number of two-point shots. Notice that the three-point shot count never needs to be found: the condition is about points, so " $2y$ points" is all we use. Read whether a condition compares counts or points.

The trap

Setting the number of three-point shots equal to the number of two-point shots, instead of equating the points they scored.

Common mistakes

Setting the number of three-point shots equal to the number of two-point shots, instead of equating the points they scored.
Solving for $y = 12$ and answering $12$ , forgetting the question asks for free throws, which is $y + 1$ .

Techniques

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