AMC 10 · 2013 · #10

Grade 7 algebra
systems-of-equationspercentagelinear-equations-two-var convert-to-algebra ↑ Prerequisites: linear-equations-two-varpercentage
📏 Medium solution 💡 2 insights
Problem
A team scores 54 points from two-point and three-point shots. They make 50% of their two-point tries and 40% of their three-point tries, and they take 50% more two-point tries than three-point tries. Find how many three-point shots they attempted.

Pick an answer.

(A)
10
(B)
15
(C)
20
(D)
25
(E)
30

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

Two unknown counts are linked by a scoring total and a ratio. Name each count with a letter, translate every sentence into an equation, and track the point units so the totals line up. Then substitution collapses it to one equation in one variable.

1STEP 1

Name the two counts

Let x be the number of two-point shots attempted and y the number of three-point shots attempted.

x = two-point attempts, y = three-point attempts
2STEP 2

Translate 50% more

"50% more than y" is y plus half of y, so x = 1.5y.

x = y + 0.5y = 1.5y
3STEP 3

Count the points

Twos give 0.5x made at 2 points each, threes give 0.4y made at 3 each, so x + 1.2y = 54.

(0.5x) · 2_x + (0.4y) · 3₁.2y = 54 → x + 1.2y = 54
4STEP 4

Substitute and solve

Swap x for 1.5y: 1.5y + 1.2y = 2.7y = 54, so y = 20 three-point attempts, choice (C).

1.5y + 1.2y = 2.7y = 54 → y = 54/2.7 = 20
Answer
20
With y = 20, x = 1.5(20) = 30. Points: made two-pointers give 0.5(30)(2) = 30, made three-pointers give 0.4(20)(3) = 24, and 30 + 24 = 54. This matches the total, and 20 is one of the listed choices.
💡Key takeaway

Name each unknown with a letter, turn every sentence into an equation, then swap one variable in for the other so only one is left to solve.

  • Name the two counts
  • Translate 50% more
  • Count the points
  • Substitute and solve