2013 AMC 10A Problem 9

P1-10~2 minPrint

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

In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?

Show solution without answering

Key insight

A three-point try averages 0.2 times 3 = 0.6 points and a two-point try averages 0.3 times 2 = 0.6, so 30 tries give 18.

Solution

The problem does not say how many shots of each kind she took. That is a hint that the answer does not depend on it. Let us see why.

Average points for one three-point try. She makes $20\%$ of them, and each make is worth $3$ . On average one try gives $0.2 \times 3 = 0.6$ points. (For example, $10$ tries give $2$ makes, which is $6$ points, or $0.6$ per try.)
Average points for one two-point try. She makes $30\%$ of them, each worth $2$ . On average one try gives $0.3 \times 2 = 0.6$ points. (Ten tries give $3$ makes, which is $6$ points.)
Every try is worth the same. Both kinds of shot give $0.6$ points per try. So it does not matter how the $30$ tries are split.
Total. $30 \times 0.6 = 18$ points.

Shenille scored $\boxed{\textbf{(B)}\ 18}$ points.

Why this works

If a question leaves out a detail, the answer probably does not depend on it. Finding the "value per try" for each kind of shot shows the reason: both are $0.6$ . Then the total is simply the number of tries times $0.6$ .

Alternative approach

Since the split does not matter, pick an easy one: all $30$ shots are three-pointers. She makes $20\%$ of $30$ , which is $6$ shots, worth $6 \times 3 = 18$ points. With algebra: if $x$ shots are three-pointers, the points are $0.6x + 0.6(30 - x) = 18$ , and the $x$ disappears.

Concise solution

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

Let $x$ be the number of three-point attempts, so there were $30 - x$ two-point attempts.

Points from three-pointers: $20\%$ of $x$ attempts succeed, each worth $3$ , giving $0.2 \cdot 3x = 0.6x$ .

Points from two-pointers: $30\%$ of $30 - x$ attempts succeed, each worth $2$ , giving $0.3 \cdot 2(30 - x) = 0.6(30 - x)$ .

Total:
$0.6x + 0.6(30 - x) = 0.6 \cdot 30 = 18.$

The $x$ cancels, so the split does not matter. Shenille scored $\boxed{\textbf{(B)}\ 18}$ points.

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
Prealgebra, Chapter 13: Data and Statistics (§13.1)