2015 AMC 10A Problem 5

P1-10~2 minPrint

Ratios, Percents & Averages Statistics & Data Set up the equation/formula and compute; no special trick needed

Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?

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Key insight

Turn each average into a total: 14 tests averaging 80 add to 1120, and 15 tests averaging 81 add to 1215. The difference, 95, is Payton's score.

Solution

An average hides a total. Turn both averages back into totals, and Payton's score is the difference.

1. Before Payton's test, $15 - 1 = 14$ tests were graded, with average $80$ . Their total is $14 \times 80 = 1120$ points.
2. After Payton's test, all $15$ tests average $81$ . The total is now $15 \times 81 = 1215$ points.
3. The only thing added between the two totals is Payton's score:
$1215 - 1120 = 95.$

The answer is $\boxed{\textbf{(E)}\ 95}$ .

Why this works

Total $=$ average $\times$ number of items. Whenever one item is added to or removed from a group, change both averages into totals and subtract. Be careful to use the right count each time: here the first average covered only $14$ tests.

Alternative approach

Think of the average rising by one point. Payton's score must first match the new average, $81$ , and then also supply one extra point to each of the other $14$ students to lift their average from $80$ to $81$ . So Payton scored $81 + 14 = 95$ .

Concise solution

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

Before Payton's test, $14$ tests were graded with average $80$ , so their total was $14 \cdot 80 = 1120$ .

After Payton's test, all $15$ tests average $81$ , so the total became $15 \cdot 81 = 1215$ .

Payton's score is the difference: $1215 - 1120 = 95$ .

The answer is $\boxed{\textbf{(E)}\ 95}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Prealgebra, Chapter 13: Data and Statistics
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)