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2015 AMC 10A Problem 5

2015 AMC 10A Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 1 of 5 · about 2 min · position P1-10

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?

A)
$81$
B)
$85$
C)
$91$
D)
$94$
E)
$95$

Answer

Key insight

Averages hide totals: 14 tests average 80 (sum 1120), 15 tests average 81 (sum 1215), and the difference 95 is Payton's score.

Solution

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 $\boxed{\textbf{(E)}\ 95}$ .

Why this works

An average is a total in disguise; multiply back to totals whenever an item is added or removed. A quicker view: adding Payton raised each of the $14$ other scores' "share" by $1$ point, so Payton must have scored $81 + 14 = 95$ , the new average plus one point for each other student.

The trap

Using 15 students for the first average (sum 1200) instead of 14, since Payton's test was not yet graded.

Common mistakes

Using $15$ students for the first average (sum $1200$ ) instead of $14$ , since Payton's test was not yet graded.
Answering $81$ , the new average, as if the added score must equal the average.

Techniques

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