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2011 AMC 10B Problem 2

2011 AMC 10B Problem 2AMC 10 Step by Step

Name: Date: September 13, 2026

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

Josanna's test scores to date are $90, 80, 70, 60,$ and $85$ . Her goal is to raise her test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?

A)
$80$
B)
$82$
C)
$85$
D)
$90$
E)
$95$

Answer

Key insight

The current average is 77, so six tests must total at least 6 times 80 = 480; the current total is 385, leaving 95.

Solution

The five scores sum to $90 + 80 + 70 + 60 + 85 = 385$ , so the current average is $385/5 = 77$ .

The goal is an average of at least $80$ over six tests, which means the six scores must total at least $6 \cdot 80 = 480$ . The sixth score must therefore be at least
$480 - 385 = 95.$

Check: $(385 + 95)/6 = 480/6 = 80$ , exactly $3$ above $77$ . So the minimum score is $\boxed{\textbf{(E)}\ 95}$ .

Why this works

Averages are easiest to handle as totals: "average of $k$ tests is $m$ " means "the sum is $km$ ". Raising the average of $n+1$ items by $d$ requires the new item to exceed the old average by $(n+1)d$ , since it must lift every one of the $n+1$ scores by $d$ . Here $77 + 6 \cdot 3 = 95$ .

The trap

Thinking a score 3 points above the current average (80) is enough, when in fact one score must pull six scores up by 3 each.

Common mistakes

Thinking a score 3 points above the current average (80) is enough, when in fact one score must pull six scores up by 3 each.
Dividing $480$ by $5$ instead of $6$ , or computing the current total incorrectly.

Techniques

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