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2018 AMC 10B Problem 2

2018 AMC 10B Problem 2AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 2 min · position P1-10

Sam drove $96$ miles in $90$ minutes. His average speed during the first $30$ minutes was $60$ mph (miles per hour), and his average speed during the second $30$ minutes was $65$ mph. What was his average speed, in mph, during the last $30$ minutes?

A)
$64$
B)
$65$
C)
$66$
D)
$67$
E)
$68$

Answer

Key insight

Half an hour at 60 mph is 30 miles and at 65 mph is 32.5 miles, leaving 33.5 miles for the last half hour, i.e. 67 mph.

Solution

Each $30$ -minute stretch is half an hour.

First half hour at $60$ mph: $60 \cdot \tfrac12 = 30$ miles.
Second half hour at $65$ mph: $65 \cdot \tfrac12 = 32.5$ miles.

That leaves $96 - 30 - 32.5 = 33.5$ miles for the final half hour, so the speed then was
$\frac{33.5}{1/2} = 67 \text{ mph}.$

The answer is $\boxed{\textbf{(D)}\ 67}$ .

Why this works

Distance is additive, so a trip in pieces is handled by computing each piece's distance with $d = rt$ and subtracting from the total. Convert every time to hours before multiplying by an mph speed.

Alternative approach

Because the three intervals have equal length, the overall average speed is the plain average of the three speeds. Overall average is $96 \div 1.5 = 64$ mph, so $(60 + 65 + v)/3 = 64$ gives $v = 67$ .

The trap

Multiplying speeds by 30 minutes instead of 0.5 hours, or averaging 60 and 65 without accounting for the 96-mile total.

Common mistakes

Multiplying speeds by 30 minutes instead of 0.5 hours, or averaging 60 and 65 without accounting for the 96-mile total.
Answering $64$ , the overall average speed, instead of the speed on the final stretch.

Techniques

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