2017 AMC 10B Problem 2

P1-10~2 minPrint

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

Sofia ran $5$ laps around the $400$ -meter track at her school. For each lap, she ran the first $100$ meters at an average speed of $4$ meters per second and the remaining $300$ meters at an average speed of $5$ meters per second. How much time did Sofia take running the $5$ laps?

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Key insight

Find the time for each part of a lap: 100 meters takes 25 seconds and 300 meters takes 60 seconds, so one lap is 85 seconds.

Solution

Speeds cannot simply be averaged, but times can be added. So find the time for each part of a lap, then add.

First $100$ meters at $4$ meters per second: $100 \div 4 = 25$ seconds.
Remaining $300$ meters at $5$ meters per second: $300 \div 5 = 60$ seconds.
One lap takes $25 + 60 = 85$ seconds.
Five laps take $5 \cdot 85 = 425$ seconds.
Change to minutes: $7$ minutes is $420$ seconds, and $425 - 420 = 5$ . So the time is $7$ minutes and $5$ seconds.

The answer is $\boxed{\textbf{(C)}\ \text{7 minutes and 5 seconds}}$ .

Why this works

Whenever a trip is made of pieces at different speeds, work out each piece's time with time $=$ distance $\div$ speed, and add the times. Leave the change from seconds to minutes for the very end, and check it, because that is where careless slips happen.

Concise solution

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

Compute the time for one lap segment by segment.

First $100$ meters at $4$ m/s: $\frac{100}{4} = 25$ seconds.
Remaining $300$ meters at $5$ m/s: $\frac{300}{5} = 60$ seconds.

One lap takes $25 + 60 = 85$ seconds, so five laps take $5 \cdot 85 = 425$ seconds. Since $425 = 7 \cdot 60 + 5$ , that is $7$ minutes and $5$ seconds.

The answer is $\boxed{\textbf{(C)}\ \text{7 minutes and 5 seconds}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)