2021 AMC Fall 10A Problem 4

P1-10~2 minPrint

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

Mr. Lopez has a choice of two routes to get to work. Route A is $6$ miles long, and his average speed along this route is $30$ miles per hour. Route B is $5$ miles long, and his average speed along this route is $40$ miles per hour, except for a $\frac{1}{2}$ -mile stretch in a school zone where his average speed is $20$ miles per hour. By how many minutes is Route B quicker than Route A?

Show solution without answering

Key insight

Find the time for each piece of each route in minutes. Route A takes 12 minutes. Route B takes 6.75 + 1.5 = 8.25 minutes. Subtract.

Solution

Use time $=$ distance $\div$ speed. The answers are in minutes, so turn each time into minutes (multiply hours by $60$ ).

Route A. $6$ miles at $30$ mph takes $\frac{6}{30} = \frac{1}{5}$ of an hour. That is $\frac{60}{5} = 12$ minutes.
Route B, school zone. $\frac12$ mile at $20$ mph takes $\frac{1}{2} \div 20 = \frac{1}{40}$ of an hour. That is $\frac{60}{40} = 1.5$ minutes.
Route B, the rest. The school zone is part of the $5$ miles, so $5 - \frac12 = 4.5$ miles are left, at $40$ mph. That takes $\frac{4.5}{40}$ of an hour, which is $\frac{4.5 \times 60}{40} = \frac{270}{40} = 6.75$ minutes.
Route B total. $1.5 + 6.75 = 8.25$ minutes.
Difference. $12 - 8.25 = 3.75$ minutes, which is $3\frac34$ minutes.

The answer is $\boxed{\textbf{(B)}\ 3\frac{3}{4}}$ .

Why this works

When the speed changes partway through a trip, find the time for each piece separately and add them. You cannot just average the two speeds. Choosing minutes at the start keeps the numbers matched to the form of the answer choices.

Concise solution

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

Work in minutes, using time $=$ distance $\div$ speed and $1$ hour $= 60$ minutes.

Route A: $\dfrac{6}{30}$ hour $= \dfrac{1}{5}$ hour $= 12$ minutes.

Route B has two legs. The school zone is $\tfrac12$ mile at $20$ mph: $\dfrac{1/2}{20} = \dfrac{1}{40}$ hour $= 1.5$ minutes. The remaining $4.5$ miles at $40$ mph: $\dfrac{4.5}{40} = \dfrac{9}{80}$ hour $= 6.75$ minutes. Total: $1.5 + 6.75 = 8.25$ minutes.

Route B is quicker by $12 - 8.25 = 3.75 = 3\tfrac{3}{4}$ minutes.

The answer is $\boxed{\textbf{(B)}\ 3\frac{3}{4}}$ .

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)