2005 AMC 10A Problem 7

P1-10~3 minPrint

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

Josh and Mike live $13$ miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?

Show solution without answering

Key insight

Josh rides twice as long at four fifths the speed, so he covers 8 fifths of Mike's distance. Mike's 5 parts plus Josh's 8 parts make 13 miles.

Solution

Distance is speed times time. The problem compares both Josh's speed and Josh's time to Mike's, so compare their distances first.

1. Josh versus Mike. Josh rode for $2$ times as long, at $\frac{4}{5}$ of the speed. His distance is
$2 \times \frac{4}{5} = \frac{8}{5}$
times Mike's distance.
2. Think in parts. If Mike rode $5$ parts, Josh rode $8$ parts. That is the same ratio $8$ to $5$ .
3. Use the total. They rode toward each other until they met, so together they covered the whole $13$ miles. That is $5 + 8 = 13$ parts, so each part is $1$ mile.
4. Mike's distance. $5$ parts, which is $5$ miles.

Check: Josh rode $8$ miles and Mike rode $5$ , a total of $13$ , and $8$ is $\frac{8}{5}$ of $5$ .

The answer is $\boxed{\textbf{(B)}\ 5}$ .

Why this works

When two people ride toward each other and meet, their distances add up to the starting gap. You never need the actual speeds or times here: because the problem gives Josh's speed and time as multiples of Mike's, their product gives the ratio of distances directly, and splitting $13$ in that ratio finishes the job.

Alternative approach

Use a letter. Call Mike's distance $m$ . Josh's distance is $\frac{8}{5}m$ , so $m + \frac{8}{5}m = 13$ . That is $\frac{13}{5}m = 13$ , so $m = 5$ .

Concise solution

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

Suppose Mike rode at rate $r$ for time $t$ , covering $rt$ miles. Josh rode for time $2t$ at rate $\frac{4}{5}r$ , so his distance is
$\frac{4}{5}r \cdot 2t = \frac{8}{5}\,rt.$
Josh's distance is $\frac{8}{5}$ of Mike's. When they meet, the two distances make up the whole $13$ miles. With $m = rt$ for Mike's distance,
$m + \frac{8}{5}m = \frac{13}{5}m = 13 \quad\Longrightarrow\quad m = 5.$

The answer is $\boxed{\textbf{(B)}\ 5}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Prealgebra, Chapter 13: Data and Statistics (§13.1)