2009 AMC 10A Problem 20

P16-20~3 minPrint

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

Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?

Show solution without answering

Key insight

The gap closes 1 km per minute, split 3 to 1, so Lauren rides 1/4 km per minute. After 5 minutes, she alone rides the last 15 km.

Solution

1. The gap shrinks by $1$ km each minute because both riders move toward each other. Andrea rides $3$ times as fast as Lauren, so split that $1$ km into $4$ equal parts: Andrea covers $3$ parts and Lauren covers $1$ part.
2. So Lauren rides $\frac{1}{4}$ km per minute and Andrea rides $\frac{3}{4}$ km per minute. (Check: $\frac14 + \frac34 = 1$ .)
3. For the first $5$ minutes the gap shrinks $1$ km each minute, so after $5$ minutes it is $20 - 5 = 15$ km.
4. Then Andrea stops. Now only Lauren closes the gap, at $\frac14$ km per minute. To cover $15$ km she needs
$15 \div \frac{1}{4} = 15 \times 4 = 60\ \text{minutes}.$
5. Total time from the start: $5 + 60 = 65$ minutes.

The answer is $\boxed{\textbf{(D)}\ 65}$ .

Why this works

When two people move toward each other, the gap shrinks at the sum of their speeds. A $3$ to $1$ speed ratio splits that sum into $\frac34$ and $\frac14$ . The trip has two stages with different closing speeds, so find the distance left when the stage changes, then use distance $=$ speed $\times$ time again.

Concise solution

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

Let Lauren's speed be $v$ km/min; Andrea's is $3v$ . Approaching each other, the gap closes at $v + 3v = 4v = 1$ km/min, so
$v = \tfrac{1}{4} \text{ km/min}, \qquad 3v = \tfrac{3}{4} \text{ km/min}.$

In the first $5$ minutes the gap shrinks by $5 \cdot 1 = 5$ km, leaving $20 - 5 = 15$ km between them.

After Andrea stops, only Lauren moves, at $\frac14$ km/min. Covering $15$ km takes
$\frac{15}{1/4} = 60 \text{ minutes}.$

Total time from the start: $5 + 60 = 65$ minutes.

The answer is $\boxed{\textbf{(D)}\ 65}$ .

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)