# Difference between revisions of "2009 AMC 10A Problems/Problem 20"

## Problem

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?

$\mathrm{(A)}\ 20 \qquad \mathrm{(B)}\ 30 \qquad \mathrm{(C)}\ 55 \qquad \mathrm{(D)}\ 65 \qquad \mathrm{(E)}\ 80$

## Solution 1

Let their speeds in kilometers per hour be $v_A$ and $v_L$. We know that $v_A=3v_L$ and that $v_A+v_L=60$. (The second equation follows from the fact that $1\mathrm km/min = 60\mathrm km/h$.) This solves to $v_A=45$ and $v_L=15$.

As the distance decreases at a rate of $1$ kilometer per minute, after $5$ minutes the distance between them will be $20-5=15$ kilometers.

From this point on, only Lauren will be riding her bike. As there are $15$ kilometers remaining and $v_L=15$, she will need exactly an hour to get to Andrea. Therefore the total time in minutes is $5+60 = \boxed{65}$.

## Solution 2

Because the speed of Andrea is 3 times as fast as Lauren and the distance between them is decreasing at a rate of 1 kilometer per minute, Andrea's speed is $\frac{3}{4} \textbf{km/min}$, and Lauren's $\frac{1}{4} \textbf{km/min}$. Therefore, after 5 minutes, Andrea will have biked $\frac{3}{4} \cdot 5 = \frac{15}{4}$km.

In all, Lauren will have to bike $20 - \frac{15}{4} = \frac{80}{4} - \frac{15}{4} = \frac{65}{4}$km. Because her speed is $\frac{1}{4} \textbf{km/min}$, the time elapsed will be $\frac{\frac{65}{4}}{\frac{1}{4}} = \boxed{\textbf{D) 65}}$

~savannahsolver