2009 AMC 10A Problem 20

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 $\boxed{\textbf{(D)}\ 65}$.

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.

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 $\boxed{\textbf{(D)}\ 65}$.