2009 AMC 10A Problems/Problem 20

Revision as of 21:56, 16 January 2016 by Kkwang (talk | contribs) (Solution)

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

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 $1\,\unit{\mathrm km/min} = 60\,\unit{\mathrm km/h}$ (Error compiling LaTeX. ! Undefined control sequence.).) 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 $5+60 = \boxed{65}$.

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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