2010 AMC 8 Problem 8

Attempt Problem 8 of the 2010 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction 12\frac{1}{2} mile in front of her. After she passes him, she can see him in her rear view mirror until he is 12\frac{1}{2} mile behind her. Emily rides at a constant rate of 1212 miles per hour, and Emerson skates at a constant rate of 88 miles per hour. For how many minutes can Emily see Emerson?
 6 \ 6
 8 \ 8
 12 \ 12
 15 \ 15
 16 \ 16
Answer: D
Concepts:relative speeddistance rate and timeunit conversion
Small hint:
During the visible interval Emily gains a total of 11 mile on Emerson.
Big hint:
Use Emily’s speed relative to Emerson’s speed.
Solution:
Let dd be how far Emily is ahead of Emerson. Emily sees Emerson if −12≤d≤12.-\dfrac 12 \leq d \leq \frac 12. Suppose at t=0,t= 0, where tt is in hours, that d=−12.d = -\frac 12. Then, d=(12−8)t−12.d = (12-8)t - \frac 12 . Since we must find where d=12,d = \frac 12, we find the time where 12=4t−12  ⟹  1=4t\frac 12 = 4t - \frac 12 \implies 1 = 4t   ⟹  t=0.25.\implies t = 0.25. Since 0.250.25 hours passed, we know that 60⋅0.25=1560\cdot0.25 = 15 minutes passed.
Therefore, the answer is D .
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