1996 AHSME Problems/Problem 13
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Problem
Sunny runs at a steady rate, and Moonbeam runs times as fast, where is a number greater than 1. If Moonbeam gives Sunny a head start of meters, how many meters must Moonbeam run to overtake Sunny?
Solution 1
If Sunny runs at a rate of for meters in minutes, then .
In that case, Moonbeam's rate is , and Moonbeam's distance is , and the amount of time is the same. Thus,
Solving each equation for , we have
Cross multiplying, we get
Solving for , we get , which leads to .
Note that is the distance that Sunny ran. Moonbeam ran meters more, for a total of . This is answer .
Solution 2
Note that is a length, while is a dimensionless constant. Thus, and cannot be added, and and are not proper answers, since they both contain .
Thus, we only concern ourselves with answers .
If is a very, very large number, then Moonbeam will have to run just over meters to reach Sunny. Or, in the language of limits:
, where is the distance Moonbeam needs to catch Sunny at the given rate ratio of .
In option , when gets large, the distance gets large. Thus, is not a valid answer.
In option , when gets large, the distance approaches , not as desired. This is not a valid answer. (In fact, this is the distance Sunny runs, which does approach as Moonbeam gets faster and faster.)
In option , when gets large, the ratio gets very close to, but remains just a tiny bit over, the number . Thus, when you multiply it by , the ratio in option gets very close to, but remains just a tiny bit over, . Thus, the best option out of all the choices is .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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