1996 AHSME Problems/Problem 13
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
If Sunny runs at a rate of for . Then the distance covered is . Now we know that Moonbeam runs times as fast than sunny, so Moonbeam runs at the rate of . Now moonbeam gave sunny a headstart of meters, so he will catch on Sunny at the rate of . At time Moon beam will catch on Sunny. Now we are asked how much in meters he have to run to catch on Sunny. That is .
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 .
Solution 3
Assume that Sunny originally runs at a unit speed, and thus Moonbeam runs at a rate of .
Choose a new reference frame where Sunny is still, and Moonbeam runs at a rate of . In this new reference frame, the distance to be run is still .
Moonbeam runs this distance in a time of
Returning to the original reference frame, if Moonbeam runs for seconds, Moonbeam will cover a distance of , which is option .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.