1996 AHSME Problems/Problem 13
Contents
[hide]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
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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