Difference between revisions of "1996 AHSME Problems/Problem 13"
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===Solution 1=== | ===Solution 1=== | ||
| − | If Sunny runs at a rate of <math>s</math> for <math>h</math>. Then the distance covered is <math>s*h</math>. Now we know that moonbeam runs <math>m</math> times as fast than sunny So moonebeam runs at the rate of <math>ms</math>. Now moonbeam gave sunny a haeadstart of <math>h</math> meters . So he will catch on Sunny at the rate of <math>s(m-1</math>)<math> . At time </math>\frac{h}{m-1}<math> Moon beam will catch on Sunny . Now we are asked how much in meters he have to run | + | If Sunny runs at a rate of <math>s</math> for <math>h</math>. Then the distance covered is <math>s*h</math>. Now we know that moonbeam runs <math>m</math> times as fast than sunny So moonebeam runs at the rate of <math>ms</math>. Now moonbeam gave sunny a haeadstart of <math>h</math> meters . So he will catch on Sunny at the rate of <math>s(m-1</math>)<math> . At time </math>\frac{h}{m-1}<math> Moon beam will catch on Sunny.Now we are asked how much in meters he have to run to catch on Sunny.That is </math>\frac{hm}{m-1}$. |
===Solution 2=== | ===Solution 2=== | ||
Revision as of 15:15, 9 June 2019
Problem
Sunny runs at a steady rate, and Moonbeam runs 
times as fast, where is a number greater than 1. If Moonbeam gives Sunny a head start of 
meters, how many meters must Moonbeam run to overtake Sunny?
Solution
Solution 1
If Sunny runs at a rate of 
for . Then the distance covered is 
. Now we know that moonbeam runs times as fast than sunny So moonebeam runs at the rate of 
. Now moonbeam gave sunny a haeadstart of meters . So he will catch on Sunny at the rate of 
)
\frac{h}{m-1}
\frac{hm}{m-1}$.
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. 
