Difference between revisions of "1996 AHSME Problems/Problem 13"
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<math> \text{(A)}\ hm\qquad\text{(B)}\ \frac{h}{h+m}\qquad\text{(C)}\ \frac{h}{m-1}\qquad\text{(D)}\ \frac{hm}{m-1}\qquad\text{(E)}\ \frac{h+m}{m-1} </math> | <math> \text{(A)}\ hm\qquad\text{(B)}\ \frac{h}{h+m}\qquad\text{(C)}\ \frac{h}{m-1}\qquad\text{(D)}\ \frac{hm}{m-1}\qquad\text{(E)}\ \frac{h+m}{m-1} </math> | ||
| − | ==Solution == | + | ==Solution 1== |
If Sunny runs at a rate of <math>s</math> for <math>x</math> meters in <math>t</math> minutes, then <math>s = \frac{x}{t}</math>. | If Sunny runs at a rate of <math>s</math> for <math>x</math> meters in <math>t</math> minutes, then <math>s = \frac{x}{t}</math>. | ||
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Solving for <math>x</math>, we get <math>msx - xs = hs</math>, which leads to <math>x = \frac{h}{m - 1}</math>. | Solving for <math>x</math>, we get <math>msx - xs = hs</math>, which leads to <math>x = \frac{h}{m - 1}</math>. | ||
| − | Note that <math>x</math> is the distance that Sunny ran. Moonbeam ran <math>h</math> meters more, for a total of <math>h + \frac{h}{m-1} = \frac{h(m-1) + h}{m-1} = \frac{hm}{m-1}</math>. This is answer <math>\boxed{D}</math>. | + | Note that <math>x</math> is the distance that Sunny ran. Moonbeam ran <math>h</math> meters more, for a total of <math>h + \frac{h}{m-1} = \frac{h(m-1) + h}{m-1} = \frac{hm}{m-1}</math>. This is answer <math>\boxed{D}</math>. |
==Solution 2== | ==Solution 2== | ||
Revision as of 10:52, 19 August 2011
Contents
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 | |
| 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 | ||
