Difference between revisions of "1996 AHSME Problems/Problem 13"

(Solution)
(Solution)
 
(3 intermediate revisions by the same user not shown)
Line 7: Line 7:
 
__TOC__
 
__TOC__
 
==Solution==
 
==Solution==
If Sunny runs at a rate of <math>s</math> for <math>h</math>. Then the distance covered is <math>sh</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 headstart of <math>h</math> meters . So he will catch  on Sunny at the rate of <math>s(m-1)</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}</math>.
+
If Sunny runs at a rate of <math>s</math> for <math>h</math>. Then the distance covered is <math>sh</math>. Now we know that Moonbeam runs <math>m</math> times as fast than Sunny, so Moonbeam runs at the rate of <math>ms</math>. Now Moonbeam gave Sunny a headstart of <math>h</math> meters, so he will catch  on Sunny at the rate of <math>s(m-1)</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}</math>.
  
 
===Solution 2===
 
===Solution 2===

Latest revision as of 19:34, 2 November 2023

Problem

Sunny runs at a steady rate, and Moonbeam runs $m$ times as fast, where $m$ is a number greater than 1. If Moonbeam gives Sunny a head start of $h$ meters, how many meters must Moonbeam run to overtake Sunny?

$\text{(A)}\ hm\qquad\text{(B)}\ \frac{h}{h+m}\qquad\text{(C)}\ \frac{h}{m-1}\qquad\text{(D)}\ \frac{hm}{m-1}\qquad\text{(E)}\ \frac{h+m}{m-1}$

Solution

If Sunny runs at a rate of $s$ for $h$. Then the distance covered is $sh$. Now we know that Moonbeam runs $m$ times as fast than Sunny, so Moonbeam runs at the rate of $ms$. Now Moonbeam gave Sunny a headstart of $h$ meters, so he will catch on Sunny at the rate of $s(m-1)$ . At time $\frac{h}{m-1}$ Moon beam will catch on Sunny. Now we are asked how much in meters he have to run to catch on Sunny. That is $\frac{hm}{m-1}$.

Solution 2

Note that $h$ is a length, while $m$ is a dimensionless constant. Thus, $h$ and $m$ cannot be added, and $B$ and $E$ are not proper answers, since they both contain $h+m$.

Thus, we only concern ourselves with answers $A, C, D$.

If $m$ is a very, very large number, then Moonbeam will have to run just over $h$ meters to reach Sunny. Or, in the language of limits:

$\lim_{m\rightarrow \infty} d(m) = h$, where $d(m)$ is the distance Moonbeam needs to catch Sunny at the given rate ratio of $m$.

In option $A$, when $m$ gets large, the distance gets large. Thus, $A$ is not a valid answer.

In option $C$, when $m$ gets large, the distance approaches $0$, not $h$ as desired. This is not a valid answer. (In fact, this is the distance Sunny runs, which does approach $0$ as Moonbeam gets faster and faster.)

In option $D$, when $m$ gets large, the ratio $\frac{m}{m-1}$ gets very close to, but remains just a tiny bit over, the number $1$. Thus, when you multiply it by $h$, the ratio in option $D$ gets very close to, but remains just a tiny bit over, $h$. Thus, the best option out of all the choices is $\boxed{D}$.

Solution 3

Assume that Sunny originally runs at a unit speed, and thus Moonbeam runs at a rate of $m$.

Choose a new reference frame where Sunny is still, and Moonbeam runs at a rate of $m-1$. In this new reference frame, the distance to be run is still $h$.

Moonbeam runs this distance $h$ in a time of $\frac{h}{m-1}$

Returning to the original reference frame, if Moonbeam runs for $\frac{h}{m-1}$ seconds, Moonbeam will cover a distance of $\frac{hm}{m-1}$, which is option $\boxed{D}$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png