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

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 $\boxed{D}$ .

@@ Line 7: / Line 7: @@
 __TOC__
 ==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 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>.
+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===

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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