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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>
  
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==Solution ==
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__TOC__
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==Solution==
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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 haeadstart 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>.
  
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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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===Solution 2===
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In that case, Moonbeam's rate is <math>ms</math>, and Moonbeam's distance is <math>x + h</math>, and the amount of time <math>t</math> is the same.  Thus, <math>ms = \frac{x + h}{t}</math>
 
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Solving each equation for <math>t</math>, we have <math>t = \frac{x}{s} = \frac{x + h}{ms}</math>
 
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Cross multiplying, we get <math>msx = xs + hs</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>.
 
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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>. 
 
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==Solution 2==
 
  
 
Note that <math>h</math> is a length, while <math>m</math> is a dimensionless constant.  Thus, <math>h</math> and <math>m</math> cannot be added, and <math>B</math> and <math>E</math> are not proper answers, since they both contain <math>h+m</math>.
 
Note that <math>h</math> is a length, while <math>m</math> is a dimensionless constant.  Thus, <math>h</math> and <math>m</math> cannot be added, and <math>B</math> and <math>E</math> are not proper answers, since they both contain <math>h+m</math>.
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In option <math>D</math>, when <math>m</math> gets large, the ratio <math>\frac{m}{m-1}</math> gets very close to, but remains just a tiny bit over, the number <math>1</math>.  Thus, when you multiply it by <math>h</math>, the ratio in option <math>D</math> gets very close to, but remains just a tiny bit over, <math>h</math>.  Thus, the best option out of all the choices is <math>\boxed{D}</math>.
 
In option <math>D</math>, when <math>m</math> gets large, the ratio <math>\frac{m}{m-1}</math> gets very close to, but remains just a tiny bit over, the number <math>1</math>.  Thus, when you multiply it by <math>h</math>, the ratio in option <math>D</math> gets very close to, but remains just a tiny bit over, <math>h</math>.  Thus, the best option out of all the choices is <math>\boxed{D}</math>.
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===Solution 3===
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Assume that Sunny originally runs at a unit speed, and thus Moonbeam runs at a rate of <math>m</math>.
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Choose a new reference frame where Sunny is still, and Moonbeam runs at a rate of <math>m-1</math>.  In this new reference frame, the distance to be run is still <math>h</math>.
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Moonbeam runs this distance <math>h</math> in a time of <math>\frac{h}{m-1}</math>
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Returning to the original reference frame, if Moonbeam runs for <math>\frac{h}{m-1}</math> seconds, Moonbeam will cover a distance of <math>\frac{hm}{m-1}</math>, which is option <math>\boxed{D}</math>.
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=12|num-a=14}}
 
{{AHSME box|year=1996|num-b=12|num-a=14}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 16:22, 9 June 2019

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 moonebeam runs at the rate of $ms$. Now moonbeam gave sunny a haeadstart 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 (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

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