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

(Solution 2)
(Solution)
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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 ==
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==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>.
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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>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 10:52, 19 August 2011

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 1

If Sunny runs at a rate of $s$ for $x$ meters in $t$ minutes, then $s = \frac{x}{t}$.

In that case, Moonbeam's rate is $ms$, and Moonbeam's distance is $x + h$, and the amount of time $t$ is the same. Thus, $ms = \frac{x + h}{t}$

Solving each equation for $t$, we have $t = \frac{x}{s} = \frac{x + h}{ms}$

Cross multiplying, we get $msx = xs + hs$

Solving for $x$, we get $msx - xs = hs$, which leads to $x = \frac{h}{m - 1}$.

Note that $x$ is the distance that Sunny ran. Moonbeam ran $h$ meters more, for a total of $h + \frac{h}{m-1} = \frac{h(m-1) + h}{m-1} = \frac{hm}{m-1}$. This is answer $\boxed{D}$.

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 $\boxed{D}$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AHSME Problems and Solutions
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