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Difference between revisions of "2008 AIME II Problems/Problem 2"

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== Problem ==
 
== Problem ==
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Rudolph bikes at a [[constant]] rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the <math>50</math>-mile mark at exactly the same time. How many minutes has it taken them?
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Rudolph bikes at a [[constant]] rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the <math>50</math>-mile mark at exactly the same time. How many minutes has it taken them?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
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== Solution ==
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== Simple Solution ==
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Let <math>r</math> be the time Rudolph takes disregarding breaks and <math>\frac{4}{3}r</math> be the time Jennifer takes disregarding breaks. We have the equation
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<cmath>r+5\left(49\right)=\frac{4}{3}r+5\left(24\right)</cmath>
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<cmath>125=\frac13r</cmath>
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<cmath>r=375.</cmath>
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Thus, the total time they take is <math>375 + 5(49) = \boxed{620}</math> minutes.
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== Solution 1==
 
Let Rudolf bike at a rate <math>r</math>, so Jennifer bikes at the rate <math>\dfrac 34r</math>. Let the time both take be <math>t</math>.  
 
Let Rudolf bike at a rate <math>r</math>, so Jennifer bikes at the rate <math>\dfrac 34r</math>. Let the time both take be <math>t</math>.  
  
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Using the formula <math>r = \frac dt</math>, since both Jennifer and Rudolf bike <math>50</math> miles,
 
Using the formula <math>r = \frac dt</math>, since both Jennifer and Rudolf bike <math>50</math> miles,
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<cmath>\begin{align}r &= \frac{50}{t-245}\\
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<center><cmath>\begin{align}r &= \frac{50}{t-245}\\
 
\frac{3}{4}r &= \frac{50}{t-120}
 
\frac{3}{4}r &= \frac{50}{t-120}
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\end{align}</cmath>
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\end{align}</cmath></center>
 
Substituting equation <math>(1)</math> into equation <math>(2)</math> and simplifying, we find  
 
Substituting equation <math>(1)</math> into equation <math>(2)</math> and simplifying, we find  
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<cmath>\begin{align*}50 \cdot \frac{3}{4(t-245)} &= 50 \cdot \frac{1}{t-120}\\
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<center><cmath>\begin{align*}50 \cdot \frac{3}{4(t-245)} &= 50 \cdot \frac{1}{t-120}\\
 
\frac{1}{3}t &= \frac{245 \cdot 4}{3} - 120\\
 
\frac{1}{3}t &= \frac{245 \cdot 4}{3} - 120\\
 
t &= \boxed{620}\ \text{minutes}
 
t &= \boxed{620}\ \text{minutes}
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\end{align*}</cmath>
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\end{align*}</cmath></center>
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==Solution 2==
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Let the total time that Jennifer and Rudolph bike, including rests, to be <math>t</math> minutes.  Furthermore, let Rudolph's biking rate be <math>r</math> so Jennifer's biking rate is <math>\frac{3}{4}r</math>. Note that Rudolf takes 49 breaks, taking <math>49\cdot 5</math> minutes, and Jennifer takes 24 breaks, taking <math>24\cdot 5</math> minutes.  Since they both reach the 50 mile mark, then by <math>d=rt</math>, the rate times time taken for Rudolph and Jennifer must equal.  Hence, we disregard the breaks from the total time taken and get the equation  <cmath>r(t-49\cdot 5)=\frac{3}{4}r(t-24\cdot 5),</cmath>yielding <math>t=\boxed{620}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 10:53, 19 June 2019

Problem

Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $50$-mile mark at exactly the same time. How many minutes has it taken them?

Simple Solution

Let $r$ be the time Rudolph takes disregarding breaks and $\frac{4}{3}r$ be the time Jennifer takes disregarding breaks. We have the equation \[r+5\left(49\right)=\frac{4}{3}r+5\left(24\right)\] \[125=\frac13r\] \[r=375.\] Thus, the total time they take is $375 + 5(49) = \boxed{620}$ minutes.

Solution 1

Let Rudolf bike at a rate $r$, so Jennifer bikes at the rate $\dfrac 34r$. Let the time both take be $t$.

Then Rudolf stops $49$ times (because the rest after he reaches the finish does not count), losing a total of $49 \cdot 5 = 245$ minutes, while Jennifer stops $24$ times, losing a total of $24 \cdot 5 = 120$ minutes. The time Rudolf and Jennifer actually take biking is then $t - 245,\, t-120$ respectively.

Using the formula $r = \frac dt$, since both Jennifer and Rudolf bike $50$ miles,

\begin{align}r &= \frac{50}{t-245}\\ \frac{3}{4}r &= \frac{50}{t-120} \end{align}

Substituting equation $(1)$ into equation $(2)$ and simplifying, we find

\begin{align*}50 \cdot \frac{3}{4(t-245)} &= 50 \cdot \frac{1}{t-120}\\ \frac{1}{3}t &= \frac{245 \cdot 4}{3} - 120\\ t &= \boxed{620}\ \text{minutes} \end{align*}

Solution 2

Let the total time that Jennifer and Rudolph bike, including rests, to be $t$ minutes. Furthermore, let Rudolph's biking rate be $r$ so Jennifer's biking rate is $\frac{3}{4}r$. Note that Rudolf takes 49 breaks, taking $49\cdot 5$ minutes, and Jennifer takes 24 breaks, taking $24\cdot 5$ minutes. Since they both reach the 50 mile mark, then by $d=rt$, the rate times time taken for Rudolph and Jennifer must equal. Hence, we disregard the breaks from the total time taken and get the equation \[r(t-49\cdot 5)=\frac{3}{4}r(t-24\cdot 5),\]yielding $t=\boxed{620}$.

See also

2008 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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