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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>
  
 
== Solution ==
 
== Solution ==

Revision as of 16:13, 18 November 2015

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?

Solution

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*}

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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