Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AIME II Problems/Problem 2"
Line 8: Line 8:
  
 
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,
−
<center><math>\begin{align}r &= \frac{50}{t-245}\\
+
<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}
−
\end{align}</math></center>
+
\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  
−
<center><math>\begin{align*}50 \cdot \frac{3}{4(t-245)} &= 50 \cdot \frac{1}{t-120}\\
+
<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}
−
\end{align*}</math></center>
+
\end{align*}</cmath></center>
  
 
== See also ==
 
== See also ==

Revision as of 02:13, 3 March 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_2&oldid=68133"