Difference between revisions of "2008 AIME II Problems/Problem 2"
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== Problem == | == 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 <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>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 17:13, 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 -mile mark at exactly the same time. How many minutes has it taken them?
Solution
Let Rudolf bike at a rate , so Jennifer bikes at the rate
. Let the time both take be
.
Then Rudolf stops times (because the rest after he reaches the finish does not count), losing a total of
minutes, while Jennifer stops
times, losing a total of
minutes. The time Rudolf and Jennifer actually take biking is then
respectively.
Using the formula , since both Jennifer and Rudolf bike
miles,
![\begin{align}r &= \frac{50}{t-245}\\ \frac{3}{4}r &= \frac{50}{t-120} \end{align}](http://latex.artofproblemsolving.com/0/0/9/009c2785ed941a2e6127ad143e7d0958eeb670b9.png)
Substituting equation into equation
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*}](http://latex.artofproblemsolving.com/9/0/3/903cba8139ad0245619f6ae90070a72bf9bb6c06.png)
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.