Difference between revisions of "2008 AIME II Problems/Problem 2"
(→Problem) |
m (→Solution 2) |
||
(4 intermediate revisions by 3 users not shown) | |||
Line 2: | Line 2: | ||
<!-- 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> | <!-- 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 == | + | == Simple Solution == |
+ | 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 | ||
+ | <cmath>r+5\left(49\right)=\frac{4}{3}r+5\left(24\right)</cmath> | ||
+ | <cmath>125=\frac13r</cmath> | ||
+ | <cmath>r=375.</cmath> | ||
+ | Thus, the total time they take is <math>375 + 5(49) = \boxed{620}</math> minutes. | ||
+ | |||
+ | == 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>. | ||
Line 16: | Line 23: | ||
t &= \boxed{620}\ \text{minutes} | t &= \boxed{620}\ \text{minutes} | ||
\end{align*}</cmath></center> | \end{align*}</cmath></center> | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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 == |
Latest revision as of 11:53, 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 -mile mark at exactly the same time. How many minutes has it taken them?
Simple Solution
Let be the time Rudolph takes disregarding breaks and
be the time Jennifer takes disregarding breaks. We have the equation
Thus, the total time they take is
minutes.
Solution 1
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)
Solution 2
Let the total time that Jennifer and Rudolph bike, including rests, to be minutes. Furthermore, let Rudolph's biking rate be
so Jennifer's biking rate is
. Note that Rudolf takes 49 breaks, taking
minutes, and Jennifer takes 24 breaks, taking
minutes. Since they both reach the 50 mile mark, then by
, 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
yielding
.
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.