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

Revision as of 10:50, 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 be $t$ minutes. Furthermore, let Rudolph's biking rate be $r$ so Jennifer's biking rate is $\frac{3}{4}r$ . Because they both reached the fifty mile mark at the same time, we don't count the rest afterward so 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 (d=50), 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 $r(t-49\cdot 5)=frac{3}{4}(t-24\cdot 5),$ yielding $t=\boxed{620}$ .

@@ Line 9: / Line 9: @@
 Thus, the total time they take is <math>375 + 5(49) = \boxed{620}</math> minutes.
-== Solution ==
+== 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>.
@@ Line 23: / Line 23: @@
 t &= \boxed{620}\ \text{minutes}
 \end{align*}</cmath></center>
+==Solution 2==
+Let the total time that Jennifer and Rudolph bike 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>. Because they both reached the fifty mile mark at the same time, we don't count the rest afterward so 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 (d=50), 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}(t-24\cdot 5),</cmath>yielding <math>t=\boxed{620}</math>.
 == 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
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