2008 AIME II Problems/Problem 2

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

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

$\begin{align*}50 \cdot \frac{3}{4(t-245)} &= 50 \cdot \frac{1}{t-120}\\ \frac{1}{3}t &= \frac{245 \cdot 4}{3} - 120\\ t &= \boxed{620}\ \text{minutes} \end{align*}$

2008 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2008 AIME II Problems/Problem 2

Contents

Problem

Simple Solution

Solution

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