2008 AIME II Problems/Problem 2

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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

$\begin{align*}50 \cdot \frac{3}{4(t-245)} &= 50 \cdot \frac{1}{t-120}\\\frac{1}{3}t &= \frac{245 \cdot 4}{3} - 120\\t...$

2008 AIME II (Problems • 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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2008 AIME II Problems/Problem 2

From AoPSWiki

Problem

Solution

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