AIME 2008 II · 第 2 题

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,

$$\begin{aligned}r &= \frac{50}{t-245}\\ \frac{3}{4}r &= \frac{50}{t-120} \end{aligned}$$ Substituting equation $(1)$ into equation $(2)$ and simplifying, we find

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

Solution 2

Let the total time that Jennifer and Rudolph bike, including rests, to be $t$ minutes. Furthermore, let Rudolph's biking rate be $r$ so Jennifer's biking rate is $\frac{3}{4}r$. Note that 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, 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

$$r(t-49\cdot 5)=\frac{3}{4}r(t-24\cdot 5),$$ yielding $t=\boxed{620}$.