1952 AMC 12 Problem 48

Question

Two cyclists,

k

miles apart, and starting at the same time, would be together in

r

hours if they traveled in the same direction, but would pass each other in

t

hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is:

\dfrac{r+t}{r-t}

\dfrac r{r-t}

\dfrac{r+t}{r}

\dfrac rt

\dfrac{r+k}{t-k}

Let the faster and slower speeds be

u

and

v

, and write the same-direction and opposite-direction relative-speed equations

Use

(u-v)r=k

and

(u+v)t=k

, then solve for

\frac{u}{v}

Let the speeds be

u\gt v.

The two meeting conditions give

u-v=\frac kr,\qquad u+v=\frac kt.

Adding and subtracting these equations yields

\begin{aligned} u&=\frac k2\left(\frac1r+\frac1t\right),\\ v&=\frac k2\left(\frac1t-\frac1r\right). \end{aligned}

Therefore

\frac uv=\frac{r+t}{r-t}.

Thus, the correct answer is A.

Problem 48 in other years