1952 AHSME Problems/Problem 48

Problem

Two cyclists, miles apart, and starting at the same time, would be together in hours if they traveled in the same direction, but would pass each other in hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is:

Solution

pair A,B,C;
A=(0,0); B=(8,0); C=(4,1);
draw((A)--(B));
label("$A$",A,S); label("$B$",B,SE); label("$k$",C,SW);

Solution 2

We have a simple formula $d = rt$. The times the problem gives us, $r$ and $t$ are kind of annoying, so I will just let $x$ and $y$ be $r$ and $t$, respectively. I will also let $r_{1}$ and $r_{2}$ being the rate of the riders, where $r_{1}>r_{2}$. We can then write our equations based off these numbers, so I got $k = (r_1-r_2)x$ and $k = (r_1+r_2)y$. Rearranging the equations and solving for $r_1$ and $r_2$, I got $r_1 = k/x + r_2$ (from first equation) and $r_1 = k/y - r_2$ (from second equation). Also, we have $r_2 = k/y - r_1$ and $r_2 = -(k/x) + r_1$. Adding the first two together and the last two together, I got $2r_1 = \frac{(xk+yk)}{xy}$ and $2r_2 = \frac{(xk - yk)}{xy}$. Finally, dividing $2r_1$ by $2r_2$ gives us $r_1/r_2 = \frac{(x+y)}{(x-y)}$. Substituting $x$ and $y$ for $r$ and $t$, I got $\boxed{\text{(A) } \frac {r + t}{r - t}}$ -DragonFish12345 with credits to @RJ5303707 for LATEX
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See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.