1952 AHSME Problems/Problem 48

Problem

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:

$\text{(A) } \frac {r + t}{r - t} \qquad \text{(B) } \frac {r}{r - t} \qquad \text{(C) } \frac {r + t}{r} \qquad \text{(D) } \frac{r}{t}\qquad \text{(E) } \frac{r+k}{t-k}$

Solution

[asy] 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); [/asy]

Solution 2

- I did this because this problem does not have any solution explained to us, rather just a diagram. This is not the official solution, it's just the way I solved it. Enjoy! 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

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 47
Followed by
Problem 49
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