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

$\frac{r+t}{r-t}$

The correct answer is Option (4) → $\frac{r+t}{r-t}$

Let the speeds of the cyclists be $v_1$​ (faster) and $v_2$​ (slower).

• When they travel in the same direction, they meet in r hours:

$v_1 - v_2 = \frac{k}{r}$​

• When they travel in opposite directions, they meet in t hours:

$v_1 + v_2 = \frac{k}{t}$​

Now,

$\frac{v_1}{v_2} = \frac{(v_1+v_2)+(v_1-v_2)}{(v_1+v_2)-(v_1-v_2)} = \frac{\frac{k}{t}+\frac{k}{r}}{\frac{k}{t}-\frac{k}{r}}$

Cancel k:

$= \frac{\frac{1}{t}+\frac{1}{r}}{\frac{1}{t}-\frac{1}{r}} = \frac{r+t}{r-t}$