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:

Let the speed of the faster cyclist be $v_1$ and the slower cyclist be $v_2$.

The two cyclists are $k$ kilometers apart initially.

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When traveling in the same direction, the faster cyclist catches up to the slower one. The relative speed is the difference of their speeds.

Distance $=$ Relative Speed $\times$ Time

$k = (v_1 - v_2) \times r$

$v_1 - v_2 = \frac{k}{r}$ ... (Equation 1)

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When traveling in opposite directions, they move towards each other. The relative speed is the sum of their speeds.

Distance $=$ Relative Speed $\times$ Time

$k = (v_1 + v_2) \times t$

$v_1 + v_2 = \frac{k}{t}$ ... (Equation 2)

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Adding Equation 1 and Equation 2:

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

$2v_1 = k\left(\frac{1}{r} + \frac{1}{t}\right)$

$2v_1 = k\left(\frac{t + r}{rt}\right)$

$v_1 = \frac{k(r + t)}{2rt}$

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Subtracting Equation 1 from Equation 2:

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

$2v_2 = k\left(\frac{1}{t} - \frac{1}{r}\right)$

$2v_2 = k\left(\frac{r - t}{rt}\right)$

$v_2 = \frac{k(r - t)}{2rt}$

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The ratio of the speed of the faster cyclist to that of the slower:

$\frac{v_1}{v_2} = \frac{\frac{k(r + t)}{2rt}}{\frac{k(r - t)}{2rt}}$

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

Therefore, the ratio is $\dfrac{r+t}{r-t}$