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:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{r+t}{r-t}\)
Moving in the same direction, the gap of \(k\) km closes at the difference of the speeds; moving towards each other, it closes at the sum of the speeds. Time = distance ÷ speed gives two equations.
Faster speed \(=a\), slower speed \(=b\)
Same direction, together in \(r\) hours: \(a-b=\frac{k}{r}\) (1)
Opposite directions, together in \(t\) hours: \(a+b=\frac{k}{t}\) (2)
Adding (1) and (2), the \(b\) terms cancel: \(\displaystyle 2a=\frac{k}{t}+\frac{k}{r}=\frac{k(r+t)}{rt}\) (3)
Subtracting (1) from (2), the \(a\) terms cancel: \(\displaystyle 2b=\frac{k}{t}-\frac{k}{r}=\frac{k(r-t)}{rt}\) (4)
Dividing (3) by (4), \(2\), \(k\) and \(rt\) cancel: \(\displaystyle \frac{a}{b}=\frac{r+t}{r-t}\)