Question

Two cyclists, ${}^{\prime }{k}^{\prime }km$ apart, and starting at the same time, would be together in $r$ hours if they travelled in the same direction, 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

• A. $\frac{r+t}{r-t}$
• B. $\frac{r}{r-t}$
• C. $\frac{r+t}{r}$
• D. $\frac{r}{t}$

Answer 1

Answer: (A)

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

Answer is A.
Let the speed of the faster and slower cyclist be u and v respectively.
Distance between both the cyclist = $k$ km.
Speed, when they travel in the same direction$=u-v$.
Time taken to cover $k$ km $=r$ hours.
As, $Distance=Speed\times time$
$k=(u-v)\times r-eqn1$.
When they travel in the opposite direction $=u+v$.
Time taken to cover $k$km$=t$hours.
$Distance=Speed\times Time$, so, $k=(u+v)\times t-eqn2$.
From equation $1$ and equation $2$,
$(u-v)\times r=(u+v)\times t$.
That is,
$\frac{u+v}{u-v}=\frac{r}{t}$
$\Rightarrow \frac{u+v+u-v}{u+v-u+v}=\frac{r+t}{r-t}$,
$\Rightarrow \frac{2u}{2v}=\frac{r+t}{r-t}$,
$\Rightarrow \frac{u}{v}=\frac{r+t}{r-t}$.
Hence, The ratio of the speed of the faster cyclist to that of the slower is $\frac{r+t}{r-t}$.