Question

The difference of the speed of the faster and slower car is $20Kmh{r}^{-1}$. If slower car is takes $1$ hour more than the faster car to travel a distance of $400Km$, find the speed of both the cars.

Answer 1

Let speed fast car be '$x$' km/hr

and slow car be '$y$' km/hr

given $(x-y)=20km/hr$

$\Rightarrow x=(20+y)km/hr$

time taken by $x$ to travel $400km=\frac{400}{x}$

time taken by $y$ to travel $400km=\frac{400}{y}$

given $\left(\frac{400}{y}\right)=\frac{400}{x}+1$

$\left[\frac{400}{y}\right]=\frac{400}{(20+y)}+1$

$1=400[\frac{-1}{(20+y)}+\frac{1}{y}]$

$\Rightarrow \frac{1}{400}=\frac{(20+y-y)}{y(20+y)}$

$\Rightarrow \left(20\right)\left(400\right)=y(20+y)$

$\Rightarrow 80\times 100=y(20+y)$

$\Rightarrow y=80km/hr$ and $x=20+y=100km/hr$

$\therefore$ speed of both cars are $80km/hr$ and $100km/hr$