Points A and B are 90 km apart from each other on a highway. A car starts from A and another from B at the same time. If they go in the same direction they meet in 9 hours and if they go in opposite directions they meet in $\frac{9}{7}$ hours. Then a speed of the car A is

Let X and Y be two cars starting from points A and B respectively. Let the speed of car X be x km/hr and that of car Y be y km/hr.

Case – I : 
When two cars move in the same directions : 
Suppose two cars meet at point Q. Then, 
                Distance travelled by car X = AQ,
                Distance travelled by car Y = BQ

It is given that two cars meet in 9 hours.
Distance travelled by car X in 9 hours = 9x km
$\Rightarrow \mathrm{AQ}=9\mathrm{x}$ 
Distance travelled by car y in 9 hours = 9y km
$\Rightarrow \mathrm{BQ}=9\mathrm{y}$

Clearly, $\mathrm{AQ}-\mathrm{BQ}=\mathrm{AB}$

$\Rightarrow 9\mathrm{x}-9\mathrm{y}=90[\because \mathrm{AB}=90\mathrm{km}]$

$\Rightarrow \mathrm{x}-\mathrm{y}=10....\left(\mathrm{i}\right)$

Case – II : 
When two cars move in opposite directions : 
Suppose two cars meet at point P. Then, 
                  Distance travelled by car X = AP,
                  Distance travelled by car Y = BP
In this case, two cars meet in  $\frac{9}{7}$ hours.
$\therefore$  Distance travelled by car X in  $\frac{9}{7}$ hours = $\frac{9}{7}$ x km
$\Rightarrow \mathrm{BP}=\frac{9}{7}\mathrm{y}$ 
Clearly, $\mathrm{AP}+\mathrm{BP}=\mathrm{AB}$
 $\begin{array}{l}\Rightarrow \frac{9}{7}\mathrm{x}+\frac{9}{7}\mathrm{y}=90\\ \Rightarrow \frac{9}{7}(\mathrm{x}+\mathrm{y})=90\\ \Rightarrow \mathrm{x}+\mathrm{y}=70\end{array}$
 
Solving equations (i) and (ii), we get  x = 40 and y = 30  
Hence, speed of car X is 40 km/hr and speed of car Y is 30 km/hr.