Points A and B are 70 …

Suppose, P and Q be the cars starting from A and B respectively and let their speeds be x km/hr and y km/hr respectively.
Case I:
When the cars P and Q move in the same direction.
Distance covered by the car P in 7 hours = 7x km
Distance covered by the car Q in 7 hours = 7y km
Let the cars meet at point M.

{tex}\therefore{/tex}{tex}AM = 7x\ km\ \ and\ \ BM = 7y\ km{/tex}
{tex}\therefore{/tex}{tex}AM - BM = AB{/tex}
{tex}\Rightarrow{/tex}{tex}7x - 7y = 70{/tex}
{tex}\Rightarrow{/tex}{tex}7(x - y) = 70{/tex}
{tex}\Rightarrow{/tex}{tex} x - y = 10{/tex}..........(i)
Case II:
When the cars P and Q move in the opposite directions.
Distance covered by the car P in 1 hour = x km
Distance covered by the car Q in 1 hour = y km
In this case, let the cars meet at the point N

{tex}\therefore{/tex} {tex}AN = x\ km\ \ and\ \ BN = y\ km{/tex}
{tex}\therefore{/tex} {tex}AN + BN = AB{/tex}
{tex}\Rightarrow{/tex}{tex}x + y = 70{/tex}........(ii)
Adding (i) and (ii), we get
{tex}2x = 80{/tex}
{tex}\Rightarrow{/tex}{tex}x = 40{/tex}
Putting {tex}x = 40{/tex} in (i), we get
{tex}40 - y = 10​​​​​​​{/tex}
{tex}\Rightarrow{/tex} {tex}y =(40 - 10) = 30{/tex}
{tex}\therefore{/tex} {tex}x = 40, y = 30{/tex}
Speed of car starting from point A = {tex}40km/hr.{/tex}
Speed of car Starting from point B = {tex}30km/hr.{/tex}