Places A and B are 80 …

Suppose x and y be two cars starting from points A and B respectively.
Let the speed of the car X be x km/hr and that of the 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 = {tex}AQ{/tex}
Distance travelled by car Y = {tex}BQ{/tex}
Distance travelled by car X in 8 hours = 8x km.
{tex}AQ = 8x{/tex}
Distance travelled by car Yin 8 hours = 8y km.
{tex}BQ = 8y{/tex}
Clearly {tex}AQ - BQ = AB{/tex}
{tex}8x - 8y = 80{/tex}
{tex}\Rightarrow {/tex} {tex}x - y = 10{/tex} ...(i)
Case II: When two cars move in opposite direction
Suppose two cars meet at point P, then,
Distance travelled by X car X = AP
Distance travelled by Y car Y = BP
we can write it as 1 hour = {tex}\frac{{20}}{{60}}{/tex}hours.
Therefore, Distance travelled by a car Y in {tex}\frac{4}{3}{/tex}hrs = {tex}\frac{4}{3}{/tex}x km.
AP + BP = AB
{tex}\frac{4}{3}x{/tex} km + {tex}\frac{4}{3} y{/tex} km = 80
{tex}\frac{4}{3}{/tex}{tex}(x + y) = 80{/tex}
{tex}( x + y ) = 80{/tex} × {tex}\frac{3}{4}{/tex}
{tex}x + y = 60{/tex} ...(ii)

By solving equations (i) and (ii), we get, {tex}x = 35.{/tex}
By substituting x = 35 in equation (ii), we get
{tex}x + y = 60{/tex}
{tex}35 + y = 60{/tex}
{tex}y = 60 - 35{/tex}
{tex}y = 25.{/tex}
Hence, speed of car X {tex}= 35\ km/hr.{/tex}
Speed of car Y ={tex} 25\ km/hr.{/tex}