Question

A pedestrian and a cyclist start simultaneously towards each other from towns $A$ and $B$ which are $40$ km apart and meet two hours after the start. Then they resumed their trips and the cyclist arrives at $7$ h $30$ mm earlier than the pedestrian arrives at $B$. Find the speeds of the pedestrian and the cyclist.

Answer 1

let speeds of pedestrian and cyclist be $x,y$

after 2 hrs they covered $2x$ and $2y$ distances respectively.

total distance is $40$ so the sum of the distances covered by them $=40$

$⟹x+y=20⟹y=20-x$

cyclist has arrived $7.5h$ earlier

$⟹\frac{40}{20-x}=7.5+\frac{40}{x}$

$⟹3x^2-8x-160=0$

$⟹x=8.75$