Let the second racer takes ‘x’ hr with speed S₂
and the first racer takes [x – (5/6)] hr with speed S₁
Given, Total distance = 50 km
Then, S₁ = 50/[x – (5/6)]
S₂ = 50/x
Total speed = S₁ + S₂
= {50/[x – (5/6)]} + [50/x]
= 50 {1/[x – (5/6)] + [1/x]}
= 50 {x + [x – (5/6)] / [x – (5/6)]x}
= 50 {[2x – (5/6)] / [x² – (5x/6)]}
= 50 {[(12x – 5)/6] / [(6x² – 5x)/6]}
= 50 {[12x – 5] / [6x² – 5x]}

As they cross each other in 1hr,
Now, Time = Distance/ Total Speed
—> 1 = 50 / (50 {[12x – 5] / [6x² – 5x]})
—> 1 = 1 / {[12x – 5] / [6x² – 5x]}
—> 1 = [6x² – 5x] / [12x – 5]
—> [12x – 5] = [6x² – 5x]
—> 6x² – 5x – 12x + 5 = 0
—> 6x² – 17x + 5 = 0
—> 6x² – 15x – 2x + 5 = 0
—> 3x(2x – 5) – 1(2x – 5) = 0
—> (2x – 5) (3x – 1) = 0
—> x = 5/2, 1/3

[Neglecting x = 1/3]
Now, We have to find the speed of the slower racer ie., second racer.
Put x = 5/2 in S₂, we get
S₂ = 50/(5/2)
= (50 * 2) / 5
= 20 km/hr
Thus, the speed of the slower racer = 20 km/hr.