let

d = 300 km the distance

v1 = the speed of Steve

t1 = the speed of Steve

v2 = the speed of Mohammad

t2 = the speed of Mohammad

Steve rides his Harley 20 km per hour faster than Mohammad rides his Yamaha

v1 = 20 + v2

Steve travels from Cranbrook to Medicine Hat in one and one forth hours less time than Mohammad

t1 = t2 - 1 1/4

t1 = t2 - 1.25

since the average rate v = d/t => d = v*t

v1*t1 = d

v2*t2 = d

if we plug v1 and t2 in the above equations we get

(20 + v2)(t2 - 1.25) = 300

v2*t2 = 300

by solving the above system we find and consider only the positive solutions

(click here to see the solutions

https://quickmath.com/webMathematica3/quickmath/equations/solve/advanced.jsp#c=solve_advancedsolveequations&v1=(20+%2B+v2)(t2+-+1.25)+%3D+300%0Av2*t2+%3D+300&v2=v2%0At2 )

v2 = 60

v1 = 20 + 60 = 80 km/h

the average speed of Steve is 80 km/h.