Let the distance ==D
Let Oscar's speed ==S
Let the usual time of arrival==T
D / [S + 10] ==T - 18 / 60,
D / [S + 20] ==T - 1/2,
D / S ==T, solve for D, S, T
S ==40 MPH - their initial speed
T ==3/2 ==1.5 hours - the usual time of arrival
D ==60 miles - distance between their house and the conference venue.