Answer:
a) on the way there, the speed = 50 m/hr
b) on the way back home, speed = 30 m/hr
Step-by-step explanation:

on the way to Disney land:
distance = d
speed = s
time = t = 6 hours
d = s × t
d = s × 6
d = 6s - - - - - (1)
On the way back:
speed decreased by 20m/hr
This means that s₂ = (s - 20)m/hr
time = t₂ = 10 hours
∴ d = s₂ × t₂
d = (s - 20) × 10
d = 10s - 200 - - - - (2)
<em>Note that the distance is the same to and from disney land, therefore, equation (1) equals equation (2)</em>
∴ 6s = 10s - 200
<em>collecting like terms:</em>
10s - 6s = 200
4s = 200
∴ s = 200 ÷ 4
s = 50 m/hr
Therefore on the way there, the car was moving at a apeed of 50m/hr
speed on the way back home (s₂) = (s - 20) m/hr
∴ s₂ = 50 - 20
s₂ = 30 m/hr
Therefore on the way back home the speed was 30 m/hr