Average speed on his return drive = 60 mph
Average Speed on his Original drive = 40 mph
Distance on his original drive = 120m
Distance on his return drive = 120m
Step-by-step explanation
We need to first complete the table.
Rate X Time = Distance
Rate Time Distance
Original drive x 3 d
Return drive x+20 2 d
But we know that: Rate X Time = Distance
x X 3 = d
⇒ 3x = d ---------------------------------------(1)
Similarly, (x+20) X 2 = d
⇒ 2(x+20) = d ---------------------------------------(2)
Substitute equation (1) in (2)
3x = 2(x + 20)
Open the parenthesis
3x = 2x + 40
Collect ;like term.
3x - 2x = 40
x = 40
Hence, the average speed on his return drive = x + 20 = 40 + 20 = 60 mph
Average Speed on his Original drive = x = 40 mph
Distance
The distance are all equal, both on his original drive and on the reurn drive.
So, let's solve for d
3x = d
Substitute x = 40 into the above.
3(40) = d
d = 120m
Hence, distance on his original drive = 120m
Distance on his return drive = 120m