The Solution:
Given the initial values as below:
[tex]\begin{gathered} \text{Let the original speed/ rate be x} \\ \text{Time(t)}=11\text{ hours} \\ So,\text{ we have the distance(d) to be:} \\ d=x\times11=11x\ldots eqn(1) \end{gathered}[/tex]
The average speed was increased by 18m/h and a return time of 8 hours only.
[tex]\begin{gathered} \text{ speed(s)=(x+18) m/h} \\ \text{ Time(t)=8 hrs} \\ So, \\ d=(x+18)(8)=8x+144\ldots eqn(2) \end{gathered}[/tex]
Equating both eqn(1) and eqn(2), we get
[tex]\begin{gathered} 11x=8x+144 \\ 11x-8x=144 \\ 3x=144 \\ \text{Dividing both sides by 3} \\ \frac{3x}{3}=\frac{144}{3} \\ \\ x=48\text{ m/h} \end{gathered}[/tex]
The distance is
[tex]d=11x=11\times48=528m[/tex]
Thus, the average speed on his return drive is:
[tex]\text{ Average speed=x}+18=48+18=66\text{ m/h}[/tex]
Therefore, the correct answer is 66 m/h
Original values:
x=48m/h
t=11 hrs
d=528m
Return Drive:
x=66m/h
t=8 hrs
d=528m
