91Ó°ÊÓ

The distance covered by Person J $x=100mi$

On her way to Grandmother's house, she drives half the distance with a speed of $40mph$and half the distance with a speed of $60mph$.

On her return journey, she drives half the time at $40mph$and half the time at $60mph$.

The total distance traveled by Person J on the way to Grandmother's house is $100mi$.

Half the distance is $\frac{100}{2}=50mi$.

The total time taken to cover$100mi$will be as follows.

$$\begin{gathered} t=\left(\frac{50}{40}\right)+\left(\frac{50}{60}\right) \\ t=2.083hours \end{gathered}$$

The average speed is given by

Average speed $=\frac{Totaldis\tan cetraveled}{Totaltimetaken}$

$$\begin{gathered} v_{avg}=\frac{100}{2.083} \\ {v}_{avg}=48mph \end{gathered}$$

Therefore, the average speed of Person J is $48mph$.

The total distance traveled by Person J on the return trip is $100mi$.

The total time t hour required to cover this distance is calculated as follows.

$$\begin{gathered} x=40\left(\frac{1}{2}t\right)+60\left(\frac{1}{2}t\right) \\ 100=40\left(\frac{1}{2}t\right)+60\left(\frac{1}{2}t\right) \\ 100=20t+30t \\ 100=50t \\ t=2hours \end{gathered}$$

The average speed of Person J on her return trip is

$$\begin{gathered} v_{avg}=\frac{100}{2} \\ {v}_{avg}=50mph \end{gathered}$$

Therefore, the average speed of Person J on her return trip is $50mph$.