hardnstrong wrote:

Where does it say that total distance is 40

:

Distance is given as x % of total distance

we only have speed, which is 40m/ph and 60m/ph

Nowhere it's said that the total distance is 40 miles. I should have written this more clearly: distance \(d\) will cancel out from the equation (edited the earlier post to clear this). So we can assume distance to be some number. I chose 40 as it's easy for calculation.

The solution with \(d\):

\(Average \ speed=\frac{distance}{total \ time}\).

Francine traveled \(x\) percent of the total distance at an average speed of 40 miles per hour --> time needed for this part of the trip: \(t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*d}{40}=\frac{dx}{40*100}\);

Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*d}{60}=\frac{(100-x)d}{60*100}\);

\(Total \ time=t_1+t_2=\frac{dx}{40*100}+\frac{(100-x)d}{60*100}=\frac{d(x+200)}{12000}\);

\(Average \ speed=\frac{distance}{total \ time}=\frac{d}{\frac{d(x+200)}{12000}}=\frac{12000}{x+200}\).

Answer: E.

Hope it's clear.

can you expain step by step(with comments) how from adding up two TIMES you get this \(\frac{d(x+200)}{12000}\)

\(Total \ time=t_1+t_2=\frac{dx}{40*100}+\frac{(100-x)d}{60*100}=\frac{d(x+200)}{12000}\);