M07 Q24

\(Speed_{Average} = \frac{Total \hspace{2} Distance}{Total \hspace{2} Time}\)

\(Speed_{Average} = \frac{Total \hspace{2} Distance}{Time \hspace{2} Taken{A \hspace{2} to \hspace{2} B}+Time \hspace{2} Taken{B \hspace{2} to \hspace{2} A}}\)

\(Speed_{Average} = 700\)

Let the distance between A to B be D.
\(Total \hspace{2} Distance = D+D=2D\)

\(Time \hspace{2} Taken{A \hspace{2} to \hspace{2} B}=12.4\)

\(Time \hspace{2} Taken{B \hspace{2} to \hspace{2} A}=10.26\)

\(Total \hspace{2} Time = 12.4+10.26=22.66\)

\(700=\frac{2D}{22.66}\)

\(D=\frac{700*22.66}{2}=7931\)

Ans: "C"
****************************************

Coming to your question:
"Average Distance= (700+v)*Time". I am not aware of any such formula.
Moreover, distance between 2 points A and B is constant. What is average distance then, I wonder!!!

700= Average Speed of the plane both ways.
Earth's speed = \(\nu_{e}\)
Plane's speed = \(\nu_{p}\)

A to B: Speed= Plane's speed-Earth's speed=\(\nu_{p}-\nu_{e}\)
B to A: Speed= Plane's speed+Earth's speed=\(\nu_{p}+\nu_{e}\)

From A to B:
\(Time=\frac{Distance}{Speed}\)
\(12.4=\frac{D}{\nu_{p}-\nu_{e}}\)

From B to A:
\(Time=\frac{Distance}{Speed}\)
\(10.26=\frac{D}{\nu_{p}+\nu_{e}}\)

Since we don't know either the speed of plane or the speed of the earth, proceeding this way would prove futile.