mirzohidjon wrote:

NewSc2 wrote:

Pinali wrote:

I don't see any reason why it can't apply here . The only difference here is we know the value of time and unknown is speed. while in the example you said time is unknown factor and we know the speed . I hope this helps.

btw is the answer ~ 9064 km approximately?

No, the answer is 700*11.33 = 7931. I went with the assumption that the plane would have spent *more* time flying the slower speed, so the actual distance would be slightly more than 7931, but this was wrong. I've edited my post to show the answer choices.

The only reasoning I can see is that weighted rates skew speed, and not time spent (e.g., if time spent walking to work was 6 hours, and walking home 4 hours -- average is 5 hours), but this question still relies on speed to calculate the distance.

Can any of experts comment on this problem. I have problem understanding the concept as well. This problem should be solved using the formula Average Distance= Speed * Time, since the time and speed are different for two trips (from A to b and from b to A because there was the speed of Earth which rotates one way), I am trying to use the formula Average Distance= (700+v)*Time, where v is the speed of the Earth's rotation as plane moves from A to B.

However, the OA solution to this problem is very straightforward, taking the average of time and multiplying it by 700.

Any help with this problem would be appreciated. Thank you.

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

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

\(Speed_{Average} = 700\)

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"

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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.

_________________

~fluke

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