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schumacher
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I am getting 2.1 hrs.

I however took the speed as (600 - 40 = 560).

What's the OA ?
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2 Hours

Speed = distance / time

time = distance / speed

relative velocity from SF -> Hawaii = 640 kmph
relative velocity from Hawaii -> SF = 560 kmph

Consider that the jet is x km from Hawaii when it has to return

at this point in time, the jet is 2400 - x km from SF

Equating the time:

(2400 - x) / 640 = x / 560

Solving for x, we get x = 1120 which implies, the jet is 1280 from SF. At 640 kmph, it would take the jet 2 hours to cover this distance.
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maaverick
2 Hours

Speed = distance / time

time = distance / speed

relative velocity from SF -> Hawaii = 640 kmph
relative velocity from Hawaii -> SF = 560 kmph

Consider that the jet is x km from Hawaii when it has to return

at this point in time, the jet is 2400 - x km from SF

Equating the time:

(2400 - x) / 640 = x / 560

Solving for x, we get x = 1120 which implies, the jet is 1280 from SF. At 640 kmph, it would take the jet 2 hours to cover this distance.


Exactly the same approach I used. The only difference was I used the 1120 as the distance already traveled and reasoned if you had to fly back you'd go at 560. So the time would be D/R=1120/560=28/14=2 hours. The same basic idea but this way saves subtracting 1120 from 2400.
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Quote:
2 Hours

Speed = distance / time

time = distance / speed

relative velocity from SF -> Hawaii = 640 kmph
relative velocity from Hawaii -> SF = 560 kmph

Consider that the jet is x km from Hawaii when it has to return

at this point in time, the jet is 2400 - x km from SF

Equating the time:

(2400 - x) / 640 = x / 560

Solving for x, we get x = 1120 which implies, the jet is 1280 from SF. At 640 kmph, it would take the jet 2 hours to cover this distance.



You are almost right. The question asks for the time already traveled, not the time that would take to complete the journey (either to Hawaii or to San Francisco)!!

Imagine that the plane had travelled t hours from Hawaii. It would have covered 640t miles (you are trying to determine at this point whether to go back or continue with the journey). Remaining distance to cover is 2400-640t miles. If you were to proceed, time it would take would be (2400-640t)/640. If you were to go back, the time that would take is 640t/560

Solving for t, we get 1.75 hours. This is the time from the take off.
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I agree with maaverick's answer. His wording in the last sentence is a bit confusing, but I think his point is if the plane leaves Hawaii 2 hours, it would be faster for the pilot to fly directly to SF than returning to Hawaii.
He does not mean the remaining time to reach SF when the plane is in the air is 2 hours.

Very nice approach from Maaverick. Correct me if I misinterpret your solution, Mavverick.
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Can someone please explain to me what exactly tailwind means, I am unsure on how r we saying that:

relative velocity from SF -> Hawaii = 640 kmph
relative velocity from Hawaii -> SF = 560 kmph
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Tailwind here means the wind blows with the same direction when the plan travels from Hawaii to SF. It will make the plane faster in the direction from Hawaii to SF, but slower in the other way around.
This is my interpretation, and pls correct me if I am wrong.



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