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Q: Algebra problem with several variables ( Answered,   0 Comments )
Subject: Algebra problem with several variables
Category: Science > Math
Asked by: downtheline-ga
List Price: $20.00
Posted: 14 Dec 2003 19:23 PST
Expires: 13 Jan 2004 19:23 PST
Question ID: 287206
I have a math problem. I've often taken the United to Kona, Hawaii
flight and they play a game called halfway to hawaii. The pilots give
you the following data: total miles to Kona from SF, average speed of
each half of the flight, headwind/tailwind for each half (which you simply
add/subtract from speed), exact takeoff and estimated landing time. What is the
algebraic formula to solve this problem?? The last time I took the
trip, the following facts resulted in a the halfway point at 10:03 AM
Hawaii time (I couldn't solve for it even with the facts!): 7:30 AM
takeoff (HI time), 5 hrs 15 minutes total flight time, 485 MPH for
first half/462 for second, 33 and 45 MPH head wind (so subtract) for
1st/2nd half. Thanks for helping as it's been driving me crazy that I
can't figure it out.
Subject: Re: Algebra problem with several variables
Answered By: endo-ga on 14 Dec 2003 20:05 PST

Thank you for your interesting question! I'd love to fly to Hawaii and
have fun answering maths questions at the same time :)

Anyways here is the calculation, which should allow you to answer what
you're looking for.

First you need to calculate the total distance covered. Lets call this
distance 'd'.

We know that the time needed to travel half of d at 452 miles per hour
(485-33) added to the time needed to travel half of d at 417 miles per
hour (462-45) totals 5.25 hours

Since time = distance / speed we have:

d x ( 1/(2 x 452) + 1/(2 x 417)) = 5.25 hours
d x ( 1/904 + 1/834) = 5.25 hours
d x (834+904)/(834x904) = 5.25 hours
d = 5.25 x 834 x 904 / (834+904) = 2277.42 miles

Half way is thus: 2277.42/2 = 1138.71 miles

Time needed to fly 1138.71 miles at 452 miles per hour:

1138.71 / 252 = 2.519 hours = 2 hours 31 minutes 9 seconds

Thus if we add this to the takeoff time of 7:30 we get 10:01. Give or
take 30 seconds from each (since the times are not given to the
second) and take into account that the speeds are approximate, then
you have your answer.

I hope this answers your question. If something is unclear of if you
need any clarifications, please do not hesitate to ask.


Request for Answer Clarification by downtheline-ga on 14 Dec 2003 21:11 PST

Hmmm. thanks. I have a couple of points of clarification. Looking at
my notes from the info we were given, the pilot also told us that the
nautical miles from SFO to Kona were 2097 (not the 2277 that you
calculated). He also mentioned that he was proving some "extraneous"
(not needed) info. Could his total miles number be a ruse?
also, I'm sure I'm just being dense but what is the significance of
the 252 in your calculation to get to 2.519 hours?
lastly, if time=distance/speed are you simply making the algebraic
leap (my math memory is fuzzy) to get to d X speed by "multiplying the
inverse" or 1/(2X452)+1/(2x417)?
thanks again. 

Clarification of Answer by endo-ga on 14 Dec 2003 21:55 PST

I think the value given by the pilot is the shortest distance between
SFO and Kona rather than the flight path the plane takes and the
distance it travels. So indeed I suppose it's a ruse. Secondly it
would be too easy otherwise. You would just do:

distance/(2 x speed) =  2097/(2x452) = 2.31 hours = 2 hours 19 minutes

Which is wrong.

Sorry the 252 is a mistake (typo), it's supposed to be 452, the speed
during the first half, 485-33 = 452. Sorry for the confusion.

Sorry about that last part. I'll rewrite that to make it clearer.

So we have time = distance / speed

distance travelled = d/2 + d/2
time needed to travel first half: d/(2x452) 
time needed to travel second half: d/(2x417)

time needed in total: d/(2x452) + d/(2x417) = d x ( 1/(2 x 452) + 1/(2 x 417)) 
(factor out d)

Then the rest is the same.

I hope this is clearer.

If you require any further calculations please do not hesitate to ask.

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