SOLUTION: I cannot come up with what formula to use. The problem is: On a 1287 mile long road trip in windy conditions, a family calculates their fuel economy to be 18 miles per gallon ove

Question 935193: I cannot come up with what formula to use. The problem is: On a 1287 mile long road trip in windy conditions, a family calculates their fuel economy to be 18 miles per gallon over the 2 days it takes to reach their destination. On their 2 day return trip, with a similar wind at their backs, their calculations show a fuel economy of 26 miles per gallon. Calculate the expected fuel economy of their vehicle in ideal (non-windy) conditions as well as the influence of the wind (w) in miles per gallon.

Found 2 solutions by Alan3354, MathLover1:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
This is not a reasonable problem.
It's easy to find the average, (18 + 26)/2 = 22 mi/gallon.
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But, wind resistance is a function on the square of the speed, and no speeds were stated, just 1287 miles in 2 days (a lot of glass time, imo).
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Whatever answer the teacher comes up with should be subjected to serious analysis and criticism.
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
Length of Road = mile.
Let the fuel economy without wind =
Let the influence of wind on fuel economy =
so while going fuel economy =
While returning the fuel economy =
so
&

plugging the value of we get
Fuel used in forward journey = gallons
Fuel used in return journey = gallons
Total fuel used in round trip = gallons.
Total fuel used in non windy condition = gallons
So the fuel used is more in windy conditions by gallons.

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