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

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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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
Length of Road = 1287 mile.
Let the fuel economy without wind =+x
Let the influence of wind on fuel economy = w
so while going fuel economy = x-w
While returning the fuel economy = x%2Bw
so x-w+=+18
&+x%2Bw+=+26
2x+=+44
x+=+22
plugging the value of x we get w+=+4
Fuel used in forward journey = 1287%2F18+=+71.5 gallons
Fuel used in return journey =+1287%2F%2826-4%29=1287%2F22+=+58.5 gallons
Total fuel used in round trip =+130 gallons.
Total fuel used in non windy condition = 1287%2A2%2F22+=+117 gallons
So the fuel used is more in windy conditions by 13 gallons.