Question

with 17 and 18

ALGEBRA1-Alternate Exam 12-continued Student's Name Student Number 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. 17. 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. 18. Compare how many gallons of fuel they used for the round trip to the expected usage in ideal (non-windy) travel conditions. Which uses more? By how many gallons? Round to the nearest hundredth, if necessary.

Answer 1

17. Let the fuel economy of the vehicle in ideal (non-windy) conditions be f miles per gallon and let the influence of the wind be w miles per gallon.

Then f-w = 18…(1) and f+w = 26…(2).

On adding the 2 equations, we get 2f = 18+26 = 44 so that f = 44/2 = 22.

Now, on substituting f = 22 in the 2^nd equation, we get 22+w = 26 so that w = 26-22 = 4.

Thus, the fuel economy of the vehicle in ideal (non-windy) conditions is expected to be 22 miles per gallon and the influence of the wind has an impact of 4 miles per gallon.

18. On the trip towards their desination, the vehicle consumed 1287/18 = 71.5 gallons of fuel and on the return trip, the vehicle consumed 1287/26 = 49.5 gallons of fuel. Therefore, during the return trip, the vehicle consumed 71.5+49.5 = 121 gallons of fuel.

In ideal (non-windy) conditions, the vehicle would consume 2*1287/22 = 117 gallons of fuel.

Thus, in windy conditions, the vehicle would consume 4 more gallons of fuel.