Math Help - % Increase

1. % Increase

I have made some modifications to my automobile to increase the fuel mileage. The EPA estimates my car's hwy fuel mileage (at 55 mph) to be about 39 mpg. In a recent road test at an average speed of 72.8 mph on the hwy (a 32% increase in velocity over the 55 mph the EPA tests at) I achieved 41.6 miles per gallon. It is estimated that the average car is 18% less efficient at that speed differential (72.8 mph - 55 mph Speed diff = 17.8 mph). With the fuel saving devices I installed, at a higher velocity, 17.8 mph faster, I achieved better fuel mileage than the EPA estimated 39 mpg at 55 mph. This is a 2.6 mpg increase over the EPA estimate while driving 72.8 mph (41.6 - 39 mpg = 2.6 mpg increase).

Considering the 18% loss in efficiency of an average car traveling 17.8 miles faster than 55 mph, can you help me solve for the actual mpg % increase I get now vs. the EPA mpg sticker equivalent? ( I can't drive 55!)

2. Re: % Increase

Originally Posted by Jamez
I have made some modifications to my automobile to increase the fuel mileage. The EPA estimates my car's hwy fuel mileage (at 55 mph) to be about 39 mpg. In a recent road test at an average speed of 72.8 mph on the hwy (a 32% increase in velocity over the 55 mph the EPA tests at) I achieved 41.6 miles per gallon. It is estimated that the average car is 18% less efficient at that speed differential (72.8 mph - 55 mph Speed diff = 17.8 mph). With the fuel saving devices I installed, at a higher velocity, 17.8 mph faster, I achieved better fuel mileage than the EPA estimated 39 mpg at 55 mph. This is a 2.6 mpg increase over the EPA estimate while driving 72.8 mph (41.6 - 39 mpg = 2.6 mpg increase).

Considering the 18% loss in efficiency of an average car traveling 17.8 miles faster than 55 mph, can you help me solve for the actual mpg % increase I get now vs. the EPA mpg sticker equivalent? ( I can't drive 55!)
The EPA says avg mpg @ 55 mph is 39 mpg.

At 72.8 mph there is a loss of 18% fuel efficiency, i.e. the estimated mpg is $(1-18\%)39=32$ mpg

Your test achieved 41.6 mpg at 72.8 mph

$\dfrac {41.6}{32}\approx 1.3$

So you achieved at 30% increase in fuel efficiency over the EPA rating.

3. Re: % Increase

Awesomeness! Thx Romsek. 1.3 x 100 -100 = 30%
Problem solved.