**More on fuel efficiency improvements and reductions in gas consumption**.

What is the amount of fuel
saved per mile driven when the fuel efficiency of a car is improved from 20 mpg
to 25 mpg?

How much does one need to
improve the fuel efficiency of a 30-mpg vehicle in order to achieve the same
reduction in fuel consumption?

**Answer**: We know that gas consumption is equal to miles driven divided by mpg.

Gallons = Miles / (Miles/Gallon)

So the gas saved per mile
driven from an improvement in fuel efficiency from 20 to 25 miles per gallon is

Reduction in Gas Consumption
= (1/20) – (1/25) or 0.05 -0.04 or 0.01 gallon.

Now assume the initial fuel
consumption for the car is 30 mpg and the reduction in gas consumption from a
one-mile trip is 0.01 gallons

Reduced Gas Consumption =
0.01 = (1/30) – (1/NFE)

Where NFE is the new fuel
efficiency.

1/NFE = (1/30) – 0.01 =
0.023333

Take the reciprocal of both
sides and get

NFE = 42.857.

We can check our answer by
observing that

(1/30) – (1/42.852) is equal to
0.01.

0.033333 – 0.023333 is equal
to 0.01.

**A Note:**This post shows that a relatively small improvement in fuel efficiency from a very fuel inefficient car can result in large reductions in gasoline. However, returns from improvement in fuel efficiency will eventually diminish. Consider a 1000-mpg car. The amount of gas needed for this car to travel 1 mile is 1/1000 or 0.001 gallon. An increase in the fuel efficiency of this car to 2000 mpg would only result in a 0.0005 reduction in gas consumed for each mile traveled.

Note also we are a long way
from 1000 mpg.

This issue was previously
looked at more informally in the post below.

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