Example

Fuel Consumption and Car Weight: An Inverse Variation Application

Problem: A car's fuel consumption (mpg) varies inversely with its weight. A car that weighs 3100 pounds gets 26 mpg on the highway. (a) Write the equation of variation. (b) What would be the fuel consumption of a car that weighs 4030 pounds?

Part (a) — Find the equation.

Let ff = fuel consumption and ww = weight. Because ff varies inversely with ww, write the inverse variation formula using these variables:

f=kwf = \frac{k}{w}

Substitute the known values f=26f = 26 and w=3100w = 3100:

26=k310026 = \frac{k}{3100}

Solve for the constant of variation by multiplying both sides by 3100:

310026=k3100 \cdot 26 = k

k=80,600k = 80{,}600

Substitute kk back into the formula:

f=80,600wf = \frac{80{,}600}{w}

Part (b) — Find ff when w=4030w = 4030.

Substitute w=4030w = 4030 into the equation:

f=80,6004030=20f = \frac{80{,}600}{4030} = 20

A car that weighs 4030 pounds would have a fuel consumption of 20 mpg. The heavier car gets fewer miles per gallon, which is consistent with the inverse relationship between weight and fuel efficiency.

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Updated 2026-04-21

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