Unit Conversions (Part 2):

Perform Unit Conversions for Fuel Economy

Another application of unit conversions is to calculate fuel economy. Fuel economy of vehicles compares the distance the vehicle can go compared to the amount of fuel it uses or the cost of the fuel. In this next video, we use unit conversion to determine the cost of a delivery. The following video will explain how to solve this example:

Video Source (08:53 mins) | Transcript

Even though our conversion factors look different, the process is the same. Before starting the problem, scan the question to find the conversion rates it gives you, then complete the steps below:

  1. Start with what you know
  2. Determine what you want to get in the end. (Figure out what the end units should be.)
  3. Determine what conversion factor(s) to use. You may sometimes need more than one.
  4. Multiply by 1 in the form of the conversion factor that cancels out the unwanted units.

Additional Resources

Practice Problems

  1. You are planning to go on a trip with some friends. Your vehicle gets 25 miles per gallon (mpg), and you want to drive from Los Angeles to San Francisco. San Francisco is 382 miles away. The current gas price is $3.75 per gallon. How much will the gas for your trip cost, in US Dollars? Round to the nearest penny (hundredth).
    25 mi = 1 gal
    $3.75 = 1 gal

  2. How much does it cost to drive 23 miles using a truck that takes 10 gallons to go 120 miles, if the current cost of gas is $3.55? Round to the nearest penny (hundredth).
    10 gal = 120 mi
    $3.55 = 1 gal

  3. Pam lives in the mountains far away from any power source. She uses an oil lamp to light her cabin at night. She goes through 3 ounces of oil each night. The current cost of oil is $9 per gallon. How much does it cost Pam to light her cabin each night in US Dollars? Use the following information to find the answer. Round to the nearest penny (hundredth).
    $9 = 1 gal
    3 oz = 1 night
    1 gal = 128 oz

  4. A truck has a fuel efficiency of 9 milliliters per 100 meters. Find the truck’s fuel efficiency in liters per kilometers using the following information.
    1 L = 1000 ml
    1 km = 1000 m

  5. How much does it cost to drive 45 kilometers using a car that takes 5 liters of petrol (or gasoline) to go 100 kilometers? The current cost of petrol is €1.50.
    Round to the nearest hundredth.
    5 L = 100 km
    €1.50 = 1 L

Solutions

  1. \(\frac{382\:\text{mi}}{1}\times\frac{1\:\text{gal}}{25\:\text{mi}}\times\frac{\$3.75}{1\:\text{gal}} = \$57.30\) (Written Solution)

  2. \(\frac{23\:\text{mi}} {1}\times\frac{10\:\text{gal}}{120\:\text{mi}}\times\frac{\$3.55}{1\:\text{gal}} = \$6.80\)

  3. \(\frac{3\:\text{oz}}{1\:\text{night}}\times\frac{1\:\text{gal}}{128\:\text{oz}}\times\frac{\$9}{1\:\text{gal}}= \$0.21\) per night, or \(\frac{\$0.21}{1\:\text{night}}\) (Written Solution)

  4. \(\frac{9\:\text{ml}}{100\:\text{m}}\times\frac{1\:\text{L}}{1000\:\text{ml}}\times\frac{1000\:\text{m}}{1\:\text{km}} = \frac{.09\:\text{L}}{1\:\text{km}} \) (Fuel efficiency is often written as liters per 100 kilometers. If we want to see it this way, we can multiply the numerator and denominator by 100.) \( \frac{.09\:\text{L}}{1\:\text {km}}\times\frac{100}{100} = \frac {9\:\text{L}}{100\:\text{km}}\)

  5. \(\frac{45\: {\text{km}}}{1}\times\frac{5\: {\text{L}}}{100\:{\text{km}}}\times\frac{€1.50}{1\: {\text{L}}} = €3.38\)