If you wanted to power the city of Lincoln, Nebraska using wind power, how many wind turbines would you need to install? Questions like these can be answered using rates and proportions.
RATES
A rate is the ratio (fraction) of two quantities.
A unit rate is a rate with a denominator of one.
Recall Reducing Fractions
The Equivalent Fractions Property states that
If [latex]a,b,c[/latex] are numbers where [latex]b\ne 0,c\ne 0[/latex], then
Your car can drive 300 miles on a tank of 15 gallons. Express this as a rate.
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Expressed as a rate, [latex]\displaystyle\frac{300\text{ miles}}{15\text{ gallons}}[/latex]. We can divide to find a unit rate:[latex]\displaystyle\frac{20\text{ miles}}{1\text{ gallon}}[/latex], which we could also write as [latex]\displaystyle{20}\frac{\text{miles}}{\text{gallon}}[/latex], or just 20 miles per gallon.
Proportion Equation
A proportion equation is an equation showing the equivalence of two rates or ratios.
For an overview on rates and proportions, using the examples on this page, view the following video.
Using Variables to represent unknowns
Recall that we can use letters we call variables to “stand in” for unknown quantities. Then we can use the properties of equality to isolate the variable on one side of the equation. Once we have accomplished that, we say that we have “solved the equation for the variable.”
In the example below, you are asked to solve the proportion (an equality given between two fractions) for the unknown value [latex]x[/latex].
Ex. Solve the proportion [latex]\dfrac{7}{3}=\dfrac{x}{15}[/latex]
We see that the variable we wish to isolate is being divided by 15. We can reverse that by multiplying on both sides by 15.
Solve the proportion [latex]\displaystyle\frac{5}{3}=\frac{x}{6}[/latex] for the unknown value x.
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This proportion is asking us to find a fraction with denominator 6 that is equivalent to the fraction[latex]\displaystyle\frac{5}{3}[/latex]. We can solve this by multiplying both sides of the equation by 6, giving [latex]\displaystyle{x}=\frac{5}{3}\cdot6=10[/latex].
Example
A map scale indicates that ½ inch on the map corresponds with 3 real miles. How many miles apart are two cities that are [latex]\displaystyle{2}\frac{1}{4}[/latex] inches apart on the map?
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We can set up a proportion by setting equal two [latex]\displaystyle\frac{\text{map inches}}{\text{real miles}}[/latex] rates, and introducing a variable, x, to represent the unknown quantity—the mile distance between the cities.
Many proportion problems can also be solved using dimensional analysis, the process of multiplying a quantity by rates to change the units.
Example
Your car can drive 300 miles on a tank of 15 gallons. How far can it drive on 40 gallons?
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We could certainly answer this question using a proportion: [latex]\displaystyle\frac{300\text{ miles}}{15\text{ gallons}}=\frac{x\text{ miles}}{40\text{ gallons}}[/latex].
However, we earlier found that 300 miles on 15 gallons gives a rate of 20 miles per gallon. If we multiply the given 40 gallon quantity by this rate, the gallons unit “cancels” and we’re left with a number of miles:
Notice if instead we were asked “how many gallons are needed to drive 50 miles?” we could answer this question by inverting the 20 mile per gallon rate so that the miles unit cancels and we’re left with gallons:
A worked example of this last question can be found in the following video.
Notice that with the miles per gallon example, if we double the miles driven, we double the gas used. Likewise, with the map distance example, if the map distance doubles, the real-life distance doubles. This is a key feature of proportional relationships, and one we must confirm before assuming two things are related proportionally.
You have likely encountered distance, rate, and time problems in the past. This is likely because they are easy to visualize and most of us have experienced them first hand. In our next example, we will solve distance, rate and time problems that will require us to change the units that the distance or time is measured in.
Example
A bicycle is traveling at 15 miles per hour. How many feet will it cover in 20 seconds?
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To answer this question, we need to convert 20 seconds into feet. If we know the speed of the bicycle in feet per second, this question would be simpler. Since we don’t, we will need to do additional unit conversions. We will need to know that 5280 ft = 1 mile. We might start by converting the 20 seconds into hours: