We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > Mathematics for the Liberal Arts

Proportions and Rates

Learning Outcomes

  • Given the part and the whole, write a percent
  • Calculate both relative and absolute change of a quantity
  • Calculate tax on a purchase
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. two wind turbines in a field of flowers and low trees

Rates

A rate is the ratio (fraction) of two quantities. A unit rate is a rate with a denominator of one.

Example

Your car can drive 300 miles on a tank of 15 gallons. Express this as a rate.

Answer: 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. https://youtu.be/aZrio6ztHKE

Example

Solve the proportion [latex]\displaystyle\frac{5}{3}=\frac{x}{6}[/latex] for the unknown value x.

Answer: 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?

Answer: 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.

[latex]\displaystyle\frac{\frac{1}{2}\text{map inch}}{3\text{ miles}}=\frac{2\frac{1}{4}\text{map inches}}{x\text{ miles}}[/latex] Multiply both sides by and rewriting the mixed number
[latex]\displaystyle\frac{\frac{1}{2}}{3}\cdot{x}=\frac{9}{4}[/latex] Multiply both sides by 3
[latex]\displaystyle\frac{1}{2}x=\frac{27}{4}[/latex] Multiply both sides by 2 (or divide by ½)
[latex]\displaystyle{x}=\frac{27}{2}=13\frac{1}{2}\text{ miles}[/latex]
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?

Answer: 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: [latex-display]\displaystyle40\text{ gallons}\cdot\frac{20\text{ miles}}{\text{gallon}}=\frac{40\text{ gallons}}{1}\cdot\frac{20\text{ miles}}{\text{gallons}}=800\text{ miles}[/latex-display] 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: [latex-display]\displaystyle{50}\text{ miles}\cdot\frac{1\text{ gallon}}{20\text{ miles}}=\frac{50\text{ miles}}{1}\cdot\frac{1\text{ gallon}}{20\text{ miles}}=\frac{50\text{ gallons}}{20}=2.5\text{ gallons}[/latex-display]

A worked example of this last question can be found in the following video. https://youtu.be/jYwi3YqP0Wk
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?

Answer: 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: [latex-display]\displaystyle{20}\text{ seconds}\cdot\frac{1\text{ minute}}{60\text{ seconds}}\cdot\frac{1\text{ hour}}{60\text{ minutes}}=\frac{1}{180}\text{ hour}[/latex-display] Now we can multiply by the 15 miles/hr [latex-display]\displaystyle\frac{1}{180}\text{ hour}\cdot\frac{15\text{ miles}}{1\text{ hour}}=\frac{1}{12}\text{ mile}[/latex-display] Now we can convert to feet [latex-display]\displaystyle\frac{1}{12}\text{ mile}\cdot\frac{5280\text{ feet}}{1\text{ mile}}=440\text{ feet}[/latex-display] We could have also done this entire calculation in one long set of products: [latex-display]\displaystyle20\text{ seconds}\cdot\frac{1\text{ minute}}{60\text{ seconds}}\cdot\frac{1\text{ hour}}{60\text{ minutes}}=\frac{15\text{ miles}}{1\text{ miles}}=\frac{5280\text{ feet}}{1\text{ mile}}=\frac{1}{180}\text{ hour}[/latex-display]

View the following video to see this problem worked through. https://youtu.be/fyOcLcIVipM

Try It

A 1000 foot spool of bare 12-gauge copper wire weighs 19.8 pounds. How much will 18 inches of the wire weigh, in ounces?
 

Example

Suppose you’re tiling the floor of a 10 ft by 10 ft room, and find that 100 tiles will be needed. How many tiles will be needed to tile the floor of a 20 ft by 20 ft room?

Answer: In this case, while the width the room has doubled, the area has quadrupled. Since the number of tiles needed corresponds with the area of the floor, not the width, 400 tiles will be needed. We could find this using a proportion based on the areas of the rooms: [latex-display]\displaystyle\frac{100\text{ tiles}}{100\text{ft}^2}=\frac{n\text{ tiles}}{400\text{ft}^2}[/latex-display]

Other quantities just don’t scale proportionally at all.

Example

Suppose a small company spends $1000 on an advertising campaign, and gains 100 new customers from it. How many new customers should they expect if they spend $10,000?

Answer: While it is tempting to say that they will gain 1000 new customers, it is likely that additional advertising will be less effective than the initial advertising. For example, if the company is a hot tub store, there are likely only a fixed number of people interested in buying a hot tub, so there might not even be 1000 people in the town who would be potential customers.

Matters of scale in this example and the previous one are explained in more detail here. https://youtu.be/-e2typcrhLE
  Sometimes when working with rates, proportions, and percents, the process can be made more challenging by the magnitude of the numbers involved. Sometimes, large numbers are just difficult to comprehend.

Examples

The 2010 U.S. military budget was $683.7 billion. To gain perspective on how much money this is, answer the following questions.
  1. What would the salary of each of the 1.4 million Walmart employees in the US be if the military budget were distributed evenly amongst them?
  2. If you distributed the military budget of 2010 evenly amongst the 300 million people who live in the US, how much money would you give to each person?
  3. If you converted the US budget into $100 bills, how long would it take you to count it out - assume it takes one second to count one $100 bill.

Answer: Here we have a very large number, about $683,700,000,000 written out. Of course, imagining a billion dollars is very difficult, so it can help to compare it to other quantities.

  1. If that amount of money was used to pay the salaries of the 1.4 million Walmart employees in the U.S., each would earn over $488,000.
  2. There are about 300 million people in the U.S. The military budget is about $2,200 per person.
  3. If you were to put $683.7 billion in $100 bills, and count out 1 per second, it would take 216 years to finish counting it.

 

Example

Compare the electricity consumption per capita in China to the rate in Japan.

Answer: To address this question, we will first need data. From the CIA[footnote]https://www.cia.gov/library/publications/the-world-factbook/rankorder/2042rank.html[/footnote] website we can find the electricity consumption in 2011 for China was 4,693,000,000,000 KWH (kilowatt-hours), or 4.693 trillion KWH, while the consumption for Japan was 859,700,000,000, or 859.7 billion KWH. To find the rate per capita (per person), we will also need the population of the two countries.   From the World Bank,[footnote]http://data.worldbank.org/indicator/SP.POP.TOTL[/footnote] we can find the population of China is 1,344,130,000, or 1.344 billion, and the population of Japan is 127,817,277, or 127.8 million. Computing the consumption per capita for each country: China: [latex]\displaystyle\frac{4,693,000,000,000\text{KWH}}{1,344,130,000\text{ people}}[/latex] ≈ 3491.5 KWH per person Japan: [latex]\displaystyle\frac{859,700,000,000\text{KWH}}{127,817,277\text{ people}}[/latex] ≈ 6726 KWH per person While China uses more than 5 times the electricity of Japan overall, because the population of Japan is so much smaller, it turns out Japan uses almost twice the electricity per person compared to China.

Working with large numbers is examined in more detail in this video. https://youtu.be/rCLh8ZvSQr8
 

Licenses & Attributions

CC licensed content, Original

CC licensed content, Shared previously

  • Problem Solving. Authored by: David Lippman. Located at: http://www.opentextbookstore.com/mathinsociety/. License: CC BY-SA: Attribution-ShareAlike.
  • wind-364996_1280. Authored by: Stevebidmead. License: CC0: No Rights Reserved.
  • Basic rates and proportions. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Proportions using dimensional analysis. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Proportions with unit conversion. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Considering how/if things scale. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Comparing quantities involving large numbers. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
  • Question ID 17454. Authored by: Lippman, David. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.