5.3: Proportion

A proportion is a statement that two ratios are equal or equivalent.

Table 5.3.1 some proportions.

Table 5.3.1: Proportion Examples
Proportion Fraction Form Read Like This
[latex]1:2=2:4[/latex] [latex]\dfrac{1}{2}=\dfrac{2}{4}[/latex] 1 is to 2 as 2 is to 4
[latex]1:4=25:100[/latex] [latex]\dfrac{1}{4}=\dfrac{25}{100}[/latex] 1 is to 4 as 25 is to 100
[latex]18:9=10:5[/latex] [latex]\dfrac{18}{9}=\dfrac{10}{5}[/latex] 18 is to 9 as 10 is to 5
[latex]15:20=3:4[/latex] [latex]\dfrac{15}{20}=\dfrac{3}{4}[/latex] 15 is to 20 as 3 is to 4

Proportions can be used to solve many math problems. You will soon learn to use proportions to solve problems involving percent. The techniques you practice in this section are important for that problem-solving work.

Problems often give incomplete information; that is, one of the terms is missing. To solve such problems, you first find the comparison or ratio that is given. It may be:

  • a quantity of one thing that is mixed with a larger quantity of something else
  • a scale of measurement given on a map, such as 1 cm on the map represents a 100 km distance on land
  • cost for a certain number of items
  • time to travel a certain distance

The problem will then give one term of the second ratio in the proportion. For example, if you have been told that 3 heads of lettuce cost $1.49, you may be asked to find the cost of 7 heads of lettuce.

The missing term is the second cost. The proportion will be:

[latex]\begin{equation}\begin{split} \dfrac{\text{number of heads of lettuce}}{\text{cost}} &= \dfrac{\text{number of heads of lettuce}}{\text{cost}} \\ \dfrac{3}{$1.49} &= \dfrac{7}{?} \\ 3:$1.49 &= 7: ? \end{split}\end{equation}[/latex]

The most important thing to remember is to keep the order of comparison the same in the first and second ratios in a proportion. If the first ratio compares time to distance, then the second ratio in the proportion must compare time to distance.

[latex]\dfrac{\text{time}}{\text{distance}}=\dfrac{\text{time}}{\text{distance}}[/latex]

Or it could be:

[latex]\dfrac{\text{distance}}{\text{time}}=\dfrac{\text{distance}}{\text{time}}[/latex]

Once you have decided on the order of comparison, it is a simple matter to write the proportion using the numbers given in the problem. Use a letter to stand for the missing term.

How would you find a missing term?

  • You can use your skills with equivalent ratios (finding higher and lower terms).
  • You can use your fraction skills of cross-multiplying and then dividing to find the missing term.

Using Equivalent Ratios to Solve Proportions

To solve a proportion problem using equivalent ratios:

  1. Decide on the order of comparison and write a ratio that describes the information given in the problem. Write a proportion using words of the items being compared in fraction form.
  2. Write two more ratios with the numbers matching the words in the first ratio. The missing term (number) can be given a letter (e.g. N).
  3. Mentally set the ratio with words (the first ratio) aside.
  4. Multiply or divide the complete ratio to find the missing term.

Example A

Use 1 teaspoon of baking powder for every 2 cups of flour. If a recipe uses 6 cups of flour, how much baking powder is needed

The missing term is the teaspoons of baking powder for 6 cups of flour. Call this term N.

Step 1:

Ratio is [latex]\dfrac{\text{baking powder}}{\text{flour}}[/latex]

Step 2:

[latex]\dfrac{\text{baking powder}}{\text{flour}}=\dfrac{1}{2}=\dfrac{\text{N}}{6}[/latex]

Step 3:

[latex]\dfrac{1}{2}=\dfrac{\text{N}}{6}[/latex]

Step 4:

[latex]\dfrac{1}{2}\times \left(\dfrac{3}{3}\right)= \dfrac{1 \times 3}{2 \times 3}=\dfrac{3}{6}[/latex], so [latex]\dfrac{1}{2}=\dfrac{3}{6}[/latex], so [latex]\text{N}=3[/latex]

Use 3 teaspoons of baking powder for 6 cups of flour.

Example B

Reports suggest that 3 out of 10 people will at some time miss work due to back pain. If a company has 1000 employees, how many can be expected to miss work due to back pain?

The missing term is the number of people out of 1000 who will miss work due to back pain. Call this term P.

Step 1:

Ratio is [latex]\dfrac{\text{people who will miss work}}{\text{all people at work}}[/latex]

Step 2:

[latex]\dfrac{\text{people who will miss work}}{\text{all the people at work}}=\dfrac{3}{10}=\dfrac{\text{P}}{1000}[/latex]

Step 3:

[latex]\dfrac{3}{10}=\dfrac{\text{P}}{1000}[/latex]

Step 4:

[latex]\dfrac{3}{10}\times\left(\dfrac{100}{100}\right)= \dfrac{3 \times 100}{10 \times 100}=\dfrac{300}{1000}[/latex], so [latex]\dfrac{3}{10}=\dfrac{300}{1000}[/latex], so [latex]\text{P}=300[/latex]

300 people out of 1000 people may miss work due to back pain.

Exercise 1

Write the ratio of the words to describe the information given.

Example: Three cups of flour to one teaspoon of yeast.

Answer: [latex]\dfrac{\text{flour}}{\text{yeast}}[/latex]

 

  1. Four parts oil, ten parts gasoline
  2. One centimetre represents 100 kilometres
  3. 100 grams for $6.89
  4. 3 eggs for each cup of milk
  5. 5 men and 7 women

Exercise 1 Answers

  1. [latex]\dfrac{\text{oil}}{\text{gasoline}}[/latex]
  2. [latex]\dfrac{\text{centimetres}}{\text{kilometres}}[/latex]
  3. [latex]\dfrac{\text{grams}}{\text{dollars}}[/latex]
  4. [latex]\dfrac{\text{eggs}}{\text{milk}}[/latex]
  5. [latex]\dfrac{\text{men}}{\text{women}}[/latex]

Exercise 2

Use equivalent ratios to find the answers.

Example: One cup of sugar and four cups of water will make great hummingbird food. How much sugar do you need for 8 cups of water?

[latex]\dfrac{\text{sugar}}{\text{water}}\longrightarrow\dfrac{1}{4}=\dfrac{\text{N}}{8}\longrightarrow\dfrac{1}{4}\times\left(\dfrac{2}{2}\right) = \dfrac{1 \times 2}{4 \times 2}=\dfrac{2}{8}\longrightarrow\text{N}=2[/latex]

 

  1. Reports show that for every 100 vehicles checked by police, 20 vehicles do not meet the safety standard. If only 50 vehicles are checked, how many would not meet the safety standard?
  2. Four litres of paint covers 24 square metres of wall. How much paint is needed to cover 72 square metres?
  3. Powdered milk uses 1 part milk powder to 3 parts water. How much powder should be added to 9 parts water?

Exercise 2 Answers

  1. 10 cars would not meet the safety standards
  2. 12 litres of paint
  3. 3 parts milk powder

Exercise 3

Use equivalent ratios to find the missing term in these proportions.

  1. [latex]3:5=\text{Y}:15[/latex]
  2. [latex]1:2=\text{P}:8[/latex]
  3. [latex]5:7=10:\text{N}[/latex]
  4. [latex]2:3=8:\text{W}[/latex]
  5. [latex]4:7=16:\text{A}[/latex]
  6. [latex]1:3=2:\text{N}[/latex]
  7. The KX 250 motorcycle (see Figure 5.3.1) uses a mixture of one part oil to 30 parts of gasoline. How much oil must be added to 3,000 mL of gasoline?
    A young man crouching down in front of a motorcycle, filling it with an oil and gasoline mixture.
    Figure 5.3.1

Exercise 3 Answers

  1. [latex]\text{Y}=9[/latex]
  2. [latex]\text{P}=4[/latex]
  3. [latex]\text{N}=14[/latex]
  4. [latex]\text{W}=12[/latex]
  5. [latex]\text{A}=28[/latex]
  6. [latex]\text{N}=6[/latex]
  7. [latex]\text{N}=100 \text{mL}[/latex]

Using Cross-Multiplication to Solve a Proportion

Review cross products:

Multiply the numerator of each fraction with the denominator of the other fraction.

[latex]\dfrac{2}{5}\rlap{\nearrow}{\searrow}\dfrac{4}{10}[/latex]

[latex]\begin{equation} \begin{split} 2\times10 & =5\times4 \\ 20 & =20 \end{split} \end{equation}[/latex]

[latex]2\times10=20[/latex] and [latex]5\times4=20[/latex]

Therefore:

[latex]\dfrac{2}{5}=\dfrac{4}{10}[/latex]

Remember: When the cross products are the same, the fractions are equivalent.

When finding the missing terms in a proportion, cross-multiplication can be used. Follow the examples carefully.

Example D

[latex]\dfrac{2}{3}=\dfrac{\text{N}}{45}[/latex]

Step 1: Cross-multiply.

[latex]\begin{equation} \begin{split} 2\times45 &= 3\times\text{N} \\ 90 &= 3\text{N} \end{split} \end{equation}[/latex]

The idea is to have the unknown term N by itself on one side of the equal sign. To do that, remember these things that you already know:

  • Division and multiplication are opposite operations
  • Whatever is done to one side of an equation or proportion must be done to the other side to keep the equation equal

Step 2: 3N means N is multiplied by 3.

To get rid of the 3, divide by 3.

You must also divide the other side of the equation by 3.

[latex]\dfrac{90}{3}=\dfrac{3\text{N}}{3}[/latex]

Step 3: Solve by reducing the [latex]\dfrac{3}{3}[/latex] and dividing 90 by 3.

[latex]\begin{equation}\begin{split} \dfrac{90}{3} &= \dfrac{3\text{N}}{3} \\ \dfrac{90}{3} &= \dfrac{1\text{N}}{1} \\ \dfrac{90}{3} &= \text{N} \\ 90\div3 &= \text{N} \\ 30 &= \text{N} \end{split}\end{equation}[/latex]

Reducing the fraction [latex]\dfrac{3\text{N}}{3}[/latex] to [latex]\dfrac{1\text{N}}{1}[/latex] to N is also called cancelling. In math, a fraction can be cancelled when the numerator and denominator are the same number. For example:

[latex]\dfrac{6\text{P}}{6}=\dfrac{1\text{P}}{1}=\text{P}[/latex]

Example E

[latex]\dfrac{6}{7}=\dfrac{24}{\text{N}}[/latex]

Step 1: Cross-multiply.

[latex]\begin{equation}\begin{split} 6\times\text{N} &= 7\times24 \\ 6\text{N} &= 168 \end{split}\end{equation}[/latex]

Step 2: Divide both sides by 6. The 6’s with the N will cancel (reduce), and the N will be alone.

[latex]\begin{equation}\begin{split} \dfrac{6\text{N}}{6} &= \dfrac{168}{6} \\ \dfrac{\cancel{6}\text{N}}{\cancel{6}} &= \dfrac{168}{6}  \\ \text{N} &= 168\div6 \\  \text{N} &= 28 \\  \dfrac{6}{7} &= \dfrac{24}{28} \end{split}\end{equation}[/latex]

Check: Use cross-multiplying.

[latex]\begin{equation}\begin{split} \text{Is }6\times28 &= 7\times24? \\ 6\times28 &= 168 \\ 7\times24 &= 168 \\ \text{the cross-product }168 &= \text{the cross-product }168 \\ \text{Yes - }6:7 &= 24:28 \end{split}\end{equation}[/latex]

Example F

[latex]\dfrac{8}{10}=\dfrac{\text{N}}{80}[/latex]

Step 1: Cross-multiply.

[latex]\begin{equation}\begin{split} 8\times80 &= 10\times\text{N} \\ 640 &= 10\text{N} \end{split}\end{equation}[/latex]

Step 2: Divide both sides by 10 so N will be alone.

[latex]\begin{equation}\begin{split} \dfrac{640}{10} &= \dfrac{10\text{N}}{10} \\ \dfrac{64\cancel{0}}{\cancel{10}} &= \dfrac{\cancel{10}\text{N}}{\cancel{10}} \\ 64 &= \text{N} \end{split}\end{equation}[/latex]

To solve a proportion problem using cross-multiplication:

  1. Write the first ratio using the information given.
  2. Write the proportion using a letter in place of the missing term. Be sure the order of comparison is the same in both the first and second ratios in your proportion.
  3. Write the proportion in the fraction form. (Try to simplify the ratio before you do all the calculations).
  4. Cross-multiply and set the cross-products equal to each other.
  5. Divide both sides of the equation by the number with the unknown term.
  6. Check by putting your answer back into the original proportion and cross-multiplying.

Exercise 4

Practise using cross-multiplying to find the missing term in these proportions.

Example:     [latex]\begin{equation}\begin{split} \dfrac{5}{8} &= \dfrac{\text{N}}{32} \\ \\ 5\times32 &= 8\times\text{N} \\ \\ 160 &= 8\text{N} \\ \\ \dfrac{160}{8} &= \dfrac{8\text{N}}{8} \\ \\ 160\div8 &= \text{N} \\ \\ 20 &= \text{N} \end{split}\end{equation}[/latex]

 

  1. [latex]\dfrac{4}{\text{N}}=\dfrac{24}{30}[/latex]
  2. [latex]\dfrac{12}{4}=\dfrac{18}{\text{x}}[/latex]
  3. [latex]\dfrac{\text{y}}{6}=\dfrac{20}{12}[/latex]
  4. [latex]4:15=8:\text{N}[/latex]
  5. [latex]\text{W}:100=6:50[/latex]

Exercise 4 Answers

  1. [latex]\text{N}=5[/latex]
  2. [latex]\text{x}=6[/latex]
  3. [latex]\text{y}=10[/latex]
  4. [latex]\text{N}=30[/latex]
  5. [latex]\text{W}=12[/latex]

The numbers in a ratio often are common fractions, decimals or mixed numbers.

Follow exactly the same steps that you have been using to solve whole number proportions. The calculations will use your skills with fractions.

Example G

[latex]2\frac{1}{4}:3=\text{N}:7[/latex]

Step 1: Rewrite the proportion.

[latex]\dfrac{2\frac{1}{4}}{3}=\dfrac{\text{N}}{7}[/latex]

Step 2: Cross-multiply.

[latex]\begin{equation}\begin{split} 2\dfrac{1}{4}\times7 &= 3\times\text{N} \\ \\ \dfrac{9}{4}\times\dfrac{7}{1} &= 3\times\text{N} \\ \\ \dfrac{63}{4} &= 3\text{N} \\ \\ \dfrac{63}{4}\div\dfrac{3}{1} &= \dfrac{3\times\text{N}}{3} &\longrightarrow \dfrac{63}{4}\times\dfrac{1}{3} &= \text{N} \\ \\ \dfrac{63}{12} &= \text{N} &\longrightarrow 5\dfrac{3}{12}\longrightarrow5\dfrac{1}{4} &= \text{N} \end{split}\end{equation}[/latex]

Exercise 5

Practise using cross-multiplying to find the missing term in these proportions.

Example:     [latex]\begin{equation}\begin{split} 6.5:5 &= 13:\text{A} \\ \\ \dfrac{6.5}{5} &= \dfrac{13}{\text{A}} \\ \\ 6.5\text{A} &= 65 \\ \\ \text{A} &= 65\div6.5 \\ \\ \text{A} &= 10 \end{split}\end{equation}[/latex]

 

  1. [latex]3\dfrac{1}{2}:2=\text{N}:8[/latex]
  2. [latex]9:6=4\dfrac{1}{2}:\text{N}[/latex]
  3. [latex]7.5:\text{B}=10:20[/latex]
  4. [latex]3.75:5=9\text{x}[/latex]
  5. [latex]4\dfrac{1}{8}:\text{A}=3:6[/latex]

Exercise 5 Answers

  1. [latex]\text{N}=14[/latex]
  2. [latex]\text{N}=3[/latex]
  3. [latex]\text{B}=15[/latex]
  4. [latex]\text{x}=12[/latex]
  5. [latex]\text{A}=8\dfrac{1}{4}[/latex] or [latex]8.25[/latex]

Exercise 6

Solve the following word problems.

  1. Joanne can walk 18 km in 3 hours. How far can she walk at the same rate in 5½ hours?
  2. The taxes on the property valued at $300000 are valued at $5000. At the same rate of taxation, what would the taxes be on the smaller lot down the street, which is valued at
    $240000?
  3. One B.C. road map has a scale of 0.5 centimetres, equal to 10 kilometres. Complete the chart in Table 5.3.2 by calculating actual driving distances in kilometres between some B.C. places.The proportions will be [latex]\dfrac{0.5}{10}=\dfrac{\text{cm given in chart}}{\text{actual distance in km}}[/latex]
    Table 5.3.2: Exercise 6c Complete the Chart
    Places in B.C. Number of cm Between Places on the Map Actual Distance in Kilometres
    Kelowna and Vernon 2.5 cm —
    Burns Lake and Vanderhoof 5.5 cm —
    TaTa Creek and Skookumchuk 0.75 cm —
    Kitimat and Terrace 3.3 cm —
  4. The directions on the lawn fertilizer say to spread 1.7 kg over 100 m2 of lawn.
    1. How much fertilizer is needed for a 130 m2 lawn?
    2. How much fertilizer for a 75 m2 lawn?

Exercise 6 Answers

  1. 33 km
  2. $4000
  3. Table 5.3.3: Exercise 6c Answers
    Places in B.C. Number of cm Between Places on the Map Actual Distance in Kilometres
    Kelowna and Vernon 2.5 cm 50 km
    Burns Lake and Vanderhoof 5.5 cm 110 km
    TaTa Creek and Skookumchuk 0.75 cm 15 km
    Kitimat and Terrace 3.3 cm 66 km
  4. Answers
    1. 2.21 kg
    2. 1.275 kg

5.3 Practice Questions

  1. Solve the following proportions.
    1. [latex]\text{N}:14=28:56[/latex]
    2. [latex]3:11=\text{N}:22[/latex]
    3. [latex]50:45=10:\text{N}[/latex]
    4. [latex]4\dfrac{1}{5}:\text{Y}=3:2[/latex]

     

  2. Complete the following exercise on map scales.
    1. Get a map of BC, a map of Canada, and a map of your city or town.
    2. Find the scale on each map (usually at the bottom) and write down the ratio of map distance to the actual distance.
    3. With another student or an instructor, calculate actual distances between places by measuring the distance on the map and working out the proportion according to the scale given. Do at least three distance calculations on each map.

    Ask your instructor to mark your work.

5.2 Practice Answers

  1. Solve these proportions.
    1. [latex]\text{N}=7[/latex]
    2. [latex]\text{N}=6[/latex]
    3. [latex]\text{N}=9[/latex]
    4. [latex]\text{Y}=2\dfrac{4}{5}[/latex] or [latex]2.8[/latex]
  2. See your instructor.

Attribution

All figures in this chapter are from Topic C: Proportion in Adult Literacy Fundamental Mathematics: Book 6 – 2nd Edition by Liz Girard and Wendy Tagami, via BCcampus.

This chapter has been adapted from Topic C: Proportion in Adult Literacy Fundamental Mathematics: Book 6 – 2nd Edition (BCcampus) by Liz Girard and Wendy Tagami (2022), which is under a CC BY 4.0 license.

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Intermediate PreAlgebra: Building Success Copyright © 2024 by Kim Tamblyn, TRU Open Press is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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