Explainer: Proportions

In this explainer, we will learn how to get an unknown value in a proportional relation using cross multiplication and how to apply that in real-life situations.

Before looking at how to solve some proportion questions, let us summarize the main features of directly proportional relationships between two quantities.

Recap: Proportional Relationships

The quantity 𝑦 is said to be directly proportional to the quantity π‘₯ when their ratio is the same in all situations. Therefore, 𝑦π‘₯=π‘˜, where π‘˜ is a constant.

This equation can be rewritten in the form 𝑦=π‘˜β‹…π‘₯. The constant π‘˜ is the coefficient of proportionality.

Equal ratios between two proportional quantities can be understood and represented on a double-line diagram in two different ways depending on the context.

In the first diagram, the 𝑦π‘₯ ratios are seen as proportions, that is, part (𝑦)-to-whole (π‘₯) ratios.

In the second diagram, the 𝑦π‘₯ ratios are seen either as part-to-part ratios or as rates, that is, as rates comparing two quantities different in nature. In this case, the coefficient of proportionality is a unit rate: π‘˜ is the unit rate that gives the value of 𝑦 when π‘₯=1. Its unit is then β€œunit of [𝑦] per unit of [π‘₯].”

Considering the equality of two 𝑦-to-π‘₯ ratios, 𝑦π‘₯=𝑦π‘₯,

we see that it can be rearranged in different ways. For instance, by multiplying both sides by π‘₯β‹…π‘₯, we get π‘₯⋅𝑦=π‘₯⋅𝑦.

From this equation, if we divide both sides by π‘₯β‹…π‘¦οŠ¨οŠ¨, we find 𝑦𝑦=π‘₯π‘₯.

We can also divide both sides of π‘₯⋅𝑦=π‘₯β‹…π‘¦οŠ¨οŠ§οŠ§οŠ¨ by π‘¦β‹…π‘¦οŠ§οŠ¨ to find that π‘₯𝑦=π‘₯𝑦,

which simply means that if 𝑦 is proportional to π‘₯, then π‘₯ is proportional to 𝑦.

The ratios π‘₯π‘₯ and π‘¦π‘¦οŠ§οŠ¨ also form a proportion. This can be represented on a double-line diagram with another two pairs of arrows, with π‘˜=π‘₯π‘₯=π‘¦π‘¦οŽ˜οŠ§οŠ¨οŠ§οŠ¨.

The graph of a proportional relationship in a coordinate plane is a straight line through the origin. The line goes through the point (1,π‘Ÿ), where π‘Ÿ is the unit rate.

Let us now go through some questions about proportional relationships. We are going to start with a purely mathematical question, without any context, to get clear ideas about equal ratios.

Example 1: Solving a Proportion Algebraically

If 7π‘₯=4278, find the value of π‘₯.

Answer

This question can be solved algebraically by multiplying both sides of the question by 78π‘₯, which gives 7Γ—78=42π‘₯; and then dividing both sides by 42, we get π‘₯=7Γ—7842=13.

If now we think about the meaning of such an equation, we find that saying that two ratios are equal can be represented by two different double-line diagrams, corresponding to two different situations.

In the first, we consider a part-to-whole ratio, called a proportion, and we are looking for the whole, π‘₯, so that 7 is a part that compares to π‘₯ in the same way as 42 compares to 78.

In the second, this equality of ratios can be understood either as an equality of part-to-part ratios or as an equality of rates, that is, rates that compare quantities different in nature (expressed in different units). Then, the question is to find the missing value π‘₯ in the pair (7,π‘₯), having the pair (42,78) and given that we have a proportional relationship.

These two situations are different if we think of a context, but they are mathematically strictly equivalent, as illustrated here.

In both situations, we find that π‘₯=7Γ—7842=786=13.

We are going now to look at a proportion question in a context.

Example 2: Solving a Speed Problem

William ran 8 km in 36 minutes. How long would it take him to run 14 km at the same speed?

  1. 63 min
  2. 20 min
  3. 83 min
  4. 22 min

Answer

In this question, we want to find how long it would take William to run 14 km at the same speed as when he ran 8 km in 36 minutes. The speed is the distance-to-time ratio. If the speed is the same, then we can write that 836=8𝑑,kmminkmmin with 𝑑 being the time we are looking for.

We can rearrange this equation by first multiplying both sides by 36𝑑, which leads to 8𝑑=14Γ—36.

Dividing both sides by 8, we get 𝑑=14Γ—368=7Γ—9=63.min

We could have solved this question by working out first the unit rate, which would be here the time in minutes it takes William to run 1 km. This is 36Γ·8=4.5/.minkm

Running 14 km would then take William 14 times this time; that is, 14Γ—4.5=63.min

The reasoning applied here can be visualized on a double-line diagram.

Our understanding of proportional relationships can be used to compare different unit rates.

Example 3: Comparing Unit Prices

The table shows what four stores are offering to their customers. Which store has the best offer?

StoreThe Digital ShopElectronicaMusiclandRecord Town
Offer10 CDs for $422 CDs for $399 CDs for $635 CDs for $37
  1. Musicland
  2. Record Town
  3. Electronica
  4. The Digital Shop

Answer

In this question, we need to work out the unit rate, that is, the price per CD, in each store. The unit rate is given by dividing the price given by the number of CDs this price is for. The results are given in the table.

StoreThe Digital ShopElectronicaMusiclandRecord Town
Unit Price$4210=$4.20/CDCD$392=$19.50/CDCD$639=$7/CDCD$375=$7.40/CDCD

We find that the best offer (the lowest unit price) is made by The Digital Shop.

Example 4: Comparing Water Consumption

The given table shows the monthly usage of water of three different families. Determine the family who uses the least amount of water per person, and find the amount.

Family NameFamily SizeWater (gal)
Jacob43,360
David22,230
James33,420
  1. The David family, 1,140 gal per person
  2. The James family, 1,115 gal per person
  3. The Jacob family, 1,140 gal per person
  4. The David family, 1,115 gal per person
  5. The Jacob family, 840 gal per person

Answer

The table gives the number of people in three different families and their respective amount of water used. We want to compare the amount of water used per person in each family. To find this amount, we simply need to divide, for each family, the given amount of water used by the number of people in it. The results are given in the table.

Family NameWater Usage per Person
Jacob3,3604=840/galpeoplegalperson
David2,2302=1,115/galpeoplegalperson
James3,4203=1,140/galpeoplegalperson

The family that uses the least amount of water per person is the Jacob family, with a water consumption rate of 840 gal/person.

Example 5: Comparing Unit Rates from a Graph

The graph shows the time it takes Benjamin (red) and Ethan (blue) to prepare different numbers of sandwiches.

  • The preparation time is proportional to the number of sandwiches to prepare. What are the unit rates (i.e., the time it takes to prepare one sandwich) for Benjamin and Ethan?
    1. Benjamin: 0.5 min/sandwich, Ethan: 0.57 min/sandwich
    2. Benjamin: 1.75 min/sandwich, Ethan: 2 min/sandwich
    3. Benjamin: 0.57 min/sandwich, Ethan: 0.5 min/sandwich
    4. Benjamin: 1 min/sandwich, Ethan: 2.33 min/sandwich
    5. Benjamin: 2 min/sandwich, Ethan: 1.75 min/sandwich
  • Who makes sandwiches faster?
    1. Benjamin
    2. Ethan

Answer

The graphs show the time needed to prepare sandwiches against the number of sandwiches for Benjamin and Ethan. To find their unit rates, we need to take two points on the graph, for instance, (40,80) for Benjamin and (40,70) for Ethan, meaning that Benjamin makes 40 sandwiches in 80 minutes and Ethan makes 40 sandwiches in 70 minutes.

From this, we find that the time it takes to make one sandwich is 8040=2/minsandwichesminsandwich for Benjamin and 7040=1.75/minsandwichesminsandwich for Ethan.

From this, we can conclude that Ethan is faster since he takes less time to make one sandwich.

Note that there is a catch here because we are used to seeing graphs plotted in the other way around, that is, showing the number of sandwiches made against time. Then, the graph that is on the top corresponds to a higher speed. Here, however, the graphs show the time it takes to make a given number of sandwiches. Therefore, the graph on the bottom corresponds to a higher speed.

Key Points

  1. The quantity 𝑦 is said to be directly proportional to the quantity π‘₯ when their ratio is the same in all situations. Therefore, 𝑦π‘₯=π‘˜, where π‘˜ is a constant. This equation can be rewritten in the form 𝑦=π‘˜β‹…π‘₯. The constant π‘˜ is the coefficient of proportionality.
  2. The equality of two ratios of 𝑦 to π‘₯ in two different situations, 𝑦π‘₯=𝑦π‘₯, can be rearranged to find many equivalent equations, such as π‘₯⋅𝑦=π‘₯β‹…π‘¦οŠ§οŠ¨οŠ¨οŠ§, π‘₯π‘₯=π‘¦π‘¦οŠ¨οŠ§οŠ¨οŠ§, and π‘₯𝑦=π‘₯π‘¦οŠ§οŠ§οŠ¨οŠ¨.
  3. The way different values of two proportional quantities are linked together (as described with the above equations) can be represented on a double-line diagram with arrows showing how the same multiplication takes us from one value to the other in the two pairs (either from bottom to top or from right to left and vice versa, as all arrows can be flipped by taking the inverse factor: if 𝑦=π‘˜β‹…π‘₯, then π‘₯=1π‘˜π‘¦οŠ§οŠ§).
  4. The graph of a proportional relationship on a coordinate plane is a straight line through the origin. The line goes through the point (1,π‘Ÿ) where π‘Ÿ is the unit rate.

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