Ratios and Proportions | Algebra

Introduction

Ratios questions and their close cousins dealing with proportions always have to do with a relationship between numbers. Since a similar relationship can also be expressed by a fraction, a decimal, or a percent, it’s important to know the differences between these equivalent expressions. Percent questions are quite frequent on the CLT and have their own dedicated module in this course. Fractions and decimals are less frequent but do occur; they will be treated in this lesson. But back to ratios: ratios are relationships between numbers that express how frequent each value is compared to the other. They are expressed as with a colon (:) as in the ratio $4 : 3$ . Proportions occur when there are two examples of the same relationship; think of a map where 1 inch equals 500 miles, and therefore by proportion would say that 3 inches equals 1,500 miles. We will present many examples of ratios and proportions below.

Approach Question

The ratio of sulfur atoms to titanium atoms in a titanium disulfide molecule is $2 : 1$ and the ratio of the atomic mass of titanium to sulfur is approximately $3 : 2$ . What percent of the atomic mass of a titanium disulfide molecule is accounted for by the titanium atom, rounded to the nearest whole percent?

A. $20%$
B. $35%$
C. $42%$
D. $58%$

Explanation

This question illustrates the form that ratio problems take on the CLT: they are typically in the form of word problems. With word problems, it’s always crucial to map the information carefully. Misreading the order of either of the ratios in this question will lead to an incorrect answer.

Two important observations for ratio questions in general are relevant here. The first is that a ratio is not the same as a fixed pair of numbers. It would difficult how important this fact is to CLT ratio problems; many test takers go wrong by interpreting a ratio of, say, $5$ blue tokens to $3$ red tokens as if there were exactly $5$ blue and $3$ red present. At the same time, we can observe that if trying an example, we actually can assume (momentarily) that the ratio numbers are absolute numbers. This will work as long as we remain aware that we are only considering one possible example out of all the possible scenarios in the question.

Here’s how these two principles work in this case. On the one hand, the $2 : 1$ ratio does not mean there are only two atoms of sulfur and one atom of titanium in the molecule (although a chemist will tell you that does turn out to be the case!). Similarly, the $3 : 2$ ratio does not mean that titanium has a mass of $3$ while sulfur has a mass of $2$ (ignoring whatever units we might be dealing with, since the question doesn’t give units). But we might as well assume all the numbers given in the ratio for our calculations, since we’re only asked to compare something about titanium and sulfur.

So let’s do that. In the first ratio, let’s assume there is one titanium atom and, in the second ratio, that titanium has a mass of $3$ . That means one atom has a mass of $1 \times 3 = 3$ . Meanwhile, if we assume that there two sulfur atoms present and that each sulfur atom has a mass of two, we’ll find a total sulfur mass of $2 \times 2 = 4$ . So the ratio of the masses of all titanium atoms present to all sulfur atoms present is $3 : 4$ .

This question, however, ratchets up the difficulty by now asking about a percent. To figure out the percent of titanium mass in the total mass, we have to think about the total ratio “parts” we have in this scenario. Three parts titanium plus four parts sulfur means seven parts total. So the proportion of titanium mass out of the total is best represented by the fraction $\frac{3}{7}$ .

Which of these percents is $\frac{3}{7}$ ? If you’ve completed the percents module you may have a sharper idea about this than you would otherwise. We could apply the percent formula, but let’s use smart test logic instead. If half of $7$ is $3.5$ , then $3$ is a little less than half of $7$ , so a little less than $50$ . The answer has to be $C$ ; choice $B$ is closer to one-third than one-half. Choice $A$ and choice $D$ are far too low and too high, respectively. (However, Choice $D$ is an excellent trap answer; make sure to use the UnCLES method and emphasize the fact that titanium comes before sulfur in the ratio. If you get that backward, you’ll think the answer is $D$ ).

The answer is C.

Definitions

Proportion: A proportion is an equation of two ratios. Because of the equals sign, we know that the relationship between the numbers in each ratio is constant. That allows us to discover the value of an unknown by cross-multiplying. Example: if you know a recipe calls for 2 cups of flour for every 3 eggs, you could find out how many cups of flour to combine with a dozen eggs with the following equation: $\frac{2}{3} = \frac{x}{12}$ .

Topics for Cross-Reference

Variations

As mentioned in the introduction, ratios are interchangeable with fractions. The successful test taker will find it natural to quickly render something like $2 : 7$ as $\frac{2}{7}$ or, if necessary, change a $\frac{3}{4}$ fraction into a $3 : 4$ ratio. This can be helpful as long as you keep in mind that, in these fractions, both parts of the fractions are parts of the ratio and the whole is not represented. To represent the whole, consider the note in Strategy Insights below.

Ratios can also be converted into decimals, but this is much rarer because decimals don’t demonstrate a relationship between two values as readily as ratios or fractions do. Decimals are most helpful when $10$ (or $100$ , or $1000$ , etc.) is the sum of the ratio parts. If you have a $3 : 7$ ratio of cats to dogs, you can say that the relatively frequency of cats in this scenario is $0.3$ because $3 + 7 = 10$ , so $10$ becomes the denominator when you are considering the total.

Strategy Insights

We will repeat two ideas from the introduction in the section on strategy because they so often appear, and so frequently confuse students, on a test like the CLT.

Repeat after us: a ratio is not the same as two fixed numbers! If you see a ratio of $9 : 4$ used to describe the relationship between the population of towns A and B, that doesn’t mean there are exactly $9$ people in A and exactly $4$ people in B!
Watch out for challenging problems in which the ratio total becomes relevant. If a bag contains $2$ red tickets for every $7$ blue tickets and you want to know your chances of picking a red ticket, you have to consider that the sum of the total ratio parts is $2 + 7 = 9$ . You have a $\frac{2}{9}$ chance of picking a red ticket.

Flashcard Fodder

The algebraic skill most helpful for proportion problems is cross-multiplication. The “cross” refers to the fact that you (twice) multiply the numerator of one fraction with the denominator of the other. You then set the products equal to each other. For a flashcard, you can represent cross-multiplication as follows:

one side of the card: $\frac{a}{b} = \frac{c}{d}$
the other side: $a d = b c$

Sample Questions

Difficulty 1

Ivy surveys her closet and calculates that the ratio of blue items of clothing to green items is $5 : 2$ . If Ivy has 10 green items of clothing in her closet, how many clothing items in the closet are blue?

A. 4
B. 10
C. 20
D. 25

(spoiler)

The answer is D. This is a great opportunity to set up a proportion: $\frac{5}{2} = \frac{B}{G}$ ; then, identifying blue as the unknown quantity and green as the known quantity of $10$ : $\frac{5}{2} = \frac{x}{10}$ . Make sure, when creating a proportion, that the same kind of element is in both numerators and the same kind is in both denominators. In this proportion, the $5$ and the $x$ correspond to blue items, whereas the $2$ and the $10$ correspond to green. If you accidentally flip a fraction, you will arrive at the wrong answer. In case you need process of elimination here, consider that you could eliminate choices $A$ and $B$ as two small; if there are more blue items than green in Ivy’s closest, then there have to be more than $10$ blue items.

Difficulty 2

After Christmas break, two new boys and three new girls join Josh’s grade. If the ratio of boys to girls in the grade was $9 : 8$ before the new students arrived, what is the ratio between boys and girls in the class after the change?

A. 5:6
B. 1:1
C. 6:5
D. It is impossible to determine from the given information.

(spoiler)

The answer is D. Be careful! We can’t simply add $2$ to the number of boys and $3$ to the number of girls in the ratio. We don’t know if the numbers $9$ and $8$ represent the actual number of boys and girls in Josh’s grade or simply the relationship between the two. For all we know, there could be $90$ boys and $80$ girls - a $9 : 8$ ratio. If we don’t know the original values, we can’t know exactly how the addition of the new boys and girls will change the situation.

Difficulty 3

Thus far in the field hockey season, Park School has scored twice as many goals as Covenant School. Meanwhile, Areopagus Academy has scored one-third as many goals as Covenant School. What is the ratio of goals scored by Park School to goals by Areopagus Academy?

A. 6:1
B. 3:2
C. 2:3
D. 1:6

(spoiler)

The answer is A. We need to carefully follow the assignments in each ratio and ensure that we don’t get anything reversed. To approach this problem algebraically, we could assign the variable $x$ to one of the schools. If we call goals scored by Park School $x$ , then Covenant School would be represented by $\frac{x}{2}$ , while Areopagus would come out as $\frac{x}{2} \times \frac{x}{3} = \frac{x}{6}$ . The ratio of $x$ to $\frac{x}{6}$ is $6 : 1$ . However, if we realize that Areopagus has the smallest number of goals, we can avoid fractions by assigning $x$ to Areopagus instead. This makes Covenant equal to $3 x$ and Park School double that, or $6 x$ . The ratio of $6 x$ to $x$ is a bit easier to understand.

If you prefer to plug in actual numbers, it helps to analyze the problem first to determine where to begin. As shown above, we should start with Areopagus and work our way upward in goals from there. If we choose a nice round number of $10$ for Areopagus, that gives us $30$ for Covenant and $60$ for Park. $60$ to $10$ is equivalent to $6 : 1$ .

Difficulty 4

In the debate elective at Veritas et Lux High School, the ratio of seniors to juniors is $4 : 3$ . Meanwhile, the ratios of juniors to sophomores is $7 : 5$ . What is the ratio of seniors to sophomores?

A. 8:5
B. 28:15
C. 21:20
D. 4:5

(spoiler)

The answer is B. There is a wonderfully simple strategy to a question like wherein one of the quantities is present in both ratios and so serves at the “link” between them. With the ratios in the form $A : B$ and $B : C$ , simply convert both to fractions and multiply the fractions, simplifying where possible. In this case, $\frac{4}{3} \times \frac{7}{5} = \frac{28}{15}$ , and we have our answer.

Why does this work? Another way of looking at this question is that we need to equalize the common element in the ratios. Since juniors appear in both ratios, let’s find the least common multiple of the numbers for juniors; the $L CM$ of $3$ and $7$ is $21$ . That means that if we make the value of juniors $21$ in both ratios, we’ll be able to compare the seniors and sophomores in an “apples to apples” scenario. The first ratio, rendered with the juniors as $21$ , now becomes $28 : 21$ . Similarly, the second ratio becomes $21 : 15$ . Comparing $28$ for the seniors to $15$ for the sophomores. confirms the multiplying fractions approach.

Difficulty 5

On a proportionally correct map of the $50$ US States, the ratio of Texas’ area to that of Alaska is approximately $6 : 11$ . Meanwhile, the ratio of Alaska’s area to that of the continental U.S. is approximately $1 : 5$ . Using these ratios, if you randomly throw a dart at this U.S. map and are guaranteed to hit the map, what are the odds that you hit Texas? (Note: ignore tiny Hawaii in imagining this situation [sorry, Hawaii!].)

A. 1:11
B. 6:55
C. 1:5
D. 5:11

(spoiler)

The answer is A. The trap answer is choice $B$ because it would appear, with Alaska being the common element linking the ratios, that we should be able to multiply $\frac{6}{11}$ by $\frac{1}{5}$ to get our answer. The problem with such an approach in this case is that the second ratio is not Alaska compared to the whole U.S. but Alaska compared to the continental U.S. To get the ratio total for the whole U.S. (ignoring Hawaii as directed), we have to add the parts of the second ratio to make $6$ . So the fractions we need to multiply are really $\frac{6}{11}$ and $\frac{1}{6}$ . The sixes cancel and we are left with $\frac{1}{11}$ or 1:11.

For Reflection

How will you approach ratios and proportions questions on test day? Write down at least three takeaways from this module.
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
What is the difference between a ratio and a proportion? Make sure you’re clear on when to use proportions: you have to have two equal ratios where you know three of the four values but need to solve for the unknown fourth.