SAT Math Skill Review: Ratios, Proportions & Variations
The concepts of ratios, proportions and variations, which build upon each other, all involve change. Ratios are used to compare two quantities. Proportions are merely statements of equality involving ratios and variations involve problems written as proportions.
While you may not realize it, you use ratios every day. For example, when you talk about a vehicle’s rate of travel being 45 mph, you are talking about the ratio of distance traveled to time traveled. Ratios are used to compare two quantities. They are typically written in one of two ways: as a fraction or with a colon (:). The ratio read as “4 to 10” could be written as 4/10 or as 4 : 10. The ratio expressing the relationship between a day and a week could be written as 1/7 or 1 : 7 because there are 7 days in a week.
Keep in mind, when you write a ratio as a fraction, it is not a fraction in the true sense of the term (i.e. part over whole).
A proportion is a mathematical statement equating two ratios. Two ratios are said to be equal if, when written in fractional form, the fractions are equivalent fractions. You can compare and solve ratios using cross multiplications. If the cross products are equal, the ratios are equal.
When solving a proportion question, you’ll want to rewrite the proportion in fractional format. Let’s take a look at an example:
Note, in the above example, you could have also solved this problem by observation if you recognized that 4/12 is simply 1/3. One-third of 6 is 2. Observations such as this will provide significant time savings on the exam, but on more difficult problems, it’s important to know the basics.
Variations deal with explaining, in mathematical language, how one quantity changes with respect to one or more other quantities. There are two primary types of variation: direct variation and inverse variation. We’ll look at these first, and then take a look at two extensions of these concepts: joint variation and combined variation.
Let’s take a look at an example of direct variation. The amount of flour needed in a bread recipe varies with the number of loaves being made. Making more loaves will require more flour; making fewer loaves will require less flour. In direct variation, the variables (amount of flour and number of loaves in our example) will move in the same direction.
Problems involving direct variation can be solved using proportions. Let’s see this in action:
If it takes 3 gallons of paint to cover 100 square feet, how many gallons of paint will be needed to cover 600 square feet?
The problem gives us the ratio of 3 gallons of paint to cover 100 square feet. We can use this information to set up our proportion:
Cross-multiply and solve for x:
When we say "y varies directly as x," we could also write:
In the above equation, k is called the constant of variation. In the paint example, the number of gallons of paint varies directly with the square footage that will be covered. The constant of variation is 3/100. You would then have:
When two variables or quantities change in opposite directions, you have inverse variation. Let’s take a look at an example: The time it takes to paint a house varies with the number of people doing the work.
In this example, the time required to paint the house varies inversely with the number of people painting. This means the more people painting the house, the less total time it will take to paint. When we say "y varies inversely with x," we can express this as:
Once again, k is the constant of variation. We can find k by rearranging the formula as k = xy. Thus, k is simply the product of the known values for the two variables. Let’s consider the following example:
A particular hotel has a custodial staff of 12 employees, and they can typically clean all of the hotel rooms in 6 hours. If four members of the custodial staff are not at work today, how long will it take the remaining custodians to clean all of the hotel rooms?
In this example, the total time taken to complete the job is inversely proportional to the number of workers. The constant of variation, k, is simply 6 • 12 = 72. We want to know the numbers of hours it will take the remaining custodians to do the job. Since 4 are absent, that leaves 8 workers (12 – 4 = 8).
We can thus solve the problem as detailed below:
The correct answer is 9 hours.
Joint variation is just an extension of direct variation. When we say that "y varies jointly with x and z" we have an example of joint variation. This can be written as:
Again, k is the constant of variation, simply the ratio of y to x and z (x times z) Here’s an example:
The variable y varies jointly with x and z. The value of y is 12 when x is 4 and z is 8. What is y when x = 6 and z = 10?
Since it is pretty clear we have a joint variation problem, we solve for k by plugging the known variables (y = 12 when x is 4 and z is 8) into our joint variation equation:
Once we have a value forwe plug this back into the equation of joint variation along with the variables we do know and solve for our unknown:
Combined variation involves both direct and inverse variation. For example, if we say "y varies directly with x and inversely with z," we are faced with combined variation. This can be written as follows:
Consider the following example:
It takes 2 hours for 3 people to paint 100 ft. of 6-ft. fencing. Assuming that each person is capable of painting at the same rate, how long will it take for 12 people to paint 1,800 ft. of the same fencing?
The total time (t) needed to paint a section of fencing is directly proportional to the length of the fence (l) and inversely proportional to the number of people (n) who are painting. (Note, we are using different letters for our variables, but it doesn’t matter!).
The steps are the same as in the other example problems. First, we determine the constant of variation, k.
Now that we have k, we plug it into the equation and solve for the missing variable:
The correct answer is 9 hours.
- There are 14 boys and 16 girls in Tyler’s class. What ratio best represents the relationship between the number of boys and the number of students in Tyler’s class?
- If the ratio of A to B is 3 : 4, and the ratio of B to C is 2 : 3, what is the value of A when C is 5?
Answers and Explanations
- The correct answer is C. It’s critical to read this question carefully as it’s easy to make a careless error on this one. The question gives you the number of boys (14) and the number of girls (16), but it’s asking you for the ratio between the number of boys (14) and the total number of students (14 + 16 = 30). Sneaky. The ratio itself could be written as a fractional format (14/30) or as 14 : 30. None of these are choices however, so you need to reduce to find the answer, choice C, which could have just as easily been written as 7 : 15. You’ll note that the choices also include options for the ratio of boys to girls (choices A and D), and the ratio of girls to the total (choice E). It’s not a tough question conceptually, but it’s an easy one to miss if you aren’t paying close enough attention!
- The correct answer is C. You are given two separate ratios with one common variable (B). You’ll need to combine them. If the ratio of A to B is 3 : 4 and B to C is 2 : 3, the common variable needs to be the same in both relationships to combine. If you multiply both B and C by 2 in the second relationship, it’s not actually changing -- 2 : 3 is the same as 4 : 6 -- but now you can combine the two relationships to have a ratio A to B to C of 3 : 4 : 6. Remove B and you have your ratio of A to C, or 3 : 6. To determine what A would be if C were 5, set it up as a proportion and cross multiply: 32/6 = x/5. Solve for the missing value, 3 • 5 = 6 • x, so 15 = 6x, and thus, x = 15/6 = 5/2.