# Ratio and Proportion Questions on the SAT

Ratios and proportions appear in many word problems on the SAT math section, so a solid understanding of this topic is essential for a successful test day. This article will address some of the fundamental characteristics of ratio and proportion questions on the SAT and their solutions. In addition, we’ll include a discussion of a related topic — direct variation — which is often confusing to students.

## Here are the topics we’ll cover:

To begin, we can answer the question, “What is a ratio?”

## What Is a Ratio?

A ratio relates two quantities. For example, if we have 5 green pencils and 6 orange pencils in a box, we can say that the ratio of green pencils to orange pencils is 5 to 6.

If, instead, the box contained 10 green pencils and 12 orange pencils, the new ratio would be 10 to 12, but we could reduce the ratio to its equivalent ratio of 5 green to 6 orange. Just as the fraction 10/12 can be reduced to 5/6, the ratio of 10 to 12 can be reduced to 5 to 6.

KEY FACT:

A ratio compares two quantities.

We can express ratios in three equivalent ways. Using the previous example, we can express the ratio of green pencils to orange pencils as any of the following:

1. green pencils to orange pencils = 5 to 6
2. green pencils : orange pencils = 5 : 6
3. (green pencils) / (orange pencils) = 5 / 6

While all three options are perfectly fine to use, when solving the problems below, we will use the fraction notation (number 3 on our list).

## Part-to-Part and Part-to-Whole Ratios

All ratios provide the following two items of information:

• how the parts of a ratio relate to each other
• how the parts of a ratio relate to the total of the parts

Returning to the pencil example, the ratio 5 to 6 that relates green pencils to orange pencils is an example of what is called a part-to-part ratio, because the green and orange pencils are each only a part of the collection.

In addition, this same information can also be used to calculate what is called a part-to-whole ratio. We have a total of 11 pencils in the box when there are 5 green pencils and 6 orange pencils. Thus, the ratio of green pencils to this total is the part-to-whole ratio 5 to 11, or 5/11. Similarly, the ratio of orange pencils to the total is the part-to-whole ratio 6 to 11, or 6/11.

KEY FACT:

When two parts are compared, we call their relationship a part-to-part ratio. When a part is compared to the entire amount, we call the relationship a part-to-whole relationship. From a part-to-part ratio, we can calculate the corresponding part-to-whole ratio.

Let’s practice with a mini-example.

### Mini Example 1: Part-to-Part and Part-to-Whole Ratios

At Polly’s Pet Daycare, the ratio of the number of dogs to the number of cats is 5 to 2. What is the part-to-part ratio of cats to dogs, and what is the part-to-whole ratio of dogs to the total number of pets at the daycare?

To determine the part-to-part ratio of cats to dogs, we must flip the order of the given ratio, 5 to 2, because in that ratio, the 5 represents the number of dogs and the 2 represents the number of cats. Thus, the ratio of cats to dogs is:

cats/dogs = 2/5

Now, we can calculate a number to represent the total number of pets at the daycare by adding the numbers of cats with the number of dogs. Thus, 5 + 2 = 7 represents the total. So, the part-to-whole ratio of the number of dogs to the total number of pets is:

dogs/total = 5/7

This part-to-whole ratio is valid regardless of how many pets are at Polly’s Pet Daycare. For example, if there are 5 dogs and 2 cats, then there are a total of 7 animals, and the ratio of cats to total pets is 2 to 7. Or, if there are 15 dogs and 6 cats, then the total number of animals is 15 + 6 = 21, and so the ratio of cats to total pets would be 6/21, which is equivalent to 2/7.

We can generalize the above to a rule. If we have a part-to-part ratio x : y, then we can create the following part-to-whole ratios representing the same situation:

1. x / (x + y)
2. y / (x + y)

KEY FACT:

If we have a part-to-part ratio of x / y, then the part-to-whole ratios are x / (x + y) and y / (x + y).

## Answering Equivalent Ratio Questions

Earlier in this article, we considered an example in which we had 11 pencils in a box and the ratio of green to orange pencils was 5 to 6. Now, let’s extend this discussion to use a given ratio to determine the quantity of a particular item using equivalent ratios.

### Mini Example 2: Using Equivalent Ratios

What if we were told that in another box, the ratio of green pencils to orange pencils was still 5 to 6, but there were 18 orange pencils in the box? In that case, could we determine the number of green pencils? Yes, we could!

Let’s let G equal the number of green pencils in this new box. Then, because we know that the ratio in the new box must still be equivalent to 5/6, we can set our old ratio equal to a new ratio in which we have G green pencils and 18 orange pencils:

(green pencils) / (orange pencils) = 5/6 = G/18

Note that when we set up the above equivalent ratios, we need to ensure they are both in the same order: green in the numerator and orange in the denominator. We now have an equation that we can use to solve for the value of G. We begin by cross-multiplying:

5/6 = G/18

(5)(18) = (6)(G)

90 = 6G

15 = G

Thus, there are 15 green pencils in the box.

Next, let’s practice with some examples of both types of ratio questions you might encounter on the SAT.

### Example 1: Equivalent Ratios

On a test, the ratio of true/false questions to multiple-choice questions is 4 to 5. If there are 30 multiple-choice questions on the test, then how many true-false questions are there?

• 9
• 11
• 22
• 24

#### Solution:

First, we are given that the part-to-part ratio of true/false questions to multiple-choice questions is 4 to 5. Next, since we are given that there are 30 multiple-choice questions on the test, we can create the following equation in which T represents the number of true/false questions:

T/30 = 4/5

Next, we cross-multiply and solve this equation to determine T:

(T)(5) = (4)(30)

5T = 120

T = 24

Thus, there are 24 true-false questions on the test.

Let’s practice with another example question.

### Example 2: Part-to-Part and Part and Part-to-Whole Ratios

If the ratio of ham sandwiches to turkey sandwiches at a deli is 8 to 3, and there are a total of 33 sandwiches of either type at the deli, then how many turkey sandwiches are at the deli?

• 9
• 11
• 22
• 24

#### Solution:

To start, we see that we have a part-to-part ratio of ham sandwiches to turkey sandwiches that is equal to 8 to 3, with 8 representing the number of ham sandwiches and 3 representing the number of turkey sandwiches.

However, we are then given the total number of sandwiches. Thus, we need to use the information from the part-to-part ratio to find a number representing the total number of sandwiches. Doing so will allow us to determine the number of turkey sandwiches.

We can express the total number of ham sandwiches and turkey sandwiches as 8 + 3 = 11, and so the part-to-whole ratio of turkey sandwiches to the total is 3/11. We are also given that the total number of sandwiches is 33. Thus, we can let the number of turkey sandwiches = T and create the following equation:

T/33 = 3/11

Solving, we have:

(T)(11) = (3)(33)

11T = 99

T = 9

There are 9 turkey sandwiches at the deli.

Now that we have taken a look at ratios, let’s look at a related concept that is often tested on the SAT: proportions.

## Proportions

A proportion is an equation that shows the equality of two ratios. Nearly all the questions about proportional relationships that you might see on the SAT will cover either real-life situations in word problems or the calculation of the lengths of similar triangles.

It is important to become familiar with common phrasing on the SAT that indicates that you will use proportions to solve a problem. These phrases are:

• “directly proportional”
• “at the same rate”
• “at this rate”

KEY FACT:

On the SAT, you will see phrases such as “directly proportional,” “at the same rate,” or “at this rate” to indicate that it is a proportion question.

Now, let’s look at a straightforward mini-example.

### Mini-Example 3: Proportions

If Martha uses 9 lemons to make 1 jug of lemonade, how many lemons does she need to make 3 jugs of lemonade?

First, we see that our two quantities are lemons and jugs. We know that the proportion of lemons to jugs is 9 to 1, or 9 / 1, and, to keep the variables straight, we include units in the ratio, saying 9 lemons / 1 jug. Our question is to calculate x, the number of lemons needed to make 3 jugs. Thus, we can say that we want to determine the equivalent ratio x lemons / 3 jugs.

Let’s set up the proportion by equating the two proportions:

9 lemons / 1 jug = x lemons / 3 jugs

We can solve for x by cross-multiplying:

(9)(3) = (1)(x)

27 = x

Thus, we see that we need 27 lemons to make 3 jugs of lemonade.

Let’s look at another proportion example question.

### Example 3: Proportions

A blue shark can swim 17.7 miles in 60 minutes. At this rate, which of the following is closest to the number of miles the shark can swim in 5 minutes?

• 0.3
• 1.5
• 11
• 89

#### Solution:

The given ratio is 17.7 miles / 60 minutes, and the equivalent ratio is x miles / 5 minutes. We can now set up the proportion and solve for x:

17.7 miles / 60 minutes = x miles / 5 minutes

(17.7)(5) = 60x

88.5 = 60x

1.475 = x

The closest answer choice is 1.5 miles.

## Direct Variation

An important type of proportion that you may see tested on the SAT is commonly presented with the phrases “direct variation” or “directly proportional to.”

Direct variation means that if y varies directly with x, then if x increases by some factor, then y will increase by the same factor. For example, if x doubles, then y will double. Or if x is halved, then y will also be halved.

KEY FACT:

Direct variation between two variables x and y means that if x increases by a particular factor, then y will increase by the same factor.

Direct variation is usually represented by the equation y = kx, where k is a positive constant, called the constant of variation or the constant of proportionality.

Let’s look at a mini-example that illustrates how to solve problems involving direct variation.

### Mini-Example 4: Direct Variation

The amount of money a company pays a worker to do a particular job varies directly with the number of hours spent on the job. If a worker spends 5 hours on a job and the total cost is $300, what would the total cost be for an 8-hour job? Since the total cost varies directly with the number of hours, we use the direct proportion formula y = kx, where y = the total cost, x = the number of hours, and k = the constant of variation. Direct variation problems are almost always 2-step problems. In the first step, we use the given information to calculate the value of k, the constant of variation. First, we must determine the value of k. We do this by substituting the given information y = 300 and x = 5 into the direct proportion formula: y = kx 300 = k(5) 300 = 5k k = 60 Now that we know the value of k, we can do the second step of the direct variation problem. We first write the direction variation formula as y = 60x. Then we can easily answer the question posed: “What is the total cost of an 8-hour job?”: y = kx y = (60)(8) y = 480 Thus, the cost of an 8-hour job is$480.

Let’s try some example questions.

### Example 4: Direct Variation

During a thunderstorm, you see the lightning before you hear the thunder. You know that the distance between you and the storm is in direct proportion to the time interval between the lightning and the thunder.

Suppose that the thunder from a storm 5,400 feet away takes 5 seconds to reach you. How far away is the storm when the time interval between the lightning and the thunder is 8 seconds?

• 5,280
• 5,400
• 8,640
• 43,200

#### Solution:

First, we note that there is a direct relationship between the distance and the time interval. Thus, we use the direct variation formula y = kx, where y = distance and x = time interval.

Next, we use the given information to determine the value of k. We are given that y  = 5,400 feet and x = 5 seconds. Plugging those values into the formula, we have:

y = kx

5,400 = k(5)

5,400 / 5 = k

1,080 = k

Now that we know that the constant of variation k is equal to 1,080, we can answer the question:  “How far away is the storm when the time interval between the lightning and the thunder is 8 seconds?”

y = kx

y = (1,080)(8)

y = 8,640

Now, let’s solve a more challenging question about direct variation.

### Example 5: Direct Variation

Hooke’s Law states that the force needed to keep a spring stretched x units beyond its natural length is directly proportional to x.

A spring has a natural length of 10 cm and a force of 40 N is required to maintain the spring stretched to a length of 15 cm. How many centimeters (beyond its natural length) can a force of 32 N stretch the spring?

• 2.5
• 3
• 3.5
• 4

#### Solution:

We use the direct variation formula y = kx to first determine the constant of variation. We are given that y = 40 and x = 5. Note that x = 15 – 10 = 5 because x is the number of units that the spring is stretched beyond its natural length.

y = kx

40 = k(5)

8 = k

Now that we know that the constant of variation k is equal to 8, we can answer the question:  “How many centimeters (beyond its natural length) can a force of 32 N stretch the spring?”

First, we know that y = 8x is the direct variation equation for this spring. We are given the force y = 32 N. Let’s plug these values into the formula:

y = kx

y = 8x

32 = 8x

x = 4

We see that a force of 32 N stretches the spring 4 cm beyond its natural length.

## Key Takeaways

This article has presented you with ratio and proportion examples. Additionally, we have introduced you to direct variation and given you some practice problems.

We now know that all ratios provide the following two pieces of information:

• how the parts of a ratio relate to each other, called the part-to-part ratio
• how the parts of a ratio relate to the total of both parts, called the part-to-whole ratio

We have learned how to calculate the part-to-whole ratio from the part-to-part ratio, and how to calculate the part-to-part ratios from the part-to-whole ratio.

We have learned that a proportion is an equation that equates two ratios. We write proportions to help us establish equivalent ratios and solve for unknown quantities.

One important proportional relationship is direct variation. It tells us that two quantities are related to each other by a constant. In direct variation, for example, if one quantity doubles, so does the other. In this case, the constant is equal to 2. We use the direct variation formula y = kx to set up direct variation problems.

We’ve covered 3 important concepts tested on the SAT math section. If you encounter SAT ratio and proportion questions, or a question on direct variation, you shouldn’t have any difficulty answering them!

## What’s Next?

While we have provided a solid introduction to ratios, proportions, and direct variation here, we teach these topics in much greater detail in the Target Test Prep SAT course.

If you would like some additional information about the digital SAT math section, read our article that discusses all the math concepts tested on the SAT.