# SAT Math Word Problems

Most students tremble with fear when they think about preparing for the Math portion of the SAT.  After all, there are 20+ topics tested, and the test requires mastery of topics ranging from arithmetic to trigonometry. Some topics cover basic math questions from algebra class, such as solving linear equations or recognizing graphs of parabolas. Solving questions such as these is tough enough, but the SAT adds a layer of difficulty by presenting word problems from a wide range of disciplines that require you to use your algebra skills to solve them.

To tackle these word problems, you must first understand the context of the word problem and then create the equations that you’ll need to solve, based on the information provided in the question.

In this article, I’ll first cover two critical algebra skills you need to master. Then I’ll provide a number of strategies for efficiently solving hard SAT word problems, as well as examples, so we can see these strategies in action and avoid common mistakes.

## Here are the topics we’ll cover:

We’ll start with solving basic linear equations, then we’ll review the substitution method of solving SAT algebra problems. Finally, we’ll use these two techniques to solve math word problem examples.

## Basic Algebra – Solving Single-Variable Linear Equations

The simplest algebra questions on the SAT are single-variable linear equations. Your goal is to solve for the value of the variable:

• 2x + 5 = x
• 4 = 5n
• 2v + 5 = 3v – 7

Note that solving single-variable equations generally requires that we combine all terms containing the variable on one side of the equation and all constant terms on the other side.

Let’s practice simplifying and solving an equation with one variable.

If 8q – 2 = 10q + 14, then what is the value of q?

First, we combine our constant terms by adding 2 to both sides:

8q – 2 + 2 = 10q + 14 + 2

8q = 10q + 16

Next, we combine the variable terms by subtracting 10q from both sides of the equation:

8q – 10q = 10q + 16 – 10q

-2q = 16

Finally, we divide both sides of the equation by -2 to isolate the variable:

-2q / -2 = 16 / -2

q = -8

KEY FACT:

When solving for a single variable, we combine like terms and isolate the variable we are solving for.

Now, let’s practice with an example.

### Example 1: Linear Equation with One Variable

If -2x – 11 = 5x + 3, then x equals which of the following?

• -5
• -2
• 0
• 2

#### Solution:

-2x – 11 = 5x + 3

-2x = 5x + 14

-7x = 14

x = -2

Now let’s turn our attention to applying algebra to solve a straightforward word problem involving sales tax.

## Sales Tax Questions

A common single-variable word problem that you might be asked on the SAT involves the calculation of sales tax. When you buy an item, you not only pay for the item itself but you also generally pay a surcharge called sales tax, typically stated as a percentage of the cost of the item. We add the sales tax to the cost of the item to determine the actual amount we will pay at the register.

The equation for purchasing an item in an area where sales tax is charged is:

Cost of Item + Sales Tax = Total

Let’s look at a basic example of the calculation of sales tax.

Laura buys a blender for $40.00 in a town where the sales tax rate is 4.5%. How much will she have to pay, including sales tax? We see that the cost of the item is$40.00, and the sales tax rate is 4.5%. Thus, the sales tax amount is 4.5% of $40.00, or 40 x 0.045. Let’s put these into the equation: Cost of Item + Sales Tax = Total 40 + (40)(0.045) = Total 40 + 1.8 = Total 41.8 = Total Thus, Laura will pay a total of$41.80: the item cost of $40.00 and the sales tax of$1.80.

TTP PRO TIP:

Know the sales tax formula: Cost of Item + Sales Tax = Total

Let’s look at a sample SAT question about sales tax.

### Sales Tax: Example 1

Alex buys a pair of shoes. The sales tax rate in his town is 6%. The register total for his purchase, including sales tax, is $115.54. How much sales tax did Alex pay? •$6.00
• $6.54 •$6.93
• $15.54 #### Solution: We set up the linear equation by noting that the total purchase price is equal to the cost of the item plus the sales tax: Total = Cost of Item + Sales Tax We are given the total and the sales tax rate. However, we don’t know the actual cost of the item, so we will let x = the cost of the item: 115.54 = x + (0.06)(x) We combine like terms on the right side of the equation and solve for x: 115.54 = 1.06x 115.54 / 1.06 = x 109 = x We see that the cost of the item is$109.00. Now we calculate the sales tax:

109 x 0.06 = 6.54

The sales tax is $6.54. Answer: B Next, let’s discuss linear equations with two variables. ## Linear Equations with Two Variables/Systems of Linear Equations Two-variable equations have not just one but two different variables: • v – u = 12 • 2x + y = 10 • 5z – 3y = 42 These are called two-variable equations because they contain two different variables. If we have a second equation that contains one or both of the variables, the two equations are called a system of linear equations. One common way to solve for the values of the variables is by using the substitution method. Let’s discuss the substitution method now. ### Using the Substitution Method to Solve a System of Linear Equations When we’re working with SAT systems of equations word problems, the essence of the substitution method is that we first isolate one variable in one of the equations. Then we substitute whatever that variable is equal to into the other equation. Let’s practice with an example. #### Substitution Method Practice 1 In the system of linear equations below, determine the value of a. b = 3a (equation 1) a + b = 12 (equation 2) ##### Solution: Looking at our two equations, we see that b is already isolated in equation 1. So, we can substitute 3a (from equation one) for b (in equation two). This gives us: a + 3a = 12 4a = 12 a = 3 #### Substitution Method Practice 2 For the system of linear equations below, what is the value of x? 3x + y = 11 (equation 1) 2x + 5y = 3 (equation 2) Solution: First, let’s isolate y in equation 1 by subtracting 3x from both sides of the equation: 3x + y = 11 (equation 1) y = 11 – 3x Now, we can substitute 11 – 3x for y in equation 2. This gives us: 2x + 5(11 – 3x) = 3 2x + 55 – 15x = 3 -13x = -52 x = 4 KEY FACT: We can use the substitution method to solve a system of linear equations. Next, let’s discuss the many topics on the SAT in which we use algebra to determine the solution. ## When Dealing With Word Problems, Translation Is Key Here are a couple of sample SAT math word problems that you might encounter: Luca buys apples, which cost$2 each, and bananas, which cost $3 each, at the market. If he buys twice as many bananas as apples and spends$48 at the market, how many bananas does he buy?

Tyrone and Greg have a total of 40 marbles. If Tyrone has 4 times the number of marbles that Greg has, how many marbles does Greg have?

The bottom line is that a word problem presents a scenario requiring us to translate the given information into an algebraic equation that we then solve.

Word problems are not just about solving equations; they are also about translating words into equations! Let’s look at some common translations:

• “Is” translates to equals (=)

Kendra is the same age as Carla

Kendra’s age = Carla’s age

• “More” translates to addition (+)

Arita has 6 more marbles than Pablo

Arita = Pablo + 6

• “Less/fewer” translates to subtraction (-)

Sammy has 3 fewer coins than Rati

Sammy = Rati 3

• “Times as many” translates to multiplication (✖)

Harold has 5 times as many newspapers as Phoebe

Harold = Phoebe ✖ 5

TTP PRO TIP:

Know the common translations of words to algebraic equations.

Before jumping into word problem practice questions, let’s discuss one point of confusion students have when translating words into equations: properly balancing the equations.

## You Must Make Sure Your Equations are Balanced

A basic principle for translating words into equations is that the equations must be balanced correctly. If they are not, you might obtain reversed values for your variables’ values, or you might get a negative answer. Let’s look at a few correct and incorrect ways to balance equations, and we’ll explain why the equations are balanced correctly.

### Balancing Equations: Example 1

Sherry has 30 more dollars than Melanie.

#### Solution:

When we are setting up this equation, we must understand that Sherry has more money than Melanie. For example, if Melanie has 20 dollars, then Sherry has 50 dollars.

So, to properly balance the equation, we must add 30 dollars to Melanie’s amount and set it equal to Sherry’s amount.

If we let S = Sherry’s money and M = Melanie’s money, we have:

S = 30 + M

### Balancing Equations: Example 2

Nala has 4 fewer toys than Frank.

#### Solution:

When considering this equation, we must understand that Nala has fewer toys than Frank.

In other words, Frank has more toys than Nala.

So, to set up a balanced equation, we must subtract 4 from Frank’s amount and set that equal to Nala’s.

If we let N = the number of Nala’s toys and F = the number of Frank’s toys, we have:

N – 4 = F

### Balancing Equations: Example 3

There are 5 times as many baseballs as tennis balls.

#### Solution:

When considering this equation, we must understand that there are more baseballs than tennis balls.

So, to set up a correct equation, we multiply the number of tennis balls by 5 and set that amount equal to the number of baseballs.

If we let B = the number of baseballs and T = the number of tennis balls, we have:

B = 5T

Now that we are familiar with how to translate, balance, and solve equations, let’s jump into the various types of general word problems you will encounter on the SAT, and how to solve them.

## “Number of Items” Questions

We have already encountered some simple examples of “number of items” questions in this article. Now, let’s attack a full-fledged problem like those you might encounter on test day.

### “Number of Items”: Example 1

Julia and Tory have a total of 30 candy bars. If Julia has 5 times as many candy bars as Tory, how many candy bars does Tory have?

• 4
• 5
• 6
• 10

#### Solution:

First, we define our variables.

Let’s let J = the number of candy bars Julia has and T = the number of candy bars Tory has.

Next, let’s create equations based on the information given in the question stem.

“Julia and Tory have a total of 30 candy bars” is translated as:

J + T = 30 (equation 1)

“Julia has 5 times as many candy bars as Tory” is translated as:

J = 5T (equation 2)

Next, we use the substitution method by substituting 5T for J in equation 1:

5T + T = 30

6T = 30

T = 5

Let’s practice one more. This time, we will add a twist.

### “Number of Items”: Example 2

Leti has 4 times as many cookies as Henrik. If Leti gives Henrik 7 cookies, Henrik will have 4 fewer cookies than Leti. How many cookies did Leti originally have?

• 6
• 8
• 12
• 24

#### Solution:

We can let L = the number of cookies Leti has and H = the number of cookies Henrik has. Next, we use the information from the question stem to create equations.

Because Leti has 4 times as many cookies as Henrik, we have:

L = 4H (equation 1)

After Leti gives Henrik 7 cookies, Leti will have L – 7 cookies. After Henrik gets the 7 cookies from Leti, he will have H + 7 cookies. He will still have 4 fewer cookies than Leti.

Thus, our second equation is:

L – 7 = (H + 7) +4

L = H + 18 (equation 2)

Next, we can use the substitution method, substituting 4H for L in equation 2:

4H = H + 18

3H = 18

H = 6

Next, let’s discuss another common type of SAT word problem: age problems.

## Age Problems

Age problems are a common type of SAT problem with a unique spin. Instead of having a straightforward translation, as we practiced above, an age problem will usually make us translate words into equations based on an age in the past, an age in the future, or even both.

For example, we might need to consider Ann’s age 10 years from now. To do this, we let Ann’s age today = A, so her age in 10 years will be A + 10.

Similarly, Ann’s age 12 years ago would be expressed as A – 12.

KEY FACT:

Age problems generally require us to compare ages in the past or future.

Now, let’s practice with an example.

### Age Problem: Example 1

Chantal is 6 years younger than Yolanda. If, in 5 years, Yolanda will be twice as old as Chantal, how old is Chantal today?

• 1
• 2
• 6
• 7

#### Solution:

Let’s let C = Chantal’s age today and Y = Yolanda’s age today. Next, let’s create some equations.

Because Chantal is 6 years younger than Yolanda, we have:

C = Y – 6 (equation 1)

Now we express what each age will be in 5 years:Chantal will be C + 5 and Yolanda will be Y + 6. The relationship between their ages in 5 years is that Yolanda will be twice Chantal’s age:

Y + 5 = 2(C + 5)

Y + 5 = 2C + 10 (equation 2)

Next, we use the substitution method and substitute Y – 6 for C in equation 2, and we have:

Y + 5 = 2(Y – 6) + 10

Y + 5 = 2Y – 12 + 10

Y + 5 = 2Y – 2

7 = Y

Yolanda is 7 years old today. Thus, Chantal is 7 – 6 = 1 year old today.

Next, let’s discuss money problems.

## Money Problems

In general, money problems on the SAT involve two different commodities, such as adult tickets and child tickets, and the revenue earned from selling them. We can create two equations, one relating the number of items and the second relating the revenue earned from the sale of the items.

We will see that the substitution method is extremely useful for solving the two equations that are created from the given information.

TTP PRO TIP:

Be prepared to use the substitution method when dealing with money problems.

### Money Problem: Example 1

At an ice cream stand, customers can buy either a small cone for $3.00 or a large cone for$4.50. Yesterday, the owner reported he had sold 720 cones, and his revenue for the day was $2,340. How many small cones did he sell? • 120 • 300 • 420 • 600 #### Solution: First, let’s define our variables. We can let the number of small cones sold = x and the number of large cones sold = y. Next, let’s create our equations. Because the owner sold 720 ice cream cones, we have: x + y = 720 (equation 1) His total revenue was$2,340 from selling x small cones for $3.00 each and y large cones for$4.50 each:

3x + 4.5y = 2,340 (equation 2)

Let’s isolate x in equation 1:

x = 720 – y

Now, let’s substitute 720 – y for x into equation 2 and solve:

3(720 – y) + 4.5y = 2,340

2,160 – 3y + 4.5y = 2,340

1.5y = 180

y = 120

Since the number of large cones sold is 120, the number of small cones sold is 720 – 120 = 600.