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Our math problem solver that lets you input a wide variety of math math problems and it will provide a step by step answer. This math solver excels at math word problems as well as a wide range of math subjects.

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Math Word Problem Solutions

Math word problems require interpreting what is being asked and simplifying that into a basic math equation. Once you have the equation you can then enter that into the problem solver as a basic math or algebra question to be correctly solved. Below are math word problem examples and their simplified forms.

Word Problem: Rachel has 17 apples. She gives some to Sarah. Sarah now has 8 apples. How many apples did Rachel give her?

Simplified Equation: 17 - x = 8

Word Problem: Rhonda has 12 marbles more than Douglas. Douglas has 6 marbles more than Bertha. Rhonda has twice as many marbles as Bertha has. How many marbles does Douglas have?

Variables: Rhonda's marbles is represented by (r), Douglas' marbles is represented by (d) and Bertha's marbles is represented by (b)

Simplified Equation: {r = d + 12, d = b + 6, r = 2 �� b}

Word Problem: if there are 40 cookies all together and Angela takes 10 and Brett takes 5 how many are left?

Simplified: 40 - 10 - 5

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Precalculus Solutions

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Calculus Solutions

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Statistics Solutions

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Solving Word Questions

With LOTS of examples!

In Algebra we often have word questions like:

Example: Sam and Alex play tennis.

On the weekend Sam played 4 more games than Alex did, and together they played 12 games.

How many games did Alex play?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

  • Read the whole thing first
  • Do a sketch if possible
  • Assign letters for the values
  • Find or work out formulas

You should also write down what is actually being asked for , so you know where you are going and when you have arrived!

Also look for key words:

Thinking Clearly

Some wording can be tricky, making it hard to think "the right way around", such as:

Example: Sam has 2 dollars less than Alex. How do we write this as an equation?

  • Let S = dollars Sam has
  • Let A = dollars Alex has

Now ... is that: S − 2 = A

or should it be: S = A − 2

or should it be: S = 2 − A

The correct answer is S = A − 2

( S − 2 = A is a common mistake, as the question is written "Sam ... 2 less ... Alex")

Example: on our street there are twice as many dogs as cats. How do we write this as an equation?

  • Let D = number of dogs
  • Let C = number of cats

Now ... is that: 2D = C

or should it be: D = 2C

Think carefully now!

The correct answer is D = 2C

( 2D = C is a common mistake, as the question is written "twice ... dogs ... cats")

Let's start with a really simple example so we see how it's done:

Example: A rectangular garden is 12m by 5m, what is its area ?

Turn the English into Algebra:

  • Use w for width of rectangle: w = 12m
  • Use h for height of rectangle: h = 5m

Formula for Area of a Rectangle : A = w × h

We are being asked for the Area.

A = w × h = 12 × 5 = 60 m 2

The area is 60 square meters .

Now let's try the example from the top of the page:

tennis

Example: Sam and Alex play Tennis. On the weekend Sam played 4 more games than Alex did, and together they played 12 games. How many games did Alex play?

  • Use S for how many games Sam played
  • Use A for how many games Alex played

We know that Sam played 4 more games than Alex, so: S = A + 4

And we know that together they played 12 games: S + A = 12

We are being asked for how many games Alex played: A

Which means that Alex played 4 games of tennis.

Check: Sam played 4 more games than Alex, so Sam played 8 games. Together they played 8 + 4 = 12 games. Yes!

A slightly harder example:

table

Example: Alex and Sam also build tables. Together they make 10 tables in 12 days. Alex working alone can make 10 in 30 days. How long would it take Sam working alone to make 10 tables?

  • Use a for Alex's work rate
  • Use s for Sam's work rate

12 days of Alex and Sam is 10 tables, so: 12a + 12s = 10

30 days of Alex alone is also 10 tables: 30a = 10

We are being asked how long it would take Sam to make 10 tables.

30a = 10 , so Alex's rate (tables per day) is: a = 10/30 = 1/3

Which means that Sam's rate is half a table a day (faster than Alex!)

So 10 tables would take Sam just 20 days.

Should Sam be paid more I wonder?

And another "substitution" example:

track

Example: Jenna is training hard to qualify for the National Games. She has a regular weekly routine, training for five hours a day on some days and 3 hours a day on the other days. She trains altogether 27 hours in a seven day week. On how many days does she train for five hours?

  • The number of "5 hour" days: d
  • The number of "3 hour" days: e

We know there are seven days in the week, so: d + e = 7

And she trains 27 hours in a week, with d 5 hour days and e 3 hour days: 5d + 3e = 27

We are being asked for how many days she trains for 5 hours: d

The number of "5 hour" days is 3

Check : She trains for 5 hours on 3 days a week, so she must train for 3 hours a day on the other 4 days of the week.

3 × 5 hours = 15 hours, plus 4 × 3 hours = 12 hours gives a total of 27 hours

Some examples from Geometry:

Example: A circle has an area of 12 mm 2 , what is its radius?

  • Use A for Area: A = 12 mm 2
  • Use r for radius

And the formula for Area is: A = π r 2

We are being asked for the radius.

We need to rearrange the formula to find the area

Example: A cube has a volume of 125 mm 3 , what is its surface area?

Make a quick sketch:

  • Use V for Volume
  • Use A for Area
  • Use s for side length of cube
  • Volume of a cube: V = s 3
  • Surface area of a cube: A = 6s 2

We are being asked for the surface area.

First work out s using the volume formula:

Now we can calculate surface area:

An example about Money:

pizza

Example: Joel works at the local pizza parlor. When he works overtime he earns 1¼ times the normal rate. One week Joel worked for 40 hours at the normal rate of pay and also worked 12 hours overtime. If Joel earned $660 altogether in that week, what is his normal rate of pay?

  • Joel's normal rate of pay: $N per hour
  • Joel works for 40 hours at $N per hour = $40N
  • When Joel does overtime he earns 1¼ times the normal rate = $1.25N per hour
  • Joel works for 12 hours at $1.25N per hour = $(12 × 1¼N) = $15N
  • And together he earned $660, so:

$40N + $(12 × 1¼N) = $660

We are being asked for Joel's normal rate of pay $N.

So Joel’s normal rate of pay is $12 per hour

Joel’s normal rate of pay is $12 per hour, so his overtime rate is 1¼ × $12 per hour = $15 per hour. So his normal pay of 40 × $12 = $480, plus his overtime pay of 12 × $15 = $180 gives us a total of $660

More about Money, with these two examples involving Compound Interest

Example: Alex puts $2000 in the bank at an annual compound interest of 11%. How much will it be worth in 3 years?

This is the compound interest formula:

So we will use these letters:

  • Present Value PV = $2,000
  • Interest Rate (as a decimal): r = 0.11
  • Number of Periods: n = 3
  • Future Value (the value we want): FV

We are being asked for the Future Value: FV

Example: Roger deposited $1,000 into a savings account. The money earned interest compounded annually at the same rate. After nine years Roger's deposit has grown to $1,551.33 What was the annual rate of interest for the savings account?

The compound interest formula:

  • Present Value PV = $1,000
  • Interest Rate (the value we want): r
  • Number of Periods: n = 9
  • Future Value: FV = $1,551.33

We are being asked for the Interest Rate: r

So the annual rate of interest is 5%

Check : $1,000 × (1.05) 9 = $1,000 × 1.55133 = $1,551.33

And an example of a Ratio question:

Example: At the start of the year the ratio of boys to girls in a class is 2 : 1 But now, half a year later, four boys have left the class and there are two new girls. The ratio of boys to girls is now 4 : 3 How many students are there altogether now?

  • Number of boys now: b
  • Number of girls now: g

The current ratio is 4 : 3

Which can be rearranged to 3b = 4g

At the start of the year there was (b + 4) boys and (g − 2) girls, and the ratio was 2 : 1

b + 4 g − 2 = 2 1

Which can be rearranged to b + 4 = 2(g − 2)

We are being asked for how many students there are altogether now: b + g

There are 12 girls !

And 3b = 4g , so b = 4g/3 = 4 × 12 / 3 = 16 , so there are 16 boys

So there are now 12 girls and 16 boys in the class, making 28 students altogether .

There are now 16 boys and 12 girls, so the ratio of boys to girls is 16 : 12 = 4 : 3 At the start of the year there were 20 boys and 10 girls, so the ratio was 20 : 10 = 2 : 1

And now for some Quadratic Equations :

Example: The product of two consecutive even integers is 168. What are the integers?

Consecutive means one after the other. And they are even , so they could be 2 and 4, or 4 and 6, etc.

We will call the smaller integer n , and so the larger integer must be n+2

And we are told the product (what we get after multiplying) is 168, so we know:

n(n + 2) = 168

We are being asked for the integers

That is a Quadratic Equation , and there are many ways to solve it. Using the Quadratic Equation Solver we get −14 and 12.

Check −14: −14(−14 + 2) = (−14)×(−12) = 168 YES

Check 12: 12(12 + 2) = 12×14 = 168 YES

So there are two solutions: −14 and −12 is one, 12 and 14 is the other.

Note: we could have also tried "guess and check":

  • We could try, say, n=10: 10(12) = 120 NO (too small)
  • Next we could try n=12: 12(14) = 168 YES

But unless we remember that multiplying two negatives make a positive we might overlook the other solution of (−14)×(−12).

Example: You are an Architect. Your client wants a room twice as long as it is wide. They also want a 3m wide veranda along the long side. Your client has 56 square meters of beautiful marble tiles to cover the whole area. What should the length of the room be?

Let's first make a sketch so we get things right!:

  • the length of the room: L
  • the width of the room: W
  • the total Area including veranda: A
  • the width of the room is half its length: W = ½L
  • the total area is the (room width + 3) times the length: A = (W+3) × L = 56

We are being asked for the length of the room: L

This is a quadratic equation , there are many ways to solve it, this time let's use factoring :

And so L = 8 or −14

There are two solutions to the quadratic equation, but only one of them is possible since the length of the room cannot be negative!

So the length of the room is 8 m

L = 8, so W = ½L = 4

So the area of the rectangle = (W+3) × L = 7 × 8 = 56

There we are ...

... I hope these examples will help you get the idea of how to handle word questions. Now how about some practice?

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Equation Word Problems Worksheets

This compilation of a meticulously drafted equation word problems worksheets is designed to get students to write and solve a variety of one-step, two-step and multi-step equations that involve integers, fractions, and decimals. These worksheets are best suited for students in grade 6 through high school. Click on the 'Free' icons to sample our handouts.

One Step Equation Word Problem Worksheets

One Step Equation Word Problem Worksheets

Read and solve this series of word problems that involve one-step equations. Apply basic operations to find the value of unknowns.

(15 Worksheets)

Two-Step Equation Word Problems: Integers

Two-Step Equation Word Problems: Integers

Interpret this set of word problems that require two-step operations to solve the equations. Each printable worksheet has five word problems ideal for 6th grade, 7th grade, and 8th grade students.

  • Download the set

Multi-Step Equation Word Problems: Integers

Multi-Step Equation Word Problems: Integers

Read each multi-step word problem in these high school pdf worksheets and set up the equation. Solve and find the value of the unknown. More than two steps are required to solve the problems.

Two-Step Equation Word Problems: Fractions and Decimals

Two-Step Equation Word Problems: Fractions and Decimals

Read each word problem and set up the two-step equation. Solve the equation and find the solution. This selection of worksheets includes both fractions and decimals.

Multi-Step Equation Word Problems: Fractions and Decimals

Multi-Step Equation Word Problems: Fractions and Decimals

Write multi-step equations that involve both fractions and decimals based on the word problems provided here. Validate your responses with our answer keys.

MCQ - Equation Word Problems

MCQ - Equation Word Problems

Pick the correct two-step equation that best matches word problems presented here. Evaluate the ability of students to solve two-step equations with this array of MCQ worksheets.

Related Worksheets

» One-Step Equation

» Two-Step Equation

» Multi-Step Equation

» Algebraic Identities

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Algebra Topics  - Introduction to Word Problems

Algebra topics  -, introduction to word problems, algebra topics introduction to word problems.

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Algebra Topics: Introduction to Word Problems

Lesson 9: introduction to word problems.

/en/algebra-topics/solving-equations/content/

What are word problems?

A word problem is a math problem written out as a short story or scenario. Basically, it describes a realistic problem and asks you to imagine how you would solve it using math. If you've ever taken a math class, you've probably solved a word problem. For instance, does this sound familiar?

Johnny has 12 apples. If he gives four to Susie, how many will he have left?

You could solve this problem by looking at the numbers and figuring out what the problem is asking you to do. In this case, you're supposed to find out how many apples Johnny has left at the end of the problem. By reading the problem, you know Johnny starts out with 12 apples. By the end, he has 4 less because he gave them away. You could write this as:

12 - 4 = 8 , so you know Johnny has 8 apples left.

Word problems in algebra

If you were able to solve this problem, you should also be able to solve algebra word problems. Yes, they involve more complicated math, but they use the same basic problem-solving skills as simpler word problems.

You can tackle any word problem by following these five steps:

  • Read through the problem carefully, and figure out what it's about.
  • Represent unknown numbers with variables.
  • Translate the rest of the problem into a mathematical expression.
  • Solve the problem.
  • Check your work.

We'll work through an algebra word problem using these steps. Here's a typical problem:

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took two days, and the van cost $360. How many miles did she drive?

It might seem complicated at first glance, but we already have all of the information we need to solve it. Let's go through it step by step.

Step 1: Read through the problem carefully.

With any problem, start by reading through the problem. As you're reading, consider:

  • What question is the problem asking?
  • What information do you already have?

Let's take a look at our problem again. What question is the problem asking? In other words, what are you trying to find out?

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

There's only one question here. We're trying to find out how many miles Jada drove . Now we need to locate any information that will help us answer this question.

There are a few important things we know that will help us figure out the total mileage Jada drove:

  • The van cost $30 per day.
  • In addition to paying a daily charge, Jada paid $0.50 per mile.
  • Jada had the van for 2 days.
  • The total cost was $360 .

Step 2: Represent unknown numbers with variables.

In algebra, you represent unknown numbers with letters called variables . (To learn more about variables, see our lesson on reading algebraic expressions .) You can use a variable in the place of any amount you don't know. Looking at our problem, do you see a quantity we should represent with a variable? It's often the number we're trying to find out.

Since we're trying to find the total number of miles Jada drove, we'll represent that amount with a variable—at least until we know it. We'll use the variable m for miles . Of course, we could use any variable, but m should be easy to remember.

Step 3: Translate the rest of the problem.

Let's take another look at the problem, with the facts we'll use to solve it highlighted.

The rate to rent a small moving van is $30 per day , plus $0.50 per mile . Jada rented a van to drive to her new home. It took 2 days , and the van cost $360 . How many miles did she drive?

We know the total cost of the van, and we know that it includes a fee for the number of days, plus another fee for the number of miles. It's $30 per day, and $0.50 per mile. A simpler way to say this would be:

$30 per day plus $0.50 per mile is $360.

If you look at this sentence and the original problem, you can see that they basically say the same thing: It cost Jada $30 per day and $0.50 per mile, and her total cost was $360 . The shorter version will be easier to translate into a mathematical expression.

Let's start by translating $30 per day . To calculate the cost of something that costs a certain amount per day, you'd multiply the per-day cost by the number of days—in other words, 30 per day could be written as 30 ⋅ days, or 30 times the number of days . (Not sure why you'd translate it this way? Check out our lesson on writing algebraic expressions .)

$30 per day and $.50 per mile is $360

$30 ⋅ day + $.50 ⋅ mile = $360

As you can see, there were a few other words we could translate into operators, so and $.50 became + $.50 , $.50 per mile became $.50 ⋅ mile , and is became = .

Next, we'll add in the numbers and variables we already know. We already know the number of days Jada drove, 2 , so we can replace that. We've also already said we'll use m to represent the number of miles, so we can replace that too. We should also take the dollar signs off of the money amounts to make them consistent with the other numbers.

30 ⋅ 2 + .5 ⋅ m = 360

Now we have our expression. All that's left to do is solve it.

Step 4: Solve the problem.

This problem will take a few steps to solve. (If you're not sure how to do the math in this section, you might want to review our lesson on simplifying expressions .) First, let's simplify the expression as much as possible. We can multiply 30 and 2, so let's go ahead and do that. We can also write .5 ⋅ m as 0.5 m .

60 + .5m = 360

Next, we need to do what we can to get the m alone on the left side of the equals sign. Once we do that, we'll know what m is equal to—in other words, it will let us know the number of miles in our word problem.

We can start by getting rid of the 60 on the left side by subtracting it from both sides .

The only thing left to get rid of is .5 . Since it's being multiplied with m , we'll do the reverse and divide both sides of the equation with it.

.5 m / .5 is m and 300 / 0.50 is 600 , so m = 600 . In other words, the answer to our problem is 600 —we now know Jada drove 600 miles.

Step 5: Check the problem.

To make sure we solved the problem correctly, we should check our work. To do this, we can use the answer we just got— 600 —and calculate backward to find another of the quantities in our problem. In other words, if our answer for Jada's distance is correct, we should be able to use it to work backward and find another value, like the total cost. Let's take another look at the problem.

According to the problem, the van costs $30 per day and $0.50 per mile. If Jada really did drive 600 miles in 2 days, she could calculate the cost like this:

$30 per day and $0.50 per mile

30 ⋅ day + .5 ⋅ mile

30 ⋅ 2 + .5 ⋅ 600

According to our math, the van would cost $360, which is exactly what the problem says. This means our solution was correct. We're done!

While some word problems will be more complicated than others, you can use these basic steps to approach any word problem. On the next page, you can try it for yourself.

Let's practice with a couple more problems. You can solve these problems the same way we solved the first one—just follow the problem-solving steps we covered earlier. For your reference, these steps are:

If you get stuck, you might want to review the problem on page 1. You can also take a look at our lesson on writing algebraic expressions for some tips on translating written words into math.

Try completing this problem on your own. When you're done, move on to the next page to check your answer and see an explanation of the steps.

A single ticket to the fair costs $8. A family pass costs $25 more than half of that. How much does a family pass cost?

Here's another problem to do on your own. As with the last problem, you can find the answer and explanation to this one on the next page.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

Problem 1 Answer

Here's Problem 1:

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

Answer: $29

Let's solve this problem step by step. We'll solve it the same way we solved the problem on page 1.

Step 1: Read through the problem carefully

The first in solving any word problem is to find out what question the problem is asking you to solve and identify the information that will help you solve it . Let's look at the problem again. The question is right there in plain sight:

So is the information we'll need to answer the question:

  • A single ticket costs $8 .
  • The family pass costs $25 more than half the price of the single ticket.

Step 2: Represent the unknown numbers with variables

The unknown number in this problem is the cost of the family pass . We'll represent it with the variable f .

Step 3: Translate the rest of the problem

Let's look at the problem again. This time, the important facts are highlighted.

A single ticket to the fair costs $8 . A family pass costs $25 more than half that . How much does a family pass cost?

In other words, we could say that the cost of a family pass equals half of $8, plus $25 . To turn this into a problem we can solve, we'll have to translate it into math. Here's how:

  • First, replace the cost of a family pass with our variable f .

f equals half of $8 plus $25

  • Next, take out the dollar signs and replace words like plus and equals with operators.

f = half of 8 + 25

  • Finally, translate the rest of the problem. Half of can be written as 1/2 times , or 1/2 ⋅ :

f = 1/2 ⋅ 8 + 25

Step 4: Solve the problem

Now all we have to do is solve our problem. Like with any problem, we can solve this one by following the order of operations.

  • f is already alone on the left side of the equation, so all we have to do is calculate the right side.
  • First, multiply 1/2 by 8 . 1/2 ⋅ 8 is 4 .
  • Next, add 4 and 25. 4 + 25 equals 29 .

That's it! f is equal to 29. In other words, the cost of a family pass is $29 .

Step 5: Check your work

Finally, let's check our work by working backward from our answer. In this case, we should be able to correctly calculate the cost of a single ticket by using the cost we calculated for the family pass. Let's look at the original problem again.

We calculated that a family pass costs $29. Our problem says the pass costs $25 more than half the cost of a single ticket. In other words, half the cost of a single ticket will be $25 less than $29.

  • We could translate this into this equation, with s standing for the cost of a single ticket.

1/2s = 29 - 25

  • Let's work on the right side first. 29 - 25 is 4 .
  • To find the value of s , we have to get it alone on the left side of the equation. This means getting rid of 1/2 . To do this, we'll multiply each side by the inverse of 1/2: 2 .

According to our math, s = 8 . In other words, if the family pass costs $29, the single ticket will cost $8. Looking at our original problem, that's correct!

So now we're sure about the answer to our problem: The cost of a family pass is $29 .

Problem 2 Answer

Here's Problem 2:

Answer: $70

Let's go through this problem one step at a time.

Start by asking what question the problem is asking you to solve and identifying the information that will help you solve it . What's the question here?

To solve the problem, you'll have to find out how much money Mo gave to charity. All the important information you need is in the problem:

  • The amount Flor donated is three times as much the amount Mo donated
  • Flor and Mo's donations add up to $280 total

The unknown number we're trying to identify in this problem is Mo's donation . We'll represent it with the variable m .

Here's the problem again. This time, the important facts are highlighted.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo . Between the two of them, they donated $280 . How much money did Mo give?

The important facts of the problem could also be expressed this way:

Mo's donation plus Flor's donation equals $280

Because we know that Flor's donation is three times as much as Mo's donation, we could go even further and say:

Mo's donation plus three times Mo's donation equals $280

We can translate this into a math problem in only a few steps. Here's how:

  • Because we've already said we'll represent the amount of Mo's donation with the variable m , let's start by replacing Mo's donation with m .

m plus three times m equals $280

  • Next, we can put in mathematical operators in place of certain words. We'll also take out the dollar sign.

m + three times m = 280

  • Finally, let's write three times mathematically. Three times m can also be written as 3 ⋅ m , or just 3 m .

m + 3m = 280

It will only take a few steps to solve this problem.

  • To get the correct answer, we'll have to get m alone on one side of the equation.
  • To start, let's add m and 3 m . That's 4 m .
  • We can get rid of the 4 next to the m by dividing both sides by 4. 4 m / 4 is m , and 280 / 4 is 70 .

We've got our answer: m = 70 . In other words, Mo donated $70 .

The answer to our problem is $70 , but we should check just to be sure. Let's look at our problem again.

If our answer is correct, $70 and three times $70 should add up to $280 .

  • We can write our new equation like this:

70 + 3 ⋅ 70 = 280

  • The order of operations calls for us to multiply first. 3 ⋅ 70 is 210.

70 + 210 = 280

  • The last step is to add 70 and 210. 70 plus 210 equals 280 .

280 is the combined cost of the tickets in our original problem. Our answer is correct : Mo gave $70 to charity.

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1.20: Word Problems for Linear Equations

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Word problems are important applications of linear equations. We start with examples of translating an English sentence or phrase into an algebraic expression.

Example 18.1

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

Solution: We call the variable \(x .\) Twice the variable is \(2 x .\) Adding \(2 x\) to 4 gives:

\[4 + 2x\nonumber\]

b) Three times a number is subtracted from 7.

Solution: Three times a number is \(3 x .\) We need to subtract \(3 x\) from 7. This means:\

\[7-3 x\nonumber\]

c) 8 less than a number.

Solution: The number is denoted by \(x .8\) less than \(x\) mean, that we need to subtract 8 from it. We get:

\[x-8\nonumber\]

For example, 8 less than 10 is \(10-8=2\).

d) Subtract \(5 p^{2}-7 p+2\) from \(3 p^{2}+4 p\) and simplify.

Solution: We need to calculate \(3 p^{2}+4 p\) minus \(5 p^{2}-7 p+2:\)

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)\nonumber\]

Simplifying this expression gives:

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)=3 p^{2}+4 p-5 p^{2}+7 p-2 =-2 p^{2}+11 p-2\nonumber\]

e) The amount of money given by \(x\) dimes and \(y\) quarters.

Solution: Each dime is worth 10 cents, so that this gives a total of \(10 x\) cents. Each quarter is worth 25 cents, so that this gives a total of \(25 y\) cents. Adding the two amounts gives a total of

\[10 x+25 y \text{ cents or } .10x + .25y \text{ dollars}\nonumber\]

Now we deal with word problems that directly describe an equation involving one variable, which we can then solve.

Example 18.2

Solve the following word problems:

a) Five times an unknown number is equal to 60. Find the number.

Solution: We translate the problem to algebra:

\[5x = 60\nonumber\]

We solve this for \(x\) :

\[x=\frac{60}{5}=12\nonumber\]

b) If 5 is subtracted from twice an unknown number, the difference is \(13 .\) Find the number.

Solution: Translating the problem into an algebraic equation gives:

\[2x − 5 = 13\nonumber\]

We solve this for \(x\). First, add 5 to both sides.

\[2x = 13 + 5, \text{ so that } 2x = 18\nonumber\]

Dividing by 2 gives \(x=\frac{18}{2}=9\).

c) A number subtracted from 9 is equal to 2 times the number. Find the number.

Solution: We translate the problem to algebra.

\[9 − x = 2x\nonumber\]

We solve this as follows. First, add \(x\) :

\[9 = 2x + x \text{ so that } 9 = 3x\nonumber\]

Then the answer is \(x=\frac{9}{3}=3\)

d) Multiply an unknown number by five is equal to adding twelve to the unknown number. Find the number.

Solution: We have the equation:

\[5x = x + 12.\nonumber\]

Subtracting \(x\) gives

\[4x = 12.\nonumber\]

Dividing both sides by 4 gives the answer: \(x=3\).

e) Adding nine to a number gives the same result as subtracting seven from three times the number. Find the number.

Solution: Adding 9 to a number is written as \(x+9,\) while subtracting 7 from three times the number is written as \(3 x-7\). We therefore get the equation:

\[x + 9 = 3x − 7.\nonumber\]

We solve for \(x\) by adding 7 on both sides of the equation:

\[x + 16 = 3x.\nonumber\]

Then we subtract \(x:\)

\[16 = 2x.\nonumber\]

After dividing by \(2,\) we obtain the answer \(x=8\)

The following word problems consider real world applications. They require to model a given situation in the form of an equation.

Example 18.3

a) Due to inflation, the price of a loaf of bread has increased by \(5 \%\). How much does the loaf of bread cost now, when its price was \(\$ 2.40\) last year?

Solution: We calculate the price increase as \(5 \% \cdot \$ 2.40 .\) We have

\[5 \% \cdot 2.40=0.05 \cdot 2.40=0.1200=0.12\nonumber\]

We must add the price increase to the old price.

\[2.40+0.12=2.52\nonumber\]

The new price is therefore \(\$ 2.52\).

b) To complete a job, three workers get paid at a rate of \(\$ 12\) per hour. If the total pay for the job was \(\$ 180,\) then how many hours did the three workers spend on the job?

Solution: We denote the number of hours by \(x\). Then the total price is calculated as the price per hour \((\$ 12)\) times the number of workers times the number of hours \((3) .\) We obtain the equation

\[12 \cdot 3 \cdot x=180\nonumber\]

Simplifying this yields

\[36 x=180\nonumber\]

Dividing by 36 gives

\[x=\frac{180}{36}=5\nonumber\]

Therefore, the three workers needed 5 hours for the job.

c) A farmer cuts a 300 foot fence into two pieces of different sizes. The longer piece should be four times as long as the shorter piece. How long are the two pieces?

\[x+4 x=300\nonumber\]

Combining the like terms on the left, we get

\[5 x=300\nonumber\]

Dividing by 5, we obtain that

\[x=\frac{300}{5}=60\nonumber\]

Therefore, the shorter piece has a length of 60 feet, while the longer piece has four times this length, that is \(4 \times 60\) feet \(=240\) feet.

d) If 4 blocks weigh 28 ounces, how many blocks weigh 70 ounces?

Solution: We denote the weight of a block by \(x .\) If 4 blocks weigh \(28,\) then a block weighs \(x=\frac{28}{4}=7\)

How many blocks weigh \(70 ?\) Well, we only need to find \(\frac{70}{7}=10 .\) So, the answer is \(10 .\)

Note You can solve this problem by setting up and solving the fractional equation \(\frac{28}{4}=\frac{70}{x}\). Solving such equations is addressed in chapter 24.

e) If a rectangle has a length that is three more than twice the width and the perimeter is 20 in, what are the dimensions of the rectangle?

Solution: We denote the width by \(x\). Then the length is \(2 x+3\). The perimeter is 20 in on one hand and \(2(\)length\()+2(\)width\()\) on the other. So we have

\[20=2 x+2(2 x+3)\nonumber\]

Distributing and collecting like terms give

\[20=6 x+6\nonumber\]

Subtracting 6 from both sides of the equation and then dividing both sides of the resulting equation by 6 gives:

\[20-6=6 x \Longrightarrow 14=6 x \Longrightarrow x=\frac{14}{6} \text { in }=\frac{7}{3} \text { in }=2 \frac{1}{3} \text { in. }\nonumber\]

f) If a circle has circumference 4in, what is its radius?

Solution: We know that \(C=2 \pi r\) where \(C\) is the circumference and \(r\) is the radius. So in this case

\[4=2 \pi r\nonumber\]

Dividing both sides by \(2 \pi\) gives

\[r=\frac{4}{2 \pi}=\frac{2}{\pi} \text { in } \approx 0.63 \mathrm{in}\nonumber\]

g) The perimeter of an equilateral triangle is 60 meters. How long is each side?

Solution: Let \(x\) equal the side of the triangle. Then the perimeter is, on the one hand, \(60,\) and on other hand \(3 x .\) So \(3 x=60\) and dividing both sides of the equation by 3 gives \(x=20\) meters.

h) If a gardener has \(\$ 600\) to spend on a fence which costs \(\$ 10\) per linear foot and the area to be fenced in is rectangular and should be twice as long as it is wide, what are the dimensions of the largest fenced in area?

Solution: The perimeter of a rectangle is \(P=2 L+2 W\). Let \(x\) be the width of the rectangle. Then the length is \(2 x .\) The perimeter is \(P=2(2 x)+2 x=6 x\). The largest perimeter is \(\$ 600 /(\$ 10 / f t)=60\) ft. So \(60=6 x\) and dividing both sides by 6 gives \(x=60 / 6=10\). So the dimensions are 10 feet by 20 feet.

i) A trapezoid has an area of 20.2 square inches with one base measuring 3.2 in and the height of 4 in. Find the length of the other base.

Solution: Let \(b\) be the length of the unknown base. The area of the trapezoid is on the one hand 20.2 square inches. On the other hand it is \(\frac{1}{2}(3.2+b) \cdot 4=\) \(6.4+2 b .\) So

\[20.2=6.4+2 b\nonumber\]

Multiplying both sides by 10 gives

\[202=64+20 b\nonumber\]

Subtracting 64 from both sides gives

\[b=\frac{138}{20}=\frac{69}{10}=6.9 \text { in }\nonumber\]

and dividing by 20 gives

Exit Problem

Write an equation and solve: A car uses 12 gallons of gas to travel 100 miles. How many gallons would be needed to travel 450 miles?

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Translating Word Problems: Keywords

Keywords Examples

The hardest thing about doing word problems is using the part where you need to take the English words and translate them into mathematics. Usually, once you get the math equation, you're fine; the actual math involved is often fairly simple. But figuring out the actual equation can seem nearly impossible. What follows is a list of hints and helps. Be advised, however: To really learn "how to do" word problems, you will need to practice, practice, practice.

How do I convert word problems into math?

  • Read the entire exercise.
  • Work in an organized manner.
  • Look for the keywords.
  • Apply your knowledge of "the real world".

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Algebra Word Problems

Step 1 in effectively translating and solving word problems is to read the problem entirely. Don't start trying to solve anything when you've only read half a sentence. Try first to get a feel for the whole problem; try first to see what information you have, and then figure out what you still need.

Step 2 is to work in an organized manner. Figure out what you need but don't have. Name things. Pick variables to stand for the unknows, clearly labelling these variables with what they stand for. Draw and label pictures neatly. Explain your reasoning as you go along. And make sure you know just exactly what the problem is actually asking for. You need to do this for two reasons:

  • Working clearly will help you think clearly, and
  • figuring out what you need will help you translate your final answer back into English.

Regarding point (a) above:

It can be really frustrating (and embarassing) to spend fifteen minutes solving a word problem on a test, only to realize at the end that you no longer have any idea what " x " stands for, so you have to do the whole problem over again. I did this on a calculus test — thank heavens it was a short test! — and, trust me, you don't want to do this to yourself. Taking fifteen seconds to label things is a better use of your time than spending fifteen minutes reworking the entire exercise!

Step 3 is to look for "key" words. Certain words indicate certain mathematica operations. Some of those words are easy. If an exercise says that one person "added" her marbles to the pile belonging to somebody else, and asks for how many marbles are now in the pile, you know that you'll be adding two numbers.

What are common keywords for word problems?

The following is a listing of most of the more-common keywords for word problems:

increased by more than combined, together total of sum, plus added to comparatives ("greater than", etc)

Subtraction:

decreased by minus, less difference between/of less than, fewer than left, left over, after save (old-fashioned term) comparatives ("smaller than", etc)

Multiplication:

of times, multiplied by product of increased/decreased by a factor of (this last type can involve both addition or subtraction and multiplication!) twice, triple, etc each ("they got three each", etc)

per, a out of ratio of, quotient of percent (divide by 100) equal pieces, split average

is, are, was, were, will be gives, yields sold for, cost

Note that "per", in "Division", means "divided by", as in "I drove 90 miles on three gallons of gas, so I got 30 miles per gallon". Also, "a" sometimes means "divided by", as in "When I tanked up, I paid $12.36 for three gallons, so the gas was $4.12 a gallon".

Warning: The "less than" construction, in "Subtraction", is backwards in the English from what it is in the math. If you need, for instance, to translate " 1.5 less than x ", the temptation is to write " 1.5 −  x ". Do not do this!

You can see how this is wrong by using this construction in a "real world" situation: Consider the statement, "He makes $1.50 an hour less than me." You do not figure his wage by subtracting your wage from $1.50 . Instead, you subtract $1.50 from your wage. So remember: the "less than" construction is backwards.

(Technically, the "greater than" construction, in "Addition", is also backwards in the math from the English. But the order in addition doesn't matter, so it's okay to add backwards, because the result will be the same either way.)

Also note that order is important in the "quotient/ratio of" and "difference between/of" constructions. If a problems says "the ratio of x and y ", it means " x divided by y ", not " y divided by x ". If the problem says "the difference of x and y ", it means " x  −  y ", not " y  −  x ".

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Some times, you'll be expected to bring your "real world" knowledge to an exercise. For instance, suppose you're told that "Shelby worked eight hours MTThF and six hours WSat". You would be expected to understand that this meant that she worked eight hours for each of the four days Monday, Tuesday, Thursday, and Friday; and six hours for each of the two days Wednesday and Saturday. Suppose you're told that Shelby earns "time and a half" for any hours she works over forty for a given week. You would be expected to know that "time and a half" means 1.5 times her base rate of pay; if her base rate is twelve dollars an hour, then she'd get 1.5 × 12 = 18 dollars for every over-time hour.

You'll be expected to know that a "dozen" is twelve; you may be expected to know that a "score" is twenty. You'll be expected to know the number of days in a year, the number of hours in a day, and other basic units of measure.

Probably the greatest source of error, though, is the use of variables without definitions. When you pick a letter to stand for something, write down explicitly what that latter is meant to stand for. Does " S " stand for "Shelby" or for "hours Shelby worked"? If the former, what does this mean, in practical terms? (And, if you can't think of any meaningful definition, then maybe you need to slow down and think a little more about what's going on in the word problem.)

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In all cases, don't be shy about using your "real world" knowledge. Sometimes you'll not feel sure of your translation of the English into a mathematical expression or equation. In these cases, try plugging in numbers. For instance, if you're not sure if you should be dividing or multiplying, try the process each way with regular numbers. For instance, suppose you're not sure if "half of (the unknown amount)" should be represented by multiplying by one-half, or by dividing by one-half. If you use numbers, you can be sure. Pick an easy number, like ten. Half of ten is five, so we're looking for the operation (that is, multiplication or division) that gives us an answer of 5 . First, let's try division:

ten divided by one-half:

10/(1/2) = (10/1)×(2/1) = 20

Well, that's clearly wrong. How about going the other way?

ten multiplied by one-half:

(10)×(1/2) = 10 ÷ 2 = 5

That's more like it! You know that half of ten is five, and now you can see which mathematical operations gets you the right value. So now you'd know that the expression you're wanting is definitely " (1/2) x ".

You have experience and knowledge; don't be afraid to apply your skills to this new context!

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REAL WORLD PROBLEMS: How to Write Equations Based on Algebra Word Problems

I know that you often sit in class and wonder, "Why am I forced to learn about equations, Algebra and variables?"

But... trust me, there are real situations where you will use your knowledge of Algebra and solving equations to solve a problem that is not school related. And... if you can't, you're going to wish that you remembered how.

It might be a time when you are trying to figure out how much you should get paid for a job, or even more important, if you were paid enough for a job that you've done. It could also be a time when you are trying to figure out if you were over charged for a bill.

This is important stuff - when it comes time to spend YOUR money - you are going to want to make sure that you are getting paid enough and not spending more than you have to.

Ok... let's put all this newly learned knowledge to work.

Click here if you need to review how to solve equations.

There are a few rules to remember when writing Algebra equations:

Writing Equations For Word Problems

  • First, you want to identify the unknown, which is your variable. What are you trying to solve for? Identify the variable: Use the statement, Let x = _____. You can replace the x with whatever variable you are using.
  • Look for key words that will help you write the equation. Highlight the key words and write an equation to match the problem.
  • The following key words will help you write equations for Algebra word problems:

subtraction

Multiplication.

double (2x)

triple (3x)

quadruple (4x)

divided into

Let's look at an example of an algebra word problem.

Example 1: Algebra Word Problems

Linda was selling tickets for the school play.  She sold 10 more adult tickets than children tickets and she sold twice as many senior tickets as children tickets.

  • Let x represent the number of children's tickets sold.
  • Write an expression to represent the number of adult tickets sold.
  • Write an expression to represent the number of senior tickets sold.
  • Adult tickets cost $5, children's tickets cost $2, and senior tickets cost $3. Linda made $700. Write an equation to represent the total ticket sales.
  • How many children's tickets were sold for the play? How many adult tickets were sold? How many senior tickets were sold?

As you can see, this problem is massive!  There are 5 questions to answer with many expressions to write. 

Algebra word problem solutions

A few notes about this problem

1. In this problem, the variable was defined for you.  Let x represent the number of children’s tickets sold tells what x stands for in this problem.  If this had not been done for you, you might have written it like this:

        Let x = the number of children’s tickets sold

2. For the first expression, I knew that 10 more adult tickets were sold.  Since more means add, my expression was x +10 .  Since the direction asked for an expression, I don’t need an equal sign.  An equation is written with an equal sign and an expression is without an equal sign.  At this point we don’t know the total number of tickets.

3. For the second expression, I knew that my key words, twice as many meant two times as many.  So my expression was 2x .

4.  We know that to find the total price we have to multiply the price of each ticket by the number of tickets.  Take note that since x + 10 is the quantity of adult tickets, you must put it in parentheses!  So, when you multiply by the price of $5 you have to distribute the 5.

5.  Once I solve for x, I know the number of children’s tickets and I can take my expressions that I wrote for #1 and substitute 50 for x to figure out how many adult and senior tickets were sold.

Where Can You Find More Algebra Word Problems to Practice?

Word problems are the most difficult type of problem to solve in math. So, where can you find quality word problems WITH a detailed solution?

The Algebra Class E-course provides a lot of practice with solving word problems for every unit! The best part is.... if you have trouble with these types of problems, you can always find a step-by-step solution to guide you through the process!

Click here for more information.

The next example shows how to identify a constant within a word problem.

Example 2 - Identifying a Constant

A cell phone company charges a monthly rate of $12.95 and $0.25 a minute per call. The bill for m minutes is $21.20.

1. Write an equation that models this situation.

2. How many minutes were charged on this bill?

solve math word problems equation

Notes For Example 2

  • $12.95 is a monthly rate. Since this is a set fee for each month, I know that this is a constant. The rate does not change; therefore, it is not associated with a variable.
  • $0.25 per minute per call requires a variable because the total amount will change based on the number of minutes. Therefore, we use the expression 0.25m
  • You must solve the equation to determine the value for m, which is the number of minutes charged.

The last example is a word problem that requires an equation with variables on both sides.

Example 3 - Equations with Variables on Both Sides

You have $60 and your sister has $120. You are saving $7 per week and your sister is saving $5 per week. How long will it be before you and your sister have the same amount of money? Write an equation and solve.

solve math word problems equation

Notes for Example 3

  • $60 and $120 are constants because this is the amount of money that they each have to begin with. This amount does not change.
  • $7 per week and $5 per week are rates. They key word "per" in this situation means to multiply.
  • The key word "same" in this problem means that I am going to set my two expressions equal to each other.
  • When we set the two expressions equal, we now have an equation with variables on both sides.
  • After solving the equation, you find that x = 30, which means that after 30 weeks, you and your sister will have the same amount of money.

I'm hoping that these three examples will help you as you solve real world problems in Algebra!

  • Solving Equations
  • Algebra Word Problems

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feature_words-1

About 25% of your total SAT Math section will be word problems, meaning you will have to create your own visuals and equations to solve for your answers. Though the actual math topics can vary, SAT word problems share a few commonalities, and we’re here to walk you through how to best solve them.

This post will be your complete guide to SAT Math word problems. We'll cover how to translate word problems into equations and diagrams, the different types of math word problems you’ll see on the test, and how to go about solving your word problems on test day.

Feature Image: Antonio Litterio /Wikimedia

What Are SAT Math Word Problems?

A word problem is any math problem based mostly or entirely on a written description. You will not be provided with an equation, diagram, or graph on a word problem and must instead use your reading skills to translate the words of the question into a workable math problem. Once you do this, you can then solve it.

You will be given word problems on the SAT Math section for a variety of reasons. For one, word problems test your reading comprehension and your ability to visualize information.

Secondly, these types of questions allow test makers to ask questions that'd be impossible to ask with just a diagram or an equation. For instance, if a math question asks you to fit as many small objects into a larger one as is possible, it'd be difficult to demonstrate and ask this with only a diagram.

Translating Math Word Problems Into Equations or Drawings

In order to translate your SAT word problems into actionable math equations you can solve, you’ll need to understand and know how to utilize some key math terms. Whenever you see these words, you can translate them into the proper mathematical action.

For instance, the word "sum" means the value when two or more items are added together. So if you need to find the sum of a and b , you’ll need to set up your equation like this: a+b.

Also, note that many mathematical actions have more than one term attached, which can be used interchangeably.

Here is a chart with all the key terms and symbols you should know for SAT Math word problems:

Now, let's look at these math terms in action using a few official examples:

body_sat_math_sample_question_1

We can solve this problem by translating the information we're given into algebra. We know the individual price of each salad and drink, and the total revenue made from selling 209 salads and drinks combined. So let's write this out in algebraic form.

We'll say that the number of salads sold = S , and the number of drinks sold = D . The problem tells us that 209 salads and drinks have been sold, which we can think of as this:

S + D = 209

Finally, we've been told that a certain number of S and D have been sold and have brought in a total revenue of 836 dollars and 50 cents. We don't know the exact numbers of S and D , but we do know how much each unit costs. Therefore, we can write this equation:

6.50 S + 2 D = 836.5

We now have two equations with the same variables ( S and D ). Since we want to know how many salads were sold, we'll need to solve for D so that we can use this information to solve for S . The first equation tells us what S and D equal when added together, but we can rearrange this to tell us what just D equals in terms of S :

Now, just subtract S from both sides to get what D equals:

D = 209 − S

Finally, plug this expression in for D into our other equation, and then solve for S :

6.50 S + 2(209 − S ) = 836.5

6.50 S + 418 − 2 S = 836.5

6.50 S − 2 S = 418.5

4.5 S = 418.5

The correct answer choice is (B) 93.

body_sat_math_sample_question_2

This word problem asks us to solve for one possible solution (it asks for "a possible amount"), so we know right away that there will be multiple correct answers.

Wyatt can husk at least 12 dozen ears of corn and at most 18 dozen ears of corn per hour. If he husks 72 dozen at a rate of 12 dozen an hour, this is equal to 72 / 12 = 6 hours. You could therefore write 6 as your final answer.

If Wyatt husks 72 dozen at a rate of 18 dozen an hour (the highest rate possible he can do), this comes out to 72 / 18 = 4 hours. You could write 4 as your final answer.

Since the minimum time it takes Wyatt is 4 hours and the maximum time is 6 hours, any number from 4 to 6 would be correct.

body_Latin

Though the hardest SAT word problems might look like Latin to you right now, practice and study will soon have you translating them into workable questions.

Typical SAT Word Problems

Word problems on the SAT can be grouped into three major categories:

  • Word problems for which you must simply set up an equation
  • Word problems for which you must solve for a specific value
  • Word problems for which you must define the meaning of a value or variable

Below, we look at each world problem type and give you examples.

Word Problem Type 1: Setting Up an Equation

This is a fairly uncommon type of SAT word problem, but you’ll generally see it at least once on the Math section. You'll also most likely see it first on the section.

For these problems, you must use the information you’re given and then set up the equation. No need to solve for the missing variable—this is as far as you need to go.

Almost always, you’ll see this type of question in the first four questions on the SAT Math section, meaning that the College Board consider these questions easy. This is due to the fact that you only have to provide the setup and not the execution.

body_sat_math_sample_question_3

To solve this problem, we'll need to know both Armand's and Tyrone's situations, so let's look at them separately:

Armand: Armand sent m text messages each hour for 5 hours, so we can write this as 5m —the number of texts he sent per hour multiplied by the total number of hours he texted.

Tyrone: Tyrone sent p text messages each hour for 4 hours, so we can write this as 4 p —the number of texts he sent per hour multiplied by the total number of hours he texted.

We now know that Armand's situation can be written algebraically as 5m , and Tyrone's can be written as 4 p . Since we're being asked for the expression that represents the total number of texts sent by Armand and Tyrone, we must add together the two expressions:

The correct answer is choice (C) 5m + 4 p

Word Problem Type 2: Solving for a Missing Value

The vast majority of SAT Math word problem questions will fall into this category. For these questions, you must both set up your equation and solve for a specific piece of information.

Most (though not all) word problem questions of this type will be scenarios or stories covering all sorts of SAT Math topics , such as averages , single-variable equations , and ratios . You almost always must have a solid understanding of the math topic in question in order to solve the word problem on the topic.

body_sat_math_sample_question_4

Let's try to think about this problem in terms of x . If Type A trees produced 20% more pears than Type B did, we can write this as an expression:

x + 0.2 x = 1.2 x = # of pears produced by Type A

In this equation, x is the number of pears produced by Type B trees. If we add 20% of x (0.2 x ) to x , we get the number of pears produced by Type A trees.

The problem tells us that Type A trees produced a total of 144 pears. Since we know that 1.2 x is equal to the number of pears produced by Type A, we can write the following equation:

1.2 x = 144

Now, all we have to do is divide both sides by 1.2 to find the number of pears produced by Type B trees:

x = 144 / 1.2

The correct answer choice is (B) 120.

You might also get a geometry problem as a word problem, which might or might not be set up with a scenario, too. Geometry questions will be presented as word problems typically because the test makers felt the problem would be too easy to solve had you been given a diagram, or because the problem would be impossible to show with a diagram. (Note that geometry makes up a very small percentage of SAT Math . )

body_SAT_word_problem_5

This is a case of a problem that is difficult to show visually, since x is not a set degree value but rather a value greater than 55; thus, it must be presented as a word problem.

Since we know that x must be an integer degree value greater than 55, let us assign it a value. In this case, let us call x 56°. (Why 56? There are other values x could be, but 56 is guaranteed to work since it's the smallest integer larger than 55. Basically, it's a safe bet!)

Now, because x = 56, the next angle in the triangle—2 x —must measure the following:

Let's make a rough (not to scale) sketch of what we know so far:

body_triangle_ex_1

Now, we know that there are 180° in a triangle , so we can find the value of y by saying this:

y = 180 − 112 − 56

One possible value for y is 12. (Other possible values are 3, 6, and 9. )

Word Problem Type 3: Explaining the Meaning of a Variable or Value

This type of problem will show up at least once. It asks you to define part of an equation provided by the word problem—generally the meaning of a specific variable or number.

body_sat_math_sample_question_6

This question might sound tricky at first, but it's actually quite simple.

We know tha t P is the number of phones Kathy has left to fix, and d is the number of days she has worked in a week. If she's worked 0 days (i.e., if we plug 0 into the equation), here's what we get:

P = 108 − 23(0)

This means that, without working any days of the week, Kathy has 108 phones to repair. The correct answer choice, therefore, is (B) Kathy starts each week with 108 phones to fix.

body_juggle

To help juggle all the various SAT word problems, let's look at the math strategies and tips at our disposal.

Disappointed with your scores? Want to improve your SAT score by 160 points?   We've written a guide about the top 5 strategies you must use to have a shot at improving your score. Download it for free now:

SAT Math Strategies for Word Problems

Though you’ll see word problems on the SAT Math section on a variety of math topics, there are still a few techniques you can apply to solve word problems as a whole.

#1: Draw It Out

Whether your problem is a geometry problem or an algebra problem, sometimes making a quick sketch of the scene can help you understand what exactly you're working with. For instance, let's look at how a picture can help you solve a word problem about a circle (specifically, a pizza):

body_sat_math_sample_question_7_2

If you often have trouble visualizing problems such as these, draw it out. We know that we're dealing with a circle since our focus is a pizza. We also know that the pizza weighs 3 pounds.

Because we'll need to solve the weight of each slice in ounces, let's first convert the total weight of our pizza from pounds into ounces. We're given the conversion (1 pound = 16 ounces), so all we have to do is multiply our 3-pound pizza by 16 to get our answer:

3 * 16 = 48 ounces (for whole pizza)

Now, let's draw a picture. First, the pizza is divided in half (not drawn to scale):

body_sat_math_sample_question_7_diagram_1

We now have two equal-sized pieces. Let's continue drawing. The problem then says that we divide each half into three equal pieces (again, not drawn to scale):

body_sat_math_sample_question_7_diagram_2

This gives us a total of six equal-sized pieces. Since we know the total weight of the pizza is 48 ounces, all we have to do is divide by 6 (the number of pieces) to get the weight (in ounces) per piece of pizza:

48 / 6 = 8 ounces per piece

The correct answer choice is (C) 8.

As for geometry problems, remember that you might get a geometry word problem written as a word problem. In this case, make your own drawing of the scene. Even a rough sketch can help you visualize the math problem and keep all your information in order.

#2: Memorize Key Terms

If you’re not used to translating English words and descriptions into mathematical equations, then SAT word problems might be difficult to wrap your head around at first. Look at the chart we gave you above so you can learn how to translate keywords into their math equivalents. This way, you can understand exactly what a problem is asking you to find and how you’re supposed to find it.

There are free SAT Math questions available online , so memorize your terms and then practice on realistic SAT word problems to make sure you’ve got your definitions down and can apply them to the actual test.

#3: Underline and/or Write Out Important Information

The key to solving a word problem is to bring together all the key pieces of given information and put them in the right places. Make sure you write out all these givens on the diagram you’ve drawn (if the problem calls for a diagram) so that all your moving pieces are in order.

One of the best ways to keep all your pieces straight is to underline your key information in the problem, and then write them out yourself before you set up your equation. So take a moment to perform this step before you zero in on solving the question.

#4: Pay Close Attention to What's Being Asked

It can be infuriating to find yourself solving for the wrong variable or writing in your given values in the wrong places. And yet this is entirely too easy to do when working with math word problems.

Make sure you pay strict attention to exactly what you’re meant to be solving for and exactly what pieces of information go where. Are you looking for the area or the perimeter? The value of x, 2x, or y?

It’s always better to double-check what you’re supposed to find before you start than to realize two minutes down the line that you have to begin solving the problem all over again.

#5: Brush Up on Any Specific Math Topic You Feel Weak In

You're likely to see both a diagram/equation problem and a word problem for almost every SAT Math topic on the test. This is why there are so many different types of word problems and why you’ll need to know the ins and outs of every SAT Math topic in order to be able to solve a word problem about it.

For example, if you don’t know how to find an average given a set of numbers, you certainly won’t know how to solve a word problem that deals with averages!

Understand that solving an SAT Math word problem is a two-step process: it requires you to both understand how word problems work and to understand the math topic in question. If you have any areas of mathematical weakness, now's a good time to brush up on them—or else SAT word problems might be trickier than you were expecting!

body_ready-1

All set? Let's go!

Test Your SAT Math Word Problem Knowledge

Finally, it's time to test your word problem know-how against real SAT Math problems:

Word Problems

1. No Calculator

body_sat_math_test_question_1

2. Calculator OK

body_sat_math_test_question_2

3. Calculator OK

body_sat_math_test_question_3

4. Calculator OK

body_sat_math_test_question_4

Answers: C, B, A, 1160

Answer Explanations

1. For this problem, we have to use the information we're given to set up an equation.

We know that Ken spent x dollars, and Paul spent 1 dollar more than Ken did. Therefore, we can write the following equation for Paul:

Ken and Paul split the bill evenly. This means that we'll have to solve for the total amount of both their sandwiches and then divide it by 2. Since Ken's sandwich cost x dollars and Paul's cost x + 1, here's what our equation looks like when we combine the two expressions:

Now, we can divide this expression by 2 to get the price each person paid:

(2 x + 1) / 2

But we're not finished yet. We know that both Ken and Paul also paid a 20% tip on their bills. As a result, we have to multiply the total cost of one bill by 0.2, and then add this tip to the bill. Algebraically, this looks like this:

( x + 0.5) + 0.2( x + 0.5)

x + 0.5 + 0.2 x + 0.1

1.2 x + 0.6

The correct answer choice is (C) 1.2 x + 0.6

2. You'll have to be familiar with statistics in order to understand what this question is asking.

Since Nick surveyed a random sample of his freshman class, we can say that this sample will accurately reflect the opinion (and thus the same percentages) as the entire freshman class.

Of the 90 freshmen sampled, 25.6% said that they wanted the Fall Festival held in October. All we have to do now is find this percentage of the entire freshmen class (which consists of 225 students) to determine how many total freshmen would prefer an October festival:

225 * 0.256 = 57.6

Since the question is asking "about how many students"—and since we obviously can't have a fraction of a person!—we'll have to round this number to the nearest answer choice available, which is 60, or answer choice (B).

3. This is one of those problems that is asking you to define a value in the equation given. It might look confusing, but don't be scared—it's actually not as difficult as it appears!

First off, we know that t represents the number of seconds passed after an object is launched upward. But what if no time has passed yet? This would mean that t = 0. Here's what happens to the equation when we plug in 0 for t :

h (0) = -16(0)2 + 110(0) + 72

h (0) = 0 + 0 + 72

As we can see, before the object is even launched, it has a height of 72 feet. This means that 72 must represent the initial height, in feet, of the object, or answer choice (A).

4. You might be tempted to draw a diagram for this problem since it's talking about a pool (rectangle), but it's actually quicker to just look at the numbers given and work from there.

We know that the pool currently holds 600 gallons of water and that water has been hosed into it at a rate of 8 gallons a minute for a total of 70 minutes.

To find the amount of water in the pool now, we'll have to first solve for the amount of water added to the pool by hose. We know that 8 gallons were added each minute for 70 minutes, so all we have to do is multiply 8 by 70:

8 * 70 = 560 gallons

This tells us that 560 gallons of water were added to our already-filled, 600-gallon pool. To find the total amount of water, then, we simply add these two numbers together:

560 + 600 = 1160

The correct answer is 1160.

body_sleepy-1

Aaaaaaaaaaand time for a nap.

Key Takeaways: Making Sense of SAT Math Word Problems

Word problems make up a significant portion of the SAT Math section, so it’s a good idea to understand how they work and how to translate the words on the page into a proper expression or equation. But this is still only half the battle.

Though you won’t know how to solve a word problem if you don’t know what a product is or how to draw a right triangle, you also won’t know how to solve a word problem about ratios if you don’t know how ratios work.

Therefore, be sure to learn not only how to approach math word problems as a whole, but also how to narrow your focus on any SAT Math topics you need help with. You can find links to all of our SAT Math topic guides here to help you in your studies.

What’s Next?

Want to brush up on SAT Math topics? Check out our individual math guides to get an overview of each and every topic on SAT Math . From polygons and slopes to probabilities and sequences , we've got you covered!

Running out of time on the SAT Math section? We have the know-how to help you beat the clock and maximize your score .

Been procrastinating on your SAT studying? Learn how you can overcome your desire to procrastinate and make a well-balanced prep plan.

Trying to get a perfect SAT score? Take a look at our guide to getting a perfect 800 on SAT Math , written by a perfect scorer.

Want to improve your SAT score by 160 points?   We have the industry's leading SAT prep program. Built by Harvard grads and SAT full scorers, the program learns your strengths and weaknesses through advanced statistics, then customizes your prep program to you so you get the most effective prep possible.   Along with more detailed lessons, you'll get thousands of practice problems organized by individual skills so you learn most effectively. We'll also give you a step-by-step program to follow so you'll never be confused about what to study next.   Check out our 5-day free trial today:

Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.

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COMMENTS

  1. Word Problems Calculator

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    Math Word Problem Solutions. Math word problems require interpreting what is being asked and simplifying that into a basic math equation. Once you have the equation you can then enter that into the problem solver as a basic math or algebra question to be correctly solved. Below are math word problem examples and their simplified forms. Word ...

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    Equation Word Problems Worksheets. This compilation of a meticulously drafted equation word problems worksheets is designed to get students to write and solve a variety of one-step, two-step and multi-step equations that involve integers, fractions, and decimals. These worksheets are best suited for students in grade 6 through high school.

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    Word Problem Calculator. Get detailed solutions to your math problems with our Word Problem step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here. Go! Symbolic mode. Text mode.

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    Two-step equations word problems. Mindy and Troy combined ate 9 pieces of the wedding cake. Mindy ate 3 pieces of cake and Troy had 1 4 of the total cake. Write an equation to determine how many pieces of cake ( c) there were in total. Find the total number of pieces of cake. Learn for free about math, art, computer programming, economics ...

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    Loo Dadoo. 10 months ago. e is a constant that is approximately equal to 2.71. Here, e is raised to the (0.4t) power, which will either increase the value of f (t) as t increases. For example, at f (1), the function equals 24 x e^ (0.4), but at f (0), the function equals 24 x e^ (0), or 24. ( 1 vote) Upvote.

  9. Linear equation word problems (video)

    David Severin. All Sal is doing in this video is finding the key features of a linear equation given in slope intercept (almost) form y=mx+b. Given Q=15+0.4t, we can switch to slope intercept form by moving right side to Q=0.4t+15 noting the y is represented by Q and x is represented by t. The slope is 0.4 and the y intercept is 15.

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    Step 4: Solve the problem. Now all we have to do is solve our problem. Like with any problem, we can solve this one by following the order of operations. f is already alone on the left side of the equation, so all we have to do is calculate the right side.

  12. Microsoft Math Solver

    Get math help in your language. Works in Spanish, Hindi, German, and more. Online math solver with free step by step solutions to algebra, calculus, and other math problems. Get help on the web or with our math app.

  13. Microsoft Math Solver

    Get math help in your language. Works in Spanish, Hindi, German, and more. Online math solver with free step by step solutions to algebra, calculus, and other math problems. Get help on the web or with our math app.

  14. 1.20: Word Problems for Linear Equations

    Solution: Translating the problem into an algebraic equation gives: 2x − 5 = 13. We solve this for x. First, add 5 to both sides. 2x = 13 + 5, so that 2x = 18. Dividing by 2 gives x = 18 2 = 9. c) A number subtracted from 9 is equal to 2 times the number. Find the number. Solution: We translate the problem to algebra.

  15. How to write word problems as equations

    For instance, suppose you wanted to use algebra to solve the following word problem: John's age is four less than twice Mary's age. If Mary is ???18???, how old is John? The first step in solving a word problem like this is to define the variables. What that means is to state the particular quantity that each variable stands for.

  16. word problem

    The Algebra Calculator is a versatile online tool designed to simplify algebraic problem-solving for users of all levels. Here's how to make the most of it: Begin by typing your algebraic expression into the above input field, or scanning the problem with your camera. After entering the equation, click the 'Go' button to generate instant solutions.

  17. How to turn word problems into math

    Algebra Word Problems. Step 1 in effectively translating and solving word problems is to read the problem entirely. Don't start trying to solve anything when you've only read half a sentence. Try first to get a feel for the whole problem; try first to see what information you have, and then figure out what you still need.

  18. Algebra Word Problems

    Identify the variable: Use the statement, Let x = _____. You can replace the x with whatever variable you are using. Look for key words that will help you write the equation. Highlight the key words and write an equation to match the problem. The following key words will help you write equations for Algebra word problems: Addition. altogether.

  19. Solve

    Differentiation. dxd (x − 5)(3x2 − 2) Integration. ∫ 01 xe−x2dx. Limits. x→−3lim x2 + 2x − 3x2 − 9. Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.

  20. Word Problems

    Teachers can use the word problem generators to create word problems for students. Once you answer a word problem the answer is shown in a proof. The equation is created and solved in detail and can be used to check your own work. The best way for you to learn to solve word problems is to practice to translating the words into math problems ...

  21. Systems of equations word problems

    Systems of equations word problems. Google Classroom. Microsoft Teams. You might need: Calculator. Malcolm and Ravi raced each other. The average of their maximum speeds was 260 km/h . If doubled, Malcolm's maximum speed would be 80 km/h more than Ravi's maximum speed. What were Malcolm's and Ravi's maximum speeds?

  22. The Complete Guide to SAT Math Word Problems

    Word Problem Type 1: Setting Up an Equation. This is a fairly uncommon type of SAT word problem, but you'll generally see it at least once on the Math section. You'll also most likely see it first on the section. For these problems, you must use the information you're given and then set up the equation.

  23. Math Equation Solver

    PEMDAS is an acronym that may help you remember order of operations for solving math equations. PEMDAS is typcially expanded into the phrase, "Please Excuse My Dear Aunt Sally." The first letter of each word in the phrase creates the PEMDAS acronym. Solve math problems with the standard mathematical order of operations, working left to right:

  24. Factoring 8*w^2-24*w+18

    Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)

  25. word problems

    Learn how to solve word problems on a number line with Symbolab's free online calculator. Find examples, tips and detailed solutions for any math topic.

  26. Simple Equations

    This lesson covers skills from the following lesson of the NCERT Math Textbook - (i) 4.4 - What equation is . Learn. One-step addition equation (Opens a modal) ... Linear equations word problems (basic) Get 3 of 4 questions to level up! Model with one-step equations and solve Get 3 of 4 questions to level up! Quiz 3.

  27. Mathway

    Free math problem solver answers your algebra homework questions with step-by-step explanations. Mathway. Visit Mathway on the web. Start 7-day free trial on the app. Start 7-day free trial on the app. Download free on Amazon. Download free in Windows Store. get Go. Algebra. Basic Math. Pre-Algebra. Algebra. Trigonometry. Precalculus.

  28. Mathway

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