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\mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
\mathrm{Kira\:went\:for\:a\:drive\:in\:her\:new\:car.\:She\:drove\:for\:142.5\:miles\:at\:a\:speed\:of\:57\:mph.\:For\:how\:many\:hours\:did\:she\:drive?}
\mathrm{The\:sum\:of\:two\:numbers\:is\:249\:.\:Twice\:the\:larger\:number\:plus\:three\:times\:the\:smaller\:number\:is\:591\:.\:Find\:the\:numbers.}
\mathrm{If\:2\:tacos\:and\:3\:drinks\:cost\:12\:and\:3\:tacos\:and\:2\:drinks\:cost\:13\:how\:much\:does\:a\:taco\:cost?}
\mathrm{You\:deposit\:3000\:in\:an\:account\:earning\:2\%\:interest\:compounded\:monthly.\:How\:much\:will\:you\:have\:in\:the\:account\:in\:15\:years?}
How do you solve word problems?
To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
How do you identify word problems in math?
Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
Is there a calculator that can solve word problems?
Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
What is an age problem?
An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

word-problems-calculator

Middle School Math Solutions – Inequalities Calculator Next up in our Getting Started maths solutions series is help with another middle school algebra topic - solving...

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Algebra Topics - Introduction to Word Problems

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

Algebra Topics: Introduction to Word Problems

Lesson 9: introduction to word problems.

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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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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:

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:

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:

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:

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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Introduction to Word Problems

These lessons, with videos, examples, solutions and worksheets, help Grade 5 students learn how to solve word problems.

Related Pages More Grade 5 Math Word Problems More Lessons for Grade 5 Math Math Worksheets

The following diagram gives some examples of word problems keywords or clue words. Scroll down the page for more examples and solutions of word problems.

Introduction to Word Problem Terms

Problem Solving Strategies

Explain the meanings of the four basic operations–addition, subtraction, multiplication and division–so that you can understand how to solve word problems correctly.

Helpful hints for solving word problems

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How do you solve word problems in math?

Master word problems with eight simple steps from a math tutor!

Author Amber Watkins

Published April 2024

Key takeaways
Students who struggle with reading, tend to struggle with understanding and solving word problems. So the best way to solve word problems in math is to become a better reader!
Mastery of word problems relies on your child’s knowledge of keywords for word problems in math and knowing what to do with them.
There are 8 simple steps each child can use to solve word problems- let’s go over these together.

Table of contents

How to solve word problems

Lesson credits

As a tutor who has seen countless math worksheets in almost every grade – I’ll tell you this: every child is going to encounter word problems in math. The key to mastery lies in how you solve them! So then, how do you solve word problems in math?

In this guide, I’ll share eight steps to solving word problems in math.

How to solve word problems in math in 8 steps

Step 1: read the word problem aloud.

For a child to understand a word problem, it needs to be read with accuracy and fluency! That is why, when I tutor children with word problems, I always emphasize the importance of reading properly.

Mastering step 1 looks like this:

Allow your child to read the word problem aloud to you.
Don’t let your child skip over or mispronounce any words.
If necessary, model how to read the word problem, then allow your child to read it again. Only after the word problem is read accurately, should you move on to step 2.

Step 2: Highlight the keywords in the word problem

The keywords for word problems in math indicate what math action should be taken. Teach your child to highlight or underline the keywords in every word problem.

Here are some of the most common keywords in math word problems:

Subtraction words – less than, minus, take away
Addition words – more than, altogether, plus, perimeter
Multiplication words – Each, per person, per item, times, area
Division words – divided by, into
Total words – in all, total, altogether

Let’s practice. Read the following word problem with your child and help them highlight or underline the main keyword, then decide which math action should be taken.

Michael has ten baseball cards. James has four baseball cards less than Michael. How many total baseball cards does James have?

The words “less than” are the keywords and they tell us to use subtraction .

Step 3: Make math symbols above keywords to decode the word problem

As I help students with word problems, I write math symbols and numbers above the keywords. This helps them to understand what the word problem is asking.

Let’s practice. Observe what I write over the keywords in the following word problem and think about how you would create a math sentence using them:

Step 4: Create a math sentence to represent the word problem

Using the previous example, let’s write a math sentence. Looking at the math symbols and numbers written above the word problem, our math sentence should be: 10 – 5 = 5 !

Each time you practice a word problem with your child, highlight keywords and write the math symbols above them. Then have your child create a math sentence to solve.

Step 5: Draw a picture to help illustrate the word problem

Pictures can be very helpful for problems that are more difficult to understand. They also are extremely helpful when the word problem involves calculating time , comparing fractions , or measurements .

Step 6: Always show your work

Help your child get into the habit of always showing their work. As a tutor, I’ve found many reasons why having students show their work is helpful:

By showing their work, they are writing the math steps repeatedly, which aids in memory
If they make any mistakes they can track where they happened
Their teacher can assess how much they understand by reviewing their work
They can participate in class discussions about their work

Step 7: When solving word problems, make sure there is always a word in your answer!

If the word problem asks: How many peaches did Lisa buy? Your child’s answer should be: Lisa bought 10 peaches .

If the word problem asks: How far did Kyle run? Your child’s answer should be: Kyle ran 20 miles .

So how do you solve a word problem in math?

Together we reviewed the eight simple steps to solve word problems. These steps included identifying keywords for word problems in math, drawing pictures, and learning to explain our answers.

Is your child ready to put these new skills to the test? Check out the best math app for some fun math word problem practice.

Parents, sign up for a DoodleMath subscription and see your child become a math wizard!

Amber Watkins

Amber is an education specialist with a degree in Early Childhood Education. She has over 12 years of experience teaching and tutoring elementary through college level math. "Knowing that my work in math education makes such an impact leaves me with an indescribable feeling of pride and joy!"

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Course: Algebra 1 > Unit 14

Finding the vertex of a parabola in standard form
Graphing quadratics: standard form
Graph quadratics in standard form

Quadratic word problem: ball

Quadratic word problems (standard form)

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Video transcript

Basic Word Problems

A word problem in algebra is the equivalent of a story problem in math. When you solved story problems in your math class you had to decide what information you had and what you needed to find out. Then you decided what operation to use. Addition was used to find a totals and subtraction was used to find changes in values.

The approach to solve problems with algebra is usually quite different. Word problems are solved by separating information from the problems into two equal groups, one for each side of an equation. Examine this problem.

We know that the sum of 15 and 12 is equal to the the total amount of fruit. As explained in the Basics of the Equation lesson, an unknown number or value is represented by a letter. The total number of pieces of fruit is unknown, so we will represent that amount with x . When the value that a particular variable will represent is determined, it is defined by writing a statement like,

Once again, the sum of 15 apples and 12 oranges is equal to the total amount of fruit. This can be used to translate the problem into an equation, like the following:

The next step is to solve this equation.

Basics of Word Problems

Now solve the equation which was created in the last step.

The answer is then rewritten as a sentence.

By using simple arithmetic, this problem probably could have been solved faster without setting up an algebra equation. But, knowing how to use an equation for this problem builds awareness of concepts which are useful, and sometimes critical to solving much harder problems. One such problem will be presented in the next example.

Examine this word problem.

Take notice, this problem has two numbers which are unknown, unlike the previous one which only had one unknown value. In order for this problem to be solved using basic algebra methods, we must set up an equation that has only one variable (such as x ).

Consecutive Integer Word Problems

The problem is shown again below for reference.

To begin solving this problem, define the variable. You do not know what the first consecutive number is, so you can call it x.

Since the numbers are consecutive, meaning one number comes right after the other, the second number must be one more than the first. So, x + 1 equals the second number.

The problem says that the sum of the two numbers is 91. This can be shown in the equation like the following:

The equation which you just wrote can be solved as follows:

In the previous problem, we determined that

The answer to the equation (shown above) must now be used to find the answer to the word problem. Go back to the top of the paper you used to solve this problem. It should contain the following work:

Since you now know that x equals 45, and that the First Consecutive Number equals x, you can show this in the work like we did below.

Since x equals 45 and the Second Consecutive Number equals x + 1 you can simply add 1 to 45, to find that the Second Consecutive Number is 46. It should be shown like the work below.

This problem is now completed. If you did all of the work correctly, it should appear as ours does below.

Sample Problem Work

Keep going to learn about what to do when you encounter more than two consecutive numbers, negative consecutives, and even or odd consecutives.

Variations of Consecutive Integer World Problems

More than 2 consecutive integers.

Sometimes you will encounter a problem which has more than two consecutive numbers, such as the one below.

You can solve this much like the previous problem. The difference is that you will have to define four numbers (instead of two), like we did below. Note: each consecutive number is found by adding 1 to the previous number.

Your equation will look like this.

Negative consecutive integers

To solve problems which involve negative consecutive numbers, it is important that you ignore the negative sign, and that you do not do anything differently.

Keep the variable x positive, as shown, so that the answer does not come out wrong.

Thus the equation will have the form

Even or Odd Consecutive Numbers

The only difference between ordinary consecutive numbers and even or odd consecutive numbers is the space between each number. The next consecutive number after 16 can be found by adding 1. The next consecutive even number can be found by adding 2. Similarly, the next consecutive odd number after 7 is 7 + 2, or 9.

Examine this problem

Since each even number is 2 away from the next, it is logical that you should define each number like the following

Leading to the equation

Word Problem Basics Resources

Looking for someone to help you with algebra? At Wyzant, connect with algebra tutors and math tutors nearby. Prefer to meet online? Find online algebra tutors or online math tutors in a couple of clicks.

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The population of endangered species is changing at a rate of P′(t) = 30 − 20t.

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