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"Work" Word Problems

Painting & Pipes Tubs & Man-Hours Unequal Times Etc.

"Work" problems usually involve situations such as two people working together to paint a house. You are usually told how long each person takes to paint a similarly-sized house, and you are asked how long it will take the two of them to paint the house when they work together.

Many of these problems are not terribly realistic — since when can two laser printers work together on printing one report? — but it's the technique that they want you to learn, not the applicability to "real life".

The method of solution for "work" problems is not obvious, so don't feel bad if you're totally lost at the moment. There is a "trick" to doing work problems: you have to think of the problem in terms of how much each person / machine / whatever does in a given unit of time . For instance:

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Suppose one painter can paint the entire house in twelve hours, and the second painter takes eight hours to paint a similarly-sized house. How long would it take the two painters together to paint the house?

To find out how much they can do together per hour , I make the necessary assumption that their labors are additive (in other words, that they never get in each other's way in any manner), and I add together what they can do individually per hour . So, per hour, their labors are:

But the exercise didn't ask me how much they can do per hour; it asked me how long they'll take to finish one whole job, working togets. So now I'll pick the variable " t " to stand for how long they take (that is, the time they take) to do the job together. Then they can do:

This gives me an expression for their combined hourly rate. I already had a numerical expression for their combined hourly rate. So, setting these two expressions equal, I get:

I can solve by flipping the equation; I get:

An hour has sixty minutes, so 0.8 of an hour has forty-eight minutes. Then:

They can complete the job together in 4 hours and 48 minutes.

The important thing to understand about the above example is that the key was in converting how long each person took to complete the task into a rate.

hours to complete job:

first painter: 12

second painter: 8

together: t

Since the unit for completion was "hours", I converted each time to an hourly rate; that is, I restated everything in terms of how much of the entire task could be completed per hour. To do this, I simply inverted each value for "hours to complete job":

completed per hour:

Then, assuming that their per-hour rates were additive, I added the portion that each could do per hour, summed them, and set this equal to the "together" rate:

adding their labor:

As you can see in the above example, "work" problems commonly create rational equations . But the equations themselves are usually pretty simple to solve.

One pipe can fill a pool 1.25 times as fast as a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

My first step is to list the times taken by each pipe to fill the pool, and how long the two pipes take together. In this case, I know the "together" time, but not the individual times. One of the pipes' times is expressed in terms of the other pipe's time, so I'll pick a variable to stand for one of these times.

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Since the faster pipe's time to completion is defined in terms of the second pipe's time, I'll pick a variable for the slower pipe's time, and then use this to create an expression for the faster pipe's time:

slow pipe: s

together: 5

Next, I'll convert all of the completion times to per-hour rates:

Then I make the necessary assumption that the pipes' contributions are additive (which is reasonable, in this case), add the two pipes' contributions, and set this equal to the combined per-hour rate:

multiplying through by 20 s (being the lowest common denominator of all the fractional terms):

20 + 25 = 4 s

45/4 = 11.25 = s

They asked me for the time of the slower pipe, so I don't need to find the time for the faster pipe. My answer is:

The slower pipe takes 11.25 hours.

Note: I could have picked a variable for the faster pipe, and then defined the time for the slower pipe in terms of this variable. If you're not sure how you'd do this, then think about it in terms of nicer numbers: If someone goes twice as fast as you, then you take twice as long as he does; if he goes three times as fast as you, then you take three times as long as him. In this case, if he goes 1.25 times as fast, then you take 1.25 times as long. So the variables could have been " f  " for the number of hours the faster pipe takes, and then the number of hours for the slower pipe would have been " 1.25 f  ".

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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.

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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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Mathematical Word Problems

Math word problems is one of the most complex parts of the elementary math curriculum since translating text into symbolic math is required to solve the problem. Because the Wolfram Language has powerful symbolic computation ability, Wolfram|Alpha can interpret basic mathematical word problems and give descriptive results.

Solve a word problem and explore related facts.

Solve a word problem:

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Mathematics LibreTexts

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?

Module 10: Linear Equations

Apply a problem-solving strategy to word problems, learning outcomes.

  • Approach word problems with a positive attitude
  • Use a problem solving strategy for word problems
  • Translate more complex word problems into algebraic expressions and equations

 Approach Word Problems with a Positive Attitude

The world is full of word problems. How much money do I need to fill the car with gas? How much should I tip the server at a restaurant? How many socks should I pack for vacation? How big a turkey do I need to buy for Thanksgiving dinner, and what time do I need to put it in the oven? If my sister and I buy our mother a present, how much will each of us pay?

Now that we can solve equations, we are ready to apply our new skills to word problems. Do you know anyone who has had negative experiences in the past with word problems? Have you ever had thoughts like the student in the cartoon below?

A cartoon image of a girl with a sad expression writing on a piece of paper is shown. There are 5 thought bubbles. They read, "I don't know whether to add, subtract, multiply, or divide!", "I don't understand word problems!", "My teachers never explained this!", "If I just skip all the word problems, I can probably still pass the class.", and "I just can't do this!".

Negative thoughts about word problems can be barriers to success.

When we feel we have no control, and continue repeating negative thoughts, we set up barriers to success. We need to calm our fears and change our negative feelings.

Start with a fresh slate and begin to think positive thoughts, like the student in the cartoon below. Read the positive thoughts and say them out loud.

A cartoon image of a girl with a confident expression holding some books is shown. There are 4 thought bubbles. They read, "while word problems were hard in the past I think I can try them now.", "I am better prepared now. I think I will begin to understand word problems.", " I think I can! I think I can!", and "It may take time, but I can begin to solve word problems.".

When it comes to word problems, a positive attitude is a big step toward success.

If we take control and believe we can be successful, we will be able to master word problems.

Think of something that you can do now but couldn’t do three years ago. Whether it’s driving a car, snowboarding, cooking a gourmet meal, or speaking a new language, you have been able to learn and master a new skill. Word problems are no different. Even if you have struggled with word problems in the past, you have acquired many new math skills that will help you succeed now!

Use a Problem-Solving Strategy for Word Problems

In earlier chapters, you translated word phrases into algebraic expressions, using some basic mathematical vocabulary and symbols. Since then, you’ve increased your math vocabulary as you learned about more algebraic procedures, and you’ve had more practice translating from words into algebra.

You have also translated word sentences into algebraic equations and solved some word problems. The word problems applied math to everyday situations. You had to restate the situation in one sentence, assign a variable, and then write an equation to solve. This method works as long as the situation is familiar to you and the math is not too complicated.

Now we’ll develop a strategy you can use to solve any word problem. This strategy will help you become successful with word problems. We’ll demonstrate the strategy as we solve the following problem.

Pete bought a shirt on sale for $[latex]18[/latex], which is one-half the original price. What was the original price of the shirt?

Solution: Step 1. Read the problem. Make sure you understand all the words and ideas. You may need to read the problem two or more times. If there are words you don’t understand, look them up in a dictionary or on the Internet.

  • In this problem, do you understand what is being discussed? Do you understand every word?

Step 2. Identify what you are looking for. It’s hard to find something if you are not sure what it is! Read the problem again and look for words that tell you what you are looking for!

  • In this problem, the words “what was the original price of the shirt” tell you what you are looking for: the original price of the shirt.

Step 3. Name what you are looking for. Choose a variable to represent that quantity. You can use any letter for the variable, but it may help to choose one that helps you remember what it represents.

  • Let [latex]p=[/latex] the original price of the shirt

Step 4. Translate into an equation. It may help to first restate the problem in one sentence, with all the important information. Then translate the sentence into an equation.

The top line reads: "18 is one half of the original price". The bottom line translates the top line from words to an algebraic equation. The word "is" translates to an equal sign. The phrase "one half" translates to "1/2". The word "of" translates to a multiplication symbol. The phrase "the original price" translates to "p". This gives the full algebraic equation "18 = 1/2 times p".

Step 6. Check the answer in the problem and make sure it makes sense.

  • We found that [latex]p=36[/latex], which means the original price was [latex]\text{\$36}[/latex]. Does [latex]\text{\$36}[/latex] make sense in the problem? Yes, because [latex]18[/latex] is one-half of [latex]36[/latex], and the shirt was on sale at half the original price.

Step 7. Answer the question with a complete sentence.

  • The problem asked “What was the original price of the shirt?” The answer to the question is: “The original price of the shirt was [latex]\text{\$36}[/latex].”

If this were a homework exercise, our work might look like this:

An example of what a student's work might look like for the problem. Let p equal the original price. 18 is one half the original price. 18 equals one half p. 2 times 18 equals 2 times one half p. 36 equals p. Check: is $36 a reasonable price for a shirt? Yes. Is 18 one half of 36? Yes. The original price of the shirt was $36.

https://ohm.lumenlearning.com/multiembedq.php?id=142694&theme=oea&iframe_resize_id=mom1

We list the steps we took to solve the previous example.

Problem-Solving Strategy

  • Read the word problem. Make sure you understand all the words and ideas. You may need to read the problem two or more times. If there are words you don’t understand, look them up in a dictionary or on the internet.
  • Identify what you are looking for.  Determine the constants and variables in the problem.  A constant is a number in the problem that is not going to change.  A variable is a number that you don’t yet know its value.
  • Name what you are looking for. Choose a letter to represent that quantity.
  • Translate words into algebraic expressions and equations.  Write an equation to represent the problem. It may be helpful to first restate the problem in one sentence before translating.
  • Solve the equation using good algebra techniques.
  • Check the answer in the problem. Make sure it makes sense.
  • Answer the question with a complete sentence.

Translate word problems into expressions

One of the first steps to solving word problems is converting an English sentence into a mathematical sentence. In the table below, words or phrases commonly associated with mathematical operators are categorized. Word problems often contain these or similar words, so it’s good to see what mathematical operators are associated with them.

Some examples follow:

  • “[latex]x\text{ is }5[/latex]” becomes [latex]x=5[/latex]
  • “Three more than a number” becomes [latex]x+3[/latex]
  • “Four less than a number” becomes [latex]x-4[/latex]
  • “Double the cost” becomes [latex]2\cdot\text{ cost }[/latex]
  • “Groceries and gas together for the week cost $250” means [latex]\text{ groceries }+\text{ gas }=250[/latex]
  • “The difference of [latex]9[/latex] and a number” becomes [latex]9-x[/latex]. Notice how [latex]9[/latex] is first in the sentence and the expression.

Let’s practice translating a few more English phrases into algebraic expressions.

Translate the table into algebraic expressions:

In this example video, we show how to translate more words into mathematical expressions.

For another review of how to translate algebraic statements into words, watch the following video.

The power of algebra is how it can help you model real situations in order to answer questions about them.  Let’s use this approach with another example.

Yash brought apples and bananas to a picnic. The number of apples was three more than twice the number of bananas. Yash brought [latex]11[/latex] apples to the picnic. How many bananas did he bring?

https://ohm.lumenlearning.com/multiembedq.php?id=142722&theme=oea&iframe_resize_id=mom2

Twenty-eight less than five times a certain number is [latex]232[/latex]. What is the number?

Following the steps provided:

  • Read and understand: we are looking for a number.
  • Constants and variables:  [latex]28[/latex] and [latex]232[/latex] are constants, “a certain number” is our variable, because we don’t know its value, and we are asked to find it. We will call it [latex]x[/latex].
  • Translate:  five times a certain number translates to [latex]5x[/latex] Twenty-eight less than five times a certain number translates to [latex]5x-28[/latex], because subtraction is built backward. “is 232” translates to “[latex]=232″[/latex] since “is” is associated with equals.
  • Write an equation:  [latex]5x-28=232[/latex]

[latex]\begin{array}{r}5x-28=232\\5x=260\\x=52\,\,\,\end{array}[/latex]

[latex]\begin{array}{r}5\left(52\right)-28=232\\5\left(52\right)=260\\260=260\end{array}[/latex]

In the video that follows, we show another example of how to translate a sentence into a mathematical expression using a problem solving method.

In the next example, we will apply our Problem-Solving Strategy to applications of percent.

Nga’s car insurance premium increased by [latex]\text{\$60}[/latex], which was [latex]\text{8%}[/latex] of the original cost. What was the original cost of the premium?

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300 Primary School Maths Word Problems: Includes Examples And Answers

Sophie bartlett.

Solving word problems at KS1 and KS2 is an essential part of the new maths curriculum. Here you can find expert guidance on how to solve maths word problems as well as examples of the many different types of word problems primary school children will encounter with links to hundreds more .

What is a word problem?

Mastery helps children to explore maths in greater depth, how to teach children to solve word problems , word problems in year 1, word problems in year 2, word problems in year 3 , word problems in year 4, word problems in year 5, word problems in year 6, place value word problems, addition and subtraction word problems, multiplication and division word problems, fraction word problems, how important are word problems when it comes to the sats , remember: the word problems can change but the maths won’t .

A word problem in maths is a maths question written as one sentence or more that requires children to apply their maths knowledge to a ‘real-life’ scenario. 

This means that children must be familiar with the vocabulary associated with the mathematical symbols they are used to, in order to make sense of the word problem. 

For example:

Importance of word problems within the national curriculum

The National Curriculum states that its mathematics curriculum “aims to ensure that all pupils:

  • become fluent in the fundamentals of mathematics, including through varied and frequent practice with increasingly complex problems over time , so that pupils develop conceptual understanding and the ability to recall and apply knowledge rapidly and accurately;
  • reason mathematically by following a line of enquiry, conjecturing relationships and generalisations, and developing an argument, justification or proof using mathematical language;
  • can solve problems by applying their mathematics to a variety of routine and non-routine problems with increasing sophistication , including breaking down problems into a series of simpler steps and persevering in seeking solutions.”

To support this schools are adopting a ‘mastery’ approach to maths

The National Centre for Excellence in the Teaching of Mathematics (NCETM) have defined “teaching for mastery”, with some aspects of this definition being: 

  • Maths teaching for mastery rejects the idea that a large proportion of people ‘just can’t do maths’.  
  • All pupils are encouraged by the belief that by working hard at maths they can succeed. 
  • Procedural fluency and conceptual understanding are developed in tandem because each supports the development of the other. 
  • Significant time is spent developing deep knowledge of the key ideas that are needed to underpin future learning. The structure and connections within the mathematics are emphasised, so that pupils develop deep learning that can be sustained.

(The Essence of Maths Teaching for Mastery, 2016)

Year 3 to 6 Rapid Reasoning Worksheet for Weeks 1-6

Download for FREE 6 weeks of Rapid Reasoning worksheets. That include six weeks of daily reasoning and problem-solving questions for years 3, 4, 5 and 6!

One of NCETM’s Five Big Ideas in Teaching for Mastery (2017) is “ Mathematical Thinking: if taught ideas are to be understood deeply, they must not merely be passively received but must be worked on by the student: thought about, reasoned with and discussed with others”.

In other words – yes, fluency in arithmetic is important; however, with this often lies the common misconception that once a child has learnt the number skills appropriate to their level/age, they should be progressed to the next level/age of number skills. 

The mastery approach encourages exploring the breadth and depth of these concepts (once fluency is secure) through reasoning and problem solving. 

See the following example:

What sort of word problems might my child encounter at school.

In Key Stage 2, there are nine ‘strands’ of maths – these are then further split into ‘sub-strands’.  For example, ‘number and place value’ is the first strand: a Year 3 sub-strand of this is to “find 10 or 100 more or less than a given number”; a Year 6 sub-strand of this is to “determine the value of each digit in numbers up to 10 million”. The table below shows how the ‘sub-strands’ are distributed across each strand and year group in KS2.

Here are two simple strategies that can be applied to many word problems before solving them.

  • What do you already know?
  • How can this problem be drawn/represented pictorially?

Let’s see how this can be applied to a word problem to help achieve the answer.

Solving a simple word problem

There are 28 pupils in a class. The teacher has 8 litres of orange juice. She pours 225 millilitres of orange juice for every pupil. How much orange juice is left over?

1. What do you already know?

  • There are 1,000ml in 1 litre
  • Pours = liquid leaving the bottle = subtraction
  • For every = multiply
  • Left over = requires subtraction at some point

2. How can this problem be drawn/represented pictorially?

The bar model is always a brilliant way of representing problems, but if you are not familiar with this, there are always other ways of drawing it out. 

Read more: What is a bar model

For example, for this question, you could draw 28 pupils (or stick man x 28) with ‘225 ml’ above each one and then a half-empty bottle with ‘8 litres’ marked at the top.

Now to put the maths to work. This is a Year 6 multi-step problem, so we need to use what we already know and what we’ve drawn to break down the steps.

Solving a more complex word problem

Mara is in a bookshop. She buys one book for £6.99 and another that costs £3.40 more than the first book. She pays using a £20 notes. What change does Mara get?

  • More than = add
  • Using decimals means I will have to line up the decimal points correctly in calculations
  • Change from money = subtract

See this example of bar modelling for this question:

Now to put the maths to work using what we already know and what we’ve drawn to break down the steps.

Mara is in a bookshop. 

She buys one book for £6.99 and another that costs £3.40 more than the first book. 1) £6.99 + (£6.99 + £3.40) = £17.38

She pays using a £20 note. What change does Mara get? 2) £20 – £17.38 = £2.62

Maths word problems for years 1 to 6

The more children learn about maths as the go through primary school, the trickier the word problems they face will become.

Below you will find some information about the types of word problems your child will be coming up against on a year by year basis, and how word problems apply to each primary year group

Throughout Year 1 a child is likely to be introduced to word problems with the help of concrete resources (pieces of physical apparatus like coins, cards, counters or number lines) to help them understand the problem.

An example of a word problem for Year 1 would be:

Chris is going to buy a cake for his mum which costs 80p. How many 20p coins would he need to do this? 

Year 2 is a continuation of Year 1 when it comes to word problems, with children still using concrete maths resources to help them understand and visualise the problems they are working on

An example of a word problem for Year 2 would be:

A class of 10 children each have 5 pencils in their pencil cases. How many pencils are there in total?

With word problems for year 3 , children will move away from using concrete resources when solving word problems, and move towards using written methods. Teachers will begin to demonstrate the four operations such as addition and subtraction word problems, multiplication and division problems too.

This is also the year in which 2-step word problems will be introduced. This is a problem which requires two individual calculations to be completed.

Year 3 word problem: Geometry properties of shape

Shaun is making 3-D shapes out of plastic straws.

At the vertices where the straws meet, he uses blobs of modelling clay to fix them together

Here are some of the shapes he makes:

One of Sean’s shapes is a cuboid. Which is it? Explain your answer.

Answer: shape B as a cuboid has 12 edges (straws) and 8 vertices (clay)

Year 3 word problem: Statistics

Year 3 are collecting pebbles. This pictogram shows the different numbers of pebbles each group finds.

Answer: a) 9   b) 3 pebbles drawn

Top tip By the time children are in Year 3 many of the word problems, even one-step story problems tend to be a variation on a multiplication problem. For this reason learning times tables becomes increasingly essential at this stage. One of the best things you can do to help with Year 3 maths at home is support your child to do this. You can also help children while they are developing this skill by providing 100 squares to help them solve these word problems.

At this stage of their primary school career, children should feel confident using the written method for each of the four operations. 

Word problems for year 4 will include a variety of problems, including 2-step problems and be children will be expected to work out the appropriate method required to solve each one. 

Year 4 word problem: Number and place value

My number has four digits and has a 7 in the hundreds place. The digit which has the highest value in my number is 2. The digit which has the lowest value in my number is 6. My number has 3 fewer tens than hundreds. What is my number?

Answer: 2,746

One and 2-step word problems continue with word problems for year 5 , but this is also the year that children will be introduced to word problems containing decimals.

These are some examples of Year 5 maths word problems .

Year 5 word problem: Fractions, decimals and percentages

Stan, Frank and Norm are washing their cars outside their houses. Stan has washed 0.5 of his car. Frank has washed 1/5 of his car. Norm has washed 5% of his car.

Who has washed the most?

Explain your answer.

Answer: Stan (he has washed 0.5 whereas Frank has only washed 0.2 and Norm 0.05)

Word problems for year 6 shift from 2-step word problems to multi-step word problems. These will include fractions, decimals, percentages and time word problems. 

Here are some examples of the types of maths word problems Year 6 will have to solve.

Year 6 word problem – Ratio and proportion

This question is from the 2018 key stage 2 SATs paper. It is worth 1 mark.

The Angel of the North is a large statue in England. It is 20 metres tall and 54 metres wide. 

Ally makes a scale model of the Angel of the North. Her model is 40 centimetres tall. How wide is her model?

Answer: 108cm

Year 6 word problem – Algebra

This question is from the 2018 KS2 SATs paper. It is worth 2 marks as there are 2 parts to the answer.

Amina is making designs with two different shapes.

She gives each shape a value.

Calculate the value of each shape.

Answer: 36 (hexagon) and 25 . 

Year 6 word problem: Measurement

This question is from the 2018 KS2 SATs paper. It is worth 3 marks as it is a multi-step problem.

Answer: 1.7 litres or 1,700ml

Topic based word problems

The following examples give you an idea of the kinds of maths word problems your child will encounter for each of the 9 strands of maths in KS2.

Place value word problem Year 5

This machine subtracts one hundredth each time the button is pressed. The starting number is 8.43. What number will the machine show if the button is pressed six times? Answer: 8.37

Download free number and place value word problems for Years 3, 4, 5 and 6

Addition and subtraction word problem Year 3

In Year 3 pupils will solve addition word problems and subtraction word problems with 2 and 3 digits.

Sam has 364 sweets. He gets given 142 more. He then gives 277 away. How many sweets is he left with? Answer: 229

Download free addition and subtraction word problems for Years 3, 4, 5 and 6

Addition word problem Year 3

Alfie thinks of a number. He subtracts 70. His new number is 12. What was the number Alfie thought of? Answer: 82

Subtraction word problem Year 6

The temperature at 7pm was 4oC. By midnight, it had dropped by 9 degrees. What was the temperature at midnight? Answer: -5oC

More here: 25 addition and subtraction word problems

Multiplication and division word problem Year 3

A baker is baking chocolate cupcakes. She melts 16 chocolate buttons to make the icing for 9 cakes. How many chocolate buttons will she need to melt to make the icing for 18 cakes? Answer: 32

Multiplication word problem Year 4

E ggs are sold in boxes of 12. The egg boxes are delivered to stores in crates. Each crate holds 9 boxes. How many eggs are in a crate? Answer: 108

Download free multiplication word problems for Years 3, 4, 5 and 6

Division word problem Year 6

A factory produces 1,692 paintbrushes every day. They are packaged into boxes of 9. How many boxes does the factory produce every day? Answer: 188

Download our free division word problems worksheets for Years 3, 4, 5 and 6.

More here: 20 multiplication word problems More here: 25 division word problems

Free resource: Use these four operations word problems to practise addition, subtraction, multiplication and division all together.

Fraction word problem Year 5

At the end of every day, a chocolate factory has 1 and 2/6 boxes of chocolates left over. How many boxes of chocolates are left over by the end of a week? Answer: 9 and 2/6 or 9 and 1/3

Download free fractions and decimals word problems worksheets for Years 3, 4, 5 and 6

More here: 28 fraction word problems

Decimals word problem Year 4 (crossover with subtraction)

Which two decimals that have a difference of 0.5? 0.2, 0.25, 0.4, 0.45, 0.6, 0.75. Answer: 0.25 and 0.75

Download free decimals and percentages word problems resources for Years 3, 4, 5 and 6

Percentage word problem Year 5

There are 350 children in a school. 50% are boys. How many boys are there? Answer: 175

Measurement word problem Year 3 (crossover with subtraction)

Lucy and Ffion both have bottles of strawberry smoothie. Each bottle contains 1 litre. Lucy drinks ½ of her bottle. Ffion drinks 300ml of her bottle. How much does each person have left in both bottles? Answer: Lucy = 500ml, Ffion = 300ml

More here: 25 percentage word problems

Money word problem Year 3

James and Lauren have different amounts of money. James has twelve 2p coins. Lauren has seven 5p coins. Who has the most money and by how much? Answer: Lauren by 11p.

More here: 25 money word problems

Area word problem Year 4

A rectangle measures 6cm by 5cm.

What is its area? Answer: 30cm2

Perimeter word problem Year 4

The swimming pool at the Sunshine Inn hotel is 20m long and 7m wide. Mary swims around the edge of the pool twice. How many metres has she swum? Answer: 108m

Ratio word problem Year 6 (crossover with measurement)

A local council has spent the day painting double yellow lines. They use 1 pot of yellow paint for every 100m of road they paint. How many pots of paint will they need to paint a 2km stretch of road? Answer: 20 pots

More here: 24 ratio word problems

Bodmas word problem Year 6

Draw a pair of brackets in one of these calculations so that they make two different answers. What are the answers?

50 – 10 × 5 =

50 – 10 × 5 =

Volume word problem Year 6

This large cuboid has been made by stacking shipping containers on a boat. Each individual shipping container has a length of 6m, a width of 4m and a height of 3m. What is the volume of the large cuboid? Answer: 864m3

In the KS1 SATs, 58% (35/60 marks) of the test is comprised of maths ‘reasoning’ (word problems). 

In KS2, this increases to 64% (70/110 marks) spread over two reasoning papers, each worth 35 marks. Considering children have, in the past, needed approximately 55-60% to reach the ‘expected standard’, it’s clear that children need regular exposure to and a solid understanding of how to solve a variety of word problems.

Children have the opportunity to practice SATs style word problems in Third Space Learning’s online one-to-one SATs revision programme. Personalised to meet the needs of each student, our programme helps to fill gaps and give students more confidence going in to the SATs exams.

It can be easy for children to get overwhelmed when they first come across word problems in KS2, but it is important that you remind them that whilst the context of the problem may be presented in a different way, the maths behind it remains the same. 

Word problems are a good way to bring maths into the real world and make maths more relevant for your child, so help them practise, or even ask them to turn the tables and make up some word problems for you to solve. 

This article while written by a teacher for teachers is also suitable for those at home supporting children with home learning . More free home learning resources are also available.

Do you have pupils who need extra support in maths? Every week Third Space Learning’s maths specialist tutors support thousands of pupils across hundreds of schools with weekly online 1-to-1 lessons and maths interventions designed to address learning gaps and boost progress. Since 2013 we’ve helped over 150,000 primary and secondary school pupils become more confident, able mathematicians. Learn more or request a personalised quote for your school to speak to us about your school’s needs and how we can help.

Subsidised one to one maths tutoring from the UK’s most affordable DfE-approved one to one tutoring provider.

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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?

how to solve word problem examples

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.

how to solve word problem examples

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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Choose Your Test

Sat / act prep online guides and tips, act math word problems: the ultimate guide.

feature_typeqriter

Though the majority of ACT math problems use diagrams or simply ask you to solve given mathematical equations, you will also see approximately 15-18 word problems on any given ACT (between 25% and 30% of the total math section). This means that knowing how best to deal with word problems will help you significantly when taking the test. Though there are many different types of ACT word problems, most of them are not nearly as difficult or cumbersome as they may appear. 

This post will be your complete guide to ACT word problems:  how to translate your word problems into equations and diagrams, the different types of word problems you’ll see on the test, and how best to go about solving your word problems for test day.

What Are Word Problems?

A word problem is any problem that is based mostly or entirely on written description and does not provide you with an equation, diagram, or graph . You must use your reading skills to translate the words of the question into a workable math problem and then solve for your information.

Word problems will show up on the test for a variety of reasons. Most of the time, these types of questions act to test your reading and visualization skills, as well act as a medium to deliver questions that would otherwise be untestable. For instance, if you must determine the number of sides of an unknown polygon based on given information, a diagram would certainly give the game away!

Translating Word Problems Into Equations or Drawings

In order to translate your word problems into actionable math equations that you can solve, you’ll need to know and utilize some key math terms. Whenever you see these words, you can translate them into the proper action. For instance, the word “product” means “the value of two or more values that have been multiplied together,” so if you need to find “the product of a and b,” you’ll need to set up your equation with $a * b$.  

Let's take a look at this in action with an example problem:

body_ACT_word_problem_10

We have two different cable companies that each have different rates for installation and different monthly fees. We are asked to find out how many months it will take for the cost for each company to be the "same," which means we must set the two rates equal.

Uptown Cable charges 120 dollars for installation plus 25 dollars a month. We do not know how many months we're working with, so we will have:

$120 + 25x$

Downtown Cable charges 60 dollars for installation and 35 dollars per month. Again, we don't know how many months we're working with, but we know they will be the same, so we will have:

And, again, because we are finding the amount of months when the cost is the "same," we must set our rates equal.  

$120 + 25x = 60 + 35x$

From here, we can solve for $x$, since it is a single variable equation . 

[Note: the final answer is G , 6 months]

body_rosetta_stone

Learning the language of ACT word problems will help you to unravel much of the mystery of these types of questions. 

Typical ACT Word Problems

ACT word problems can be grouped into two major categories: word problems where you must simply set up an equation and word problems in which you must solve for a specific piece of information.

Word Problem Type 1: Setting Up an Equation

This is the less common type of word problem on the test, but you’ll generally see it at least once or twice. You'll also usually see this type of word problem first . For this type of question, you must use the given information to set up the equation, even though you don’t need to solve for the missing variable.

Almost always, you’ll see this type of question in the first ten questions on the test, meaning that the ACT test-makers consider them fairly “easy.” This is due to the fact that you only have to provide the set-up and not the execution.

body_ACT_word_problem_2

We consider a “profit” to be any money that is gained, so we must always subtract our costs from our earnings. We know that Jones had to invest 10 million starting capital, so he is only making a profit if he has earned more than 10 million dollars. This means we can eliminate answer choices C, D, and E, as they do not account for this 10 million.

Now each boat costs Jones 7,000 dollars to make and he sells them for 20,000. This means that he earns a profit of:

$20,000 - 7,000$

$13,000$ per boat.

If $x$ represents our number of boats, then our final equation will be:

$13,000x - 10,000,000$

Our final answer is A , $13,000x - 10,000,000$

Word Problem Type 2: Solving for Your Information

Other than the few set-up word questions you’ll see, the rest of your ACT 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 ACT math topics , including averages , single variable equations , and probabilities , among others. 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_ACT_word_problem_1

This question is a rare example of a time in which not every piece of given information is needed to solve the problem. For most ACT word questions, all your given information will come into play at some point, but this is not the case here (though you can use all of your information, should you so choose).

For example, we are told that 25% of a given set of jelly beans are red. 25% translates to $1/4$ because 25% is the same as $25/100$ (or $1/4$). If we are being asked to find how many jelly beans are NOT red, then we know it would be $3/4$ because 100% is the same as 1, and 1 - $1/4$ = $3/4$.

So we didn’t need to know that there were 400 jellybeans to know that our final answer is H, $3/4$.

Alternatively, we could use all of our given information and find 25% of 400 in order to find the remaining jelly beans.

$400 * {1/4}$ or $400/4$

If 100 jellybeans are red, then 400 - 100 = 300 jelly beans are NOT red. This means that the not-red jelly beans make up,

$3/4$ of the total number of jelly beans.

Again, our final answer is H, $3/4$

You might also be given a geometry problem as a word problem, which may or may not be set up with a scenario as well.

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.

body_ACT_word_problem_5

The test-makers didn’t give us a diagram, so let's make ourselves one and fill it in with what we know so far.

body_parallelogram_ex_1

We know from our studies of parallelograms  that opposite side pairs will be equal, so we know that the opposite side of our given will also be 12.

body_parallelogram_ex_2

Now we can use this information to subtract from our total perimeter.

$72 - 12 - 12$

Again, opposite sides will be equal and we know that the sum of the two remaining sides will be 48. This means that each remaining side will be:

Now we have four sides in the pairings of 12 and  24.

Our final answer is C , 12, 12, 24, 24.

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Now, how do we put our knowledge to its best effect? Let's take a look.

ACT Math Strategies for Your Word Problems

Though you’ll see word problems on a myriad of different types of ACT math topics, there are still a few techniques you can apply to solve your 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 ratio/division problem :

body_ACT_word_problem_13

Let's start by first drawing our sandwich and Jerome's portion of it.

body_sandwich_1

Now let's divvy off Kevin's portion and, by the remainder, Seth's as well. 

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By seeing the problem visually, we can see that the ratio of Jerome's share, to Kevin's, to Seth's will go in descending order of size. This let's us eliminate answer choices A, B, and C, and leaves us with answer choices D and E.

Just by drawing it out and using process of elimination, and without knowing anything else about ratios, we have a 50-50 shot of guessing the right answer. And, again, without knowing anything else about fractions or ratios, we can make an educated guess between the two options. Since Jerome's share doesn't look twice as large as Kevin's, our answer is probably not E. 

This leaves us with our final answer D , 3:2:1. 

[Note: for a breakdown on how to solve this problem using fractions and ratios instead of using a diagram and educated guessing, check out our guide to ACT fractions and ratios .]

As for geometry problems, remember — you’re often given a word problem as a word problem because it would be too simple to solve had you had a diagram to work with from the get-go . So take back the advantage and draw the picture yourself. Even a quick and dirty sketch can help you visualize the problem much easier than you can in your head and help keep all your information clear.

#2: Memorize Important Terms

If you’re not used to translating English words into mathematical equations, then ACT word problems can sound like so much nonsense and leave you floundering to set up the proper equation. Look to the chart and learn how to translate your keywords into their math equivalents. Doing so will help you to understand exactly what the problem is asking you to find.

There are free ACT math questions available online , so memorize your terms and then practice on real ACT word problems to make sure you’ve got your definitions down and can apply them to real problems.

#3: Underline and Write Out the Key Information

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

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

#4: Pay Close Attention to Exactly  What Is Being Asked of You

Little is more frustrating than 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 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 $x$ or $x + y$? Better to make sure before you start what you’re supposed to find than realize two minutes down the line that you have to solve the problem all over again.

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

You are likely to see both diagram/equation problems and word problems for any given ACT math topic on the test. Many of the topics can swing either way, which is why there are so many different types of word problems and why you’ll need to know the ins and outs of any particular math topic in order to solve its corresponding word problem. For example, if you don’t know how to properly set up a system of equations problem, you will have a difficult time of it when presented with a word problem on the topic.

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

body_thinking

All set? Time to shine!

Test Your Knowledge

Now to put your word problem know-how to the test with real ACT math problems. 

body_ACT_word_problem_3

Answers: K, C, A, E

Answer Explanations:

1) First, let us make a sketch of what we have, just so we can keep our measurements straight. We know we have two triangles, one smaller than the other, and the hypotenuse of the smaller triangle is 5. 

body_Ratios

Now our triangles are in a ratio of 2:5, so if the hypotenuse of the smaller triangle is 5, we can find the hypotenuse of the larger triangle by setting them up in a proportion . 

$2/5 = 5/x$

Our final answer is K , 12.5.

2) Because we are dealing with a hypothetical number that is increasing and decreasing based on percentage, we can solve this problem in one of two ways--by using algebra or by plugging in our own numbers. 

Solving Method 1: Algebra

If we assign our hypothetical number as $x$, we can say that $x$ is increased by 25% by saying:

$x + 0.25x$

Which gives us:

Now, we can decrease this value by 20% by saying:

$1.25x - (1.25x * 0.2)$

$1.25x - 0.25x$

This leaves us with:

$1x$ or 100% of our original number. 

Our final answer is C, 100%. 

Solving Method 2: Plugging in Numbers

Alternatively, we can use the same basic process, but make it a little simpler by using numbers instead of variables. 

Let's say our original number is 100. (Why 100? Why not! Our number can literally be anything and 100 is an easy number to work with.)

So if we need to increase 100 by 25%, we first need to find 25% of 100 and then add that to 100. 

$100 + (0.25)100$

Now we need to decrease this value by 20%, so we would say:

$125 - (0.2)125$

We are left with the same number we started with, which means we are left with 100% of the number we started with. 

Again, our final answer is C, 100%. 

3) Let's first begin by drawing a picture of our scene. We know that one vertex of the square is at (3, 0), so we can mark it on a coordinate plane. 

body_vertex_1

Now, we are told that each side of the square is 3 cm long. To make life simple, we can start by marking all the possible vertexes attached to our known vertex at (3, 0) straight up, down, and side to side. If no answers match, we can then look to vertexes at different angles. 

body_vertex_2

Our possible vertexes are:

(0, 0), (6, 0), 3, 3) and (3, -3)

One of our possible vertexes is at (6, 0 and this matches one of our answer choices, so we can stop here. 

Our final answer is A, (6, 0). 

4) We are told that Ms. Lopez throws out the lowest test score and then averages the remaining scores. Because Victor's scores are already in ascending order, we can throw out the first score of 62. 

Now to find the average of the remaining 4 scores, let us add them together and then divide by the number of scores. 

$(78 + 83 + 84 + 93)/4$

Our final answer is E, 84.5. 

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The Take-Aways

Word problems comprise a significant portion of the ACT, so it’s a good idea to understand how they work and how to translate the words into a proper equation . But remember that translating your word problems is still only half the battle.

You must also supplement this knowledge of how to solve word problems with a solid understanding of the math topic in question. For example, it won’t do a lot of good if you can translate a probability word problem if you don’t understand exactly how probabilities work. So be sure to not only learn how to approach your word problems, but also hone your focus on any math topics  you feel you need to improve upon. You can find links to all of our ACT math topic guides here to help your studies.

What’s Next?  

Want to brush up on any of your other math topics?  Check out our  individual math guides  to get the walk-through on each and every  topic on the ACT math test .  

Trying to stop procrastinating?   Learn  how to get over your desire to procrastinate  and make a well-balanced study plan. 

Running out of time on the ACT math section?  We'll teach you how to beat the clock and maximize your ACT math score . 

Trying to get a perfect score?  Check out our  guide to getting a perfect 36 on ACT math , written by a perfect-scorer. 

  

Want to improve your ACT score by 4 points? 

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Our program is entirely online, and it customizes what you study to your strengths and weaknesses . If you liked this Math lesson, you'll love our program.  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.

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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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Word Problem Practice Questions with Answer Key

how to solve word problem examples

  • Posted by Brian Stocker
  • Date February 13, 2019
  • Comments 11 comments

Problem Solving – Word Problems

Word problems are mathematical problems using everyday language and real-world situations.  Some information is given and one or more pieces or information (variables) are missing.  You must understand the given information, identify the mathematical operations necessary to solve the problem, and then carry out those operations to obtain the missing information or variables.

The Biggest Tip!

Tackling word problems is much easier if you have a systematic approach which we outline below.

Here is the biggest tip for word problems practice. Practice regularly and systematically. Sounds simple and easy right? Yes it is, and yes it really does work.   Word problems are a way of thinking and require you to translate a real world problem into mathematical terms.

Some math instructors go so far as to say that learning how to think mathematically is the main reason for teaching word problems. So what do we mean by Practice regularly and systematically? Studying word problems and math in general requires a logical and mathematical frame of mind. The only way that you can get this is by practicing regularly, which means everyday.

How to Solve Word Problems

Types of Word Problems

Most Common Word Problem Mistakes on a Test

It is critical that you practice word problems  everyday for the 5 days before the exam as a bare minimum.  If you practice and miss a day, you have lost the mathematical frame of mind and the benefit of your previous practice is pretty much gone. Anyone who has done any number of math tests will agree – you have to practice everyday.

See Also Algebra Word Problems

Effective problem-solving skills are essential in many areas of life, from academia to the workplace and beyond. Developing the ability to solve word problems requires practice and patience, as well as a strong understanding of basic mathematical concepts and operations.

Try a FREE Quiz

Word problem practice questions.

1. A box contains 7 black pencils and 28 blue ones. What is the ratio between the black and blue pens?

a. 1:4 b. 2:7 c. 1:8 d. 1:9

2. The manager of a weaving factory estimates that if 10 machines run at 100% efficiency for 8 hours, they will produce 1450 meters of cloth. Due to some tech¬nical problems, 4 machines run of 95% efficiency and the remaining 6 at 90% efficiency. How many meters of cloth can these machines will produce in 8 hours?

a. 1334 meters b. 1310 meters c. 1300 meters d. 1285 meters

3. In a local election at polling station A, 945 voters cast their vote out of 1270 registered voters. At poll¬ing station B, 860 cast their vote out of 1050 regis¬tered voters and at station C, 1210 cast their vote out of 1440 registered voters. What is the total turnout from all three polling stations?

a. 70% b. 74% c. 76% d. 80%

4. If Lynn can type a page in p minutes, what portion of the page can she do in 5 minutes?

a. p/5 b. p – 5 c. p + 5 d. 5/p

5. If Sally can paint a house in 4 hours, and John can paint the same house in 6 hours, how long will it take for both to paint a house?

a. 2 hours and 24 minutes b. 3 hours and 12 minutes c. 3 hours and 44 minutes d. 4 hours and 10 minutes

6. Employees of a discount appliance store receive an additional 20% off the lowest price on any item. If an employee purchases a dishwasher during a 15% off sale, how much will he pay if the dishwasher originally cost $450?

a. $280.90 b. $287.00 c. $292.50 d. $306.00

7. The sale price of a car is $12,590, which is 20% off the original price. What is the original price?

a. $14,310.40 b. $14,990.90 c. $15,108.00 d. $15,737.50

8. Richard gives ‘s’ amount of salary to each of his ‘n’ employees weekly. If he has ‘x’ amount of money, how many days he can employ these ‘n’ employees.

a. sx/7n b. 7x/nx c. nx/7s d. 7x/ns

9. A distributor purchased 550 kilograms of potatoes for $165. He distributed these at a rate of $6.4 per 20 kilograms to 15 shops, $3.4 per 10 kilograms to 12 shops and the remainder at $1.8 per 5 kilo¬grams. If his total distribution cost is $10, what will his profit be?

a. $10.40 b. $8.60 c. $14.90 d. $23.40

10. How much pay does Mr. Johnson receive if he gives half of his pay to his family, $250 to his land¬lord, and has exactly 3/7 of his pay left over?

a. $3600 b. $3500 c. $2800 d. $175042

11. The cost of waterproofing canvas is .50 a square yard. What’s the total cost for waterproofing a canvas truck cover that is 15’ x 24’?

a. $18.00 b. $6.67 c. $180.00 d. $20.00

1. A The ratio between black and blue pens is 7 to 28 or 7:28. Bring to the lowest terms by dividing both sides by 7 gives 1:4.

2. A At 100% efficiency 1 machine produces 1450/10 = 145 m of cloth. At 95% efficiency, 4 machines produce 4 * 145 * 95/100 = 551 m of cloth. At 90% efficiency, 6 machines produce 6 * 145 * 90/100 = 783 m of cloth. Total cloth produced by all 10 machines = 551 + 783 = 1334 m Since the information provided and the question are based on 8 hours, we did not need to use time to reach the answer.

The turnout at polling station A is 945 out of 1270 registered voters. So, the percentage turnout at station A is:

(945/1270) × 100% = 74.41%

The turnout at polling station B is 860 out of 1050 registered voters. So, the percentage turnout at station B is:

(860/1050) × 100% = 81.90%

The turnout at polling station C is 1210 out of 1440 registered voters. So, the percentage turnout at station C is:

(1210/1440) × 100% = 84.03%

To find the total turnout from all three polling stations, we need to add up the total number of voters and the total number of registered voters from all three stations:

Total number of voters = 945 + 860 + 1210 = 3015

Total number of registered voters = 1270 + 1050 + 1440 = 3760

The overall percentage turnout is:

(3015/3760) × 100% = 80.12%

Therefore, the total turnout from all three polling stations is 80.12% — rounding to 80%.

4. D This is a simple direct proportion problem: If Lynn can type 1 page in p minutes, then she can type x pages in 5 minutes We do cross multiplication: x * p = 5 * 1 Then, x = 5/p

5. A This is an inverse ratio problem. 1/x = 1/a + 1/b where a is the time Sally can paint a house, b is the time John can paint a house, x is the time Sally and John can together paint a house. So, 1/x = 1/4 + 1/6 … We use the least common multiple in the denominator that is 24: 1/x = 6/24 + 4/24 1/x = 10/24 x = 24/10 x = 2.4 hours. In other words; 2 hours + 0.4 hours = 2 hours + 0.4•60 minutes = 2 hours 24 minutes

The original price of the dishwasher is $450. During a 15% off sale, the price of the dishwasher will be reduced by:

15% of $450 = 0.15 x $450 = $67.50

So the sale price of the dishwasher will be:

$450 – $67.50 = $382.50

As an employee, the person receives an additional 20% off the lowest price, which is $382.50. We can calculate the additional discount as:

20% of $382.50 = 0.20 x $382.50 = $76.50

So the final price that the employee will pay for the dishwasher is:

$382.50 – $76.50 = $306.00

Therefore, the employee will pay $306.00 for the dishwasher.

7. D Original price = x, 80/100 = 12590/X, 80X = 1259000, X = 15,737.50.

8. D We are given that each of the n employees earns s amount of salary weekly. This means that one employee earns s salary weekly. So; Richard has ‘ns’ amount of money to employ n employees for a week. We are asked to find the number of days n employees can be employed with x amount of money. We can do simple direct proportion: If Richard can employ n employees for 7 days with ‘ns’ amount of money, Richard can employ n employees for y days with x amount of money … y is the number of days we need to find. We can do cross multiplication: y = (x * 7)/(ns) y = 7x/ns

9. B The distribution is done at three different rates and in three different amounts: $6.4 per 20 kilograms to 15 shops … 20•15 = 300 kilograms distributed $3.4 per 10 kilograms to 12 shops … 10•12 = 120 kilograms distributed 550 – (300 + 120) = 550 – 420 = 130 kilograms left. This 50 amount is distributed in 5 kilogram portions. So, this means that there are 130/5 = 26 shops. $1.8 per 130 kilograms. We need to find the amount he earned overall these distributions. $6.4 per 20 kilograms : 6.4•15 = $96 for 300 kilograms $3.4 per 10 kilograms : 3.4 *12 = $40.8 for 120 kilograms $1.8 per 5 kilograms : 1.8 * 26 = $46.8 for 130 kilograms So, he earned 96 + 40.8 + 46.8 = $ 183.6 The total distribution cost is given as $10 The profit is found by: Money earned – money spent … It is important to remember that he bought 550 kilograms of potatoes for $165 at the beginning: Profit = 183.6 – 10 – 165 = $8.6

10. B We check the fractions taking place in the question. We see that there is a “half” (that is 1/2) and 3/7. So, we multiply the denominators of these fractions to decide how to name the total money. We say that Mr. Johnson has 14x at the beginning; he gives half of this, meaning 7x, to his family. $250 to his landlord. He has 3/7 of his money left. 3/7 of 14x is equal to: 14x * (3/7) = 6x So, Spent money is: 7x + 250 Unspent money is: 6×51 Total money is: 14x Write an equation: total money = spent money + unspent money 14x = 7x + 250 + 6x 14x – 7x – 6x = 250 x = 250 We are asked to find the total money that is 14x: 14x = 14 * 250 = $3500

11. D First calculate total square feet, which is 15 * 24 = 360 ft 2 . Next, convert this value to square yards, (1 yards 2 = 9 ft 2 ) which is 360/9 = 40 yards 2 . At $0.50 per square yard, the total cost is 40 * 0.50 = $20.

  • Not reading the problem carefully and thoroughly, so that you either misunderstand or solve the problem incorrectly.
  • Not identifying the important information in the problem, such as the quantities, units, and the operation to be performed.
  • Not translating the information in the problem into mathematical language and equations.
  • Not checking the units of measure and making sure they match your final answer.
  • Not double-checking the answer to ensure it makes sense.
  • Not understanding the underlying mathematical concept or operation the problem is asking for.
  • Not using estimation or approximations as a tool to check the reasonableness of your answer.

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11 comments.

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Will we need to know units and their conversions such as yards to feet? Should we memorise those?

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are we allowed to use a calculator? To expect someone to complete these in their head is absurd especially with a time limit. The second question requires multiplication by decimals, which would be okay if you got a whole number but you dont, you get a fraction and the only way to get it to 551 is by then multiplying that number by 4. Doubt anyone would be required to do these kind of calculations in a real world scenario especially unaided and under time constraints.

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Hi Depends on the test – what test are you studying for?

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is this preparation for the CAAT level C???

These questions vary in subject and difficulty level to give students practice on different types of questions for different types of tests. They are not specific to one test or one level.

Yes the LEAST common multiple of 6 and 4 is 12 – i did it with 24 – it will give the same answer no matter which way you do it. Good point though – perhaps for simplicity sake 12 would be better.

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Are these questions appeared on the Cbest?

The are the same TYPE of questions – not exactly

' src=

sorry for the message above, i like your site and i have won 1st place in an exam due to this site

' src=

I used Chat GPT. Solved every one…. perfectly. I’m still dumb as a rock though.

Oh Really? You may want to check that again!! It gave me wrong answers and weird steps to calculate for 2 of them!

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Solving word problem chart

1. Understand the Problem by Paraphrasing

2. identify key information and variables, 3. translate words into mathematical symbols, 4. break down the problem into manageable parts, 5. draw diagrams or visual representations, 6. use estimation to predict answers, 7. apply logical reasoning for unknown variables, 8. leverage similar problems as templates, 9. check answers in the context of the problem, 10. reflect and learn from mistakes.

Have you ever observed the look of confusion on a student’s face when they encounter a math word problem ? It’s a common sight in classrooms worldwide, underscoring the need for effective strategies for solving math word problems . The main hurdle in solving math word problems is not just the math itself but understanding how to translate the words into mathematical equations that can be solved.

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Generic advice like “read the problem carefully” or “practice more” often falls short in addressing students’ specific difficulties with word problems. Students need targeted math word problem strategies that address the root of their struggles head-on. 

A Guide on Steps to Solving Word Problems: 10 Strategies 

One of the first steps in tackling a math word problem is to make sure your students understand what the problem is asking. Encourage them to paraphrase the problem in their own words. This means they rewrite the problem using simpler language or break it down into more digestible parts. Paraphrasing helps students grasp the concept and focus on the problem’s core elements without getting lost in the complex wording.

Original Problem: “If a farmer has 15 apples and gives away 8, how many does he have left?”

Paraphrased: “A farmer had some apples. He gave some away. Now, how many apples does he have?”

This paraphrasing helps students identify the main action (giving away apples) and what they need to find out (how many apples are left).

Play these subtraction word problem games in the classroom for free:

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Students often get overwhelmed by the details in word problems. Teach them to identify key information and variables essential for solving the problem. This includes numbers , operations ( addition , subtraction , multiplication , division ), and what the question is asking them to find. Highlighting or underlining can be very effective here. This visual differentiation can help students focus on what’s important, ignoring irrelevant details.

  • Encourage students to underline numbers and circle keywords that indicate operations (like ‘total’ for addition and ‘left’ for subtraction).
  • Teach them to write down what they’re solving for, such as “Find: Total apples left.”

Problem: “A classroom has 24 students. If 6 more students joined the class, how many students are there in total?”

Key Information:

  • Original number of students (24)
  • Students joined (6)
  • Looking for the total number of students

Here are some fun addition word problems that your students can play for free:

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The transition from the language of word problems to the language of mathematics is a critical skill. Teach your students to convert words into mathematical symbols and equations. This step is about recognizing keywords and phrases corresponding to mathematical operations and expressions .

Common Translations:

  • “Total,” “sum,” “combined” → Addition (+)
  • “Difference,” “less than,” “remain” → Subtraction (−)
  • “Times,” “product of” → Multiplication (×)
  • “Divided by,” “quotient of” → Division (÷)
  • “Equals” → Equals sign (=)

Problem: “If one book costs $5, how much would 4 books cost?”

Translation: The word “costs” indicates a multiplication operation because we find the total cost of multiple items. Therefore, the equation is 4 × 5 = $20

Complex math word problems can often overwhelm students. Incorporating math strategies for problem solving, such as teaching them to break down the problem into smaller, more manageable parts, is a powerful approach to overcome this challenge. This means looking at the problem step by step rather than simultaneously trying to solve it. Breaking it down helps students focus on one aspect of the problem at a time, making finding the solution more straightforward.

Problem: “John has twice as many apples as Sarah. If Sarah has 5 apples, how many apples do they have together?”

Steps to Break Down the Problem:

Find out how many apples John has: Since John has twice as many apples as Sarah, and Sarah has 5, John has 5 × 2 = 10

Calculate the total number of apples: Add Sarah’s apples to John’s to find the total,  5 + 10 = 15

By splitting the problem into two parts, students can solve it without getting confused by all the details at once.

Explore these fun multiplication word problem games:

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Diagrams and visual representations can be incredibly helpful for students, especially when dealing with spatial or quantity relationships in word problems. Encourage students to draw simple sketches or diagrams to represent the problem visually. This can include drawing bars for comparison, shapes for geometry problems, or even a simple distribution to better understand division or multiplication problems .

Problem: “A garden is 3 times as long as it is wide. If the width is 4 meters, how long is the garden?”

Visual Representation: Draw a rectangle and label the width as 4 meters. Then, sketch the length to represent it as three times the width visually, helping students see that the length is 4 × 3 = 12

Estimation is a valuable skill in solving math word problems, as it allows students to predict the answer’s ballpark figure before solving it precisely. Teaching students to use estimation can help them check their answers for reasonableness and avoid common mistakes.

Problem: “If a book costs $4.95 and you buy 3 books, approximately how much will you spend?”

Estimation Strategy: Round $4.95 to the nearest dollar ($5) and multiply by the number of books (3), so 5 × 3 = 15. Hence, the estimated total cost is about $15.

Estimation helps students understand whether their final answer is plausible, providing a quick way to check their work against a rough calculation.

Check out these fun estimation and prediction word problem worksheets that can be of great help:

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When students encounter problems with unknown variables, it’s crucial to introduce them to logical reasoning. This strategy involves using the information in the problem to deduce the value of unknown variables logically. One of the most effective strategies for solving math word problems is working backward from the desired outcome. This means starting with the result and thinking about the steps leading to that result, which can be particularly useful in algebraic problems.

Problem: “A number added to three times itself equals 32. What is the number?”

Working Backward:

Let the unknown number be x.

The equation based on the problem is  x + 3x = 32

Solve for x by simplifying the equation to 4x=32, then dividing by 4 to find x=8.

By working backward, students can more easily connect the dots between the unknown variable and the information provided.

Practicing problems of similar structure can help students recognize patterns and apply known strategies to new situations. Encourage them to leverage similar problems as templates, analyzing how a solved problem’s strategy can apply to a new one. Creating a personal “problem bank”—a collection of solved problems—can be a valuable reference tool, helping students see the commonalities between different problems and reinforcing the strategies that work.

Suppose students have solved a problem about dividing a set of items among a group of people. In that case, they can use that strategy when encountering a similar problem, even if it’s about dividing money or sharing work equally.

It’s essential for students to learn the habit of checking their answers within the context of the problem to ensure their solutions make sense. This step involves going back to the original problem statement after solving it to verify that the answer fits logically with the given information. Providing a checklist for this process can help students systematically review their answers.

Checklist for Reviewing Answers:

  • Re-read the problem: Ensure the question was understood correctly.
  • Compare with the original problem: Does the answer make sense given the scenario?
  • Use estimation: Does the precise answer align with an earlier estimation?
  • Substitute back: If applicable, plug the answer into the problem to see if it works.

Problem: “If you divide 24 apples among 4 children, how many apples does each child get?”

After solving, students should check that they understood the problem (dividing apples equally).

Their answer (6 apples per child) fits logically with the number of apples and children.

Their estimation aligns with the actual calculation.

Substituting back 4×6=24 confirms the answer is correct.

Teaching students to apply logical reasoning, leverage solved problems as templates, and check their answers in context equips them with a robust toolkit for tackling math word problems efficiently and effectively.

One of the most effective ways for students to improve their problem-solving skills is by reflecting on their errors, especially with math word problems. Using word problem worksheets is one of the most effective strategies for solving word problems, and practicing word problems as it fosters a more thoughtful and reflective approach to problem-solving

These worksheets can provide a variety of problems that challenge students in different ways, allowing them to encounter and work through common pitfalls in a controlled setting. After completing a worksheet, students can review their answers, identify any mistakes, and then reflect on them in their mistake journal. This practice reinforces mathematical concepts and improves their math problem solving strategies over time.

3 Additional Tips for Enhancing Word Problem-Solving Skills

Before we dive into the importance of reflecting on mistakes, here are a few impactful tips to enhance students’ word problem-solving skills further:

1. Utilize Online Word Problem Games

A word problem game

Incorporate online games that focus on math word problems into your teaching. These interactive platforms make learning fun and engaging, allowing students to practice in a dynamic environment. Games can offer instant feedback and adaptive challenges, catering to individual learning speeds and styles.

Here are some word problem games that you can use for free:

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2. Practice Regularly with Diverse Problems

Word problem worksheet

Consistent practice with a wide range of word problems helps students become familiar with different questions and mathematical concepts. This exposure is crucial for building confidence and proficiency.

Start Practicing Word Problems with these Printable Word Problem Worksheets:

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3. Encourage Group Work

Solving word problems in groups allows students to share strategies and learn from each other. A collaborative approach is one of the best strategies for solving math word problems that can unveil multiple methods for tackling the same problem, enriching students’ problem-solving toolkit.

Conclusion 

Mastering math word problems is a journey of small steps. Encourage your students to practice regularly, stay curious, and learn from their mistakes. These strategies for solving math word problems are stepping stones to turning challenges into achievements. Keep it simple, and watch your students grow their confidence and skills, one problem at a time.

Frequently Asked Questions (FAQs)

How can i help my students stay motivated when solving math word problems.

Encourage small victories and use engaging tools like online games to make practice fun and rewarding.

What's the best way to teach beginners word problems?

Begin with simple problems that integrate everyday scenarios to make the connection between math and real-life clear and relatable.

How often should students practice math word problems?

Regular, daily practice with various problems helps build confidence and problem-solving skills over time.

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Solving Inequality Word Questions

(You might like to read Introduction to Inequalities and Solving Inequalities first.)

In Algebra we have "inequality" questions like:

soccer teams

Sam and Alex play in the same soccer team. Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals. What are the possible number of goals Alex scored?

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 needed
  • Assign letters for the values
  • Find or work out formulas

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

The best way to learn this is by example, so let's try our first example:

Assign Letters:

  • the number of goals Alex scored: A
  • the number of goals Sam scored: S

We know that Alex scored 3 more goals than Sam did, so: A = S + 3

And we know that together they scored less than 9 goals: S + A < 9

We are being asked for how many goals Alex might have scored: A

Sam scored less than 3 goals, which means that Sam could have scored 0, 1 or 2 goals.

Alex scored 3 more goals than Sam did, so Alex could have scored 3, 4, or 5 goals .

  • When S = 0, then A = 3 and S + A = 3, and 3 < 9 is correct
  • When S = 1, then A = 4 and S + A = 5, and 5 < 9 is correct
  • When S = 2, then A = 5 and S + A = 7, and 7 < 9 is correct
  • (But when S = 3, then A = 6 and S + A = 9, and 9 < 9 is incorrect)

Lots More Examples!

pups

Example: Of 8 pups, there are more girls than boys. How many girl pups could there be?

  • the number of girls: g
  • the number of boys: b

We know that there are 8 pups, so: g + b = 8, which can be rearranged to

We also know there are more girls than boys, so:

We are being asked for the number of girl pups: g

So there could be 5, 6, 7 or 8 girl pups.

Could there be 8 girl pups? Then there would be no boys at all, and the question isn't clear on that point (sometimes questions are like that).

  • When g = 8, then b = 0 and g > b is correct (but is b = 0 allowed?)
  • When g = 7, then b = 1 and g > b is correct
  • When g = 6, then b = 2 and g > b is correct
  • When g = 5, then b = 3 and g > b is correct
  • (But if g = 4, then b = 4 and g > b is incorrect)

A speedy example:

bike

Example: Joe enters a race where he has to cycle and run. He cycles a distance of 25 km, and then runs for 20 km. His average running speed is half of his average cycling speed. Joe completes the race in less than 2½ hours, what can we say about his average speeds?

  • Average running speed: s
  • So average cycling speed: 2s
  • Speed = Distance Time
  • Which can be rearranged to: Time = Distance Speed

We are being asked for his average speeds: s and 2s

The race is divided into two parts:

  • Distance = 25 km
  • Average speed = 2s km/h
  • So Time = Distance Average Speed = 25 2s hours
  • Distance = 20 km
  • Average speed = s km/h
  • So Time = Distance Average Speed = 20 s hours

Joe completes the race in less than 2½ hours

  • The total time < 2½
  • 25 2s + 20 s < 2½

So his average speed running is greater than 13 km/h and his average speed cycling is greater than 26 km/h

In this example we get to use two inequalities at once:

ball throw

Example: The velocity v m/s of a ball thrown directly up in the air is given by v = 20 − 10t , where t is the time in seconds. At what times will the velocity be between 10 m/s and 15 m/s?

  • velocity in m/s: v
  • the time in seconds: t
  • v = 20 − 10t

We are being asked for the time t when v is between 5 and 15 m/s:

So the velocity is between 10 m/s and 15 m/s between 0.5 and 1 second after.

And a reasonably hard example to finish with:

Example: A rectangular room fits at least 7 tables that each have 1 square meter of surface area. The perimeter of the room is 16 m. What could the width and length of the room be?

Make a sketch: we don't know the size of the tables, only their area, they may fit perfectly or not!

  • the length of the room: L
  • the width of the room: W

The formula for the perimeter is 2(W + L) , and we know it is 16 m

  • 2(W + L) = 16
  • L = 8 − W

We also know the area of a rectangle is the width times the length: Area = W × L

And the area must be greater than or equal to 7:

  • W × L ≥ 7

We are being asked for the possible values of W and L

Let's solve:

This is a quadratic inequality. It can be solved many way, here we will solve it by completing the square :

Yes we have two inequalities, because 3 2 = 9 AND (−3) 2 = 9

So the width must be between 1 m and 7 m (inclusive) and the length is 8−width .

  • Say W = 1, then L = 8−1 = 7, and A = 1 x 7 = 7 m 2 (fits exactly 7 tables)
  • Say W = 0.9 (less than 1), then L = 7.1, and A = 0.9 x 7.1 = 6.39 m 2 (7 won't fit)
  • Say W = 1.1 (just above 1), then L = 6.9, and A = 1.1 x 6.9 = 7.59 m 2 (7 fit easily)
  • Likewise for W around 7 m

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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.

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  • Middle School Math Solutions – Equation Calculator Welcome to our new "Getting Started" math solutions series. Over the next few weeks, we'll be showing how Symbolab... Read More

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Linear Equation Word Problems - Examples & Practice - Expii

Linear equation word problems - examples & practice, explanations (3), (video) slope-intercept word problems.

by cpfaffinator

how to solve word problem examples

This video by cpfaffinator explains how to translate word problems into variables while working through a couple of examples.

(Note: the answer at 3:34 should be y=0.50x+3 instead of y=0.5x+b .)

When writing linear equations (also called linear models) from word problems, you need to know what the x and y variables refer to, as well as what the slope and y-intercept are. Here are some tricks to help translate :

  • The x variable. Usually, you'll be asked to find an equation in terms of some factor. This will usually be whatever affects the output of the equation.
  • The y variable. The y variable is usually whatever you're trying to find in the problem. It'll depend on the x variable.
  • The slope (m). The slope will be some sort of rate or other change over time.
  • The y-intercept (b). Look for a starting point or extra fees—something that will be added (or subtracted) on , no matter what the x variable is.

Using these parts, we can write the word problem in slope intercept form , y=mx+b, and solve.

A city parking garage charges a flat rate of $3.00 for parking 2 hours or less, and $0.50 per hour for each additional hour. Write a linear model that gives the total charge in terms of additional hours parked.

First up, we want the equation, or model, to give us the total charge. This will be what the y variable represents. y=total charge for parking The total charge depends on how many additional hours you've parked. This is the x variables. x=additional hours parked

Next, we need to find the slope, m. If we look at the slope-intercept equation , y=mx+b, the slope is multiplied with the input x. So, we want to find some rate in our problem that goes with the number of additional hours parked . In the problem, we have $0.50 per hour for each additional hour. The number of hours is being multiplied by the price of 50 cents every hour. m=0.50

The y-intercept is the starting point. In other words, if we parked our car in the garage for 0 hours, what would the price be? The problem says the garage *charges a flat rate of $3.00. A flat rate means that no matter what we're being charged 3 dollars. b=3

Plugging these into the slope-intercept equation will give us our model. y=mx+by=0.50x+3

Six 2 foot tall pine trees were planted during the school's observation of Earth Awareness Week in 1990. The trees have grown at an average rate of 34 foot per year. Write a linear model that gives the height of the trees in terms of the number of years since they were planted.

What does the y variable represent in our equation?

Years since the trees were planted

Number of trees planted

Radius of trees' roots

Height of the trees

Related Lessons

Translate to a linear equation.

Sometimes, the trick to solving a word problem will be to translate it into a linear equation. To clue you in, linear equation word problems usually involve some sort of rate of change , or steady increase (or decrease) based on a single variable. If you see the word rate , or even "per" or "each" , it's a safe bet that a word problem is calling for a linear equation.

There are a couple steps when translating from a word problem to a linear equation. Review them, then we'll work through an example.

  • Find the y variable, or output. What is the thing you're trying to find? This will often be a price, or an amount of time, or something else countable that depends on other things.
  • Find the x variable, or input. What is affecting the price, or amount of time, etc.?
  • Find the slope. What's the rate in the problem?
  • Find the y-intercept. Is there anything that's added or subtracted on top of the rate, no matter what what x is?
  • Plug all the numbers you know into y=mx+b !
Wally and Cobb are starting a catering business. They rent a kitchen for $350 a month, and charge $75 for each event they cater. If they cater 12 events in a month, how much do they profit?

First, let's find the y variable. What are we trying to find in this problem?

The name of the business

how to solve word problem examples

Word Problems with Linear Relationships

When you're faced with a linear word problem, it can feel daunting to work out a solution. The good news is that there are few simple steps that you can take to tackle linear word problems.

how to solve word problem examples

Image source: by Anusha Rahman

For future linear word problems, you can use these same steps to tackle the problem!

‘A nightmare’: Special counsel’s assessment of Biden’s mental fitness triggers Democratic panic

WASHINGTON — President Joe Biden sidestepped any criminal charges as the investigation into his handling of classified documents concluded, but the political blowback from the special counsel’s report Thursday could prove even more devastating, reinforcing impressions that he is too old and impaired to hold the highest office.

Special counsel Robert Hur’s portrait of a man who couldn’t remember when he served as Barack Obama’s vice president, or the year when his beloved son Beau died, dealt a blow to Biden’s argument that he is still sharp and fit enough to serve another four-year term.

In deciding not to charge Biden with any crimes, the special counsel wrote that in a potential trial, “Mr. Biden would likely present himself to a jury, as he did during our interview with him, as a sympathetic, well-meaning, elderly man with a poor memory.”

It was tough enough for Biden to reassure voters about his health before Hur’s report hit like a thunderclap Thursday afternoon, prompting members of his own party to question whether he could remain the nominee in November.

“It’s a nightmare,” said a Democratic House member who asked to speak anonymously to provide a frank assessment, adding that “it weakens President Biden electorally, and Donald Trump would be a disaster and an authoritarian.”

“For Democrats, we’re in a grim situation.”

Biden wasted little time before attempting to minimize the fallout. He held an unexpected exchange with reporters in the White House on Thursday night, in which he disputed Hur's assessment of his mental acuity.

Biden grew emotional when invoking the part of the report addressing the date of his son's death.

"How in the hell dare you raise that?" Biden said. "Frankly, when I was asked the question I thought to myself, 'It wasn't any of their damn business.' "

‘Beyond devastating’

Polling has long shown that age looms as Biden’s greatest liability in his expected rematch with Trump. A January poll by NBC News found that 76% of voters have major or moderate concerns about Biden’s mental and physical health.

“It’s been a problem since way before this ever happened,” said a longtime Democratic operative who noted that when focus groups are asked to apply one word to Biden, it is often “old.”

Just this week, Biden twice referred to conversations he’s had as president with foreign leaders who’ve long since died. In his remarks Thursday night defending his competency, while talking about the war in Gaza, he referred to Egyptian President Abdel Fattah el-Sissi as being the head of Mexico. White House press aides have downplayed such lapses as the sort of mistake anyone in public life can make.

The Hur report strips away the defenses that Biden’s press operation has used to protect him and raises fresh doubts about whether Biden is up to the rigors of the presidency, Democratic strategists said in interviews.

“This is beyond devastating,” said another Democratic operative, speaking on condition of anonymity to talk candidly about Biden’s shortcomings. “It confirms every doubt and concern that voters have. If the only reason they didn’t charge him is because he’s too old to be charged, then how can he be president of the United States?”

Asked if Hur’s report changes the calculus for Democrats who expect Biden to be the party’s nominee, this person said: “How the f--- does it not?”

Another Biden ally called it “the worst day of his presidency.”

“I think he needs to show us this is a demonstrably false characterization of him and that he has what it takes to win and govern.”

Biden has overwhelmingly won the first primary contests — notching victories in New Hampshire, South Carolina and Nevada. It would be virtually impossible for anyone else to challenge him at this point; the deadline has passed in more than 30 states to get on primary ballots.

Some of the president’s allies were quick to defend him. They pointed to the timing of the interview with the special counsel — days after Hamas’ attack on Israel, which had captured much of the president’s focus. Others said that in their own dealings with Biden, he shows no sign of infirmity.

“He did so well in this discussion with members,” Rep. Susan Wild, D-Pa., told NBC News after seeing the president on Thursday. “He’s very sharp, no memory issues, and his only stumbling is when he trips over words consistent with his lifelong speech impediment.”

‘Prejudicial language’

Though Biden was fortunate to escape indictment, the special counsel report may give Trump additional fodder as he fights charges for allegedly mishandling classified records at his Mar-a-Lago social club. Republicans are already accusing Biden of benefiting from a double standard . Trump will likely brandish the Hur report as proof that Biden has “weaponized” the Justice Department for political advantage.

What’s more, Democrats will now be hard-pressed to capitalize on Trump’s indictment over retaining classified records. Before Hur’s report came out, Democrats argued that the two cases were very different. Whereas Trump failed to turn over classified records even after he was asked to do so, Biden willingly cooperated with authorities and relinquished all the material he had, Biden allies had argued.

“The public understands the essential difference between presidents or vice presidents like Joe Biden who occasionally behaved in sloppy ways with respect to where they were taking documents, and a president like Trump, who deliberately makes off with hundreds of classified government documents and then hides them and refuses to return them,” Rep. Jamie Raskin, D-Md., said on Wednesday, before the report was released. (Trump has denied any wrongdoing.)

Now, the distinctions may be harder for Biden allies to draw, given that Hur wrote that there was evidence Biden “willfully retained and disclosed classified material after his vice presidency when he was a private citizen.”

The report mentions an instance in February 2017, when he was no longer vice president, when Biden read notes containing classified information “nearly verbatim” to a ghostwriter helping him with his book, “Promise Me, Dad.”

Storage of sensitive government secrets was haphazard. The report describes certain classified records involving the war in Afghanistan in Biden’s Delaware garage inside a “badly damaged box surrounded by household detritus.”

Before the report was released, Biden aides had been bracing for a finding that he had simply been careless in his treatment of classified records, a person familiar with the White House’s thinking said.

The political fallout from the report, though, is likely to be “worse,” this person said. What will stick in people’s minds is what Hur said about Biden’s memory, the person added.

Biden’s lawyers disputed the report’s description of Biden’s forgetfulness.

“We do not believe that the report’s treatment of President Biden’s memory is accurate or appropriate,” two of his lawyers wrote in a letter to Hur. “The report uses highly prejudicial language to describe a commonplace occurrence among witnesses: a lack of recall of years-old events.”

In the hours after the report was released, people close to the Biden campaign rolled out a different rebuttal. Jim Messina, who ran Obama’s 2012 re-election campaign, wrote on X, the site formerly known as Twitter, that Hur is a Republican who “knew exactly how his swipes could hurt Biden politically.”

That’s a familiar argument. Trump has also claimed that law enforcement is trying to sway the election, meaning both sides are now claiming victimization at the hands of partisan prosecutors.

“Hur knew exactly what he was doing here,” Stephanie Cutter, a veteran Democratic operative, wrote on X. “To provide political cover for himself for not prosecuting, he gratuitously leveled a personal (not legal) charge against the president that he absolutely knows is a gift to Trump. And, guess what we are all talking about?”

how to solve word problem examples

Peter Nicholas is a senior national political reporter for NBC News.

IMAGES

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