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Tricky and interesting geometry word problems

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Chapter 4: Inequalities

4.5 Geometric Word Problems

It is common to run into geometry-based word problems that look at either the interior angles, perimeter, or area of shapes. When looking at interior angles, the sum of the angles of any polygon can be found by taking the number of sides, subtracting 2, and then multiplying the result by 180°. In other words:

[latex]\text{sum of interior angles} = 180^{\circ} \times (\text{number of sides} - 2)[/latex]

This means the interior angles of a triangle add up to 180° × (3 − 2), or 180°. Any four-sided polygon will have interior angles adding to 180° × (4 − 2), or 360°. A chart can be made of these:

[latex]\begin{array}{rrrrrr} \text{3 sides:}&180^{\circ}&\times&(3-2)&=&180^{\circ} \\ \text{4 sides:}&180^{\circ}&\times&(4-2)&=&360^{\circ} \\ \text{5 sides:}&180^{\circ}&\times&(5-2)&=&540^{\circ} \\ \text{6 sides:}&180^{\circ}&\times&(6-2)&=&720^{\circ} \\ \text{7 sides:}&180^{\circ}&\times&(7-2)&=&900^{\circ} \\ \text{8 sides:}&180^{\circ}&\times&(8-2)&=&1080^{\circ} \\ \end{array}[/latex]

Example 4.5.1

The second angle [latex](A_2)[/latex] of a triangle is double the first [latex](A_1).[/latex] The third angle [latex](A_3)[/latex] is 40° less than the first [latex](A_1).[/latex] Find the three angles.

The relationships described in equation form are as follows:

[latex]A_2 = 2A_1 \text{ and } A_3 = A_1 - 40^{\circ}[/latex]

Because the shape in question is a triangle, the interior angles add up to 180°. Therefore:

[latex]A_1 + A_2 + A_3 = 180^{\circ}[/latex]

Which can be simplified using substitutions:

[latex]A_1 + (2A_1) + (A_1 - 40^{\circ}) = 180^{\circ}[/latex]

Which leaves:

[latex]\begin{array}{rrrrrrrrrrr} 2A_1&+&A_1&+&A_1&-&40^{\circ}&=&180^{\circ}&&&\\ &&&&4A_1&-&40^{\circ}&=&180^{\circ}&&\\ \\ &&&&&&4A_1&=&180^{\circ}&+&40^{\circ}\\ \\ &&&&&&A_1&=&\dfrac{220^{\circ}}{4}&\text{or}&55^{\circ} \end{array}[/latex]

This means [latex]A_2 = 2 (55^{\circ})[/latex] or 110° and [latex]A_3 = 55^{\circ}-40^{\circ}[/latex] or 15°.

Another common geometry word problem involves perimeter, or the distance around an object. For example, consider a rectangle, for which [latex]\text{perimeter} = 2l + 2w.[/latex]

Example 4.5.2

If the length of a rectangle is 5 m less than twice the width, and the perimeter is 44 m long, find its length and width.

[latex]L = 2W - 5 \text{ and } P = 44[/latex]

For a rectangle, the perimeter is defined by:

[latex]P = 2 W + 2 L[/latex]

Substituting for [latex]L[/latex] and the value for the perimeter yields:

[latex]44 = 2W + 2 (2W - 5)[/latex]

Which simplifies to:

[latex]44 = 2W + 4W - 10[/latex]

Further simplify to find the length and width:

[latex]\begin{array}{rrrrlrrrr} 44&+&10&=&6W&&&& \\ \\ &&54&=&6W&&&& \\ \\ &&W&=&\dfrac{54}{6}&\text{or}&9&& \\ \\ &\text{So}&L&=&2(9)&-&5&\text{or}&13 \\ \end{array}[/latex]

The width is 9 m and the length is 13 m.

Other common geometric problems are:

Example 4.5.3

A 15 m cable is cut into two pieces such that the first piece is four times larger than the second. Find the length of each piece.

[latex]P_1 + P_2 = 15 \text{ and } P_1 = 4P_2[/latex]

Combining these yields:

[latex]\begin{array}{rrrrrrr} 4P_2&+&P_2&=&15&& \\ \\ &&5P_2&=&15&& \\ \\ &&P_2&=&\dfrac{15}{5}&\text{or}&3 \end{array}[/latex]

This means that [latex]P_2 =[/latex] 3 m and [latex]P_1 = 4 (3),[/latex] or 12 m.

For questions 1 to 8, write the formula defining each relation. Do not solve.

  • The length of a rectangle is 3 cm less than double the width, and the perimeter is 54 cm.
  • The length of a rectangle is 8 cm less than double its width, and the perimeter is 64 cm.
  • The length of a rectangle is 4 cm more than double its width, and the perimeter is 32 cm.
  • The first angle of a triangle is twice as large as the second and 10° larger than the third.
  • The first angle of a triangle is half as large as the second and 20° larger than the third.
  • The sum of the first and second angles of a triangle is half the amount of the third angle.
  • A 140 cm cable is cut into two pieces. The first piece is five times as long as the second.
  • A 48 m piece of hose is to be cut into two pieces such that the second piece is 5 m longer than the first.

For questions 9 to 18, write and solve the equation describing each relationship.

  • The second angle of a triangle is the same size as the first angle. The third angle is 12° larger than the first angle. How large are the angles?
  • Two angles of a triangle are the same size. The third angle is 12° smaller than the first angle. Find the measure of the angles.
  • Two angles of a triangle are the same size. The third angle is three times as large as the first. How large are the angles?
  • The second angle of a triangle is twice as large as the first. The measure of the third angle is 20° greater than the first. How large are the angles?
  • Find the dimensions of a rectangle if the perimeter is 150 cm and the length is 15 cm greater than the width.
  • If the perimeter of a rectangle is 304 cm and the length is 40 cm longer than the width, find the length and width.
  • Find the length and width of a rectangular garden if the perimeter is 152 m and the width is 22 m less than the length.
  • If the perimeter of a rectangle is 280 m and the width is 26 m less than the length, find its length and width.
  • A lab technician cuts a 12 cm piece of tubing into two pieces such that one piece is two times longer than the other. How long are the pieces?
  • An electrician cuts a 30 m piece of cable into two pieces. One piece is 2 m longer than the other. How long are the pieces?

Answer Key 4.5

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Praxis Core Math

Course: praxis core math   >   unit 1.

  • Algebraic properties | Lesson
  • Algebraic properties | Worked example
  • Solution procedures | Lesson
  • Solution procedures | Worked example
  • Equivalent expressions | Lesson
  • Equivalent expressions | Worked example
  • Creating expressions and equations | Lesson
  • Creating expressions and equations | Worked example

Algebraic word problems | Lesson

  • Algebraic word problems | Worked example
  • Linear equations | Lesson
  • Linear equations | Worked example
  • Quadratic equations | Lesson
  • Quadratic equations | Worked example

What are algebraic word problems?

What skills are needed.

  • Translating sentences to equations
  • Solving linear equations with one variable
  • Evaluating algebraic expressions
  • Solving problems using Venn diagrams

How do we solve algebraic word problems?

  • Define a variable.
  • Write an equation using the variable.
  • Solve the equation.
  • If the variable is not the answer to the word problem, use the variable to calculate the answer.

What's a Venn diagram?

  • 7 + 10 − 13 = 4 ‍   brought both food and drinks.
  • 7 − 4 = 3 ‍   brought only food.
  • 10 − 4 = 6 ‍   brought only drinks.
  • Your answer should be
  • an integer, like 6 ‍  
  • a simplified proper fraction, like 3 / 5 ‍  
  • a simplified improper fraction, like 7 / 4 ‍  
  • a mixed number, like 1   3 / 4 ‍  
  • an exact decimal, like 0.75 ‍  
  • a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  
  • (Choice A)   $ 4 ‍   A $ 4 ‍  
  • (Choice B)   $ 5 ‍   B $ 5 ‍  
  • (Choice C)   $ 9 ‍   C $ 9 ‍  
  • (Choice D)   $ 14 ‍   D $ 14 ‍  
  • (Choice E)   $ 20 ‍   E $ 20 ‍  
  • (Choice A)   10 ‍   A 10 ‍  
  • (Choice B)   12 ‍   B 12 ‍  
  • (Choice C)   24 ‍   C 24 ‍  
  • (Choice D)   30 ‍   D 30 ‍  
  • (Choice E)   32 ‍   E 32 ‍  
  • (Choice A)   4 ‍   A 4 ‍  
  • (Choice B)   10 ‍   B 10 ‍  
  • (Choice C)   14 ‍   C 14 ‍  
  • (Choice D)   18 ‍   D 18 ‍  
  • (Choice E)   22 ‍   E 22 ‍  

Things to remember

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how to solve word problems in geometry

How do you solve word problems in math?

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

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Author Amber Watkins

how to solve word problems in geometry

Published April 2024

how to solve word problems in geometry

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

Table of contents

  • How to solve word problems

Lesson credits

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

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

How to solve word problems in math in 8 steps

Step 1: read the word problem aloud.

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

Mastering step 1 looks like this:

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

Step 2: Highlight the keywords in the word problem

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

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

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

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

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

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

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

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

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

how to solve word problems in geometry

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

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

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

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

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

Step 6: Always show your work

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

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

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

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

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

So how do you solve a word problem in math?

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

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

how to solve word problems in geometry

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

how to solve word problems in geometry

Amber Watkins

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

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Solving Word Questions

With LOTS of examples!

In Algebra we often have word questions like:

Example: Sam and Alex play tennis.

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

How many games did Alex play?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

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

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

Also look for key words:

Thinking Clearly

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

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

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

Now ... is that: S − 2 = A

or should it be: S = A − 2

or should it be: S = 2 − A

The correct answer is S = A − 2

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

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

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

Now ... is that: 2D = C

or should it be: D = 2C

Think carefully now!

The correct answer is D = 2C

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

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

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

Turn the English into Algebra:

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

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

We are being asked for the Area.

A = w × h = 12 × 5 = 60 m 2

The area is 60 square meters .

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

tennis

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

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

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

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

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

Which means that Alex played 4 games of tennis.

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

A slightly harder example:

table

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

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

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

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

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

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

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

So 10 tables would take Sam just 20 days.

Should Sam be paid more I wonder?

And another "substitution" example:

track

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

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

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

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

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

The number of "5 hour" days is 3

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

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

Some examples from Geometry:

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

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

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

We are being asked for the radius.

We need to rearrange the formula to find the area

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

Make a quick sketch:

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

We are being asked for the surface area.

First work out s using the volume formula:

Now we can calculate surface area:

An example about Money:

pizza

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

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

$40N + $(12 × 1¼N) = $660

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

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

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

More about Money, with these two examples involving Compound Interest

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

This is the compound interest formula:

So we will use these letters:

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

We are being asked for the Future Value: FV

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

The compound interest formula:

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

We are being asked for the Interest Rate: r

So the annual rate of interest is 5%

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

And an example of a Ratio question:

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

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

The current ratio is 4 : 3

Which can be rearranged to 3b = 4g

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

b + 4 g − 2 = 2 1

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

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

There are 12 girls !

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

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

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

And now for some Quadratic Equations :

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

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

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

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

n(n + 2) = 168

We are being asked for the integers

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

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

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

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

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

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

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

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

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

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

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

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

And so L = 8 or −14

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

So the length of the room is 8 m

L = 8, so W = ½L = 4

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

There we are ...

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

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10 Best Strategies for Solving Math Word Problems

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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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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How To Solve Algebra Word Problems?

Many students find solving algebra word problems difficult. The best way to approach word problems is to “divide and conquer”. Break the problem down into smaller bits and solve each bit at a time.

Related Pages Basic Algebra Combining Like Terms Solving Equations More Algebra Lessons

Step 1: Translate the problem into equations with variables

First, we need to translate the word problem into equation(s) with variables. Then, we need to solve the equation(s) to find the solution(s) to the word problems.

Translating words to equations

How to recognize some common types of algebra word problems and how to solve them step by step.

The following shows how to approach the common types of algebra word problems.

Age Problems usually compare the ages of people. They may involve a single person , comparing his/her age in the past, present or future. They may also compare the ages involving more than one person .

Average Problems involve the computations for arithmetic mean or weighted average of different quantities. Another common type of average problems is the average speed computation.

Coin Problems deal with items with denominated values. Similar word problems are Stamp Problems and Ticket Problems .

Consecutive Integer Problems deal with consecutive numbers. The number sequences may be Even or Odd , or some other simple number sequences.

Digit Problems involve the relationship and manipulation of digits in numbers.

Distance Problems involve the distance an object travels at a rate over a period of time. We can have objects that Travel at Different Rates , objects that Travel in Different Directions or we may need to find the distance Given the Total Time

Fraction Problems involve fractions or parts of a whole.

Geometry Word Problems deal with geometric figures and angles described in words. This include geometry word problems Involving Perimeters , Involving Areas and Involving Angles

Integer Problems involve numerical representations of word problems. The integer word problems may Involve 2 Unknowns or may Involve More Than 2 Unknowns

Interest Problems involve calculations of simple interest. Lever Problems deal with the lever principle described in word problems. Lever problem may involve 2 Objects or More than 2 Objects

Mixture Problems involve items or quantities of different values that are mixed together. This involve Adding to a Solution , Removing from a Solution , Replacing a Solution ,or Mixing Items of Different Values

Motion Word Problems are word problems that uses the distance, rate and time formula.

Number Sequence Problems use number sequences in the construction of word problems. You may be asked to find the Value of a Particular Term or the Pattern of a Sequence

Proportion Problems involve proportional and inversely proportional relationships of various quantities.

Ratio Problems require you to relate quantities of different items in certain known ratios, or work out the ratios given certain quantities. This could be Two-Term Ratios or Three-Term Ratios

Symbol Problems

Variation Word Problems may consist of Direct Variation Problems , Inverse Variation Problems or Joint Variation Problems

Work Problems involve different people doing work together at different rates. This may be for Two Persons , More Than Two Persons or Pipes Filling up a Tank

For more Algebra Word Problems and Algebra techniques, go to our Algebra page

Step 2: Solving the equations - finding the values of the variables for the equations

Simplifying the equation Isolate the variable (Transposition) Addition Method (Opposite-Coefficient Method) Substitution Method Factoring Quadratic Equations Quadratic Formula

Algebra Word Problems with examples, videos and step-by-step solutions

Age Problems Age Problems: Single Person More Than One Person Age Word Problems 1 Age Word Problems 2

Arithmetic Mean Weighted Average Average Speed

Problems involving Coins Stamp Problems Ticket Problems Coin Word Problems 1 Coin Word Problems 2

Consecutive Integers 1 Consecutive Integers 2 Consecutive Odd Integer Consecutive Even Integer Consecutive Integer Problems Even & Odd Word Problems

Simple Digit Problems Convert Digits to Numbers & Interchanging Digits Digits Word Problems

Distance Problems Travel at Different Rates Travel in Different Directions Given Total Time

Fraction Problems Percent Word Problems

Geometry Word Problems Involving Perimeters Involving Areas Involving Angles Angle Word Problems Perimeter Word Problems 1 Perimeter Word Problems 2 Perimeter Word Problems 3 Area Word Problems Area of Rectangle Word Problems Area of Triangle Word Problems Area of Parallelogram Word Problems Volume Word Problems Volume Word Problems Geometry Word Problems using Quadratic Equations Pythagorean Theorem Word Problems Trigonometry Word Problems

Involving 2 Unknowns More Than 2 Unknowns Integer Word Problems 1 Integer Word Problems 2

Interest Word Problems Simple Interest Word Problems Compound Interest Word Problems 1 Compound Interest Word Problems 2 Investment Word Problems

Involving 2 or More Objects

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  • ISBN-10 007134652X
  • ISBN-13 978-0071346528
  • Edition 1st
  • Publisher McGraw-Hill
  • Publication date November 11, 1999
  • Language English
  • Dimensions 5.5 x 0.43 x 8.4 inches
  • Print length 153 pages
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  • Publisher ‏ : ‎ McGraw-Hill; 1st edition (November 11, 1999)
  • Language ‏ : ‎ English
  • Paperback ‏ : ‎ 153 pages
  • ISBN-10 ‏ : ‎ 007134652X
  • ISBN-13 ‏ : ‎ 978-0071346528
  • Item Weight ‏ : ‎ 2.31 pounds
  • Dimensions ‏ : ‎ 5.5 x 0.43 x 8.4 inches
  • #668 in Mathematics Reference (Books)
  • #1,056 in Geometry
  • #3,104 in Geometry & Topology (Books)

About the author

Dawn b. sova.

DAWN SOVA (1949 -) is an Edgar Award winner born in Passaic, New Jersey, to first-generation Polish and Czech parents and raised in Garfield, New Jersey. She earned a doctorate from Drew University and taught for 28 years. She is the author of 22 books. In 1997, "Agatha Christie, A-to-Z" was nominated by the Mystery Writers of America for Best Non-Fiction work. In 2002, Sova won the MWA Best Non-Fiction Award for "Edgar Allan Poe A-to-Z." Both works have been featured in questions on the television show "Jeopardy."

Sova has spoken extensively on issues of intellectual freedom and published four books on the topic: "Banned Plays," "Forbidden Films," "Banned Books: Literature Suppressed on Social Grounds," and "Banned Books: Literature Suppressed on Sexual Grounds." The last two books have gone into third editions. She is also co-author of "100 Banned Books," a Book-of-the-Month Club Main Selection, and of "120 Banned Books," an entry in the 2001 Book Lovers’ Calendar. Her books have been translated into Spanish, Polish, Czechoslovakian, Italian, Korean, Chinese and Japan and published in ten countries. She has been privileged to serve as guest speaker for regional, state and local libraries; booksellers’ conferences and education associations on issues related to the numerous and continuing challenges to intellectual freedom.

In addition to her work related to censorship, she has published 16 books on a variety of topics from "Sex and the Single Mother," to "Solving Word Problems in Geometry" and "The Encyclopedia of Mistresses."

The author is currently writer-in-residence at her son’s Herbert Grand Hotel in Kingfield, Maine.

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How to Solve Angle Measurements Word Problems

Hey future geometrists! Today, we're going on a journey into the world of angles. Are you curious about how to figure out the measurement of an angle in a triangle or a circle just from a story or a problem? Well, you're in the right place. We're about to make solving angle measurements word problems as easy and exciting as solving a detective's mystery. Ready for some fun with angles? Let's dive in!

How to Solve Angle Measurements Word Problems

let’s start with some basic concepts about angles.

In fourth grade, students usually learn about four main types of angles:

  • Right Angles: This is an angle of exactly \(90^°\). It looks like the corner of a square.
  • Acute Angles: This is any angle less than \(90^°\) but more than \(0^°\).
  • Obtuse Angles: This is any angle greater than \(90^°\) but less than \(180^°\).
  • Straight Angles: This is an angle of exactly \(180^°\). It looks like a straight line.

The word problems typically involve understanding and applying these concepts to solve them.

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Mastering Grade 4 Math The Ultimate Step by Step Guide to Acing 4th Grade Math

A step-by-step guide to solving angle measurements word problems.

Here’s a simple step-by-step guide to help 4th graders solve word problems involving angle measurements. This guide assumes that they’re already familiar with basic types of angles (acute, right, obtuse, and straight).

Step 1: Understand the Problem

Read the problem carefully. What are you asked to find? What information is given to you? You might want to underline or highlight the important information.

Step 2: Represent the Problem

Try to represent the problem visually if possible. Draw the angles, lines, or shapes mentioned in the problem. This will give you a clearer idea of what you’re working with.

Step 3: Recall What You Know

Remember the properties of angles you’ve learned. For instance, the angles in a triangle add up to 180 degrees, a straight angle is 180 degrees, a right angle is 90 degrees, and a complete rotation is 360 degrees.

A Perfect Book for Grade 4 Math Word Problems!

Mastering Grade 4 Math Word Problems The Ultimate Guide to Tackling 4th Grade Math Word Problems

Step 4: set up your equations.

Use the information given in the problem and what you know about angles to set up equations. The equations will depend on the details of the problem.

Step 5: Solve the Equations

Solve the equations you’ve set up. This might involve simple addition or subtraction, or more complex operations like multiplication or division.

Step 6: Check Your Work

Once you’ve found a solution, go back to the problem. Does your answer make sense in the context of the problem? Is it a reasonable measure for an angle? If you’re unsure, check your work or try solving the problem a different way.

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Mastering grade 5 math the ultimate step by step guide to acing 5th grade math, mastering grade 3 math the ultimate step by step guide to acing 3rd grade math, mastering grade 5 math word problems the ultimate guide to tackling 5th grade math word problems, mastering grade 2 math word problems the ultimate guide to tackling 2nd grade math word problems.

by: Effortless Math Team about 12 months ago (category: Articles )

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

Simplified Equation: 17 - x = 8

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

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

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

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

Simplified: 40 - 10 - 5

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Teaching math word problems without keywords? Yes, you can! In this blog post, I share how I use operation situations to help with math word problems.

“Who can tell me a word that tells us to add when we solve a word problem?” 

Fifteen hands shot into the air, some waving that oh-so-familiar “pick-me” wave. 

One by one, students shared their words . . . 

“in all”

“all together”

“increase”

“add”

“sum”

“total”

“plus”

“combine”

“join”

“together”

“both”

“Wow! You all came up with an amazing list of words, but what about subtraction? What words signal a subtraction problem?” 

The students raised their hands to respond.

“deduct”

“decrease”

“how many more”

“take away”

“how many less” 

“less than”

“minus”

“remain”

“difference”

Once the teacher recorded the words on chart paper, the students divided themselves into groups of two and completed a set of addition and subtraction task cards. 

While the students created an impressive list of words, I wondered if all the students would be able to remember them when solving word problems. 

I’m thinking, no! 😕

So, what’s wrong with using math keywords ?

An Alternative To Keywords

If we can’t expect students to recognize these signal words to know what to do, how do we help with math word problems? 

Great question– I can answer it with two simple words– operation situations. 

Now, don’t get me wrong, I have been guilty of expecting my students to pick up on these keywords when solving problems too; however, a conference session I attended many years ago changed my way of thinking. 

At the time, I taught fourth grade and dedicated a portion of my math instruction to help my students develop word problem analysis skills with a word problem of the day program. 

As we read each day’s word problem, I noticed something interesting . . . each problem fell in a specific category for each operation.

I couldn’t believe it! It was like magic!

In the beginning, I referred to the set of operation situations I learned about at the summer conference session. 

They included: 

  • Addition: part-part-whole, joining
  • Subtraction: something leaves; comparing, missing part
  • Multiplication: groups of something 
  • Division: how many groups; how many in each group

This picture shows how the operation situations are used to solve math word problems.

Using Operation Situations To Help With Math Word Problems

Each time we encountered a new word problem, I asked the students to describe the event(s) happening in the word problem. From there, we discussed what action (operation) was taking place. 

From the beginning, most students could connect the picture they created in their heads with an operation. Then, I helped them connect the situation illustrated in the story problem with the more formal language of the corresponding operation situation. 

Consider the following problem: 

The Lions finished the 400-meter freestyle relay race in 3.83 minutes. The Ravens finished in 3.72 minutes. The Knights finished in 4.05 minutes. What is the difference between the finish times for the first and third place teams?

For this problem, students might say we are looking for the difference between the finish time for the Ravens (first place) and the Knights (third place) relay race team. Students might also say we need to find out how much faster the first-place team (the Ravens) finished the race than the third-place team (the Knights). 

During this discussion, if not mentioned by a student, I would add we need to compare the Ravens time to the Knights time and state, “when we compare two numbers, we determine the difference between the two quantities”. I might even follow with a visual to help the students understand what it means to find the difference between two numbers.  (See the image below.)

This is an illustration of the action in the problem.

Okay. I know what you’re thinking . . . what about multiple-step problems?

That’s a great question. Let’s consider another problem. 

When Brandon and Kelly arrive at the BrightStar campground, they begin preparing dessert to eat after their campfire cookout. There are eight people on the camping trip. Brandon and Kelly prepare three smores for each person. Each smore has one marshmallow. If the bag of marshmallows they used had 4 dozen marshmallows, how many marshmallows are leftover?

For this problem, students will call it a subtraction situation because they need to determine the leftover amount. I would revoice their thinking and call this a “something leaves” situation because they started with a bag of 4 dozen marshmallows, then some left, and there is now a specific amount remaining.

The only missing piece of information from this situation is the quantity removed. So we go back to the problem to look for clues. 

Looking for Clues

When we return to the problem, we see there are eight people on the camping trip. We also know Brandon and Kelly prepare three smores for each person and each smore has one marshmallow. So, to determine how many marshmallows Brandon and Kelly need, we use the “groups of something” situation– which means to multiply 8 people x 3 smores (marshmallows). 

Hopefully, students will also notice the conversion. It’s not necessarily an operation situation we need to flesh out, but students need to acknowledge it. After working through the pieces of information needed to solve the problem, students can articulate they need to subtract the number of marshmallows used (8 groups of 3) from the total in the bag (4 dozen).  

Wait . . . there’s a keyword here.

Right now, you might be saying to yourself, “This problem has a keyword, so why is all of this necessary?” 

Because, this problem could have been written without the use of the word “leftover.” In fact, it could say, “If the bag of marshmallows they used had 4 dozen marshmallows, how many marshmallows were in the bag when Brandon and Kelly put the supplies away after dessert.” Notice there isn’t a keyword here, so some students might feel discouraged before even starting. 

The Bottom Line

We want our students to engage with story problems so they can see the story in their heads. Can you see Brandon, Kelly, Brenda, Dylan, Andrea, Steve, Donna, and David around the campfire after their high school graduation making smores? (Yes, you guessed it . . . I’m an 80s child and loved watching 90210!) 

When students can visualize the situation, they can make sense of the actions taking place, such as someone counting the number of marshmallows they need and removing them from the bag. After recognizing the action, we can easily determine the quantity remaining. 

More Operation Situations

I’m sure you’re wondering how I went from eight situations to the seventeen situations on the operation situations printable I shared in my keywords for math problems blog post .

During the first year I used this strategy to help my students with math word problems, my students and I noticed some situations didn’t quite fit, such as money situations where we know the amount spent and the change received but need to know the amount given to the cashier. 

These types of problems made me curious to look for more situations I could use to address all of the types of problems we encountered; however, these initial eight situations were a great starting place and helped my students become proficient at solving math word problems for a great number of years. 

Ready to use the operation situations to help with math word problems? Download a free set of posters using the form below!

What strategies do you use to help with math word problems? Respond in the comments below.

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Shametria Routt Banks

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I appreciate this resource – and was led here via your email subscription. To get this resource, am I subscribing a second time to the same blog post emails? I’m confused.

Great question! No, you are not subscribing a second time, but the system will tag you to show you downloaded the resource. That helps me know what our teacher community wants to know more about. Nothing will change in terms of the emails– you won’t receive double the number. I hope this helps. If you have any other questions, please email me at [email protected] . Thank you!

~ Shametria

I am a teacher of students with visual impairments & am working with a 4th grade student (braille reader) who struggles with determining what operations are needed in word problems – esp. those doozies that require doing two. I am excited to take a different approach. I don’t want to spend valuable time teaching him operation words when I could be teaching him how to think more holistically. Generalization is often a problem for these students because they have such limited experience. Thank you for sharing your teaching knowledge.

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A Math Word Problem Framework That Fosters Conceptual Thinking

This strategy for selecting and teaching word problems guides students to develop their understanding of math concepts.

Photo of middle school teacher with students

Word problems in mathematics are a powerful tool for helping students make sense of and reason with mathematical concepts. Many students, however, struggle with word problems because of the various cognitive demands. As districtwide STEAM professional development specialists, we’ve spent a lot of time focusing on supporting our colleagues and students to ensure their success with word problems. We found that selecting the right word problems, as well as focusing on conceptual understanding rather than procedural knowledge, provides our students with real growth.

As our thinking evolved, we began to instill a routine that supports teaching students to solve with grit by putting them in the driver’s seat of the thinking. Below you’ll find the routine that we’ve found successful in helping students overcome the challenges of solving word problems.

Not all word problems are created equal

Prior to any instruction, we always consider the quality of the task for teaching and learning. In our process, we use word problems as the path to mathematics instruction. When selecting the mathematical tasks for students, we always consider the following questions:

  • Does the task align with the learning goals and standards?
  • Will the task engage and challenge students at an appropriate level, providing both a sense of accomplishment and further opportunities for growth?
  • Is the task open or closed? Open tasks provide multiple pathways to foster a deeper understanding of mathematical concepts and skills. Closed tasks can still provide a deep understanding of mathematical concepts and skills if the task requires a high level of cognitive demand. 
  • Does the task encourage critical thinking and problem-solving skills?
  • Will the task allow students to see the relevance of mathematics to real-world situations?
  • Does the task promote creativity and encourage students to make connections between mathematical concepts and other areas of their lives?

If we can answer yes to as many of these questions as possible, we can be assured that our tasks are rich. There are further insights for rich math tasks on NRICH and sample tasks on Illustrative Mathematics and K-5 Math Teaching Resources .

Developing conceptual understanding

Once we’ve selected the rich math tasks, developing conceptual understanding becomes our instructional focus. We present students with Numberless Word Problems and simultaneously use a word problem framework to focus on analysis of the text and to build conceptual understanding, rather than just memorization of formulas and procedures. 

  • First we remove all of the numbers and have students read the problem focusing on who or what the problem is about; they visualize and connect the scenario to their lives and experiences. 
  • Next we have our students rewrite the question as a statement to ensure that they understand the questions.
  • Then we have our students read the problem again and have them think analytically. They ask themselves these questions: Are there parts? Is there a whole? Are things joining or separating? Is there a comparison? 
  • Once that’s completed, we reveal the numbers in the problem. We have the students read the problem again to determine if they have enough information to develop a model and translate it into an equation that can be solved.
  • After they’ve solved their equation, we have students compare it against their model to check their answer.  

Collaboration and workspace are key to building the thinking

To build the thinking necessary in the math classroom , we have students work in visibly random collaborative groups (random groups of three for grades 3 through 12, random groups of two for grades 1 and 2). With random groupings, we’ve found that students don’t enter their groups with predetermined roles, and all students contribute to the thinking.

For reluctant learners, we make sure these students serve as the scribe within the group documenting each member’s contribution. We also make sure to use nonpermanent vertical workspaces (whiteboards, windows [using dry-erase markers], large adhesive-backed chart paper, etc.). The vertical workspace provides accessibility for our diverse learners and promotes problem-solving because our students break down complex problems into smaller, manageable steps. The vertical workspaces also provide a visually appealing and organized way for our students to show their work.  We’ve witnessed how these workspaces help hold their attention and improve their focus on the task at hand.

Facilitate and provide feedback to move the thinking along

As students grapple with the task, the teacher floats among the collaborative groups, facilitates conversations, and gives the students feedback. Students are encouraged to look at the work of other groups or to provide a second strategy or model to support their thinking. Students take ownership and make sense of the problem, attempt solutions, and try to support their thinking with models, equations, charts, graphs, words, etc. They work through the problem collaboratively, justifying their work in their small group. In essence, they’re constructing their knowledge and preparing to share their work with the rest of the class. 

Word problems are a powerful tool for teaching math concepts to students. They offer a practical and relatable approach to problem-solving, enabling students to understand the relevance of math in real-life situations. Through word problems, students learn to apply mathematical principles and logical reasoning to solve complex problems. 

Moreover, word problems also enhance critical thinking, analytical skills, and decision-making abilities. Incorporating word problems into math lessons is an effective way to make math engaging, meaningful, and applicable to everyday life.

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How to Solve Word Problems in Algebra

Last Updated: December 19, 2022 Fact Checked

This article was co-authored by Daron Cam . Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary's College. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 71,651 times.

You can solve many real world problems with the help of math. In order to familiarize students with these kinds of problems, teachers include word problems in their math curriculum. However, word problems can present a real challenge if you don't know how to break them down and find the numbers underneath the story. Solving word problems is an art of transforming the words and sentences into mathematical expressions and then applying conventional algebraic techniques to solve the problem.

Assessing the Problem

Step 1 Read the problem carefully.

  • For example, you might have the following problem: Jane went to a book shop and bought a book. While at the store Jane found a second interesting book and bought it for $80. The price of the second book was $10 less than three times the price of he first book. What was the price of the first book?
  • In this problem, you are asked to find the price of the first book Jane purchased.

Step 3 Summarize what you know, and what you need to know.

  • For example, you know that Jane bought two books. You know that the second book was $80. You also know that the second book cost $10 less than 3 times the price of the first book. You don't know the price of the first book.

Step 4 Assign variables to the unknown quantities.

  • Multiplication keywords include times, of, and f actor. [9] X Research source
  • Division keywords include per, out of, and percent. [10] X Research source
  • Addition keywords include some, more, and together. [11] X Research source
  • Subtraction keywords include difference, fewer, and decreased. [12] X Research source

Finding the Solution

Step 1 Write an equation.

Completing a Sample Problem

Step 1 Solve the following problem.

  • Robyn and Billy run a lemonade stand. They are giving all the money that they make to a cat shelter. They will combine their profits from selling lemonade with their tips. They sell cups of lemonade for 75 cents. Their mom and dad have agreed to double whatever amount they receive in tips. Write an equation that describes the amount of money Robyn and Billy will give to the shelter.

Step 2 Read the problem carefully and determine what you are asked to find.

  • Since you are combining their profits and tips, you will be adding two terms. So, x = __ + __.

.75c

Expert Q&A

Daron Cam

  • Word problems can have more than one unknown and more the one variable. Thanks Helpful 2 Not Helpful 1
  • The number of variables is always equal to the number of unknowns. Thanks Helpful 1 Not Helpful 0
  • While solving word problems you should always read every sentence carefully and try to extract all the numerical information. Thanks Helpful 1 Not Helpful 0

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  • ↑ Daron Cam. Academic Tutor. Expert Interview. 29 May 2020.
  • ↑ http://www.purplemath.com/modules/translat.htm
  • ↑ https://www.mathsisfun.com/algebra/word-questions-solving.html
  • ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut8_probsol.htm
  • ↑ http://www.virtualnerd.com/algebra-1/algebra-foundations/word-problem-equation-writing.php
  • ↑ https://www.khanacademy.org/test-prep/praxis-math/praxis-math-lessons/praxis-math-algebra/a/gtp--praxis-math--article--algebraic-word-problems--lesson

About This Article

Daron Cam

To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Next, summarize what information you know and what you need to know. Then, assign variables to the unknown quantities. For example, if you know that Jane bought 2 books, and the second book cost $80, which was $10 less than 3 times the price of the first book, assign x to the price of the 1st book. Use this information to write your equation, which is 80 = 3x - 10. To learn how to solve an equation with multiple variables, keep reading! Did this summary help you? Yes No

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Learners read word problem, write an equation and solve using subtraction or one to one matching.

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Algebraic Expression Word Problems and Answers

Algebraic Expression Word Problems and Answers

Whether you’re taking the UPCAT , Civil Service Exam , NMAT , or other written examinations, word problems will always be part of the Mathematics subtest. They’re among the most frequently asked questions and also one of the most challenging.

This section will provide some tips on approaching these word problems using algebra. In particular, we will solve number problems, age problems, work problems, rate problems, and so on.

Hopefully, you can answer word problems with ease after reading this chapter.

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Table of Contents

General tips in solving word problems using algebra.

Here are some quick tips to help you improve your problem-solving skills in mathematics:

  • Read the problem carefully and determine what is being asked . You must know what you must solve because it will lead you to the process you must perform to answer the problem.
  • Identify the given and define them . Identify all quantities that are provided in the problem. Ensure you know these quantities and how they relate to the problem.
  • Set up a mathematical expression that represents the problem . Your mathematical expression can be an equation or inequality . You must make the correct expression to obtain the precise answer.
  • Solve the mathematical expression you have made . Ensure that you know the techniques on how to solve equations and inequalities. If you are not confident enough with your arithmetic and algebra of equations and inequalities, practicing them before attempting to answer these word problems is highly suggested. 

Keep the above-mentioned things in mind as we discuss how to solve different word problems. 

Number Problems

Number problems are word problems that ask you to determine a specific number or quantity using the descriptions given in the problem. 

To solve number problems, you should be familiar with some keywords used in algebra. These keywords, such as sum , increased by , decreased by , of , and ratio of allow you to determine what mathematical operation/s is involved in the problem.

For instance, a number problem containing the statement “the sum of a number and 5” gives a clue that the problem involves adding numbers.

Let us try to solve some number problems:

Sample Problem 1

Twice a number increased by five is 125. What is the number?

The problem above is asking for an unknown number.

Let x be that unknown number.

Recall that the keyword “twice” means “double” or a certain number is multiplied by 2. Thus, twice the unknown number is 2x .

Twice that unknown number increased by five can be represented by 2x + 5 .

Therefore, our equation would be 2x + 5 = 125 . If we solve for x in this equation, we can determine the unknown number in the problem.

Solving the equation:

2x + 5 = 125

2x  = -5 + 125 Transposition method

(2x)/2 = 120/2 Dividing both sides of the equation by 2

Since x represents the unknown number in the problem, and we have computed that x = 60 , the unknown number is 60.

Sample Problem 2

When a number is increased by its reciprocal, the result is 2. Determine the number.

The problem is asking for an unknown number.

As stated in the problem, if the number is increased by its reciprocal, we will obtain 2. The reciprocal of x is 1/x. Thus, our equation would be:

x + 1/x = 2

To solve this equation, we must first eliminate the denominator. This can be achieved by multiplying both sides of the equation by the Least Common Denominator (LCD) , which is x :

x(x + 1/x) = x(2)

x 2 + 1  = 2x

Notice that after we eliminate the denominator, the resulting equation is quadratic . We can write it in standard form as follows:

x 2 + 1 = 2x

x 2 – 2x + 1 = 0 Transposition method

Now, let us solve for x in x 2 – 2x + 1 = 0 to determine the unknown number:

x 2 – 2x + 1 = 0 

(x – 1)(x – 1) = 0 By factoring

x – 1 = 0 x – 1 = 0 Set each factor to 0

x = 1 x = 1

Based on our solution, the unknown number in the problem is 1 . 

You can also verify that the sum of 1 and its reciprocal is 2. The reciprocal of 1 is just 1. Hence, 1 + 1 = 2. 

Sample Problem 3

The sum of four consecutive positive whole numbers is 78. Determine the largest number among the consecutive numbers.

Again, this problem is asking for an unknown number. 

Consecutive numbers are numbers that follow each other continuously in order from smallest to largest. For example, 8, 9, 10, and 11 are consecutive numbers.

Now, the problem asks us to find four consecutive positive whole numbers whose sum is 78. 

Let x represent the smallest or the first number of these four consecutive numbers.

x – smallest/first number

Since the numbers are consecutive, the second number of these four consecutive numbers must be x + 1 since you have to add 1 to the first number (which is x ) to obtain the next number.

x + 1 – second number

Now, we represent the third number as x + 2 since we have to add 2 to the first number ( which is x ) to obtain the third number.

x + 2 – third number

Lastly, we represent the fourth number or the largest number as x + 3 since we have to add 3 to the first number (which is x ) to obtain the fourth or the largest number.

x + 3 – fourth number

So far, we have the following to represent the four consecutive numbers:

x – first number

Recall that the sum of these four consecutive numbers is 78. Thus, we have this equation:

first number + second number + third number + fourth number = 78

x + (x + 1) + (x + 2) + (x + 3) = 78

We can simplify the equation above as follows:

4x + 6 = 78

Let us solve for the value of x in the equation above:

4x = -6 + 78 Transposition method

(4x)/4 = 72/4 Dividing both sides of the equation by 4

Since x represents the smallest or the first number and we have obtained x = 18 , the first number should be 18. It follows that the second number is 19, the third number is 20, and the fourth or the largest number is 21. 

Therefore, the answer to the problem is 21 .

Age Problems

One of the most commonly asked questions in mathematics proficiency exams involves age problems. These problems require you to determine the specific age of a person using details given in the problem.

Age problems are usually solvable using your intuition alone and without the need to apply algebraic concepts. However, relying on your intuition is not a reliable way to solve age problems and can be time-consuming. Furthermore, solutions obtained through this process tend to be unorganized, making it difficult to answer the given problem.

Using algebra, you can use a more systematic approach to solving age problems in less time.

Letty is three times as old as Bert. The sum of Bert’s and Letty’s ages is 48. How old is Letty ?

The problem is asking for the age of Letty, given that she is three times older than Bert.

Let x be Bert’s age.

Since Letty is three times as old as Bert, we can write Letty’s age as 3x .

The sum of Bert’s and Letty’s ages is 48. Thus, our equation would be:

Bert’s age + Letty’s age = 48

x + 3x = 48

Using the equation, we have to solve for x to determine Bert’s age:

(4x)/4 = 48/4 Dividing both sides of the equation by 4

Since x represents Bert’s age, then Bert is 12 years old. Now, we know that Letty is three times older than Bert. Therefore, Letty’s age is 3(12) = 36.

Letty is 36 years old. 

This year, Albert is 11 years older than Isaac. Three years ago, the sum of their ages was 67. Determine how old Isaac and Albert are this year.

The problem is asking for Albert’s and Isaac’s ages in the current year.

Let x be Isaac’s age in the present year.

The problem states that Albert is 11 years older this year than Isaac. Hence, we can represent Albert’s age as x + 11 this year.

This means that three years ago, Isaac’s age can be represented by x – 3 . On the other hand, Albert’s age can be represented by x + 11 – 3 = x + 8 .

To summarize:

The sum of the ages of Albert and Isaac in the past is 67. Hence, we can set up our equation as follows:

Albert’s age (past) + Isaac’s age (past) = 67

(x – 3) + (x + 8) = 67

2x + 5 = 67

Solving for x in the equation above:

2x = -5 + 67 Transposition method

2x/2 = 62/2 Dividing both sides of the equation by 2

Since x represents Isaac’s age in the present, then Isaac is 31 years old in the present. Meanwhile, since Albert is 11 years older than Isaac in the present, Albert is 31 + 11 = 42 years old.

Hence, Isaac is 31 years old, while Albert is 42 years old .

Five years from now, the sum of Richard’s age and Mike’s age will be 21. In the current year, Richard’s age is five years higher than twice Mike’s age. Determine Mike’s age in the present.

The problem is asking for Mike’s age in the present.

It implies that Mike is younger than Richard. Hence, we let x be Mike’s age in the present.

Richard’s age is 5 years higher than twice Mike’s age. Twice of Mike’s age can be represented by 2x . Thus, Richard’s age in the present can be represented by 2x + 5 .

Five years from now, Mike’s age will be x + 5 while Richard’s age will be 2x + 10 .

The problem states that in the future (5 years from now), Mike’s and Richard’s ages will sum up to 21.

Thus, we have this equation:

Mike’s age (future) + Richard’s age (future) = 21

(x + 5) + (2x + 10) = 21

Combining like terms in the equation above:

3x + 15 = 21

3x = -15 + 21 Transposition method

(3x)/3 = 6/3 Dividing both sides of the equation by 3

Looking back at our table of Richard’s and Mike’s ages, x represents Mike’s age in the present. Therefore, Mike is two years old at present .

Sample Problem 4

This year, Ravel is 12 years younger than Claude. Twelve years ago, Claude’s age was five times Ravel’s age. Determine how old Claude is in the present.

The problem is asking for Claude’s age in the present.

Ravel seems younger than Claude, so we let x be Ravel’s age in the present.

Ravel is 12 years younger than Claude. This also means that Claude is 12 years older than Ravel. Hence, let x + 12 represent Claude’s age.

Ravel’s age 12 years ago can be represented by x – 12 . On the other hand, Claude’s age can be represented by x + 12 – 12 = x .

The problem states that 12 years ago, Claude’s age was five times Ravel’s age. 

Claude’s age = 5(Ravel’s age)

x = 5(x – 12) 

x = 5x – 60 Distributive Property

Let us solve the equation above:

x = 5x – 60

x – 5x = -60 Transposition method

(-4x)/-4 = (-60)/(-4) Dividing both sides by -4

Looking back at the table, x represents Ravel’s present age while x + 12 represents Claude’s present age. Using the value of x we have obtained from the solution (which is x = 15) :

(15) + 12 = 27

Thus, Claude is 27 years old today .

Work Problems

Solving a work problem involves figuring out how long a person or group will take to finish a particular task (or work) using the details provided in the problem.

We can use a guiding formula or equation when solving work problems. This equation tells us that the sum of the reciprocal of the time it takes for A and B to finish a task is equal to the reciprocal of the time it takes for A and B to work together to finish the same task.

algebra word problems 1

Let us try to solve some work problems to see how this equation/formula works:

Jacques can finish cleaning a garage in 2 hours. On the other hand, Kath can finish cleaning the same garage in 5 hours. Suppose Jacques and Kath work together to clean the same garage; how long would it take them to accomplish it?

The problem is asking for the time it would take for Jacques and Kath to finish cleaning the garage if they work together.

Jacques can finish cleaning the garage in 2 hours.

Meanwhile, Kath can finish cleaning the garage in 5 hours.

We let t represent the time it would take for Jacques and Kath to finish cleaning the garage if they worked together.

Recall that in a work problem, we can obtain the time it would take for two persons to finish a certain task by solving this equation:

algebra word problems 2

Therefore, to solve the problem, we can use the same equation/formula:

algebra word problems 3

Using the values given in the problem:

algebra word problems 4

Solving for the value of t :

½ + ⅕ = 1/t 

To make the equation above much easier to solve, we multiply both sides of the inequality by the LCD (the LCD is 10t):

10t(½ + ⅕) = 10t(1/t) 

5t + 2t = 10

Now, let us solve the resulting linear equation above to determine the value of t :

(7t)/7 = 10/7 Dividing both sides of the equation by 7

Based on our computation above, it would take 10/7 or 1 3/7 hours to completely clean the garage if Jacques and Kath decide to work together.

There are two pipes in a large water tank: Pipe C and Pipe D. Pipe C puts water into the tank while Pipe D spills water. If only Pipe C is opened, it will fill the tank in 2 hours. Meanwhile,  if only Pipe D is opened, it will empty the tank in 3 hours. Suppose that for a particular purpose, both Pipe C and Pipe D are opened simultaneously; how many hours will it take for the tank to be full?

The problem is asking how long it would take for Pipe C to fill the tank, given that Pipe D is also open.

algebra word problems 5

We know Pipe C puts water into the tank while Pipe D spills water. This means that we should modify our formula for the work problem as follows:

algebra word problems 6

We made the operation in the formula a subtraction sign because Pipe D is spilling water, so it would take Pipe C longer to fill the tank.

Let t be the time it would take for the tank to be full given that Pipe C and Pipe D are opened at the same time.

It takes 2 hours for Pipe C to fill the tank and 3 hours for Pipe D to empty it. Inputting these values into the equation:

½  – ⅓ = 1/t 

Now, let us solve for the value of t:

½ – ⅓ = 1/t

We multiply both sides of the equation by 6t to remove the denominators in the equation:

6t(½ – ⅓) = 6t(1/t)

3t – 2t = 6

Based on our computation above, it would take 6 hours for the water tank to be full if both Pipe C and Pipe D were opened simultaneously.

Rate Problems

Rate problems refer to word problems that involve moving objects. Most of the rate problems will ask you about the distance a particular moving object covers (distance), how long it would take for a moving object to reach a certain point (time), or how fast the moving object is (rate or speed). 

Since we are dealing with a moving object’s distance, rate, and time, we must know the distance formula for moving objects.

Distance Formula

Distance = Rate x Time

The distance formula tells us that a moving object (e.g., car, bus, airplane, cyclist, etc.) will cover a distance equivalent to the product of the rate or speed of the object and the time or the duration of the object’s movement.

For example, if a car travels at a rate of 60 kph for 2 hours, the distance that the car will cover is:

Distance = 60 x 2

Distance = 120 km

Remember the distance formula because we will use it a lot to solve rate problems.

Cars A and B leave two different cities 400 km apart. Car A and B move toward each other at 60 and 50 kph, respectively. How many hours will it take for these two cars to meet or pass each other?

Given their respective rates, the problem is asking how many hours it will take for Cars A and B to meet.

It is helpful to create an illustration for this problem:

algebra word problems 7

Based on our image above, the total distance covered by the cars equals 400 km. 

Let t be the number of hours it will take for the cars to meet or pass each other.

We can create a table so that we can summarize the details of the given situation.

Based on our table above, the respective distances covered by both cars are 60t and 50t.

Again, as shown in the illustration, the total distance they will have covered (the sum of their respective distances at time t) by the time the two cars meet is 400 km.

Thus, our equation would be:

60t + 50t = 400

(110t)/110 = 400/110

t = 40/11 or 3.64 

Hence, the cars will meet after about 3.64 hours .

Fred and Ludwig love cycling. One day they decided to test how long it would take to be 35 km apart if they started at a certain point and moved away from each other. Fred moves at a rate of 15 kph while Ludwig moves at 20 kph. Compute how long it would take for them to be 35 km apart.

Given their respective rates, The problem asks for the number of hours it will take for Fred and Ludwig to be 35 km apart.

If we try to illustrate the problem above, it will look like this:

algebra word problems 8

It is seen that the total distance covered by Fred and Ludwig is equal to 35 km if they are already 35 km apart from each other.

Given their respective rates, let t represent the time it takes for Ludwig and Fred to be 35 km apart.

Based on our table above, the respective distances covered by Fred and Ludwig to be 35 km apart are 15t and 20t.

The sum of the distances covered by Fred and Ludwig when they are 35 km apart is also equal to 35 km. Thus, we have this equation:

15t + 20t = 35

Given their respective rates, it will only take 1 hour for Fred and Ludwig to be 35 km apart.

Other Word Problems

To enrich your arsenal of problem-solving techniques, let us try to answer some word problems that don’t belong to any of the categories we have discussed.

Problem 1: A bag in a department store has been tagged with a 20% discount. If you buy the bag with the discount, you only have to pay ₱960. What is the original price of the bag?

We are tasked to find the original price of the bag.

Let x be the bag’s original price (i.e., the bag’s price without the discount).

The discount amount (i.e., the amount to be deducted from the original price) can be represented by 0.20x .

The problem states that if the bag was bought with a discount, you only have to pay ₱960. Hence, ₱960 is the discounted price of the bag.

Original Price – Discount Amount = Discounted Price

Hence, our equation would be:

x – 0.20x = 960

0.80x = 960

Dividing both sides of the equation by 0.80:

(0.80x)/0.80 = 960/0.80

Therefore, the original price of the bag (price without the discount) is ₱1200.

Problem 2: You are planning to buy flower pots from an online shopping site as a gift to your mother. Each flower pot costs ₱350, and you have to pay a fixed amount of ₱100 as a delivery fee. What is the maximum number of flower pots you can buy, given that you only have a ₱1000 budget?

The problem is asking for the largest number of flower pots your budget of ₱1000 can buy.

Let x be the number of flower pots.

Then, let 350x represent the total cost of buying x flower pots.

Since there’s a fixed service fee of ₱100, the total amount you must pay to the online shopping site would be 350x + 100 .

Now, 350x + 100 must be less than or equal to 1000 since our budget is only ₱1000. In other words, the total amount we need to pay to the online shopping site should not exceed 1000.

Hence, we have this inequality:

350x + 100 ≤ 1000

Let us solve the inequality above:

350x ≤ -100 + 1000

(350x)/350 ≤ 900/350

x ≤ 900/350

We write our answers in decimals to make our interpretation easier to understand.

The solution set of the inequality is x ≤ 2.57 or the set of all numbers less than or equal to 2.57. This means that the number of flower pots you can buy with your ₱1000 budget can be any number less than 2.57. 

However, we are looking for the maximum number of flower pots we can buy. Note that the number of flower pots should be a whole number (a fractional or decimal number of flower pots does not make sense). Since our solution set is x ≤ 2.57 or the set of all numbers less than or equal to 2.57 and the largest whole number in this set is 2, the maximum number of flower pots we can buy is 2.

The answer to this problem is two flower pots .

Problem 3: Vince bought a shirt from a supermarket in his town. He also bought a headset that cost ₱700, lower than twice the price of the shirt. The total amount Vince paid for the shirt and the headset is ₱1100. How much is the shirt that Vince bought?

We are tasked to find the price of the shirt that Vince bought.

Let x be the price of the shirt.

Since the headset price is ₱700 lower than twice the price of the shirt, we can represent the headset price as 2x – 700 .

The total amount that Vince paid for the shirt and headset was ₱1100:

Price of the shirt + Price of the headset = 1100

x + (2x – 700) = 1100

3x – 700 = 1100

3x = 700 + 1100 Transposition method

(3x)/3 = 1800/3 Dividing both sides of the equation by 3

Hence, the price of the shirt is ₱600 .

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Return to the main article:  The Ultimate Basic Math Reviewer

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Written by Jewel Kyle Fabula

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how to solve word problems in geometry

Jewel Kyle Fabula

Jewel Kyle Fabula is a Bachelor of Science in Economics student at the University of the Philippines Diliman. His passion for learning mathematics developed as he competed in some mathematics competitions during his Junior High School years. He loves cats, playing video games, and listening to music.

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