problem solving algebra word problems

Solving Word Questions

With LOTS of examples!

In Algebra we often have word questions like:

Example: Sam and Alex play tennis.

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

How many games did Alex play?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

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

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

Also look for key words:

Thinking Clearly

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

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

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

Now ... is that: S − 2 = A

or should it be: S = A − 2

or should it be: S = 2 − A

The correct answer is S = A − 2

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

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

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

Now ... is that: 2D = C

or should it be: D = 2C

Think carefully now!

The correct answer is D = 2C

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

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

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

Turn the English into Algebra:

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

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

We are being asked for the Area.

A = w × h = 12 × 5 = 60 m 2

The area is 60 square meters .

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

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

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

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

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

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

Which means that Alex played 4 games of tennis.

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

A slightly harder example:

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

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

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

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

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

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

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

So 10 tables would take Sam just 20 days.

Should Sam be paid more I wonder?

And another "substitution" example:

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

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

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

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

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

The number of "5 hour" days is 3

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

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

Some examples from Geometry:

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

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

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

We are being asked for the radius.

We need to rearrange the formula to find the area

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

Make a quick sketch:

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

We are being asked for the surface area.

First work out s using the volume formula:

Now we can calculate surface area:

An example about Money:

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

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

$40N + $(12 × 1¼N) = $660

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

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

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

More about Money, with these two examples involving Compound Interest

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

This is the compound interest formula:

So we will use these letters:

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

We are being asked for the Future Value: FV

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

The compound interest formula:

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

We are being asked for the Interest Rate: r

So the annual rate of interest is 5%

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

And an example of a Ratio question:

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

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

The current ratio is 4 : 3

Which can be rearranged to 3b = 4g

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

b + 4 g − 2 = 2 1

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

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

There are 12 girls !

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

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

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

And now for some Quadratic Equations :

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

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

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

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

n(n + 2) = 168

We are being asked for the integers

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

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

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

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

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

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

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

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

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

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

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

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

And so L = 8 or −14

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

So the length of the room is 8 m

L = 8, so W = ½L = 4

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

There we are ...

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

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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 71,324 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

• 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.

• 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.

• 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

Completing a Sample 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.

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

Expert Q&A

• 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

You Might Also Like

• ↑ 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

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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Algebra Word Problems Worksheets

In algebra word problems worksheets, we will talk about algebra, which is a branch of mathematics dealing with symbols and the rules for manipulating these symbols. They represent quantities without fixed values, known as variables. Solving algebraic word problems requires us to combine our ability to create and solve equations. Translating verbal descriptions into algebraic expressions is an essential initial step in solving word problems.

Benefits of Algebra Word Problems Worksheets

We use algebra in our everyday life without even realising it. Solving algebraic word problems worksheets help kids relate and understand the relevance of algebra in the real world. Algebra finds its way while cooking, measuring ingredients, sports, finance, professional advancement etc. They help in logical thinking and help students to break down a problem and then find its solution.

Download Algebra Word Problems Worksheet PDFs

These math worksheets should be practiced regularly and are free to download in PDF formats.

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• Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Expanded Form Mean, Median & Mode
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• Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
• Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
• Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
• Linear Algebra Matrices Vectors
• Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
• Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
• Physics Mechanics
• Chemistry Chemical Reactions Chemical Properties
• Finance Simple Interest Compound Interest Present Value Future Value
• Economics Point of Diminishing Return
• Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time Volume
• Pre Algebra
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Math word problems require interpreting what is being asked and simplifying that into a basic math equation. Once you have the equation you can then enter that into the problem solver as a basic math or algebra question to be correctly solved. Below are math word problem examples and their simplified forms.

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

Simplified Equation: 17 - x = 8

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

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

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

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

Simplified: 40 - 10 - 5

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100 hard word problems in algebra 

Find below a wide variety of hard word problems in algebra. Most tricky and tough algebra word problems are covered here.  If you can solve these, you can probably solve any algebra problems. Teachers! Feel free to select from this list and give them to your students to see if they have mastered how to solve tough algebra problems. Find out below how you can print these problems. You can also purchase a solution if needed.

1.  The cost of petrol rises by 2 cents a liter. last week a man bought 20 liters at the old price. This week he bought 10 liters at the new price. Altogether, the petrol costs $9.20. What was the old price for 1 liter?

2.  Teachers divided students into groups of 3. Each group of 3 wrote a report that had 9 pictures in it. The students used 585 pictures altogether. How many students were there in all?

3. Vera and Vikki are sisters. Vera is 4 years old and Vikki is 13 years old. What age will each sister be when Vikki is twice as old as Vera?

4.  A can do a work in 14 days and working together A and B can do the same work in 10 days. In what time can B alone do the work?

5. 7 workers can make 210 pairs of cup in 6 days. How many workers are required to make 450 pairs of cup in 10 days?

6. Ten years ago the ratio between the ages of Mohan and Suman was 3:5. 11 years hence it will be 11:16. What is the present age of Mohan?

7. The ratio of girls to boys in class is 9 to 7 and there are 80 students in the class. How many girls are in the class?

8.   One ounce of solution X contains only ingredients a and b in a ratio of 2:3. One ounce of solution Y contains only ingredients a and b in a ratio of 1:2. If solution Z is created by mixing solutions X and Y in a ratio of 3:11, then 2520 ounces of solution Z contains how many ounces of a?

9.   This week Bob puts gas in his truck when the tank was about half empty. Five days later, bob puts gas again when the tank was about three fourths full. If Bob Bought 24 gallons of gas, how many gallons does the tank hold?

10. A commercial airplane flying with a speed of 700 mi/h is detected 1000 miles away with a radar. Half an hour later an interceptor plane flying with a speed of 800 mi/h is dispatched. How long will it take the interceptor plane to meet with the other plane?

11. There are 40 pigs and chickens in a farmyard. Joseph counted 100 legs in all. How many pigs and how many chickens are there?

12. The top of a box is a rectangle with a perimeter of 72 inches. If the box is 8 inches high, what dimensions will give the maximum volume?

13. You are raising money for a charity. Someone made a fixed donation of 500. Then, you require each participant to make a pledge of 25 dollars. What is the minimum amount of money raised if there are 224 participants.

14. The sum of two positive numbers is 4 and the sum of their squares is 28. What are the two numbers?

15. Flying against the jet stream, a jet travels 1880 mi in 4 hours. Flying with the jet stream, the same jet travels 5820 mi in 6 hours. What is the rate of the jet in still air and what is the rate of the jet stream?

16.  Jenna and her friend, Khalil, are having a contest to see who can save the most money. Jenna has already saved $110 and every week she saves an additional $20. Khalil has already saved $80 and every week he saves an additional $25. Let x represent the number of weeks and y represent the total amount of money saved. Determine in how many weeks Jenna and Khalil will have the same amount of money.

17. The sum of three consecutive terms of a geometric sequence is 104 and their product is 13824.find the terms.

18. The sum of the first and last of four consecutive odd integers is 52.  What are the four integers?

19. A health club charges a one-time initiation fee and a monthly fee. John  paid 100 dollars for 2 months of membership. However, Peter paid 200 for 6 months of membership. How much will Sylvia pay for 1 year of membership?

20. The sum of two positive numbers is 4 and the sum of their cubes is 28. What is the product of the two numbers?

21. A man selling computer parts realizes that when he sells 16 computer parts, his earning is $1700. When he sells 56 computer parts, his earning is $4300. What will the earning be if the man sells 30 computers parts?

22. A man has 15 coins in his pockets. These coins are dimes and quarters that add up 2.4 dollars. How many quarters and how many dimes does the man have?

23. The lengths of the sides of a triangle are in the ratio 4:3:5. Find the lengths of the sides if the perimeter is 18 inches.

24. The ratio of base to height of a equilateral triangle is 3:4. If the area of the triangle is 6, what is the perimeter of the triangle?

25. The percent return rate of a growth fund, income fund, and money market are 10%, 7%, and 5% respectively. Suppose you have 3200 to invest and you want to put twice as much in the growth fund as in the money market to maximize your return. How should you invest to get a return of 250 dollars in 1 year?

26. A shark was caught whose tail weighted 200 pounds. The head of the shark weighted as much as its tail plus half its body. Its body weighted as much as its head and tail. What is the weight of the shark?

27. The square root of a number plus two is the same as the number. What is the number?

28. Suppose you have a coupon worth 6 dollars off any item at a mall. You go to a store at the mall that offers a 20% discount. What do you need to do to save the most money?

29. Suppose your grades on three math exams are 80, 93, and 91. What grade do you need on your next exam to have at least a 90 average on the four exams?

30. Peter has a photograph that is 5 inches wide and 6 inches long. She enlarged each side by the same amount. By how much was the photograph enlarged if the new area is 182 square inches?

31. The cost to produce a book is 1200 to get started plus 9 dollars per book. The book sells for 15 dollars each. How many books must be sold to make a profit?

32. Store A sells CDs for 2 dollars each if you pay a one-time fee of 104 dollars. Store B offers 12 free CDs and charges 10 dollars for each additional CD. How many CD must you buy so it will cost the same under both plans?

33. When 4 is added to two numbers, the ratio is 5:6. When 4 is subtracted from the two numbers, the ratio is 1:2. Find the two numbers.

34. A store owner wants to sell 200 pounds of pistachios and walnuts mixed together. Walnuts cost 4 dollars per pound and pistachio cost 6 dollars per pounds. How many pounds of each type of nuts should be mixed if the store owner will charge 5 dollars for the mixture?

35. A cereal box manufacturer makes 32-ounce boxes of cereal. In a perfect world, the box will be 32-ounce every time is made. However, since the world is not perfect, they allow a difference of 0.06 ounce. Find the range of acceptable size for the cereal box.

36. A man weighing 600 kg has been losing 3.12% of his weight each month with some heavy exercises and eating the right food. What will the man weigh after 20 months?

37. An object is thrown into the air at a height of 60 feet. After 1 second and 2 second, the object is 88 feet and 84 feet in the air respectively. What is the initial speed of the object?

38. A transit is 200 feet from the base of a building. There is man standing on top of the building. The angles of elevation from the top and bottom of the man are 45 degrees and 44 degrees. What is the height of the man?

39. A lemonade consists of 6% of lemon juice and a strawberry juice consists of 15% pure fruit juice. How much of each kind should be mixed together to get a 4 Liters of a 10% concentration of fruit juice?

40. Ellen can wash her car in 60 minutes. Her older sister Sarah can do the same job in 45 minutes. How long will it take if they wash the car together?

41. A plane flies 500 mi/h. The plane can travel 1100 miles with the wind in the same amount of time as it travels 900 miles against the wind. What is the speed of the wind?

42. A company produces boxes that are 5 feet long, 4 feet wide, and 3 feet high. The company wants to increase each dimension by the same amount so that the new volume is twice as big. How much is the increase in dimension?

43. James invested half of his money in land, a tenth in stock, and a twentieth in saving bonds. Then, he put the remaining 21000 in a CD. How much money did James saved or invested?

44. The motherboards for a desktop computer can be manufactured for 50 dollars each. The development cost is 250000. The first 20 motherboard are samples and will not be sold. How many salable motherboards will yield an average cost of 6325 dollars?

45. How much of a 70% orange juice drink must be mixed with 44 gallons of a 20% orange juice drink to obtain a mixture that is 50% orange juice?

46. A company sells nuts in bulk quantities. When bought in bulk, peanuts sell  for $1.20 per pound, almonds for $ 2.20 per pound, and cashews for $3.20 per pound. Suppose a specialty shop wants a mixture of 280 pounds that will cost $2.59 per pound. Find the number of pounds of each type of nut if the sum of the number of pounds of almonds and cashews is three times the number of pounds of peanuts. Round your answers to the nearest pound.

47. A Basketball player has successfully made 36 of his last 48 free throws. Find the number of consecutive free throws the player needs to increase his success rate to 80%.

48. John can wash cars 3 times as fast as his son Erick. Working together, they need to wash 30 cars in 6 hours. How many hours will it takes each of them working alone?

49. In a college, about 36% of student are under 20 years old and 15% are over 40 years old. What is the probability that a student chosen at random is under 20 years old or over 40 years?

50. Twice a number plus the square root of the number is twelve minus the square root of the number.

51. The light intensity, I , of a light bulb varies inversely as the square of the distance from the bulb. A a distance of 3 meters from the bulb, I = 1.5 W/m^2 . What is the light intensity at a distance of 2 meters from the bulb?

52. The lengths of two sides of a triangle are 2 and 6. find the range of values for the possible lengths of the third side.

53. Find three consecutive integers such that one half of their sum is between 15 and 21.

54. After you open a book, you notice that the product of the two page numbers on the facing pages is 650. What are the two page numbers?

55. Suppose you start with a number. You multiply the number by 3, add 7, divide by ½, subtract 5, and then divide by 12. The result is 5. What number did you start with?

56. You have 156 feet of fencing to enclose a rectangular garden. You want the garden to be 5 times as long as it is wide. Find the dimensions of the garden.

57. The amount of water a dripping faucet wastes water varies directly with the amount of time the faucet drips. If the faucet drips 2 cups of water every 6 minutes, find out how long it will take the faucet to drip 10.6465 liters of water.

58. A washer costs 25% more than a dryer. If the store clerk gave a 10% discount for the dryer and a 20% discount for the washer, how much is the washer before the discount if you paid 1900 dollars.

59. Baking a tray of blueberry muffins takes 4 cups of milk and 3 cups of wheat flour. A tray of pumpkin muffins takes 2 cups of milk and 3 cups of wheat flour. A baker has 16 cups of milk and 15 cups of wheat flour. You make 3 dollars profit per tray of blueberry muffins and 2 dollars profit per tray of pumpkin muffins. How many trays of each type of muffins should you make to maximize profit?

60. A company found that -2p + 1000 models the number of TVs sold per month where p can be set as low as 200 or as high as 300. How can the company maximize the revenue?

More hard word problems in algebra

61. Your friends say that he has $2.40 in equal numbers of quarters, dimes, and nickels. How many of each coin does he have?

62. I am a two-digit number whose digit in the tenth place is 1 less than twice the digit in the ones place. When the digit in the tenth place is divided by the digit in the ones place, The quotient is 1 and the remainder is 4. What number am I?

63. A two-digit number is formed by randomly selecting from the digits 2, 4, 5, and 7 without replacement. What is the probability that a two-digit number contains a 2 or a 7?

64. Suppose you interview 30 females and 20 males at your school to find out who among them are using an electric toothbrush. Your survey revealed that only 2 males use an electric toothbrush while 6 females use it. What is the probability that a respondent did not use an electric toothbrush given that the respondent is a female?

65. An employer pays 15 dollars per hour plus an extra 5 dollars per hour for every hour worked beyond 8 hours up to a maximum daily wage of 220 dollars. Find a piecewise function that models this situation.

66. Divide me by 7, the remainder is 5. Divide me by 3, the remainder is 1 and my quotient is 2 less than 3 times my previous quotient. What number am I?

67. A company making luggage have these requirements to follow. The length is 15 inches greater than the depth and the sum of length, width, and depth may not exceed 50 inches. What is the maximum value for the depth if the manufacturer will only use whole numbers?

68. To make an open box, a man cuts equal squares from each corner of a sheet of metal that is 12 inches wide and 16 inches long. Find an expression for the volume in terms of x.

69. Ten candidates are running for president, vice-president, and secretary in the students government. You may vote for as many as 3 candidates. In how many ways can you vote for 3 or fewer candidates?

70. The half-life of a medication prescribed by a doctor is 6 hours. How many mg of this medication is left after 78 hours if the doctor prescribed 100 mg?

71. Suppose you roll a red number cube and a yellow number cube. Find P(red 2, yellow 2) and the probability to get any matching pairs of numbers.

72. A movie theater in a small town usually open its doors 3 days in a row and then closes the next day for maintenance. Another movie theater 3 miles away open 4 days in a row and then closes the next day for the same reason. Suppose both movie theaters are closed today and today is Wednesday, when is the next time they will both be closed again on the same day?

73. An investor invests 5000 dollars at 10% and the rest at 5%. How much was invested at 5% if the yield is one-fifth of the amount invested at at 10%?

74. 20000 students took a standardized math test. The scores on the test are normally distributed, with a mean score of 85 and a standard deviation of 5. About how many students scored between 90 and 95?

75. A satellite, located 2400 km above Earth’s surface, is in circular orbit around the earth. If it takes the satellite 3 hours to complete 1 orbit, how far is the satellite after 1 hour?

76. In a group of 10 people, what is the probability that at least two people in the group have the same birthday?

77. During a fundraising for cancer at a gala, everybody shakes hands with everyone else in the room before the event and after the event is finished. If n people attended the gala, how many different handshakes occur?

78. Two cubes have side lengths that are equal to 2x and 4x. How many times greater than the surface of the small cube is the surface area of the large cube?

79. Suppose you have a job in a restaurant that pays $8 per hour. You also have a job at Walmart that pays $10 per hour. You want to earn at least 200 per week. However, you want to work no more than 25 hours per week . Show 3 different ways you could work at each job.

80. Two company offer tutoring services. Company A realizes that when they tutor for 3 hours, they make 45 dollars. When they tutor for 7 hours they make 105 hours. Company B realizes that when they tutor for 2 hours, they make 34 dollars. When they tutor for 6 hours they make 102 hours. Assuming that the number of hours students sign for tutoring is the same for both company, which company will generate more revenue?

81. You want to fence a rectangular area for kids in the backyard. To save on fences, you will use the back of your house as one of the four sides. Find the possible dimensions if the house is 60 feet wide and you want to use at least 160 feet of fencing.

82. When a number in increased by 20%, the result is the same when it is decreased by 10% plus 12. What is the number?

83. The average of three numbers is 47. The biggest number is five more than twice the smallest. The range is 35. What are the three numbers?

84. The percent of increase of a number from its original amount to 36 is 80%. What is the original amount of the number?

85. When Peter drives to work, he averages 45 miles per hour because of traffic. On the way back home, he averages 60 miles per hour because traffic is not as bad. The total travel time is 2 hours. How far is Peter’s house from work?

86. An advertising company takes 20% from all revenue that it generates for its affiliates. If the affiliates were paid 15200 dollars this month, how much revenue did the advertising company generate this month?

87. A company’s revenue can be modeled with a quadratic equation. The company noticed that when they sell either 2 or or 12 items, the revenue is 0. How much is the revenue when they sell 20 items?

88. A ball bounced 4 times, reaching three-fourths of its previous height with each bounce. After the fourth bounce, the ball reached a height of 25 cm. How high was the ball when it was dropped?

89. A rental company charges 40 dollars per day plus $0.30 per mile. You rent a car and drop it off 4 days later. How many miles did you drive the car if you paid 325.5 dollars which included a 5% sales tax?

90. Two students leave school at the same time and travel in opposite directions along the same road. One walk at a rate of 3 mi/h. The other bikes at a rate of 8.5 mi/h. After how long will they be 23 miles part?

91. Brown has the same number of brothers as sisters. His sister Sylvia as twice as many brothers as sisters. How many children are in the family?

92. Ethan has the same number of male classmates as female classmates. His classmate Olivia has three-fourths as many female classmates as male classmates. How many students are in the class?

93. Noah wants to share a certain amount of money with 10 people. However, at the last minute, he is thinking about decreasing the amount by 20 so he can keep 20 for himself and share the money with only 5 people. How much money is Noah trying to share if each person still gets the same amount?

94. The square root of me plus the square root of me is me. Who am I?

95. A cash drawer contains 160 bills, all 10s and 50s. If the total value of the 10s and 50s is $1,760. How many of each type of bill are in the drawer?

96. You want to make 28 grams of protein snack mix made with peanuts and granola. Peanuts contain 7 grams of protein per ounce and granola contain 3 grams of protein per ounce. How many ounces of granola should you use for 1 ounce of peanuts?

97. The length of a rectangular prism is quadrupled, the width is doubled, and the height is cut in half. If V is the volume of the rectangular prism before the modification, express the volume after the modification in terms of V.

98. A car rental has CD players in 85% of its cars. The CD players are randomly distributed throughout the fleet of cars. If a person rents 4 cars, what is the probability that at least 3 of them will have CD players?

99. Jacob’s hourly wage is 4 times as much as Noah. When Jacob got a raise of 2 dollars, Noah accepted a new position that pays him 2 dollars less per hour. Jacob now earns 5 times as much money as Noah. How much money do they make per hour after Jacob got the raise?

You own a catering business that makes specialty cakes. Your company has decided to create three types of cakes. To create these cakes, it takes a team that consists of a decorator, a baker, and a design consultant. Cake A takes the decorator 9 hours, the baker 6 hours, and the design consultant 1 hour to complete. Cake B takes the decorator 10 hours, the baker 4 hours, and the design consultant 2 hours. Cake C takes the decorator 12 hours, the baker 4 hour, and the design consultant 1 hour. Without hiring additional employees, there are 398 decorator hours available, 164 baker hours available, and 58 design consultant hours available. How many of each type of cake can be created?

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Algebra Topics  - Distance Word Problems

Algebra topics  -, distance word problems, algebra topics distance word problems.

Algebra Topics: Distance Word Problems

Lesson 10: distance word problems.

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What are distance word problems?

Distance word problems are a common type of algebra word problems. They involve a scenario in which you need to figure out how fast , how far , or how long one or more objects have traveled. These are often called train problems because one of the most famous types of distance problems involves finding out when two trains heading toward each other cross paths.

In this lesson, you'll learn how to solve train problems and a few other common types of distance problems. But first, let's look at some basic principles that apply to any distance problem.

The basics of distance problems

There are three basic aspects to movement and travel: distance , rate , and time . To understand the difference among these, think about the last time you drove somewhere.

The distance is how far you traveled. The rate is how fast you traveled. The time is how long the trip took.

The relationship among these things can be described by this formula:

distance = rate x time d = rt

In other words, the distance you drove is equal to the rate at which you drove times the amount of time you drove. For an example of how this would work in real life, just imagine your last trip was like this:

• You drove 25 miles—that's the distance .
• You drove an average of 50 mph—that's the rate .
• The drive took you 30 minutes, or 0 .5 hours—that's the time .

According to the formula, if we multiply the rate and time , the product should be our distance.

And it is! We drove 50 mph for 0.5 hours—and 50 ⋅ 0.5 equals 25 , which is our distance.

What if we drove 60 mph instead of 50? How far could we drive in 30 minutes? We could use the same formula to figure this out.

60 ⋅ 0.5 is 30, so our distance would be 30 miles.

Solving distance problems

When you solve any distance problem, you'll have to do what we just did—use the formula to find distance , rate , or time . Let's try another simple problem.

On his day off, Lee took a trip to the zoo. He drove an average speed of 65 mph, and it took him two-and-a-half hours to get from his house to the zoo. How far is the zoo from his house?

First, we should identify the information we know. Remember, we're looking for any information about distance, rate, or time. According to the problem:

• The rate is 65 mph.
• The time is two-and-a-half hours, or 2.5 hours.
• The distance is unknown—it's what we're trying to find.

You could picture Lee's trip with a diagram like this:

This diagram is a start to understanding this problem, but we still have to figure out what to do with the numbers for distance , rate , and time . To keep track of the information in the problem, we'll set up a table. (This might seem excessive now, but it's a good habit for even simple problems and can make solving complicated problems much easier.) Here's what our table looks like:

We can put this information into our formula: distance = rate ⋅ time .

We can use the distance = rate ⋅ time formula to find the distance Lee traveled.

The formula d = rt looks like this when we plug in the numbers from the problem. The unknown distance is represented with the variable d .

d = 65 ⋅ 2.5

To find d , all we have to do is multiply 65 and 2.5. 65 ⋅ 2.5 equals 162.5 .

We have an answer to our problem: d = 162.5. In other words, the distance Lee drove from his house to the zoo is 162.5 miles.

Be careful to use the same units of measurement for rate and time. It's possible to multiply 65 miles per hour by 2.5 hours because they use the same unit: an hour . However, what if the time had been written in a different unit, like in minutes ? In that case, you'd have to convert the time into hours so it would use the same unit as the rate.

Solving for rate and time

In the problem we just solved we calculated for distance , but you can use the d = rt formula to solve for rate and time too. For example, take a look at this problem:

After work, Janae walked in her neighborhood for a half hour. She walked a mile-and-a-half total. What was her average speed in miles per hour?

We can picture Janae's walk as something like this:

And we can set up the information from the problem we know like this:

The table is repeating the facts we already know from the problem. Janae walked one-and-a-half miles or 1.5 miles in a half hour, or 0.5 hours.

As always, we start with our formula. Next, we'll fill in the formula with the information from our table.

The rate is represented by r because we don't yet know how fast Janae was walking. Since we're solving for r , we'll have to get it alone on one side of the equation.

1.5 = r ⋅ 0.5

Our equation calls for r to be multiplied by 0.5, so we can get r alone on one side of the equation by dividing both sides by 0.5: 1.5 / 0.5 = 3 .

r = 3 , so 3 is the answer to our problem. Janae walked 3 miles per hour.

In the problems on this page, we solved for distance and rate of travel, but you can also use the travel equation to solve for time . You can even use it to solve certain problems where you're trying to figure out the distance, rate, or time of two or more moving objects. We'll look at problems like this on the next few pages.

Two-part and round-trip problems

Do you know how to solve this problem?

Bill took a trip to see a friend. His friend lives 225 miles away. He drove in town at an average of 30 mph, then he drove on the interstate at an average of 70 mph. The trip took three-and-a-half hours total. How far did Bill drive on the interstate?

This problem is a classic two-part trip problem because it's asking you to find information about one part of a two-part trip. This problem might seem complicated, but don't be intimidated!

You can solve it using the same tools we used to solve the simpler problems on the first page:

• The travel equation d = rt
• A table to keep track of important information

Let's start with the table . Take another look at the problem. This time, the information relating to distance , rate , and time has been underlined.

Bill took a trip to see a friend. His friend lives 225 miles away. He drove in town at an average of 30 mph , then he drove on the interstate at an average of 70 mph . The trip took three-and-a-half hours total. How far did Bill drive on the interstate?

If you tried to fill in the table the way we did on the last page, you might have noticed a problem: There's too much information. For instance, the problem contains two rates— 30 mph and 70 mph . To include all of this information, let's create a table with an extra row. The top row of numbers and variables will be labeled in town , and the bottom row will be labeled interstate .

We filled in the rates, but what about the distance and time ? If you look back at the problem, you'll see that these are the total figures, meaning they include both the time in town and on the interstate. So the total distance is 225 . This means this is true:

Interstate distance + in-town distance = Total distance

Together, the interstate distance and in-town distance are equal to the total distance. See?

In any case, we're trying to find out how far Bill drove on the interstate , so let's represent this number with d . If the interstate distance is d , it means the in-town distance is a number that equals the total, 225 , when added to d . In other words, it's equal to 225 - d .

We can fill in our chart like this:

We can use the same technique to fill in the time column. The total time is 3.5 hours . If we say the time on the interstate is t , then the remaining time in town is equal to 3.5 - t . We can fill in the rest of our chart.

Now we can work on solving the problem. The main difference between the problems on the first page and this problem is that this problem involves two equations. Here's the one for in-town travel :

225 - d = 30 ⋅ (3.5 - t)

And here's the one for interstate travel :

If you tried to solve either of these on its own, you might have found it impossible: since each equation contains two unknown variables, they can't be solved on their own. Try for yourself. If you work either equation on its own, you won't be able to find a numerical value for d . In order to find the value of d , we'll also have to know the value of t .

We can find the value of t in both problems by combining them. Let's take another look at our travel equation for interstate travel.

While we don't know the numerical value of d , this equation does tell us that d is equal to 70 t .

Since 70 t and d are equal , we can replace d with 70 t . Substituting 70 t for d in our equation for interstate travel won't help us find the value of t —all it tells us is that 70 t is equal to itself, which we already knew.

But what about our other equation, the one for in-town travel?

When we replace the d in that equation with 70 t , the equation suddenly gets much easier to solve.

225 - 70t = 30 ⋅ (3.5 - t)

Our new equation might look more complicated, but it's actually something we can solve. This is because it only has one variable: t . Once we find t , we can use it to calculate the value of d —and find the answer to our problem.

To simplify this equation and find the value of t , we'll have to get the t alone on one side of the equals sign. We'll also have to simplify the right side as much as possible.

Let's start with the right side: 30 times (3.5 - t ) is 105 - 30 t .

225 - 70t = 105 - 30t

Next, let's cancel out the 225 next to 70 t . To do this, we'll subtract 225 from both sides. On the right side, it means subtracting 225 from 105 . 105 - 225 is -120 .

- 70t = -120 - 30t

Our next step is to group like terms—remember, our eventual goal is to have t on the left side of the equals sign and a number on the right . We'll cancel out the -30 t on the right side by adding 30 t to both sides. On the right side, we'll add it to -70 t . -70 t + 30 t is -40 t .

- 40t = -120

Finally, to get t on its own, we'll divide each side by its coefficient: -40. -120 / - 40 is 3 .

So t is equal to 3 . In other words, the time Bill traveled on the interstate is equal to 3 hours . Remember, we're ultimately trying to find the distanc e Bill traveled on the interstate. Let's look at the interstate row of our chart again and see if we have enough information to find out.

It looks like we do. Now that we're only missing one variable, we should be able to find its value pretty quickly.

To find the distance, we'll use the travel formula distance = rate ⋅ time .

We now know that Bill traveled on the interstate for 3 hours at 70 mph , so we can fill in this information.

d = 3 ⋅ 70

Finally, we finished simplifying the right side of the equation. 3 ⋅ 70 is 210 .

So d = 210 . We have the answer to our problem! The distance is 210 . In other words, Bill drove 210 miles on the interstate.

Solving a round-trip problem

It might have seemed like it took a long time to solve the first problem. The more practice you get with these problems, the quicker they'll go. Let's try a similar problem. This one is called a round-trip problem because it describes a round trip—a trip that includes a return journey. Even though the trip described in this problem is slightly different from the one in our first problem, you should be able to solve it the same way. Let's take a look:

Eva drove to work at an average speed of 36 mph. On the way home, she hit traffic and only drove an average of 27 mph. Her total time in the car was 1 hour and 45 minutes, or 1.75 hours. How far does Eva live from work?

If you're having trouble understanding this problem, you might want to visualize Eva's commute like this:

As always, let's start by filling in a table with the important information. We'll make a row with information about her trip to work and from work .

1.75 - t to describe the trip from work. (Remember, the total travel time is 1.75 hours , so the time to work and from work should equal 1.75 .)

From our table, we can write two equations:

• The trip to work can be represented as d = 36 t .
• The trip from work can be represented as d = 27 (1.75 - t ) .

In both equations, d represents the total distance. From the diagram, you can see that these two equations are equal to each other—after all, Eva drives the same distance to and from work .

Just like with the last problem we solved, we can solve this one by combining the two equations.

We'll start with our equation for the trip from work .

d = 27 (1.75 - t)

Next, we'll substitute in the value of d from our to work equation, d = 36 t . Since the value of d is 36 t , we can replace any occurrence of d with 36 t .

36t = 27 (1.75 - t)

Now, let's simplify the right side. 27 ⋅(1.75 - t ) is 47.25 .

36t = 47.25 - 27t

Next, we'll cancel out -27 t by adding 27 t to both sides of the equation. 36 t + 27 t is 63 t .

63t = 47.25

Finally, we can get t on its own by dividing both sides by its coefficient: 63 . 47.25 / 63 is .75 .

t is equal to .75 . In other words, the time it took Eva to drive to work is .75 hours . Now that we know the value of t , we'll be able to can find the distance to Eva's work.

If you guessed that we were going to use the travel equation again, you were right. We now know the value of two out of the three variables, which means we know enough to solve our problem.

First, let's fill in the values we know. We'll work with the numbers for the trip to work . We already knew the rate : 36 . And we just learned the time : .75 .

d = 36 ⋅ .75

Now all we have to do is simplify the equation: 36 ⋅ .75 = 27 .

d is equal to 27 . In other words, the distance to Eva's work is 27 miles . Our problem is solved.

Intersecting distance problems

An intersecting distance problem is one where two things are moving toward each other. Here's a typical problem:

Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading to Springfield at the same time a train leaves Springfield heading to Pawnee. One train is moving at a speed of 45 mph, and the other is moving 60 mph. How long will they travel before they meet?

This problem is asking you to calculate how long it will take these two trains moving toward each other to cross paths. This might seem confusing at first. Even though it's a real-world situation, it can be difficult to imagine distance and motion abstractly. This diagram might help you get a sense of what this situation looks like:

If you're still confused, don't worry! You can solve this problem the same way you solved the two-part problems on the last page. You'll just need a chart and the travel formula .

Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading toward Springfield at the same time a train leaves Springfield heading toward Pawnee. One train is moving at a speed of 45 mph , and the other is moving 60 mph . How long will they travel before they meet?

Let's start by filling in our chart. Here's the problem again, this time with the important information underlined. We can start by filling in the most obvious information: rate . The problem gives us the speed of each train. We'll label them fast train and slow train . The fast train goes 60 mph . The slow train goes only 45 mph .

We can also put this information into a table:

We don't know the distance each train travels to meet the other yet—we just know the total distance. In order to meet, the trains will cover a combined distance equal to the total distance. As you can see in this diagram, this is true no matter how far each train travels.

This means that—just like last time—we'll represent the distance of one with d and the distance of the other with the total minus d. So the distance for the fast train will be d , and the distance for the slow train will be 420 - d .

Because we're looking for the time both trains travel before they meet, the time will be the same for both trains. We can represent it with t .

The table gives us two equations: d = 60 t and 420 - d = 45 t . Just like we did with the two-part problems, we can combine these two equations.

The equation for the fast train isn't solvable on its own, but it does tell us that d is equal to 60 t .

The other equation, which describes the slow train, can't be solved alone either. However, we can replace the d with its value from the first equation.

420 - d = 45t

Because we know that d is equal to 60 t , we can replace the d in this equation with 60 t . Now we have an equation we can solve.

420 - 60t = 45t

To solve this equation, we'll need to get t and its coefficients on one side of the equals sign and any other numbers on the other. We can start by canceling out the -60 t on the left by adding 60 t to both sides. 45 t + 60 t is 105 t .

Now we just need to get rid of the coefficient next to t . We can do this by dividing both sides by 105 . 420 / 105 is 4 .

t = 4 . In other words, the time it takes the trains to meet is 4 hours . Our problem is solved!

If you want to be sure of your answer, you can check it by using the distance equation with t equal to 4 . For our fast train, the equation would be d = 60 ⋅ 4 . 60 ⋅ 4 is 240 , so the distance our fast train traveled would be 240 miles. For our slow train, the equation would be d = 45 ⋅ 4 . 45 ⋅ 4 is 180 , so the distance traveled by the slow train is 180 miles . Remember how we said the distance the slow train and fast train travel should equal the total distance? 240 miles + 180 miles equals 420 miles , which is the total distance from our problem. Our answer is correct.

Practice problem 1

Here's another intersecting distance problem. It's similar to the one we just solved. See if you can solve it on your own. When you're finished, scroll down to see the answer and an explanation.

Jon and Dani live 270 miles apart. One day, they decided to drive toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove an average of 70 mph. How long did they drive before they met up?

Problem 1 answer

Here's practice problem 1:

Jon and Dani live 270 miles apart. One day, they decided to drive toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove 70 mph. How long did they drive before they met up?

Answer: 2 hours .

Let's solve this problem like we solved the others. First, try making the chart. It should look like this:

Here's how we filled in the chart:

• Distance: Together, Dani and Jon will have covered the total distance between them by the time they meet up. That's 270 . Jon's distance is represented by d , so Dani's distance is 270 - d .
• Rate: The problem tells us Dani and Jon's speeds. Dani drives 65 mph , and Jon drives 70 mph .
• Time: Because Jon and Dani drive the same amount of time before they meet up, both of their travel times are represented by t .

Now we have two equations. The equation for Jon's travel is d = 65 t . The equation for Dani's travel is 270 - d = 70 t . To solve this problem, we'll need to combine them.

The equation for Jon tells us that d is equal to 65 t . This means we can combine the two equations by replacing the d in Dani's equation with 65 t .

270 - 65t = 70t

Let's get t on one side of the equation and a number on the other. The first step to doing this is to get rid of -65 t on the left side. We'll cancel it out by adding 65 t to both sides: 70 t + 65 t is 135 t .

All that's left to do is to get rid of the 135 next to the t . We can do this by dividing both sides by 135 : 270 / 135 is 2 .

That's it. t is equal to 2 . We have the answer to our problem: Dani and Jon drove 2 hours before they met up.

Overtaking distance problems

The final type of distance problem we'll discuss in this lesson is a problem in which one moving object overtakes —or passes —another. Here's a typical overtaking problem:

The Hill family and the Platter family are going on a road trip. The Hills left 3 hours before the Platters, but the Platters drive an average of 15 mph faster. If it takes the Platter family 13 hours to catch up with the Hill family, how fast are the Hills driving?

You can picture the moment the Platter family left for the road trip a little like this:

The problem tells us that the Platter family will catch up with the Hill family in 13 hours and asks us to use this information to find the Hill family's rate . Like some of the other problems we've solved in this lesson, it might not seem like we have enough information to solve this problem—but we do. Let's start making our chart. The distance can be d for both the Hills and the Platters—when the Platters catch up with the Hills, both families will have driven the exact same distance.

Filling in the rate and time will require a little more thought. We don't know the rate for either family—remember, that's what we're trying to find out. However, we do know that the Platters drove 15 mph faster than the Hills. This means if the Hill family's rate is r , the Platter family's rate would be r + 15 .

Now all that's left is the time. We know it took the Platters 13 hours to catch up with the Hills. However, remember that the Hills left 3 hours earlier than the Platters—which means when the Platters caught up, they'd been driving 3 hours more than the Platters. 13 + 3 is 16 , so we know the Hills had been driving 16 hours by the time the Platters caught up with them.

Our chart gives us two equations. The Hill family's trip can be described by d = r ⋅ 16 . The equation for the Platter family's trip is d = ( r + 15) ⋅ 13 . Just like with our other problems, we can combine these equations by replacing a variable in one of them.

The Hill family equation already has the value of d equal to r ⋅ 16. So we'll replace the d in the Platter equation with r ⋅ 16 . This way, it will be an equation we can solve.

r ⋅ 16 = (r + 15) ⋅ 13

First, let's simplify the right side: r ⋅ 16 is 16 r .

16r = (r + 15) ⋅ 13

Next, we'll simplify the right side and multiply ( r + 15) by 13 .

16r = 13r + 195

We can get both r and their coefficients on the left side by subtracting 13 r from 16 r : 16 r - 13 r is 3 r .

Now all that's left to do is get rid of the 3 next to the r . To do this, we'll divide both sides by 3: 195 / 3 is 65 .

So there's our answer: r = 65. The Hill family drove an average of 65 mph .

You can solve any overtaking problem the same way we solved this one. Just remember to pay special attention when you're setting up your chart. Just like the Hill family did in this problem, the person or vehicle who started moving first will always have a greater travel time.

Practice problem 2

Try solving this problem. It's similar to the problem we just solved. When you're finished, scroll down to see the answer and an explanation.

A train moving 60 mph leaves the station at noon. An hour later, a train moving 80 mph leaves heading the same direction on a parallel track. What time does the second train catch up to the first?

Problem 2 answer

Here's practice problem 2:

Answer: 4 p.m.

To solve this problem, start by making a chart. Here's how it should look:

Here's an explanation of the chart:

• Distance: Both trains will have traveled the same distance by the time the fast train catches up with the slow one, so the distance for both is d .
• Rate: The problem tells us how fast each train was going. The fast train has a rate of 80 mph , and the slow train has a rate of 60 mph .
• Time: We'll use t to represent the fast train's travel time before it catches up. Because the slow train started an hour before the fast one, it will have been traveling one hour more by the time the fast train catches up. It's t + 1 .

Now we have two equations. The equation for the fast train is d = 80 t . The equation for the slow train is d = 60 ( t + 1) . To solve this problem, we'll need to combine the equations.

The equation for the fast train says d is equal to 80 t . This means we can combine the two equations by replacing the d in the slow train's equation with 80 t .

80t = 60 (t + 1)

First, let's simplify the right side of the equation: 60 ⋅ ( t + 1) is 60 t + 60 .

80t = 60t + 60

To solve the equation, we'll have to get t on one side of the equals sign and a number on the other. We can get rid of 60 t on the right side by subtracting 60 t from both sides: 80 t - 60 t is 20 t .

Finally, we can get rid of the 20 next to t by dividing both sides by 20 . 60 divided by 20 is 3 .

So t is equal to 3 . The fast train traveled for 3 hours . However, it's not the answer to our problem. Let's look at the original problem again. Pay attention to the last sentence, which is the question we're trying to answer.

Our problem doesn't ask how long either of the trains traveled. It asks what time the second train catches up with the first.

The problem tells us that the slow train left at noon and the fast one left an hour later. This means the fast train left at 1 p.m . From our equations, we know the fast train traveled 3 hours . 1 + 3 is 4 , so the fast train caught up with the slow one at 4 p.m . The answer to the problem is 4 p.m .

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Solving Word Problems in Algebra Inequality Word Problems

How are you with solving word problems in Algebra? Are you ready to dive into the "real world" of inequalities? I know that solving word problems in Algebra is probably not your favorite, but there's no point in learning the skill if you don't apply it.

I promise to make this as easy as possible. Pay close attention to the key words given below, as this will help you to write the inequality. Once the inequality is written, you can solve the inequality using the skills you learned in our past lessons.

I've tried to provide you with examples that could pertain to your life and come in handy one day. Think about others ways you might use inequalities in real world problems. I'd love to hear about them if you do!

Before we look at the examples let's go over some of the rules and key words for solving word problems in Algebra (or any math class).

Word Problem Solving Strategies

• Read through the entire problem.
• Highlight the important information and key words that you need to solve the problem.
• Identify your variables.
• Write the equation or inequality.
• Write your answer in a complete sentence.
• Check or justify your answer.

I know it always helps too, if you have key words that help you to write the equation or inequality. Here are a few key words that we associate with inequalities! Keep these handy as a reference.

Inequality Key Words

• at least - means greater than or equal to
• no more than - means less than or equal to
• more than - means greater than
• less than - means less than

Ok... let's put it into action and look at our examples.

Example 1: Inequality Word Problems

Keith has $500 in a savings account at the beginning of the summer. He wants to have at least $200 in the account by the end of the summer. He withdraws $25 each week for food, clothes, and movie tickets.

• Write an inequality that represents Keith's situation.
• How many weeks can Keith withdraw money from his account? Justify your answer.

Step 1: Highlight the important information in this problem.

Note:  At least is a key word that notes that this problem must be written as an inequality.

Step 2 : Identify your variable. What don't you know? The question verifies that you don't know how many weeks.

Let w = the number of weeks

Step 3: Write your inequality.

500 - 25w > 200

I know you are saying, "How did you get that inequality?"

I know the "at least" part is tricky. You would probably think that at least means less than.

But... he wants the amount in his account to be at least $200 which means $200 or greater. So, we must use the greater than or equal to symbol.

Step 4 : Solve the inequality.

The number of weeks that Keith can withdraw money from his account is 12 weeks or less.

Step 5: Justify (prove your answer mathematically).

I'm going to prove that the largest number of weeks is 12 by substituting 12 into the inequality for w. You could also substitute any number less than 12.

Since 200 is equal to 200, my answer is correct. Any more than 12 weeks and his account balance would be less than $200.  Any number of weeks less than 12 and his account would stay above $200.

That wasn't too bad, was it? Let's take a look at another example.

Example 2: More Inequality Word Problems

Yellow Cab Taxi charges a $1.75 flat rat e in addition to $0.65 per mile . Katie has no more than $10 to spend on a ride.

• Write an inequality that represents Katie's situation.
• How many miles can Katie travel without exceeding her budget? Justify your answer.

Note:  No more than are key words that note that this problem must be written as an inequality.

Step 2 : Identify your variable. What don't you know? The question verifies that you don't know the number of miles Katie can travel.

Let m = the number of miles

Step 3:  Write the inequality.

0.65m + 1.75 < 10

Are you thinking, "How did you write that inequality?"

The "no more than" can also be tricky. "No more than" means that you can't have more than something, so that means you must have less than!

Step 4: Solve the inequality.

Since this is a real world problem and taxi's usually charge by the mile, we can say that Katie can travel 12 miles or less before reaching her limit of $10.

Are you ready to try some on your own now? Yes... of course you are! Click here to move onto the word problem practice problems.

Take a look at the questions that other students have submitted:

Quite a complicated problem about perimeter and area of a rectangle

This is a toughie.... a compound inequalities word problem

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Fraction word prob.

Fraction word problems

Here you will learn about fraction word problems, including solving math word problems within a real-world context involving adding fractions, subtracting fractions, multiplying fractions, and dividing fractions.

Students will first learn about fraction word problems as part of number and operations—fractions in 4 th grade.

What are fraction word problems?

Fraction word problems are math word problems involving fractions that require students to use problem-solving skills within the context of a real-world situation.

To solve a fraction word problem, you must understand the context of the word problem, what the unknown information is, and what operation is needed to solve it. Fraction word problems may require addition, subtraction, multiplication, or division of fractions.

After determining what operation is needed to solve the problem, you can apply the rules of adding, subtracting, multiplying, or dividing fractions to find the solution.

For example,

Natalie is baking 2 different batches of cookies. One batch needs \cfrac{3}{4} cup of sugar and the other batch needs \cfrac{2}{4} cup of sugar. How much sugar is needed to bake both batches of cookies?

You can follow these steps to solve the problem:

Step-by-step guide: Adding and subtracting fractions

Step-by-step guide: Adding fractions

Step-by-step guide: Subtracting fractions

Step-by-step guide: Multiplying and dividing fractions

Step-by-step guide: Multiplying fractions

Step-by-step guide: Dividing fractions

Common Core State Standards

How does this relate to 4 th grade math to 6 th grade math?

• Grade 4: Number and Operations—Fractions (4.NF.B.3d) Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem.
• Grade 4: Number and Operations—Fractions (4.NF.B.4c) Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. For example, if each person at a party will eat \cfrac{3}{8} of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie?
• Grade 5: Number and Operations—Fractions (5.NF.A.2) Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result \cfrac{2}{5}+\cfrac{1}{2}=\cfrac{3}{7} by observing that \cfrac{3}{7}<\cfrac{1}{2} .
• Grade 5: Number and Operations—Fractions (5.NF.B.6) Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
• Grade 5: Number and Operations—Fractions (5.NF.B.7c) Solve real world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions, e.g., by using visual fraction models and equations to represent the problem. For example, how much chocolate will each person get if 3 people share \cfrac{1}{2} \: lb of chocolate equally? How many \cfrac{1}{3} cup servings are in 2 cups of raisins?
• Grade 6: The Number System (6.NS.A.1) Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem. For example, create a story context for \cfrac{2}{3} \div \cfrac{4}{5} and use a visual fraction model to show the quotient; use the relationship between multiplication and division to explain that \cfrac{2}{3} \div \cfrac{4}{5}=\cfrac{8}{9} because \cfrac{3}{4} of \cfrac{8}{9} is \cfrac{2}{3}. (In general, \cfrac{a}{b} \div \cfrac{c}{d}=\cfrac{a d}{b c} \, ) How much chocolate will each person get if 3 people share \cfrac{1}{2} \: lb of chocolate equally? How many \cfrac{3}{4} cup servings are in \cfrac{2}{3} of a cup of yogurt? How wide is a rectangular strip of land with length \cfrac{3}{4} \: m and area \cfrac{1}{2} \: m^2?

[FREE] Fraction Operations Worksheet (Grade 4 to 6)

Use this quiz to check your grade 4 to 6 students’ understanding of fraction operations. 10+ questions with answers covering a range of 4th to 6th grade fraction operations topics to identify areas of strength and support!

How to solve fraction word problems

In order to solve fraction word problems:

Determine what operation is needed to solve.

Write an equation.

Solve the equation.

State your answer in a sentence.

Fraction word problem examples

Example 1: adding fractions (like denominators).

Julia ate \cfrac{3}{8} of a pizza and her brother ate \cfrac{2}{8} of the same pizza. How much of the pizza did they eat altogether?

The problem states how much pizza Julia ate and how much her brother ate. You need to find how much pizza Julia and her brother ate altogether , which means you need to add.

2 Write an equation.

3 Solve the equation.

To add fractions with like denominators, add the numerators and keep the denominators the same.

4 State your answer in a sentence.

The last step is to go back to the word problem and write a sentence to clearly say what the solution represents in the context of the problem.

Julia and her brother ate \cfrac{5}{8} of the pizza altogether.

Example 2: adding fractions (unlike denominators)

Tim ran \cfrac{5}{6} of a mile in the morning and \cfrac{1}{3} of a mile in the afternoon. How far did Tim run in total?

The problem states how far Tim ran in the morning and how far he ran in the afternoon. You need to find how far Tim ran in total , which means you need to add.

To add fractions with unlike denominators, first find a common denominator and then change the fractions accordingly before adding.

\cfrac{5}{6}+\cfrac{1}{3}= \, ?

The least common multiple of 6 and 3 is 6, so 6 can be the common denominator.

That means \cfrac{1}{3} will need to be changed so that its denominator is 6. To do this, multiply the numerator and the denominator by 2.

\cfrac{1 \times 2}{3 \times 2}=\cfrac{2}{6}

Now you can add the fractions and simplify the answer.

\cfrac{5}{6}+\cfrac{2}{6}=\cfrac{7}{6}=1 \cfrac{1}{6}

Tim ran a total of 1 \cfrac{1}{6} miles.

Example 3: subtracting fractions (like denominators)

Pia walked \cfrac{4}{7} of a mile to the park and \cfrac{3}{7} of a mile back home. How much farther did she walk to the park than back home?

The problem states how far Pia walked to the park and how far she walked home. Since you need to find the difference ( how much farther ) between the two distances, you need to subtract.

To subtract fractions with like denominators, subtract the numerators and keep the denominators the same.

\cfrac{4}{7}-\cfrac{3}{7}=\cfrac{1}{7}

Pia walked \cfrac{1}{7} of a mile farther to the park than back home.

Example 4: subtracting fractions (unlike denominators)

Henry bought \cfrac{7}{8} pound of beef from the grocery store. He used \cfrac{1}{3} of a pound of beef to make a hamburger. How much of the beef does he have left?

The problem states how much beef Henry started with and how much he used. Since you need to find how much he has left , you need to subtract.

To subtract fractions with unlike denominators, first find a common denominator and then change the fractions accordingly before subtracting.

\cfrac{7}{8}-\cfrac{1}{3}= \, ?

The least common multiple of 8 and 3 is 24, so 24 can be the common denominator.

That means both fractions will need to be changed so that their denominator is 24.

To do this, multiply the numerator and the denominator of each fraction by the same number so that it results in a denominator of 24. This will give you an equivalent fraction for each fraction in the problem.

\begin{aligned}&\cfrac{7 \times 3}{8 \times 3}=\cfrac{21}{24} \\\\ &\cfrac{1 \times 8}{3 \times 8}=\cfrac{8}{24} \end{aligned}

Now you can subtract the fractions.

\cfrac{21}{24}-\cfrac{8}{24}=\cfrac{13}{24}

Henry has \cfrac{13}{24} of a pound of beef left.

Example 5: multiplying fractions

Andre has \cfrac{3}{4} of a candy bar left. He gives \cfrac{1}{2} of the remaining bit of the candy bar to his sister. What fraction of the whole candy bar does Andre have left now?

It could be challenging to determine the operation needed for this problem; many students may automatically assume it is subtraction since you need to find how much of the candy bar is left.

However, since you know Andre started with a fraction of the candy bar and you need to find a fraction OF a fraction, you need to multiply.

The difference here is that Andre did NOT give his sister \cfrac{1}{2} of the candy bar, but he gave her \cfrac{1}{2} of \cfrac{3}{4} of a candy bar.

To solve the word problem, you can ask, “What is \cfrac{1}{2} of \cfrac{3}{4}? ” and set up the equation accordingly. Think of the multiplication sign as meaning “of.”

\cfrac{1}{2} \times \cfrac{3}{4}= \, ?

To multiply fractions, multiply the numerators and multiply the denominators.

\cfrac{1}{2} \times \cfrac{3}{4}=\cfrac{3}{8}

Andre gave \cfrac{1}{2} of \cfrac{3}{4} of a candy bar to his sister, which means he has \cfrac{1}{2} of \cfrac{3}{4} left. Therefore, Andre has \cfrac{3}{8} of the whole candy bar left.

Example 6: dividing fractions

Nia has \cfrac{7}{8} cup of trail mix. How many \cfrac{1}{4} cup servings can she make?

The problem states the total amount of trail mix Nia has and asks how many servings can be made from it.

To solve, you need to divide the total amount of trail mix (which is \cfrac{7}{8} cup) by the amount in each serving ( \cfrac{1}{4} cup) to find out how many servings she can make.

To divide fractions, multiply the dividend by the reciprocal of the divisor.

\begin{aligned}& \cfrac{7}{8} \div \cfrac{1}{4}= \, ? \\\\ & \downarrow \downarrow \downarrow \\\\ &\cfrac{7}{8} \times \cfrac{4}{1}=\cfrac{28}{8} \end{aligned}

You can simplify \cfrac{28}{8} to \cfrac{7}{2} and then 3 \cfrac{1}{2}.

Nia can make 3 \cfrac{1}{2} cup servings.

Teaching tips for fraction word problems

• Encourage students to look for key words to help determine the operation needed to solve the problem. For example, subtracting fractions word problems might ask students to find “how much is left” or “how much more” one fraction is than another.
• Provide students with an answer key to word problem worksheets to allow them to obtain immediate feedback on their solutions. Encourage students to attempt the problems independently first, then check their answers against the key to identify any mistakes and learn from them. This helps reinforce problem-solving skills and confidence.
• Be sure to incorporate real-world situations into your math lessons. Doing so allows students to better understand the relevance of fractions in everyday life.
• As students progress and build a strong foundational understanding of one-step fraction word problems, provide them with multi-step word problems that involve more than one operation to solve.
• Take note that students will not divide a fraction by a fraction as shown above until 6 th grade (middle school), but they will divide a unit fraction by a whole number and a whole number by a fraction in 5 th grade (elementary school), where the same mathematical rules apply to solving.
• There are many alternatives you can use in place of printable math worksheets to make practicing fraction word problems more engaging. Some examples are online math games and digital workbooks.

Easy mistakes to make

• Misinterpreting the problem Misreading or misunderstanding the word problem can lead to solving for the wrong quantity or using the wrong operation.
• Not finding common denominators When adding or subtracting fractions with unlike denominators, students may forget to find a common denominator, leading to an incorrect answer.
• Forgetting to simplify Unless a problem specifically says not to simplify, fractional answers should always be written in simplest form.

Related fractions operations lessons

• Fractions operations
• Multiplicative inverse
• Reciprocal math
• Fractions as divisions

Practice fraction word problem questions

1. Malia spent \cfrac{5}{6} of an hour studying for a math test. Then she spent \cfrac{1}{3} of an hour reading. How much longer did she spend studying for her math test than reading?

Malia spent \cfrac{1}{2} of an hour longer studying for her math test than reading.

Malia spent \cfrac{5}{18} of an hour longer studying for her math test than reading.

Malia spent \cfrac{1}{2} of an hour longer reading than studying for her math test.

Malia spent 1 \cfrac{1}{6} of an hour longer studying for her math test than reading.

To find the difference between the amount of time Malia spent studying for her math test than reading, you need to subtract. Since the fractions have unlike denominators, you need to find a common denominator first.

You can use 6 as the common denominator, so \cfrac{1}{3} becomes \cfrac{3}{6}. Then you can subtract.

\cfrac{3}{6} can then be simplified to \cfrac{1}{2}.

Finally, you need to choose the answer that correctly answers the question within the context of the situation. Therefore, the correct answer is “Malia spent \cfrac{1}{2} of an hour longer studying for her math test than reading.”

2. A square garden is \cfrac{3}{4} of a meter wide and \cfrac{8}{9} of a meter long. What is its area?

The area of the garden is 1\cfrac{23}{36} square meters.

The area of the garden is \cfrac{27}{32} square meters.

The area of the garden is \cfrac{2}{3} square meters.

The perimeter of the garden is \cfrac{2}{3} meters.

To find the area of a square, you multiply the length and width. So to solve, you multiply the fractional lengths by mulitplying the numerators and multiplying the denominators.

\cfrac{24}{36} can be simplified to \cfrac{2}{3}. 

Therefore, the correct answer is “The area of the garden is \cfrac{2}{3} square meters.”

3. Zoe ate \cfrac{3}{8} of a small cake. Liam ate \cfrac{1}{8} of the same cake. How much more of the cake did Zoe eat than Liam?

Zoe ate \cfrac{3}{64} more of the cake than Liam.

Zoe ate \cfrac{1}{4} more of the cake than Liam.

Zoe ate \cfrac{1}{8} more of the cake than Liam.

Liam ate \cfrac{1}{4} more of the cake than Zoe.

To find how much more cake Zoe ate than Liam, you subtract. Since the fractions have the same denominator, you subtract the numerators and keep the denominator the same.

\cfrac{2}{8} can be simplified to \cfrac{1}{4}. 

Therefore, the correct answer is “Zoe ate \cfrac{1}{4} more of the cake than Liam.”

4. Lila poured \cfrac{11}{12} cup of pineapple and \cfrac{2}{3} cup of mango juice in a bottle. How many cups of juice did she pour into the bottle altogether?

Lila poured 1 \cfrac{7}{12} cups of juice in the bottle altogether.

Lila poured \cfrac{1}{4} cups of juice in the bottle altogether.

Lila poured \cfrac{11}{18} cups of juice in the bottle altogether.

Lila poured 1 \cfrac{3}{8} cups of juice in the bottle altogether.

To find the total amount of juice that Lila poured into the bottle, you need to add. Since the fractions have unlike denominators, you need to find a common denominator first.

You can use 12 as the common denominator, so \cfrac{2}{3} becomes \cfrac{8}{12}.  Then you can add.

\cfrac{19}{12} can be simplified to 1 \cfrac{7}{12}. 

Therefore, the correct answer is “Lila poured 1 \cfrac{7}{12} cups of juice in the bottle altogether.”

5. Killian used \cfrac{9}{10} of a gallon of paint to paint his living room and \cfrac{7}{10} of a gallon to paint his bedroom. How much paint did Killian use in all?

Killian used \cfrac{2}{10} gallons of paint in all.

Killian used \cfrac{1}{5} gallons of paint in all.

Killian used \cfrac{63}{100} gallons of paint in all.

Killian used 1 \cfrac{3}{5} gallons of paint in all.

To find the total amount of paint Killian used, you add the amount he used for the living room and the amount he used for the kitchen. Since the fractions have the same denominator, you add the numerators and keep the denominators the same.

\cfrac{16}{10} can be simplified to 1 \cfrac{6}{10} and then further simplified to 1 \cfrac{3}{5}.

Therefore, the correct answer is “Killian used 1 \cfrac{3}{5} gallons of paint in all.”

6. Evan pours \cfrac{4}{5} of a liter of orange juice evenly among some cups.

He put \cfrac{1}{10} of a liter into each cup. How many cups did Evan fill?

Evan filled \cfrac{2}{25} cups.

Evan filled 8 cups.

Evan filled \cfrac{9}{10} cups.

Evan filled 7 cups.

To find the number of cups Evan filled, you need to divide the total amount of orange juice by the amount being poured into each cup. To divide fractions, you mulitply the first fraction (the dividend) by the reciprocal of the second fraction (the divisor).

\cfrac{40}{5} can be simplifed to 8.

Therefore, the correct answer is “Evan filled 8 cups.”

Fraction word problems FAQs

Fraction word problems are math word problems involving fractions that require students to use problem-solving skills within the context of a real-world situation. Fraction word problems may involve addition, subtraction, multiplication, or division of fractions.

To solve fraction word problems, first you need to determine the operation. Then you can write an equation and solve the equation based on the arithmetic rules for that operation.

Fraction word problems and decimal word problems are similar because they both involve solving math problems within real-world contexts. Both types of problems require understanding the problem, determining the operation needed to solve it (addition, subtraction, multiplication, division), and solving it based on the arithmetic rules for that operation.

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