Word Problems on Sets

Word problems on sets are solved here to get the basic ideas how to use the  properties of union and intersection of sets.

Solved basic word problems on sets:

1. Let A and B be two finite sets such that n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, find n(A ∩ B).

Solution:  Using the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B).  then n(A ∩ B) = n(A) + n(B) - n(A ∪ B)                       = 20 + 28 - 36                       = 48 - 36                       = 12 

2.  If n(A - B) = 18, n(A ∪ B) = 70 and n(A ∩ B) = 25, then find n(B).

Solution:  Using the formula n(A∪B) = n(A - B) + n(A ∩ B) + n(B - A)                                   70 = 18 + 25 + n(B - A)                                   70 = 43 + n(B - A)                           n(B - A) = 70 - 43                           n(B - A) = 27  Now n(B) = n(A ∩ B) + n(B - A)                 = 25 + 27                 = 52 

Different types on word problems on sets:

3.  In a group of 60 people, 27 like cold drinks and 42 like hot drinks and each person likes at least one of the two drinks. How many like both coffee and tea? 

Solution:  Let A = Set of people who like cold drinks.       B = Set of people who like hot drinks.  Given   (A ∪ B) = 60            n(A) = 27       n(B) = 42 then;

n(A ∩ B) = n(A) + n(B) - n(A ∪ B)              = 27 + 42 - 60              = 69 - 60 = 9              = 9  Therefore, 9 people like both tea and coffee. 

4.  There are 35 students in art class and 57 students in dance class. Find the number of students who are either in art class or in dance class.

•  When two classes meet at different hours and 12 students are enrolled in both activities.  •  When two classes meet at the same hour.  Solution:  n(A) = 35,       n(B) = 57,       n(A ∩ B) = 12  (Let A be the set of students in art class.  B be the set of students in dance class.)  (i) When 2 classes meet at different hours n(A ∪ B) = n(A) + n(B) - n(A ∩ B)                                                                            = 35 + 57 - 12                                                                            = 92 - 12                                                                            = 80  (ii) When two classes meet at the same hour, A∩B = ∅ n (A ∪ B) = n(A) + n(B) - n(A ∩ B)                                                                                                = n(A) + n(B)                                                                                                = 35 + 57                                                                                                = 92

Further concept to solve word problems on sets:

5. In a group of 100 persons, 72 people can speak English and 43 can speak French. How many can speak English only? How many can speak French only and how many can speak both English and French?

Solution: Let A be the set of people who speak English. B be the set of people who speak French. A - B be the set of people who speak English and not French. B - A be the set of people who speak French and not English. A ∩ B be the set of people who speak both French and English. Given, n(A) = 72       n(B) = 43       n(A ∪ B) = 100 Now, n(A ∩ B) = n(A) + n(B) - n(A ∪ B)                      = 72 + 43 - 100                      = 115 - 100                      = 15 Therefore, Number of persons who speak both French and English = 15 n(A) = n(A - B) + n(A ∩ B) ⇒ n(A - B) = n(A) - n(A ∩ B)                 = 72 - 15                 = 57 and n(B - A) = n(B) - n(A ∩ B)                    = 43 - 15                    = 28 Therefore, Number of people speaking English only = 57 Number of people speaking French only = 28

Word problems on sets using the different properties (Union & Intersection):

6. In a competition, a school awarded medals in different categories. 36 medals in dance, 12 medals in dramatics and 18 medals in music. If these medals went to a total of 45 persons and only 4 persons got medals in all the three categories, how many received medals in exactly two of these categories?

Solution: Let A = set of persons who got medals in dance. B = set of persons who got medals in dramatics. C = set of persons who got medals in music. Given, n(A) = 36                              n(B) = 12       n(C) = 18 n(A ∪ B ∪ C) = 45       n(A ∩ B ∩ C) = 4 We know that number of elements belonging to exactly two of the three sets A, B, C = n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3n(A ∩ B ∩ C) = n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3 × 4       ……..(i) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C) Therefore, n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C) From (i) required number = n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C) - 12 = 36 + 12 + 18 + 4 - 45 - 12 = 70 - 57 = 13

Apply set operations to solve the word problems on sets:

7. Each student in a class of 40 plays at least one indoor game chess, carrom and scrabble. 18 play chess, 20 play scrabble and 27 play carrom. 7 play chess and scrabble, 12 play scrabble and carrom and 4 play chess, carrom and scrabble. Find the number of students who play (i) chess and carrom. (ii) chess, carrom but not scrabble.

Solution: Let A be the set of students who play chess B be the set of students who play scrabble C be the set of students who play carrom Therefore, We are given n(A ∪ B ∪ C) = 40, n(A) = 18,         n(B) = 20         n(C) = 27, n(A ∩ B) = 7,     n(C ∩ B) = 12    n(A ∩ B ∩ C) = 4 We have n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C) Therefore, 40 = 18 + 20 + 27 - 7 - 12 - n(C ∩ A) + 4 40 = 69 – 19 - n(C ∩ A) 40 = 50 - n(C ∩ A) n(C ∩ A) = 50 - 40 n(C ∩ A) = 10 Therefore, Number of students who play chess and carrom are 10. Also, number of students who play chess, carrom and not scrabble. = n(C ∩ A) - n(A ∩ B ∩ C) = 10 – 4 = 6

Therefore, we learned how to solve different types of word problems on sets without using Venn diagram.

● Set Theory

● Sets Theory

● Representation of a Set

● Types of Sets

● Finite Sets and Infinite Sets

● Power Set

● Problems on Union of Sets

● Problems on Intersection of Sets

● Difference of two Sets

● Complement of a Set

● Problems on Complement of a Set

● Problems on Operation on Sets

● Word Problems on Sets

● Venn Diagrams in Different Situations

● Relationship in Sets using Venn Diagram

● Union of Sets using Venn Diagram

● Intersection of Sets using Venn Diagram

● Disjoint of Sets using Venn Diagram

● Difference of Sets using Venn Diagram

● Examples on Venn Diagram

8th Grade Math Practice From Word Problems on Sets to HOME PAGE

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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 .

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

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I wrote ____ in the place of total marbles since that is what the problem is asking for (the unknown).

All of this may look oversimplified, but helping children to see the underlying relationship between the quantities is important. Consider now this problem:

Example: Jenny and Kenny together have 37 marbles, and Kenny has 15. How many does Jenny have? Many teachers might try to explain this as a subtraction problem, but in the most fundamental level it is about addition! It still talks about two people having certain amount of marbles together . The relationship between the quantities is the SAME as above, so we still need to write an addition equation. Relationship: Jenny's marbles  +  Kenny's marbles  =  Total marbles Equation: _____ + 15 = 37 Then, we can solve the equation ____ + 15 = 37  by subtracting. Using this kind of approach in the elementary grades will help children to set up equations in algebra story problems later.
Example : Jenny, Kenny, and Penny together have 51 marbles. Kenny has double as many marbles as Jenny has, and Penny has 12. How many does Jenny have? The relationship between the quantities is the same, so it is solved the same way: by writing an addition equation. However, we need to denote the number of Jenny's and Kenny's marbles with something. Jenny's marbles are unknown, so we can denote that with the variable n . Then Kenny has 2 n marbles. Relationship: Jenny's marbles  +  Kenny's marbles  +  Penny's marbles  =  Total marbles Equation: n + 2 n  +  12 = 51
Example: Jane is on page 79 of her book. The book has 254 pages. How many pages does she still have to read? This time the word " still " clues us in to an additive relationship where one of the addends is missing. You can initially write an empty line for what is not known, and later replace that with a variable. pages already read  +  pages still to read  =  pages total + = This equation is of course is then solved by subtraction, but it is better if you view it as an addition situation and write an addition equation for it.
Example:   The number of hours that were left in the day was one-third of the number of hours already passed. How many hours were left in the day? (From Grade 5 word problems for kids ) Can you see the general principle governing this problem?  It talks about the hours of the day where some hours already passed and some hours left. This, of course, points to addition once again: we have one part of the day, another part, and a total. The only quantity we know is the total hours for the day. We don't know the hours already passed nor the hours left, so initially you can use two empty lines in the equation that shows the basic relationship between the quantities: hours already passed  +  hours left = total hours = Then, the information in the first sentence gives us another relationship: "The number of hours that were left in the day was one-third of the number of hours already passed." We don't know the amount of hours passed nor the hours left. So let's use the variable p for the hours passed. Then we can write an expression involving p for the hours left, because "hours left is one-third of the hours passed," or hours left  =  1/3 p Then writing 1/3 p for the "hours left" in the first equation will give us: hours already passed + hours left = total hours p + 1/3 p = 24 This can be solved using basic algebra or by guess & check.

Subtraction word problems

One situation that indicates subtraction is difference or  how many/much more . However, the presence of the word "more" can indicate either an addition or subtraction, so be careful.

Example:   Ted read 17 pages today, and Fred read 28. How many more pages did Fred read? The solution is of course 28 − 17 = 11, but it's not enough to simply announce that – children need also to understand that difference is the result of a subtraction and tells the answer to how many more . Relationship:    Pages Fred read  −  pages Ted read = difference Equation: 28  −  17 = __ Example:   Greg has 17 more marbles than Jack. Jack has 15. How many does Greg have? Here the word more has a different meaning. This problem is not about the difference. The question asks how many does Greg have – not what is the difference in the amounts of marbles. It simply states Greg has 17 more compared to Jack, so here the word more simply indicates addition: Greg has as many as Jack AND 17 more, so Greg has 15 + 17 marbles.
Example: The mass of the Great Pyramid is 557t greater than that of the Leaning Tower of Pisa. Stone Henge has a mass of 2695t which is 95t less than the Leaning Tower of Pisa. There once was a Greater Pyramid which had a mass twice that of the Great Pyramid. What was the mass of the Greater Pyramid? (From Grade 5 word problems for kids ) Each of the first three sentences give information that can be translated into an equation. The question is not about how many more so it's not about difference. One thing being greater than another implies you add. One thing being less than another implies you subtract. And one thing being twice something indicates multiplying by 2. When I read this problem, I could immediately see that I could write equations from the different sentences in the problem, but I couldn't see the answer right away. I figured that after writing the equations I would see some way forwad; probably one equation is solved and gives an answer to another equation. The first sentence says, "The mass of the Great Pyramid is 557t greater than that of the Leaning Tower of Pisa". What are the quantities and the relationship between them here? mass of Great Pyramid = mass of the Leaning Tower of Pisa + 557t The second sentence says "Stonehenge has a mass of 2695t which is 95t less than the Leaning Tower of Pisa."  Here it gives you a relationship similar to the one above, and it actually spells out the mass of Stonehenge. It's like two separate pieces of information: "Stonehenge weighs 95t less than the tower.  Stonehenge weighs 2695t."  Less means you subtract. If you have trouble deciding which is subtracted from which, you can think in your mind which is heavier: Stonehenge or the tower?   either      mass of Stonehenge = mass of tower − 95t or mass of tower = mass of Stonehenge − 95t Now since the mass of Stonehenge is given, you can solve this equation, and from that knowledge you can solve the first equation, and from that go on to the mass of the " Greater Pyramid ".

If the teacher just jumps directly to the number sentences when solving word problems, then the students won't see the step that happens in the mind before that. The quantities and the relationship between them have to be made clear and written down before fiddling with the actual numbers. Finding this relationship should be the most important part of the word problems.  One could even omit the actual calculations and concentrate just finding the quantities and relationships.

Problem of Helen's hair length

Problem.   Helen has 2 inches of hair cut off each time she goes to the hair salon. If h equals the length of hair before she cuts it and c equals the length of hair after she cuts it, which equation would you use to find the length of Helen's hair after she visit the hair salon? a. h = 2 − c      c. c = h − 2 b. c = 2 − h      d. h = c − 2

Solution.   Ignoring the letters c and h for now, what are the quantities?  What principle or relationship is there between them? Which possibility of the ones listed below is right?  Which do you take away from which?

SIMPLE, isn't it??  In the original problem, the equations are given with the help of h and c instead of the long phrases "hair length before cutting" and "hair length after cutting". You can substitute the c , h , and 2 into the relationships above, and then match the equations (1) - (4) with the equations (a) through (d).

Helping students to write the algebraic equations

One idea that came to mind is to go through the examples above, and more, based on the typical word problems in the math books, and then turn the whole thing around and have students do exercises such as:

money earned − money spent on this - money spent on that = money left

original price − discount percent x original price = discounted price

money earned each month − expenses/taxes each month = money to use each month AND money to use each month × number of months = money to use over a period of time

speed × time = distance AND distance from A to B + distance from B to C = distance from A to C

Why are math word problems SO difficult for elementary school children? Hint: it has to do with a "recipe" that many math lessons follow.

The do's and don'ts of teaching problem solving in math General advice on how you can teach problem solving in elementary, middle, and high school math.

How I Teach Word Problems by Andre Toom (PDF) This article is written by a Russian who immigrated to US and noticed how COLLEGE LEVEL students have difficulties even with the simplest word problems! He describes his ideas on how to fill in the gap formed when students haven't learned how to solve word problems in earlier education.

A list of websites focusing on word problems and problem solving Use these sites to find good word problems to solve. Most are free!

When solving word problems, students must first decide what quantity represents x and then must write all the other quuantities in terms of x. I teach the students to set up arrows according to the language in the problem. All arrows point to x. Example. Harry had 10 less toys than Mark. Sue has twice as many toys as Harry. Set up arrows: Sue--- Harry---Mark Therefore Mark is x, Harry is x-10 and Sue is 2(x-10). Students find this very helpful. Sandy Denny
My idea is the math teacher might teach and understand the students at the same time and everyone would have a sense of humor. So I think she/he will know if the students are listening or not, when after the class, talk to the student and ask what is wrong. Don't hurt the student's feelings. lorence

Practice makes perfect. Practice math at IXL.com

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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 69,129 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

how do you solve a word problem in a set

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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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Praxis Core Math

Course: praxis core math   >   unit 1.

  • Rational number operations | Lesson
  • Rational number operations | Worked example
  • Ratios and proportions | Lesson
  • Ratios and proportions | Worked example
  • Percentages | Lesson
  • Percentages | Worked example
  • Rates | Lesson
  • Rates | Worked example
  • Naming and ordering numbers | Lesson
  • Naming and ordering numbers | Worked example
  • Number concepts | Lesson
  • Number concepts | Worked example
  • Counterexamples | Lesson
  • Counterexamples | Worked example

Pre-algebra word problems | Lesson

  • Pre-algebra word problems | Worked example
  • Unit reasoning | Lesson
  • Unit reasoning | Worked example

What are pre-algebra word problems?

What skills are tested.

  • Solving real world problems by identifying relevant numbers and operations
  • Identifying contexts where the answer must be rounded up or down

How do we solve pre-algebra word problems?

  • Identify the quantity to be calculated.
  • Determine the operations and numbers needed to calculate the desired quantity.
  • Write and evaluate the expressions to get the desired quantity.
  • If the context requires rounding, round the answer.
  • (Choice A)   7 ‍   A 7 ‍  
  • (Choice B)   6 ‍   B 6 ‍  
  • (Choice C)   5 ‍   C 5 ‍  
  • (Choice D)   4 ‍   D 4 ‍  
  • (Choice E)   3 ‍   E 3 ‍  
  • (Choice A)   82.2 ‍   pounds A 82.2 ‍   pounds
  • (Choice B)   161.4 ‍   pounds B 161.4 ‍   pounds
  • (Choice C)   211.3 ‍   pounds C 211.3 ‍   pounds
  • (Choice D)   272.2 ‍   pounds D 272.2 ‍   pounds
  • (Choice E)   289.8 ‍   pounds E 289.8 ‍   pounds
  • 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)   5 ‍   A 5 ‍  
  • (Choice B)   10 ‍   B 10 ‍  
  • (Choice C)   24 ‍   C 24 ‍  
  • (Choice D)   25 ‍   D 25 ‍  
  • (Choice E)   30 ‍   E 30 ‍  

Things to remember

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Introduction to Sets

Forget everything you know about numbers.

In fact, forget you even know what a number is.

This is where mathematics starts.

Instead of math with numbers, we will now think about math with "things".

What is a set? Well, simply put, it's a collection .

First we specify a common property among "things" (we define this word later) and then we gather up all the "things" that have this common property.

For example, the items you wear: hat, shirt, jacket, pants, and so on.

I'm sure you could come up with at least a hundred.

This is known as a set .

So it is just things grouped together with a certain property in common.

There is a fairly simple notation for sets. We simply list each element (or "member") separated by a comma, and then put some curly brackets around the whole thing:

The curly brackets { } are sometimes called "set brackets" or "braces".

This is the notation for the two previous examples:

{socks, shoes, watches, shirts, ...} {index, middle, ring, pinky}

Notice how the first example has the "..." (three dots together).

The three dots ... are called an ellipsis, and mean "continue on".

So that means the first example continues on ... for infinity.

(OK, there isn't really an infinite amount of things you could wear, but I'm not entirely sure about that! After an hour of thinking of different things, I'm still not sure. So let's just say it is infinite for this example.)

  • The first set {socks, shoes, watches, shirts, ...} we call an infinite set ,
  • the second set {index, middle, ring, pinky} we call a finite set .

But sometimes the "..." can be used in the middle to save writing long lists:

Example: the set of letters:

{a, b, c, ..., x, y, z}

In this case it is a finite set (there are only 26 letters, right?)

Numerical Sets

So what does this have to do with mathematics? When we define a set, all we have to specify is a common characteristic. Who says we can't do so with numbers?

And so on. We can come up with all different types of sets.

We can also define a set by its properties, such as {x|x>0} which means "the set of all x's, such that x is greater than 0", see Set-Builder Notation to learn more.

And we can have sets of numbers that have no common property, they are just defined that way. For example:

Are all sets that I just randomly banged on my keyboard to produce.

Why are Sets Important?

Sets are the fundamental property of mathematics. Now as a word of warning, sets, by themselves, seem pretty pointless. But it's only when we apply sets in different situations do they become the powerful building block of mathematics that they are.

Math can get amazingly complicated quite fast. Graph Theory, Abstract Algebra, Real Analysis, Complex Analysis, Linear Algebra, Number Theory, and the list goes on. But there is one thing that all of these share in common: Sets .

Universal Set

Some more notation.

Now you don't have to listen to the standard, you can use something like m to represent a set without breaking any mathematical laws (watch out, you can get π years in math jail for dividing by 0), but this notation is pretty nice and easy to follow, so why not?

element symbol

Two sets are equal if they have precisely the same members. Now, at first glance they may not seem equal, so we may have to examine them closely!

Example: Are A and B equal where:

  • A is the set whose members are the first four positive whole numbers
  • B = {4, 2, 1, 3}

Let's check. They both contain 1. They both contain 2. And 3, And 4. And we have checked every element of both sets, so: Yes, they are equal!

And the equals sign (=) is used to show equality, so we write:

Example: Are these sets equal?

  • A is {1, 2, 3}
  • B is {3, 1, 2}

Yes, they are equal!

They both contain exactly the members 1, 2 and 3.

It doesn't matter where each member appears, so long as it is there.

When we define a set, if we take pieces of that set, we can form what is called a subset .

Example: the set {1, 2, 3, 4, 5}

A subset of this is {1, 2, 3}. Another subset is {3, 4} or even another is {1}, etc.

But {1, 6} is not a subset, since it has an element (6) which is not in the parent set.

In general:

A is a subset of B if and only if every element of A is in B.

So let's use this definition in some examples.

Example: Is A a subset of B, where A = {1, 3, 4} and B = {1, 4, 3, 2}?

1 is in A, and 1 is in B as well. So far so good.

3 is in A and 3 is also in B.

4 is in A, and 4 is in B.

That's all the elements of A, and every single one is in B, so we're done.

Yes, A is a subset of B

Note that 2 is in B, but 2 is not in A. But remember, that doesn't matter, we only look at the elements in A.

Let's try a harder example.

Example: Let A be all multiples of 4 and B be all multiples of 2 . Is A a subset of B? And is B a subset of A?

Well, we can't check every element in these sets, because they have an infinite number of elements. So we need to get an idea of what the elements look like in each, and then compare them.

The sets are:

  • A = {..., −8, −4, 0, 4, 8, ...}
  • B = {..., −8, −6, −4, −2, 0, 2, 4, 6, 8, ...}

By pairing off members of the two sets, we can see that every member of A is also a member of B, but not every member of B is a member of A:

A is a subset of B, but B is not a subset of A

Proper Subsets

If we look at the defintion of subsets and let our mind wander a bit, we come to a weird conclusion.

Let A be a set. Is every element of A in A ?

Well, umm, yes of course , right?

So that means that A is a subset of A . It is a subset of itself!

This doesn't seem very proper , does it? If we want our subsets to be proper we introduce (what else but) proper subsets :

A is a proper subset of B if and only if every element of A is also in B, and there exists at least one element in B that is not in A.

This little piece at the end is there to make sure that A is not a proper subset of itself: we say that B must have at least one extra element.

{1, 2, 3} is a subset of {1, 2, 3}, but is not a proper subset of {1, 2, 3}.

{1, 2, 3} is a proper subset of {1, 2, 3, 4} because the element 4 is not in the first set.

Notice that when A is a proper subset of B then it is also a subset of B.

Even More Notation

subset symbol

Empty (or Null) Set

This is probably the weirdest thing about sets.

keys on guitar

As an example, think of the set of piano keys on a guitar.

"But wait!" you say, "There are no piano keys on a guitar!"

And right you are. It is a set with no elements .

This is known as the Empty Set (or Null Set).There aren't any elements in it. Not one. Zero.

null set

Or by {} (a set with no elements)

Some other examples of the empty set are the set of countries south of the south pole .

So what's so weird about the empty set? Well, that part comes next.

Empty Set and Subsets

So let's go back to our definition of subsets. We have a set A. We won't define it any more than that, it could be any set. Is the empty set a subset of A?

Going back to our definition of subsets, if every element in the empty set is also in A, then the empty set is a subset of A . But what if we have no elements?

It takes an introduction to logic to understand this, but this statement is one that is "vacuously" or "trivially" true.

A good way to think about it is: we can't find any elements in the empty set that aren't in A , so it must be that all elements in the empty set are in A.

So the answer to the posed question is a resounding yes .

The empty set is a subset of every set, including the empty set itself.

No, not the order of the elements. In sets it does not matter what order the elements are in .

Example: {1,2,3,4} is the same set as {3,1,4,2}

When we say order in sets we mean the size of the set .

Another (better) name for this is cardinality .

A finite set has finite order (or cardinality). An infinite set has infinite order (or cardinality).

For finite sets the order (or cardinality) is the number of elements .

Example: {10, 20, 30, 40} has an order of 4.

For infinite sets, all we can say is that the order is infinite. Oddly enough, we can say with sets that some infinities are larger than others, but this is a more advanced topic in sets.

Arg! Not more notation!

Nah, just kidding. No more notation.

PrepScholar

Choose Your Test

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

feature_typeqriter

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

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

What Are Word Problems?

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

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

Translating Word Problems Into Equations or Drawings

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

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

body_ACT_word_problem_10

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

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

$120 + 25x$

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

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

$120 + 25x = 60 + 35x$

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

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

body_rosetta_stone

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

Typical ACT Word Problems

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

Word Problem Type 1: Setting Up an Equation

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

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

body_ACT_word_problem_2

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

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

$20,000 - 7,000$

$13,000$ per boat.

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

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

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

Word Problem Type 2: Solving for Your Information

Other than the few set-up word questions you’ll see, the rest of your ACT word problem questions will fall into this category. For these questions, you must both set up your equation and solve for a specific piece of information.

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

body_ACT_word_problem_1

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

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

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

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

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

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

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

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

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

Geometry questions will be presented as word problems typically because the test-makers felt the problem would be too easy to solve had you been given a diagram.

body_ACT_word_problem_5

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

body_parallelogram_ex_1

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

body_parallelogram_ex_2

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

$72 - 12 - 12$

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

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

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

body_strategy-7

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

ACT Math Strategies for Your Word Problems

Though you’ll see word problems on a myriad of different types of ACT math topics, there are still a few techniques you can apply to solve your word problems as a whole.

#1: Draw It Out

Whether your problem is a geometry problem or an algebra problem, sometimes making a quick sketch of the scene can help you understand what, exactly, you're working with . For instance, let's look at how a picture can help you solve a ratio/division problem :

body_ACT_word_problem_13

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

body_sandwich_1

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

body_sandwich_2-1

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

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

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

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

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

#2: Memorize Important Terms

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

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

#3: Underline and Write Out the Key Information

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

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

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

Little is more frustrating than solving for the wrong variable or writing in your given values in the wrong places. And yet this is entirely too easy to do when working with word problems.

Make sure you pay strict attention to exactly what you’re meant to be solving for and exactly what pieces of information go where. Are you looking for the area or the perimeter? The value of $x$ or $x + y$? Better to make sure before you start what you’re supposed to find than realize two minutes down the line that you have to solve the problem all over again.

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

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

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

body_thinking

All set? Time to shine!

Test Your Knowledge

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

body_ACT_word_problem_3

Answers: K, C, A, E

Answer Explanations:

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

body_Ratios

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

$2/5 = 5/x$

Our final answer is K , 12.5.

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

Solving Method 1: Algebra

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

$x + 0.25x$

Which gives us:

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

$1.25x - (1.25x * 0.2)$

$1.25x - 0.25x$

This leaves us with:

$1x$ or 100% of our original number. 

Our final answer is C, 100%. 

Solving Method 2: Plugging in Numbers

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

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

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

$100 + (0.25)100$

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

$125 - (0.2)125$

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

Again, our final answer is C, 100%. 

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

body_vertex_1

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

body_vertex_2

Our possible vertexes are:

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

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

Our final answer is A, (6, 0). 

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

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

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

Our final answer is E, 84.5. 

body_clapping

The Take-Aways

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

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

What’s Next?  

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

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

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

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

  

Want to improve your ACT score by 4 points? 

Check out our best-in-class online ACT prep program . We guarantee your money back if you don't improve your ACT score by 4 points or more.

Our program is entirely online, and it customizes what you study to your strengths and weaknesses . If you liked this Math lesson, you'll love our program.  Along with more detailed lessons, you'll get thousands of practice problems organized by individual skills so you learn most effectively. We'll also give you a step-by-step program to follow so you'll never be confused about what to study next.

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

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"Mixture" Word Problems

Explanation Examples

What are "mixture" word problems?

Mixture word problems are exercises which involve creating a mixture from two or more different things, and then determining some quantity (such as percentage, price, number of liters, etc) of the resulting mixture. There will always be a "rate" of some sort, such as miles per hour or cost per pound.

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Mixture Word Problems

Here is an example:

Your school is holding a family-friendly event this weekend. Students have been pre-selling tickets to the event; adult tickets are $5.00, and child tickets (for kids six years old and under) are $2.50 . From past experience, you expect about 13,000 people to attend the event.

You consult with your student ticket-sellers, and discover that they have not been keeping track of how many child tickets they have sold. The tickets are identical, until the ticket-seller punches a hole in the ticket, indicating that it is a child ticket.

But they don't remember how many holes they've punched. They only know that they've sold 548 tickets for $2460 . How much revenue from each of child and adult tickets can you expect?

To solve this, we need to figure out the ratio of tickets that have already been sold. If we work methodically, we can find the answer.

Let A stand for the number of adult tickets pre-sold. Since a total of 548 tickets have been sold, then the number of children tickets pre-sold thus far must be 548 − A .

this construction is important! When you've got a variable that stands for part of whatever it is that you're working with, then the amount that is left for the other part of whatever you're working with is found by subtracting the variable from the total. That is, (the total amount) less (the amount that is being represented by the variable) is (the amount that is left for ther other amount). This "how much is left" construction is something you will need to understand and use.

Since each adult ticket cost $5.00 , then 5A stands for the revenue brought in from the adult tickets pre-sold; likewise, 2.5(548 − A) stands for the revenue brought in from the child tickets. (Note: The per-ticket cost is the "rate" for this exercise.)

Organizing this information in a grid, we get:

From the last column, we get (total $ from the adult tickets) plus (total $ from the child tickets) is (the total $ so far), or, as an equation:

5A + 2.5((548 − A) = 2,460

5A + 1,370 − 2.5A = 2,460

1,370 + 2.5A = 2,460

2.5A = 1,090

A = (1,090)/(2.5) = 436

So 436 adult tickets were pre-sold, so the number of child tickets pre-sold is:

C = 548 − 436 = 112

So 112 child tickets were pre-sold.

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We expect about 13,000 people in total to attend this event. We have a ratio of adult pre-sold tickets to total pre-sold tickets. Assuming that the pre-sold rate (or ratio) is representative of the total number of adult tickets, we can set up a proportion (of pre-sold adult tickets to total pre-sold tickets), using a variable for the unknown total number of adult tickets expected to be sold:

436/548 = x /13,000

[(436)(13,000)]/548 = x

10,343.0656934... = x

This works out to about 10,343 adult tickets. The remaining 2,657 tickets, of the expected total of 13,000 , will be child tickets. Then the expected total ticket revenue is given by:

5(10,343) + 2.5(2,657) = 58,357.5

So the expected total revenues from ticket sales is $58,357.50 .

Let's try another one. This time, suppose you work in a lab. You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?

Let w stand for the number of liters of the weaker 10% solution. Since the total number of liters is going to be 10 , then the number of liters left, after the first w liters have been poured, will be 10 −  w liters needed of the 30% solution.

(The clear labeling of the variable is important. While I picked w to stand for the w eaker acid, I might not remember this by the end of the exercise. If I don't label, I might not be able correctly to interpret my answer in the end.)

For mixture problems, it is often very helpful to do a grid, so let's do that here:

When the problem is set up like this, we can usually use the last column to write our equation. In this case, the liters of acid within the 10% solution, plus the liters of acid within the 30% solution, must add up to the liters of acid within the 15% solution. So, adding down the right-hand column, setting the inputs equal to the mixed output, we get:

0.10 w + 0.30(10 − w ) = 1.5

0.10 w + 3 − 0.30 w = 1.5

3 − 0.2 w = 1.5

1.5 = 0.2 w

(1.5)/(0.2) = w = 7.5

Looking back and confirming that the variable stands for the number of liters of the weaker acid, we see that we would need 7.5 liters of 10% acid . This means that we would need another 10 − 7.5 = 2.5 liters of the stronger 30% acid .

(If you think about it, this makes sense. Fifteen percent is closer to 10% than it is to 30% , so we ought to need more 10% solution in our mix.)

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Strategies for Solving Word Problems – Math

how do you solve a word problem in a set

It’s one thing to solve a math equation when all of the numbers are given to you but with word problems, when you start adding reading to the mix, that’s when it gets especially tricky.

The simple addition of those words ramps up the difficulty (and sometimes the math anxiety) by about 100!

How can you help your students become confident word problem solvers? By teaching your students to solve word problems in a step by step, organized way, you will give them the tools they need to solve word problems in a much more effective way.

Here are the seven strategies I use to help students solve word problems.

1. read the entire word problem.

Before students look for keywords and try to figure out what to do, they need to slow down a bit and read the whole word problem once (and even better, twice). This helps kids get the bigger picture to be able to understand it a little better too.

2. Think About the Word Problem

Students need to ask themselves three questions every time they are faced with a word problem. These questions will help them to set up a plan for solving the problem.

Here are the questions:

A. what exactly is the question.

What is the problem asking? Often times, curriculum writers include extra information in the problem for seemingly no good reason, except maybe to train kids to ignore that extraneous information (grrrr!). Students need to be able to stay focused, ignore those extra details, and find out what the real question is in a particular problem.

B. What do I need in order to find the answer?

Students need to narrow it down, even more, to figure out what is needed to solve the problem, whether it’s adding, subtracting, multiplying, dividing, or some combination of those. They’ll need a general idea of which information will be used (or not used) and what they’ll be doing.

This is where key words become very helpful. When students learn to recognize that certain words mean to add (like in all, altogether, combined ), while others mean to subtract, multiply, or to divide, it helps them decide how to proceed a little better

Here’s a Key Words Chart I like to use for teaching word problems. The handout could be copied at a smaller size and glued into interactive math notebooks. It could be placed in math folders or in binders under the math section if your students use binders.

One year I made huge math signs (addition, subtraction, multiplication, and divide symbols) and wrote the keywords around the symbols. These served as a permanent reminder of keywords for word problems in the classroom.

If you’d like to download this FREE Key Words handout, click here:

how do you solve a word problem in a set

C. What information do I already have?

This is where students will focus in on the numbers which will be used to solve the problem.

3. Write on the Word Problem

This step reinforces the thinking which took place in step number two. Students use a pencil or colored pencils to notate information on worksheets (not books of course, unless they’re consumable). There are lots of ways to do this, but here’s what I like to do:

  • Circle any numbers you’ll use.
  • Lightly cross out any information you don’t need.
  • Underline the phrase or sentence which tells exactly what you’ll need to find.

4. Draw a Simple Picture and Label It

Drawing pictures using simple shapes like squares, circles, and rectangles help students visualize problems. Adding numbers or names as labels help too.

For example, if the word problem says that there were five boxes and each box had 4 apples in it, kids can draw five squares with the number four in each square. Instantly, kids can see the answer so much more easily!

5. Estimate the Answer Before Solving

Having a general idea of a ballpark answer for the problem lets students know if their actual answer is reasonable or not. This quick, rough estimate is a good math habit to get into. It helps students really think about their answer’s accuracy when the problem is finally solved.

6. Check Your Work When Done

This strategy goes along with the fifth strategy. One of the phrases I constantly use during math time is, Is your answer reasonable ? I want students to do more than to be number crunchers but to really think about what those numbers mean.

Also, when students get into the habit of checking work, they are more apt to catch careless mistakes, which are often the root of incorrect answers.

7. Practice Word Problems Often

Just like it takes practice to learn to play the clarinet, to dribble a ball in soccer, and to draw realistically, it takes practice to become a master word problem solver.

When students practice word problems, often several things happen. Word problems become less scary (no, really).

They start to notice similarities in types of problems and are able to more quickly understand how to solve them. They will gain confidence even when dealing with new types of word problems, knowing that they have successfully solved many word problems in the past.

If you’re looking for some word problem task cards, I have quite a few of them for 3rd – 5th graders.

This 3rd grade math task cards bundle has word problems in almost every one of its 30 task card sets..

There are also specific sets that are dedicated to word problems and two-step word problems too. I love these because there’s a task card set for every standard.

CLICK HERE to take a look at 3rd grade:

3rd Grade Math Task Cards Mega Bundle | 3rd Grade Math Centers Bundle

This 4th Grade Math Task Cards Bundle also has lots of word problems in almost every single of its 30 task card sets. These cards are perfect for centers, whole class, and for one on one.

CLICK HERE to see 4th grade:

th Grade 960 Math Task Cards Mega Bundle | 4th Grade Math Centers

This 5th Grade Math Task Cards Bundle is also loaded with word problems to give your students focused practice.

CLICK HERE to take a look at 5th grade:

5th Grade Math Task Cards Mega Bundle - 5th Grade Math Centers

Want to try a FREE set of math task cards to see what you think?

3rd Grade: Rounding Whole Numbers Task Cards

4th Grade: Convert Fractions and Decimals Task Cards

5th Grade: Read, Write, and Compare Decimals Task Cards

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how do you solve a word problem in a set

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FREE RESOURCE LIBRARY (3)

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IMAGES

  1. Solving Word Problems Chart Grade 2-8

    how do you solve a word problem in a set

  2. How to Solve a Word Problem Using Combinations: Example #1 of 2. [HD]

    how do you solve a word problem in a set

  3. Steps to Solve a Word Problem Chart by Fab and Fun in First

    how do you solve a word problem in a set

  4. The 5 Steps of Problem Solving

    how do you solve a word problem in a set

  5. How to Solve a Wordy Math Problem (with Pictures)

    how do you solve a word problem in a set

  6. steps to solve a word problem elementary

    how do you solve a word problem in a set

VIDEO

  1. Linear equations, framing equations for word problems

  2. Word problems

  3. HOW TO SOLVE WORD-PROBLEM RELATED QUESTION

  4. Work word problem || How to solve

  5. @PakeezaYousaf Multiplication#word problem#How to solve word problem

  6. how to solve word problem || इबारती सवालों को कैसे हल करें || @mathsgurushrikant5753

COMMENTS

  1. Word Problems on Sets

    Word problems on sets are solved here to get the basic ideas how to use the properties of union and intersection of sets. Solved basic word problems on sets: 1. Let A and B be two finite sets such that n (A) = 20, n (B) = 28 and n (A ∪ B) = 36, find n (A ∩ B). Solution: Using the formula n (A ∪ B) = n (A) + n (B) - n (A ∩ B).

  2. Solving Word Problems with Sets Using a Venn Diagram

    How do we solve word problems involving sets? Here's how we can do that using a Venn diagram.Let's talk about that in this new #MathMondays video.Watch the o...

  3. Algebraic word problems

    To solve an algebraic word problem: 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. It's important for us to keep in mind how we define our variables.

  4. Word Problems on Sets

    In this, you will understand how to Solve Sets Word Problems using Venn Diagrams easily. If you need help on different concepts of Sets refer to Set Theory and learn the representation of a set, types of sets, etc. Check out the Solved Examples provided and learn how to solve related problems during your work. 1.

  5. Sets Word Problems

    Simple and clear explanation on how to analyze and solve word problems involving sets using venn diagrams.

  6. Algebra Topics: Introduction to Word Problems

    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?

  7. Word Problems Calculator

    Calculus Functions Trigonometry Statistics Full pad Examples Frequently Asked Questions (FAQ) How do you solve word problems? To solve word problems start by reading the problem carefully and understanding what it's asking.

  8. Solving Word Problems With Venn Diagrams Three Sets

    Solving Word Problems With Venn Diagrams Three Sets Steve Crow 57.6K subscribers Subscribe Subscribed 197K views 5 years ago This video shows how to solve applications using Venn Diagrams....

  9. How to set up algebraic equations to match word problems

    In very simple word problems that relationship usually involves just one of the four basic operations. Then in algebra, there may be more quantities and more operations between them. Examples of addition word problems Example. Jenny has 7 marbles and Kenny has 5. How many do they have together?

  10. How to Solve Word Problems in Algebra

    1 Read the problem carefully. [1] A common setback when trying to solve algebra word problems is assuming what the question is asking before you read the entire problem. In order to be successful in solving a word problem, you need to read the whole problem in order to assess what information is provided, and what information is missing. [2] 2

  11. Pre-algebra word problems

    Solving pre-algebra word problems requires us to create expressions and evaluate them. To solve a pre-algebra word problem: Identify the quantity to be calculated. Determine the operations and numbers needed to calculate the desired quantity. Write and evaluate the expressions to get the desired quantity.

  12. The Complete Guide to SAT Math Word Problems

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

  13. How to write word problems as equations

    How to write word problems as equations — Krista King Math | Online math help Word problems can seem to be tricky at first. What is the problem actually asking you to do? There are certain phrases that always mean the same operation in math. The table below will help you learn common phrases in math and what operations they represent.

  14. "Work" Word Problems

    As you can see in the above example, "work" problems commonly create rational equations. But the equations themselves are usually pretty simple to solve. One pipe can fill a pool 1.25 times as fast as a second pipe. When both pipes are opened, they fill the pool in five hours.

  15. Solving Word Problems: Steps & Examples

    Step 1: Visualize the Problem The first step is to visualize the problem. See if you can picture what is going on. Draw pictures if that will help you. Pinpoint or highlight the important...

  16. How to solve Algebra word problems

    Watch on In this lesson you will learn how to set up and solve ratio and proportion word problems. For word problems, the best thing to do is to look at a few examples but first let's review a few vocabulary terms. Ratio: A ratio is a comparison of two items and it is often written as a fraction.

  17. How to Solve Union Sets Word Problem

    http://ItsMyAcademy.com/Set-Theory/ For List of Set Theory Tutorial videos. To solve union of set related word problem we have to use either formula or venn ...

  18. The 4 Steps to Solving Word Problems

    Generally, solving a word problem involves four easy steps: Read through the problem and set up a word equation — that is, an equation that contains words as well as numbers. Plug in numbers in place of words wherever possible to set up a regular math equation. Use math to solve the equation. Answer the question the problem asks.

  19. Introduction to Sets

    First we specify a common property among "things" (we define this word later) and then we gather up all the "things" that have this common property. For example, the items you wear: hat, shirt, jacket, pants, and so on. I'm sure you could come up with at least a hundred. This is known as a set.

  20. ACT Math Word Problems: The Ultimate Guide

    ACT word problems can be grouped into two major categories: word problems where you must simply set up an equation and word problems in which you must solve for a specific piece of information. Word Problem Type 1: Setting Up an Equation. This is the less common type of word problem on the test, but you'll generally see it at least once or twice.

  21. Algebraic Sentences Word Problems

    Almost always, the word "is" in an algebraic sentence denotes the symbol of equality. In our example above, the algebraic sentence, " Five more than twice a number is forty-three ", is translated and written into its equation form: [latex]2x + 5 = 43 [/latex]. But before we delve into solving word problems that involve algebraic ...

  22. How do you set up and solve mixture word problems?

    mixture. 10. 0.15. 0.15 (10) = 1.5. When the problem is set up like this, we can usually use the last column to write our equation. In this case, the liters of acid within the 10% solution, plus the liters of acid within the 30% solution, must add up to the liters of acid within the 15% solution.

  23. Strategies for Solving Word Problems

    1. Read the Entire Word Problem Before students look for keywords and try to figure out what to do, they need to slow down a bit and read the whole word problem once (and even better, twice). This helps kids get the bigger picture to be able to understand it a little better too. 2. Think About the Word Problem