What is the number word problems

What are number word problems in Algebra?

The purpose of number word problems is to give you practice in translating back and forth from words to numbers and vice versa.

For some people these are fun number games, but they do appear on tests and in math classes from time to time, so it’s good to be comfortable with them.

In a number word problem you’re given information about a pair or group of numbers and you usually need to translate the information into equations to solve for the numbers.

Some helpful vocabulary:

Consecutive numbers are numbers that are in order, like ???4??? and ???5???.

Consecutive even numbers then would be even numbers that are in order, like ???4??? and ???6???.

Consecutive odd numbers are odd numbers that are in order, like ???5??? and ???7???.

And of course “sum” means add, “difference” means subtract, and “product” means multiply. Remember that place value can also be used in these problems.

How to solve number word problems

Take the course

Want to learn more about Algebra 2? I have a step-by-step course for that.

Examples of number word problems

Example

The sum of two consecutive integers is ???25???. Find the numbers.

“Consecutive” means numbers are in order.

Let ???X??? be the first number. Then the next integer would be ???X+1???.

Let’s write the sum.

???X+(X+1)=25???

Now let’s solve for ???X???.

???X+X+1=25???

???2X+1=25???

???2X+1-1=25-1???

???2X=24???

???frac{2X}{2}=frac{24}{2}???

???X=12???

The first integer is ???12???, the next one is ???12+1=13???. Therefore, the integers are ???12??? and ???13???. We can double check the answer and see that ???12+13=25???.

Let’s look at another style of problem.

Example

If you add the digits of a certain two digit number, the sum is ???17???. Reversing the two digits gives a number ???9??? smaller than the original number. What is the original number?

Let ???T??? be the tens digit and ???U??? be the units digit in the original number. Then the sum of the digits is ???T+U=17???.

The value of the original number is

???10T+U???

Reversing the digits gives us a number whose value is

???10U+T???

The second number is ???9??? smaller than the original number so we can write

???text{Original number}-9=text{Second number}???

That is

???10T+U-9=10U+T???

???10T-T+U-10U-9=0???

???9T-9U-9=0???

Dividing through by ???9??? gives

???T-U-1=0???

???T-U=1???

Add the sum of the digits equation, ???T+U=17???, to this one.

???T-U+(T+U)=1+(17)???

???T+T-U+U=1+17???

???2T=18???

???T=9???

Plug ???T=9??? into ???T+U=17??? to find ???U???.

???9+U=17???

???U=8???

The original number is ???98???.

Get access to the complete Algebra 2 course

Learn math

February 11, 2021

math, learn online, online course, online math, algebra, algebra 2, algebra ii, word problems, number word problems, consecutive integers, consecutive even integers, consecutive odd integers, sum of integers, sum of consecutive integers, reversing the digits, adding the digits

Chapter 3: Graphing

Number-based word problems can be very confusing, and it takes practice to convert a word-based sentence into a mathematical equation. The best strategy to solve these problems is to identify keywords that can be pulled out of a sentence and use them to set up an algebraic equation.

Variables that are to be solved for are often written as “a number,” “an unknown,” or “a value.”

“Equal” is generally represented by the words “is,” “was,” “will be,” or “are.”

Addition is often stated as “more than,” “the sum of,” “added to,” “increased by,” “plus,” “all,” or “total.” Addition statements are quite often written backwards. An example of this is “three more than an unknown number,” which is written as [latex]x + 3.[/latex]

Subtraction is often written as “less than,” “minus,” “decreased by,” “reduced by,” “subtracted from,” or “the difference of.” Subtraction statements are quite often written backwards. An example of this is “three less than an unknown number,” which is written as [latex]x — 3.[/latex]

Multiplication can be seen in written problems with the words “times,” “the product of,” or “multiplied by.”

Division is generally found by a statement such as “divided by,” “the quotient of,” or “per.”

28 less than five times a certain number is 232. What is the number?

• 28 less means that it is subtracted from the unknown number (write this as −28)
• five times an unknown number is written as [latex]5x[/latex]
• is 232 means it equals 232 (write this as = 232)

Putting these pieces together and solving gives:

[latex]begin{array}{rrrrrr} 5x&-&28&=&232& \ &+&28&&+28& \ hline &&5x&=&260& \ \ &&x&=&dfrac{260}{5}&text{or }52 end{array}[/latex]

Fifteen more than three times a number is the same as nine less than six times the number. What is the number?

• Fifteen more than three times a number is [latex]3x + 15[/latex] or [latex]15 + 3x[/latex]
• is means =
• nine less than six times the number is [latex]6x-9[/latex]

Putting these parts together gives:

[latex]begin{array}{rrrrrrr} 3x&+&15&=&6x&-&9 \ -6x&-&15&=&-6x&-&15 \ hline &&-3x&=&-24&& \ \ &&x&=&dfrac{-24}{-3}&text{or }8& \ end{array}[/latex]

Another type of number problem involves consecutive integers, consecutive odd integers, or consecutive even integers. Consecutive integers are numbers that come one after the other, such as 3, 4, 5, 6, 7. The equation that relates consecutive integers is:

[latex]x, x + 1, x + 2, x + 3, x + 4[/latex]

Consecutive odd integers and consecutive even integers both share the same equation, since every second number must be skipped to remain either odd (such as 3, 5, 7, 9) or even (2, 4, 6, 8). The equation that is used to represent consecutive odd or even integers is:

[latex]x, x + 2, x + 4, x + 6, x + 8[/latex]

The sum of three consecutive integers is 93. What are the integers?

The relationships described in equation form are as follows:

[latex]x + x + 1 + x + 2 = 93[/latex]

Which reduces to:

[latex]begin{array}{rrrrrr} 3x&+&3&=&93& \ &-&3&&-3& \ hline &&3x&=&90& \ \ &&x&=&dfrac{90}{3}&text{or }30 \ end{array}[/latex]

This means that the three consecutive integers are 30, 31, and 32.

The sum of three consecutive even integers is 246. What are the integers?

The relationships described in equation form are as follows:

[latex]x + x + 2 + x + 4 = 246[/latex]

Which reduces to:

[latex]begin{array}{rrrrrr} 3x&+&6&=&246& \ &-&6&&-6& \ hline &&3x&=&240& \ \ &&x&=&dfrac{240}{3}&text{ or }80 \ end{array}[/latex]

This means that the three consecutive even integers are 80, 82, and 84.

Questions

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

1. Five more than twice an unknown number is 25.
2. Twelve more than 4 times an unknown number is 36.
3. Three times an unknown number decreased by 8 is 22.
4. Six times an unknown number less 8 is 22.
5. When an unknown number is decreased by 8, the difference is half the unknown number.
6. When an unknown number is decreased by 4, the difference is half the unknown number.
7. The sum of three consecutive integers is 21.
8. The sum of the first two of three odd consecutive integers, less the third, is 5.

For questions 9 to 16, write and solve the equation describing each relationship.

1. When five is added to three times a certain number, the result is 17. What is the number?
2. If five is subtracted from three times a certain number, the result is 10. What is the number?
3. Sixty more than nine times a number is the same as two less than ten times the number. What is the number?
4. Eleven less than seven times a number is five more than six times the number. Find the number.
5. The sum of three consecutive integers is 108. What are the integers?
6. The sum of three consecutive integers is −126. What are the integers?
7. Find three consecutive integers such that the sum of the first, twice the second, and three times the third is −76.
8. Find three consecutive odd integers such that the sum of the first, two times the second, and three times the third is 70.

Answer Key 3.7

Related Pages
Different types of Digit Word Problems
Integer Problems
Integers

Convert Digits to Number

If the problem involves interchanging of the digits in the integer then you would need to
convert from the digits to numbers and vice versa. To convert the digits to numbers, we
need to multiply with the digit with the place value of the digit.

For example, the value of the number formed by the digit 4 in the ten’s place and the digit 3 in
the one’s place is 4 × 10 + 3 × 1

This type of digit problems is shown in the following example

Interchanging Of Digits Problems

Example:
The sum of the digits of a two-digit number is 11. If we interchange the digits then the new
number formed is 45 less than the original. Find the original number.

Solution:
Step 1:
Assign variables
Let x = one’s digit and t = ten’s digit

Sentence: The sum of the digits of a two-digit number is 11.
x + t = 11

Isolate variable x
x = 11 – t             (equation 1)

Step 2:
Convert digits to number
Original number = t × 10 + x
Interchanged number = x × 10 + t
Sentence: If we interchange the digits then the new number formed is 45 less than the original.
Interchanged = Original – 45
x × 10 + t = t × 10 + x – 45
10x + t = 10t + x – 45
10x – x + t = 10t – 45 (–x to both sides)
10x – x = 10t – t – 45 (– t to both sides)
10x – x + 45 = 10t – t (+ 45 to both sides)
10t – t = 10x – x + 45 (Rewrite equation with t on the left hand side)

Combine like terms
10t – t = 10x – x + 45
9t = 9x + 45            (equation 2)

Substitute equation 1 into equation 2
9t = 9(11 – t) + 45
9t = 99 – 9t + 45

Isolate variable t
9t + 9t = 99 + 45
18t = 144

The ten’s digit is 8. The one’s digit is 11 – 8 = 3

Answer: The number is 83.

Videos

How to use Systems of Equations to solve Reversing Digits Word Problem?

Reversing digits word problem
Problem:
The sum of two digits of a 2-digit number is 11. Reversing the digits increase the number by 45.
What is the number?

Example:
The sum of two digits of a 2-digit number is 13. Reversing the digits increase the number by 45.
What is the number?

• Show Video Lesson

Example:
The sum of the digits of a two digit number is 9. When the digits are reversed the new number is 9
less than three times the original.

• Show Video Lesson

Examples:

1. The sum of the digits of a 2-digit number is 10. The tens digit is 4 times the ones digit. Find the number.

2. A 2-digit number is 10 times the sum of its digits. The tens digit is 2 greater than the units digit.
   Find the number.

3. The sum of the digits of a 2-digit number is 14. If 18 is added to the number, the result is the number
   with its digits reversed. Find the original number.

4. The sum of the digits of a 2 digit number is 9. If the digits are reversed, the new number is 9 less
   than 3 times the original number. Find the original number.

• Show Video Lesson

Try the free Mathway calculator and
problem solver below to practice various math topics. Try the given examples, or type in your own
problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

/en/algebra-topics/solving-equations/content/

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

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:

1. Read through the problem carefully, and figure out what it’s about.
2. Represent unknown numbers with variables.
3. Translate the rest of the problem into a mathematical expression.
4. Solve the problem.
5. 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.

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?

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 ⋅ day + $.50 ⋅ mile = $360

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 0.5m.

30 ⋅ 2 + .5 ⋅ m = 360

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.

.5m / .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.

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?

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

60 + 300

360

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.

Practice!

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:

1. Read through the problem carefully, and figure out what it’s about.
2. Represent unknown numbers with variables.
3. Translate the rest of the problem into a mathematical expression.
4. Solve the problem.
5. Check your work.

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.

Problem 1

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?

Problem 2

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:

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

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:

1. First, replace the cost of a family pass with our variable f.

f equals half of $8 plus $25

2. Next, take out the dollar signs and replace words like plus and equals with operators.

f = half of 8 + 25

3. 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.

1. f is already alone on the left side of the equation, so all we have to do is calculate the right side.

f = 1/2 ⋅ 8 + 25

2. First, multiply 1/2 by 8. 1/2 ⋅ 8 is 4.

f = 4 + 25

3. Next, add 4 and 25. 4 + 25 equals 29 .

f = 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.

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

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.

1. We could translate this into this equation, with s standing for the cost of a single ticket.

1/2s = 29 — 25

2. Let’s work on the right side first. 29 — 25 is 4.

1/2s = 4

3. 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.

s = 8

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!

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

So now we’re sure about the answer to our problem: The cost of a family pass is $29.

Problem 2 Answer

Here’s Problem 2:

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?

Answer: $70

Let’s go through this problem one step at a time.

Step 1: Read through the problem carefully

Start by asking what question the problem is asking you to solve and identifying the information that will help you solve it. What’s the question here?

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?

To solve the problem, you’ll have to find out how much money Mo gave to charity. All the important information you need is in the problem:

• The amount Flor donated is three times as much the amount Mo donated
• Flor and Mo’s donations add up to $280 total

Step 2: Represent the unknown numbers with variables

The unknown number we’re trying to identify in this problem is Mo’s donation. We’ll represent it with the variable m.

Step 3: Translate the rest of the problem

Here’s the problem again. This time, the important facts are highlighted.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

The important facts of the problem could also be expressed this way:

Mo’s donation plus Flor’s donation equals $280

Because we know that Flor’s donation is three times as much as Mo’s donation, we could go even further and say:

Mo’s donation plus three times Mo’s donation equals $280

We can translate this into a math problem in only a few steps. Here’s how:

1. Because we’ve already said we’ll represent the amount of Mo’s donation with the variable m, let’s start by replacing Mo’s donation with m.

m plus three times m equals $280

2. Next, we can put in mathematical operators in place of certain words. We’ll also take out the dollar sign.

m + three times m = 280

3. Finally, let’s write three times mathematically. Three times m can also be written as 3 ⋅ m, or just 3m.

m + 3m = 280

Step 4: Solve the problem

It will only take a few steps to solve this problem.

1. To get the correct answer, we’ll have to get m alone on one side of the equation.

m + 3m = 280

2. To start, let’s add m and 3m. That’s 4m.

4m = 280

3. We can get rid of the 4 next to the m by dividing both sides by 4. 4m / 4 is m, and 280 / 4 is 70.

m = 70.

We’ve got our answer: m = 70. In other words, Mo donated $70.

Step 5: Check your work

The answer to our problem is $70, but we should check just to be sure. Let’s look at our problem again.

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?

If our answer is correct, $70 and three times $70 should add up to $280.

1. We can write our new equation like this:

70 + 3 ⋅ 70 = 280

2. The order of operations calls for us to multiply first. 3 ⋅ 70 is 210.

70 + 210 = 280

3. The last step is to add 70 and 210. 70 plus 210 equals 280.

280 = 280

280 is the combined cost of the tickets in our original problem. Our answer is correct: Mo gave $70 to charity.

/en/algebra-topics/distance-word-problems/content/

This is the first post in the Word Problem Solving Series.  This series is intended for middle school students.  In this post, we are going to discuss how to solve number word problems in Algebra.

The that you should observer in this series translating words into algebraic expressions and equations. Prior to the discussion of word problems, you have already learned how to solve equations, so I will not show the solution in details.

Problem 1

The sum of two numbers is . The larger number is more than the smaller. What are the two numbers?

Solution

Suppose is a number, the larger number, which is more than it is . Therefore, if a number, then is the larger number more than it.  If we add the two numbers

the sum is . The sentence “The sum of two numbers is ,”  tells us that that the sum equals . Therefore, we set up the equation,

.

Simplifying, we have , which gives us . So, the smaller number is and the larger number is .

Checking the Answer

We can check if our solution is correct by verifying the answer.  Is the larger number more than the smaller? Yes. Is the sum of the two numbers . Yes, . So, we are correct.

Problem 2

One number is twice the other number. Their sum is . What are the two numbers?

Solution

If a number is , twice that number is . So, if the smaller number in the problem is , the larger number is or .  Again in the problem, it says that “their sum is ” which means that if we add and , their sum is equal to .

So, we set up the equation

.

This gives us  which gives us .  The larger number is .

Checking the Answer

Is the larger number twice the smaller? Yes. Is the sum . Yes. Therefore, we are correct.

Problem 3

The sum of three consecutive numbers is . What are the numbers?

Solution

Consecutive numbers are numbers in uninterrupted succession. For example, are five consecutive numbers.

Suppose the consecutive numbers are , , and , we can see that is the smaller. If is the smaller, then is the next number, and is the largest number.  So, if is the smaller number, is the next number, and is the largest number. If we add the three numbers

their sum is . That is,

.

Simplifying, we have . Subtracting 3 from both sides, we have . Dividing both sides by 3, we have . So the three consecutive numbers are , and .

Checking the Answer 

Are the three numbers consecutive? Yes. Is their sum .  Yes, .

From the worked examples on number problems above, we have learned that the word “is” is most of the time synonymous to the phrase “is equal to.”

When kids begin to learn early math operations like addition and subtraction, the focus revolves around solving each problem. Think about your own experience with math: many years ago, classrooms around the country drilled students with timed multiplication tests. The objective was to solve each problem accurately as quickly as possible. Even today, and without such an emphasis on drilling, kids are taught to solve a problem and move on to the next one to practice math skills in short order. The solution just might be to tackle numberless word problems!

It’s no wonder, then, why math word problems continue to confuse little learners. Most often, a child will simply skim the problem, looking for the numbers and quickly attempt to solve the problem just to move on to the next one. Sometimes, when kids skim, but can’t figure out what the question is asking quickly enough, students give up and become frustrated, claiming that word problems are too difficult. 

At this point, you might be wondering how in the world word problems can exist without the main components: the numbers! Isn’t the point of a word problem to solve it? While word problems aim to be solved, by removing the numbers, children can slow down and understand what the problem is asking the student to do. This helps kids understand the context and math processes it will take to solve it when the numbers are included. 

---

For more practice with word problems and more, download our Talented and Gifted app for games, teacher-led videos and instruction tailored for early learners! Sign up for the first month of learning with Talented and Gifted!

---

If you’re still on the fence, let’s discover more about how numberless word problems work before exploring the steps parents and teachers can take to introduce this strategy to kids.  

How do Numberless Word Problems Work? 

Numberless word problems are exactly as they sound—numberless! That is to say that the numbers are temporarily removed from existing word problems that your child has not yet attempted to solve. Numberless problems can be prepared ahead of time and given to kids on worksheets or notecards. Kids will be asked to analyze the words in the problem to figure out what type of math operation might be used to solve it if there were numbers in the problem. To do this, the parent or teacher must follow several steps to introduce the concept, and to make sure the strategy helps kids slow down and understand what each problem is asking. 

As a result, you’ll find that your child might become more confident in solving word problems when the numbers are reintroduced. Because the numbers are taken out, so is the stress of trying to solve the problem. Kids are able to slow down and analyze the problem in a way that seeks to find meaning in the words rather than simply pulling out numbers to solve. This makes a child more flexible in their thinking because they are now analyzing it for which mathematical operation or process to use instead of simply getting it done. Instead of worrying about getting the answer wrong, kids are now thinking about the concept without fear of being incorrect. 

Steps to Introduce Kids to Numberless Word Problems 

Numberless word problems can turn the tables for a student struggling with word problems in math and can be introduced to kids as young as 1st through 3rd grade. To use this strategy effectively, you’ll need to follow a few steps to ensure that its helpful for your child. 

Step 1- Select word problems and remove the numbers

Feel free to browse the internet for word problems that are grade appropriate for your little learner, or purchase pre-made numberless word problems from online resources like Teachers Pay Teachers, or through educators or teaching supply stores. You can even make word problems yourself without ever including the numbers to make them sound seamless as your child reads.  

Before presenting the problem to your child, also remove the final question below the story that sets up the problem. This is important because your child will soon make a prediction as to what the potential question may be. 

Step 2- After reading through the problem, ask your child what he or she notices about it

Tell your child that he or she is going to be a question “detective”! It’s his or her job to read through the problem and figure out what information is already given and known. For instance, find all the facts contained within the problem and write them down. Make a simple chart that lists attributes about facts from the problem as opposed to ideas of what type of math they see or predict based upon what information is given in the problem. 

Step 3- It’s time to brainstorm! Make a prediction of what question will be asked

Don’t be surprised if your little mathematician is stumped at first! You might be faced with a blank stare or confused look on your child’s face when you first ask what question will be asked following the word problem they just read. Remind them that the final math question was removed or is missing, and it’s their job as a detective to figure it out. Encourage your child to look back through the problem and the chart to come to a conclusion. Based upon the facts of the problem, steer your child to forming an idea of what type of problem might be asked based on the data collected. 

After brainstorming a few possibilities and writing down the potential questions on paper, reveal the actual question to see if your child had gotten it correct! Help your child restate the problem and write it down on the paper. 

Step 4- Identify the information needed to make the problem solvable

Obviously, numberless problems leave out the necessary ingredients needed to solve the problem. Using critical thinking skills, your child can figure this information out individually by analyzing the word problem and choosing numbers for the problem. Don’t worry if the numbers he or she chooses aren’t the original numbers from the problem! Simply let your child choose and solve the problem. 

Let’s use the following word numberless word problem as an example:

---

There were  many students  in Ms. Smith’s class.

A few more students joined her class in spring. Now there were a lot of kids  in Ms. Smith’s Class.

How many students were added to Ms. Smith’s class?

---

After the question is revealed to students, the next step is to determine which words in the problem need to be replaced by numbers. Using the example above, ask your child to determine the first piece of information that would make the problem solvable. If your child struggles, start from the first sentence, “There were many students in Ms. Smith’s class”. Point out to your child that we don’t know how many students were in her class; we only know that there were “many”. In order to make the problem solvable, we would need to replace the word many with a number. We then would have something more like, “There were 21 students in Ms. Smith’s class”

After this first tidbit of information, it might be easier for your child to fill in the blanks. The next step would be to determine how many students were in Ms. Smith’s class after more students enrolled. If there were 26 students in Ms. Smith’s class after more students joined, the problem would task children with solving the number of kids that were added to the class. 

The final problem would look something like this: 

---

There were 21 students in Ms. Smith’s class.

A few more students joined her class in spring. Now there were 26 kids in Ms. Smith’s Class.

How many students were added to Ms. Smith’s class?

---

Step 5- Solve the problem using a strategy of your child’s choosing

Encourage your child to solve the problem using a self-selected strategy that he or she is familiar with and is comfortable in using. Some ideas include writing the problem down on a small white board with a dry erase marker, using fingers, or counters. Try not to solve the problem for your child, but rather direct your child to choose a technique to solve the problem on their own. Watch as your child gets to work! Once your child is ready with an answer, ask him or her to reveal the answer to you and to explain the strategy used to solve it. 

At first glance, many parents and teachers would look at numberless word problems and think that they’re impossible or unhelpful as a strategy to help struggling students solve word problems. But given more thought, numberless problems can help kids understand the math processes at work behind the scenes of a word problem. By slowing kids down and directing their attention to the words on the page, instead of merely pulling out the numbers to solve quickly, we can help kids look at word problems in a whole new light. When kids are taught to understand the math rather than simply solving it, they boost overall math skills and gain confidence in their work! 

---

For more practice with word problems, early math, and even skills found across the curriculum, be sure to check out our ample collection of math worksheets your child’s continued success in problem solving and beyond! 

We will learn how to solve
different types of word problems on digits and numbers.

1. The number of digits used in numbering each page of a book of 150 pages is

(a) 342

(b) 348

(c) 328

(d) 322

Solution:

Required number of digits = (9 × 1) + (90 × 2) + (150 — 99) × 3

                                                = 9 + 180 + 153

                                                = 342

Answer: (a)

Note: Number of 1
digit numbers = 9

Number of 2 digit numbers = 90

2. If the digits
of a two-digit number are interchanged, the number so formed is 18 more than
the original number. The sum of digits is 8. What is the thrice value of
original number?

(a) 159

(b) 78

(c) 105

(d) None

Solution:

Let, the unit’s and ten’s digit be ‘a’ and ‘b’ respectively.

Therefore, a + b = 8 and a — b = 18 ÷ 9 = 2

Since, a = (8 + 2) ÷ 2 = 5 and b = 5 — 2= 3

Therefore, the required value is 3 × 35 = 105

Answer: (c)

3. A two-digit
number is eight times the sum of its digits. If 45 is subtracted from the
numbers, the position of digits get interchanged. The number is:

(a) 56

(b) 83

(c) 72

(d) None

Solution:

Two digits differ by = 45 ÷ 9 = 5 and the number is divisible
by 8.

Now according the options,

The required number is 72

Answer: (c)

Note: The
difference of a two digit number and the number obtained by writing it in
reverse order is equal to nine times of difference of digits.

4. Find the total
numbers of three-digit numbers, which are exactly divisible by six?

(a) 340

(b) 126

(c) 190

(d) 150

Solution:

Total number of 3-digit numbers from 100 to 999 is 900.

Therefore, the least 3-digit number divisible by 6 is 6 × 17
= 102 and greatest number is 6 × 166 = 996

Therefore, required number of numbers = 166 — 17 + 1 = 150

Answer: (d)

5. A mathematics book
containing 20 pages. One sheet is missing among them. Sum of the page numbers
of the remaining pages of the mathematics book is 195. The numbers written on
both the sides of the missing sheet of the mathematics book must be

(a) 9, 10

(b) 11, 12

(c) 5, 6

(d) 7, 8

Solution:

Sum of the page numbers of the book = 1 + 2 + 3 +
……………. + 19 + 20 = (20 × 21)/2 = 210

The sum of page numbers of missing sheet = 210 — 195 = 15 =
(7 +

Therefore, required page number of missing sheet = 7 and 8.

Answer: (d)

Problem 1 : 

The sum of 5 times a number and 8 is 48. Find the number. 

Solution : 

5x + 8  =  48

Subtract 8 from both sides. 

(5x + — 8  =  48 — 8

5x  =  40

Divide both sides by 5.

5x/5  =  40/5

x  =  8

So, the number is 8.

Problem 2 :

A number consists of three digits of which the middle one is zero  and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297. Find the number. 

Solution :

Let «x0y» be the required three digit number. (As per the given information, middle digit is zero)

«The sum of the other digits is 9» —-> x + y = 9 ——(1)

«Interchanging the first and third digits» ———> y0x

From the information given in the question, we can have

y0x — x0y  =  297

(100y + x)  —  (100x + y)  =  297

100y + x — 100x -y = 297

-99x + 99y = 297

-x + y = 3 ———(2)

Solving (1) & (2), we get  x = 3 and y = 6

So,

x0y = 306

So, the required number is 306.

Problem 3 :

Of two numbers, 1/5th of a the greater equal to 1/3rd of the smaller and their sum is 16. Find the numbers.  

Solution :

Let «x» and «y» be the required two numbers such that

x > y.

From the point «1/5th of a the greater equal to 1/3rd of the smaller», we have

(1/5)x  =  (1/3)y

3x  =  5y

3x — 5 y  =  0 ———(1)

From the point «their sum is 16», we have

x + y = 16 ———-(1)

Solving (1) and (2), we get x = 10 and y = 6.

So, the two numbers are 10 and 6.

Problem 4 :

A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed. Find the number. 

Solution :

Let «xy» be the required number between 10 and 100. (Two digit number)

«A number between 10 and 100 is five times the sum of its digits»

From the information above, we have

xy  =  5(x+y)

10x + y   =  5x + 5y

5x — 4y  =  0 ——-(1)

«If 9 be added to it the digits are reversed»

From the information above, we have

xy + 9  =  yx  

10x + y + 9  =  10y + x

9x — 9y  =  -9

x —  y  =  -1 ———(2)

Solving (1) and (2), we get x = 4 and y = 5.

So, the required number is 45.

Problem 5 :

One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the numbers.

Solution :

So, we have   x  =  3y + 2 ———(1)

Given : 4 times the smaller number exceeds the greater by 5

So, we have  4y = x + 5 ———(2)

Plugging  (1) in (2), we get  

4y  =  3y + 2 + 5

4y  =  3y + 7

4y  =  3y + 7

y  =  7

Plugging y  =  7 in (1), we get   x  =  3(7) + 2

Therefore x  =  23

So, the two numbers are 23 and 7. 

Problem 6 :

A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

Solution :

So, we have

xy  =  7(x + y)

10x + y  =  7x + 7y

10x + y  =  7x + 7y

3x — 6y  =  0

x — 2y  =  0 ———(1)

Given : The number formed by reversing the digits is 18 less than the given number

So, we have

xy  —  yx  =  18 

(10x + y) — (10y + x)  =  18

10x + y — 10y — x  =  18

9x — 9y  =  18

x — y  =  2 ———(2)

Solving (1) and (2), we get x = 4 and y = 2

xy  =  42

So, the required number is 42. 

Problem 7 : 

If a number when divided by 296 gives a remainder 75, find the remainder when 37 divides the same number.

Solution : 

x  =  37 × 8k + 37 × 2 + 1 

x  =  37(8k + 2) + 1 

So, the remainder is ‘1’ when the number ‘x’ is divided by 37.

Problem 8 : 

Find the least number which when divided by 35, leaves a remainder 25, when divided by 45, leaves a remainder 35 and when divided by 55, leaves a remainder 45.

Solution : 

For each divisor and corresponding remainder, we have to find the difference. 

35 — 25  =  10 

45 — 35  =  10 

55 — 45  =  10 

we get the difference 10 (for all divisors and corresponding remainders) 

Now we have to find the L.C.M of (35,45,55) and subtract the difference from the L.C.M. 

L.C.M of (35, 45, 55)  =  3465 

So, the required least number is 

=  3465 — 10

=  3455

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com