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How to Solve Math Problems

Last Updated: April 15, 2024 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 591,153 times.

Although math problems may be solved in different ways, there is a general method of visualizing, approaching and solving math problems that may help you to solve even the most difficult problem. Using these strategies can also help you to improve your math skills overall. Keep reading to learn about some of these math problem solving strategies.

Understanding the Problem

• Draw a Venn diagram. A Venn diagram shows the relationships among the numbers in your problem. Venn diagrams can be especially helpful with word problems.
• Draw a graph or chart.
• Arrange the components of the problem on a line.
• Draw simple shapes to represent more complex features of the problem.

Developing a Plan

Solving the Problem

Joseph Meyer

When doing practice problems, promptly check to see if your answers are correct. Use worksheets that provide answer keys for instant feedback. Discuss answers with a classmate or find explanations online. Immediate feedback will help you correct your mistakes, avoid bad habits, and advance your learning more quickly.

Expert Q&A

• Seek help from your teacher or a math tutor if you get stuck or if you have tried multiple strategies without success. Your teacher or a math tutor may be able to easily identify what is wrong and help you to understand how to correct it. Thanks Helpful 0 Not Helpful 0
• Keep practicing sums and diagrams. Go through the concept your class notes regularly. Write down your understanding of the methods and utilize it. Thanks Helpful 0 Not Helpful 0

You Might Also Like

• ↑ Daron Cam. Math Tutor. Expert Interview. 29 May 2020.
• ↑ http://www.interventioncentral.org/academic-interventions/math/math-problem-solving-combining-cognitive-metacognitive-strategies
• ↑ http://tutorial.math.lamar.edu/Extras/StudyMath/ProblemSolving.aspx
• ↑ https://math.berkeley.edu/~gmelvin/polya.pdf

About This Article

To solve a math problem, try rewriting the problem in your own words so it's easier to solve. You can also make a drawing of the problem to help you figure out what it's asking you to do. If you're still completely stuck, try solving a different problem that's similar but easier and then use the same steps to solve the harder problem. Even if you can't figure out how to solve it, try to make an educated guess instead of leaving the question blank. To learn how to come up with a solid plan to use to help you solve a math problem, scroll down! Did this summary help you? Yes No

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Simple Algebra Problems – Easy Exercises with Solutions for Beginners

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Understanding Algebraic Expressions

Breaking down algebra problems, solving algebraic equations, tackling algebra word problems, types of algebraic equations, algebra for different grades.

For instance, solving the equation (3x = 7) for (x) helps us understand how to isolate the variable to find its value.

I always find it fascinating how algebra serves as the foundation for more advanced topics in mathematics and science. Starting with basic problems such as ( $(x-1)^2 = [4\sqrt{(x-4)}]^2$ ) allows us to grasp key concepts and build the skills necessary for tackling more complex challenges.

So whether you’re refreshing your algebra skills or just beginning to explore this mathematical language, let’s dive into some examples and solutions to demystify the subject. Trust me, with a bit of practice, you’ll see algebra not just as a series of problems, but as a powerful tool that helps us solve everyday puzzles.

Simple Algebra Problems and Strategies

When I approach simple algebra problems, one of the first things I do is identify the variable.

The variable is like a placeholder for a number that I’m trying to find—a mystery I’m keen to solve. Typically represented by letters like ( x ) or ( y ), variables allow me to translate real-world situations into algebraic expressions and equations.

An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like ( x ) or ( y )), and operators (like add, subtract, multiply, and divide). For example, ( 4x + 7 ) is an algebraic expression where ( x ) is the variable and the numbers ( 4 ) and ( 7 ) are terms. It’s important to manipulate these properly to maintain the equation’s balance.

Solving algebra problems often starts with simplifying expressions. Here’s a simple method to follow:

• Combine like terms : Terms that have the same variable can be combined. For instance, ( 3x + 4x = 7x ).
• Isolate the variable : Move the variable to one side of the equation. If the equation is ( 2x + 5 = 13 ), my job is to get ( x ) by itself by subtracting ( 5 ) from both sides, giving me ( 2x = 8 ).

With algebraic equations, the goal is to solve for the variable by performing the same operation on both sides. Here’s a table with an example:

Algebra word problems require translating sentences into equations. If a word problem says “I have six less than twice the number of apples than Bob,” and Bob has ( b ) apples, then I’d write the expression as ( 2b – 6 ).

Understanding these strategies helps me tackle basic algebra problems efficiently. Remember, practice makes perfect, and each problem is an opportunity to improve.

In algebra, we encounter a variety of equation types and each serves a unique role in problem-solving. Here, I’ll brief you about some typical forms.

Linear Equations : These are the simplest form, where the highest power of the variable is one. They take the general form ( ax + b = 0 ), where ( a ) and ( b ) are constants, and ( x ) is the variable. For example, ( 2x + 3 = 0 ) is a linear equation.

Polynomial Equations : Unlike for linear equations, polynomial equations can have variables raised to higher powers. The general form of a polynomial equation is ( $a_nx^n + a_{n-1}x^{n-1} + … + a_2x^2 + a_1x + a_0 = 0$ ). In this equation, ( n ) is the highest power, and ( $a_n$ ), ( $a_{n-1} $), …, ( $a_0$ ) represent the coefficients which can be any real number.

• Binomial Equations : They are a specific type of polynomial where there are exactly two terms. Like ($ x^2 – 4 $), which is also the difference of squares, a common format encountered in factoring.

To understand how equations can be solved by factoring, consider the quadratic equation ( $x^2$ – 5x + 6 = 0 ). I can factor this into ( (x-2)(x-3) = 0 ), which allows me to find the roots of the equation.

Here’s how some equations look when classified by degree:

Remember, identification and proper handling of these equations are essential in algebra as they form the basis for complex problem-solving.

In my experience with algebra, I’ve found that the journey begins as early as the 6th grade, where students get their first taste of this fascinating subject with the introduction of variables representing an unknown quantity.

I’ve created worksheets and activities aimed specifically at making this early transition engaging and educational.

6th Grade :

Moving forward, the complexity of algebraic problems increases:

7th and 8th Grades :

• Mastery of negative numbers: students practice operations like ( -3 – 4 ) or ( -5 $\times$ 2 ).
• Exploring the rules of basic arithmetic operations with negative numbers.
• Worksheets often contain numeric and literal expressions that help solidify their concepts.

Advanced topics like linear algebra are typically reserved for higher education. However, the solid foundation set in these early grades is crucial. I’ve developed materials to encourage students to understand and enjoy algebra’s logic and structure.

Remember, algebra is a tool that helps us quantify and solve problems, both numerical and abstract. My goal is to make learning these concepts, from numbers to numeric operations, as accessible as possible, while always maintaining a friendly approach to education.

I’ve walked through various simple algebra problems to help establish a foundational understanding of algebraic concepts. Through practice, you’ll find that these problems become more intuitive, allowing you to tackle more complex equations with confidence.

Remember, the key steps in solving any algebra problem include:

• Identifying variables and what they represent.
• Setting up the equation that reflects the problem statement.
• Applying algebraic rules such as the distributive property ($a(b + c) = ab + ac$), combining like terms, and inverse operations.
• Checking your solutions by substituting them back into the original equations to ensure they work.

As you continue to engage with algebra, consistently revisiting these steps will deepen your understanding and increase your proficiency. Don’t get discouraged by mistakes; they’re an important part of the learning process.

I hope that the straightforward problems I’ve presented have made algebra feel more manageable and a little less daunting. Happy solving!

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How to Solve Math Problems Faster: 15 Techniques to Show Students

Written by Marcus Guido

• Teaching Strategies

“Test time. No calculators.”

You’ll intimidate many students by saying this, but teaching techniques to solve math problems with ease and speed can make it less daunting.

This can also  make math more rewarding . Instead of relying on calculators, students learn strategies that can improve their concentration and estimation skills while building number sense. And, while there are educators who  oppose math “tricks”  for valid reasons, proponents point to benefits such as increased confidence to handle difficult problems.

Here are 15 techniques to show students,  helping them solve math problems faster:

Addition and Subtraction

1. two-step addition.

Many students struggle when learning to add integers of three digits or higher together, but changing the process’s steps can make it easier.

The first step is to  add what’s easy.  The second step is to  add the rest.

Let’s say students must find the sum of 393 and 89. They should quickly see that adding 7 onto 393 will equal 400 — an easier number to work with. To balance the equation, they can then subtract 7 from 89.

Broken down, the process is:

• (393 + 7) + (89 – 7)

With this fast technique, big numbers won’t look as scary now.

2. Two-Step Subtraction

There’s a similar method for subtraction.

Remove what’s easy. Then remove what’s left.

Suppose students must find the difference of 567 and 153. Most will feel that 500 is a simpler number than 567. So, they just have to take away 67 from the minuend — 567 — and the subtrahend — 153 — before solving the equation.

Here’s the process:

• (567 – 67) – (153 – 67)

Instead of two complex numbers, students will only have to tackle one.

3. Subtracting from 1,000

You can  give students confidence  to handle four-digit integers with this fast technique.

To subtract a number from 1,000, subtract that number’s first two digits from 9. Then, subtract the final digit from 10.

Let’s say students must solve 1,000 – 438.  Here are the steps:

This also applies to 10,000, 100,000 and other integers that follow this pattern.

Multiplication and Division

4. doubling and halving.

When students have to multiply two integers, they can speed up the process when one is an even number. They just need to  halve the even number and double the other number.

Students can stop the process when they can no longer halve the even integer, or when the equation becomes manageable.

Using 33 x 48 as an example,  here’s the process:

The only prerequisite is understanding the 2 times table.

5. Multiplying by Powers of 2

This tactic is a speedy variation of doubling and halving.

It simplifies multiplication if a number in the equation is a power of 2, meaning it works for 2, 4, 8, 16 and so on.

Here’s what to do:  For each power of 2 that makes up that number, double the other number.

For example, 9 x 16 is the same thing as 9 x (2 x 2 x 2 x 2) or 9 x 24. Students can therefore double 9 four times to reach the answer:

Unlike doubling and halving, this technique demands an understanding of exponents along with a strong command of the 2 times table.

6. Multiplying by 9

For most students, multiplying by 9 — or 99, 999 and any number that follows this pattern — is difficult compared with multiplying by a power of 10.

But there’s an easy tactic to solve this issue, and  it has two parts.

First, students round up the 9  to 10. Second, after solving the new equation, they subtract the number they just multiplied by 10 from the answer.

For example, 67 x 9 will lead to the same answer as 67 x 10 – 67. Following the order of operations will give a result of 603. Similarly, 67 x 99 is the same as 67 x 100 – 67.

Despite more steps, altering the equation this way is usually faster.

7. Multiplying by 11

There’s an easier way for multiplying two-digit integers by 11.

Let’s say students must find the product of 11 x 34.

The idea is to put a space between the digits, making it 3_4. Then, add the two digits together and put the sum in the space.

The answer is 374.

What happens if the sum is two digits? Students would put the second digit in the space and add 1 to the digit to the left of the space.  For example:

It’s multiplication without having to multiply.

8. Multiplying Even Numbers by 5

This technique only requires basic division skills.

There are two steps,  and 5 x 6 serves as an example. First, divide the number being multiplied by 5 — which is 6 — in half. Second, add 0 to the right of number.

The result is 30, which is the correct answer.

It’s an ideal, easy technique for students mastering the 5 times table.

9. Multiplying Odd Numbers by 5

This is another time-saving tactic that works well when teaching students the 5 times table.

This one has three steps,  which 5 x 7 exemplifies.

First, subtract 1 from the number being multiplied by 5, making it an even number. Second, cut that number in half — from 6 to 3 in this instance. Third, add 5 to the right of the number.

The answer is 35.

Who needs a calculator?

10. Squaring a Two-Digit Number that Ends with 1

Squaring a high two-digit number can be tedious, but there’s a shortcut if 1 is the second digit.

There are four steps to this shortcut,  which 812 exemplifies:

• Subtract 1 from the integer: 81 – 1 = 80
• Square the integer, which is now an easier number: 80 x 80 = 6,400
• Add the integer with the resulting square twice: 6,400 + 80 + 80 = 6,560
• Add 1: 6,560 + 1 = 6,561

This work-around eliminates the difficulty surrounding the second digit, allowing students to work with multiples of 10.

11. Squaring a Two-Digit Numbers that Ends with 5

Squaring numbers ending in 5 is easier, as there are  only two parts of the process.

First, students will always make 25 the product’s last digits.

Second, to determine the product’s first digits, students must multiply the number’s first digit — 9, for example — by the integer that’s one higher — 10, in this case.

So, students would solve 952 by designating 25 as the last two digits. They would then multiply 9 x 10 to receive 90. Putting these numbers together, the  result is 9,025.

Just like that, a hard problem becomes easy multiplication for many students.

12. Calculating Percentages

Cross-multiplication is an  important skill  to develop, but there’s an easier way to calculate percentages.

For example, if students want to know what 65% of 175 is, they can multiply the numbers together and move the decimal place two digits to the left.

The result is 113.75, which is indeed the correct answer.

This shortcut is a useful timesaver on tests and quizzes.

13. Balancing Averages

To determine the average among a set of numbers, students can balance them instead of using a complex formula.

Suppose a student wants to volunteer for an average of 10 hours a week over a period of four weeks. In the first three weeks, the student worked for 10, 12 and 14 hours.

To determine the number of hours required in the fourth week, the student must  add how much he or she surpassed or missed the target average  in the other weeks:

• 14 hours – 10 hours = 4 hours
• 12 – 10 = 2
• 10 – 10 = 0
• 4 hours + 2 hours + 0 hours = 6 hours

To learn the number of hours for the final week, the student must  subtract the sum from the target average:

• 10 hours – 6 hours = 4 hours

With practice, this method may not even require pencil and paper. That’s how easy it is. 

Word Problems

14. identifying buzzwords.

Students who struggle to translate  word problems  into equations will benefit from learning how to spot buzzwords — phrases that indicate specific actions.

This isn’t a trick. It’s a tactic.

Teach students to look for these buzzwords,  and what skill they align with in most contexts:

Be sure to include buzzwords that typically appear in their textbooks (or other classroom  math books ), as well as ones you use on tests and assignments.

As a result, they should have an  easier time processing word problems .

15. Creating Sub-Questions

For complex word problems, show students how to dissect the question by answering three specific sub-questions.

Each student should ask him or herself:

• What am I looking for?  — Students should read the question over and over, looking for buzzwords and identifying important details.
• What information do I need?  — Students should determine which facts, figures and variables they need to solve the question. For example, if they determine the question is rooted in subtraction, they need the minuend and subtrahend.
• What information do I have?  — Students should be able to create the core equation using the information in the word problem, after determining which details are important.

These sub-questions help students avoid overload.

Instead of writing and analyzing each detail of the question, they’ll be able to identify key information. If you identify students who are struggling with these, you can use  peer learning  as needed.  

For more fresh approaches to teaching math in your classroom, consider treating your students to a range of  fun math activities .

Final Thoughts About these Ways to Solve Math Problems Faster

Showing these 15 techniques to students can give them the  confidence to tackle tough questions .

They’re also  mental math  exercises, helping them build skills related to focus, logic and critical thinking.

A rewarding class equals an  engaging class . That’s an easy equation to remember.

> Create or log into your teacher account on Prodigy  — a free, adaptive math game that adjusts content to accommodate player trouble spots and learning speeds. Aligned to US and Canadian curricula, it’s loved by more than 500,000 teachers and 15 million students.

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Introduction to Algebra

Algebra is great fun - you get to solve puzzles!

What is the missing number?

OK, the answer is 6, right? Because 6 − 2 = 4 . Easy stuff.

Well, in Algebra we don't use blank boxes, we use a letter (usually an x or y, but any letter is fine). So we write:

It is really that simple. The letter (in this case an x) just means "we don't know this yet", and is often called the unknown or the variable .

And when we solve it we write:

Why Use a Letter?

• it is easier to write "x" than drawing empty boxes (and easier to say "x" than "the empty box").
• if there are several empty boxes (several "unknowns") we can use a different letter for each one.

So x is simply better than having an empty box. We aren't trying to make words with it!

And it doesn't have to be x , it could be y or w ... or any letter or symbol we like.

How to Solve

Algebra is just like a puzzle where we start with something like "x − 2 = 4" and we want to end up with something like "x = 6".

But instead of saying " obviously x=6", use this neat step-by-step approach:

• Work out what to remove to get "x = ..."
• Remove it by doing the opposite (adding is the opposite of subtracting)
• Do that to both sides

Here is an example:

To remove it, do the opposite , in this case add 2

Do it to both sides

Which is ...

Why did we add 2 to both sides?

To "keep the balance"....

Just remember this:

See this in action at the Algebra Balance Animation .

Another Puzzle

What we want is an answer like "x = ...", but the +5 is in the way of that! We can cancel out the +5 with a −5 (because 5−5=0)

Have a Try Yourself

Now practice on this Simple Algebra Worksheet and then check your answers. Try to use the steps we have shown you here, rather than just guessing!

Try the questions below, then read Introduction to Algebra - Multiplication

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You don’t have to be a math teacher to know that a lot of students—and likely a lot of parents (it’s been awhile!)—are intimidated by math problems, especially if they involve large numbers. Learning techniques on how to do math quickly can help students develop greater confidence in math , improve math skills and understanding, and excel in advanced courses.

If it’s your job to teach those, here’s a great refresher.

Fast math tricks infographic

10 tricks for doing fast math

Here are 10 fast math strategies students (and adults!) can use to do math in their heads. Once these strategies are mastered, students should be able to accurately and confidently solve math problems that they once feared solving.

1. Adding large numbers

Adding large numbers just in your head can be difficult. This method shows how to simplify this process by making all the numbers a multiple of 10. Here is an example:

While these numbers are hard to contend with, rounding them up will make them more manageable. So, 644 becomes 650 and 238 becomes 240.

Now, add 650 and 240 together. The total is 890. To find the answer to the original equation, it must be determined how much we added to the numbers to round them up.

650 – 644 = 6 and 240 – 238 = 2

Now, add 6 and 2 together for a total of 8

To find the answer to the original equation, 8 must be subtracted from the 890.

890 – 8 = 882

So the answer to 644 +238 is 882.

2. Subtracting from 1,000

Here’s a basic rule to subtract a large number from 1,000: Subtract every number except the last from 9 and subtract the final number from 10

For example:

1,000 – 556

Step 1: Subtract 5 from 9 = 4

Step 2: Subtract 5 from 9 = 4

Step 3: Subtract 6 from 10 = 4

The answer is 444.

3. Multiplying 5 times any number

When multiplying the number 5 by an even number, there is a quick way to find the answer.

For example, 5 x 4 =

• Step 1: Take the number being multiplied by 5 and cut it in half, this makes the number 4 become the number 2.
• Step 2: Add a zero to the number to find the answer. In this case, the answer is 20.

When multiplying an odd number times 5, the formula is a bit different.

For instance, consider 5 x 3.

• Step 1: Subtract one from the number being multiplied by 5, in this instance the number 3 becomes the number 2.
• Step 2: Now halve the number 2, which makes it the number 1. Make 5 the last digit. The number produced is 15, which is the answer.

4. Division tricks

Here’s a quick way to know when a number can be evenly divided by these certain numbers:

• 10 if the number ends in 0
• 9 when the digits are added together and the total is evenly divisible by 9
• 8 if the last three digits are evenly divisible by 8 or are 000
• 6 if it is an even number and when the digits are added together the answer is evenly divisible by 3
• 5 if it ends in a 0 or 5
• 4 if it ends in 00 or a two digit number that is evenly divisible by 4
• 3 when the digits are added together and the result is evenly divisible by the number 3
• 2 if it ends in 0, 2, 4, 6, or 8

5. Multiplying by 9

This is an easy method that is helpful for multiplying any number by 9. Here is how it works:

Let’s use the example of 9 x 3.

Step 1 : Subtract 1 from the number that is being multiplied by 9.

3 – 1 = 2

The number 2 is the first number in the answer to the equation.

Step 2 : Subtract that number from the number 9.

9 – 2 = 7

The number 7 is the second number in the answer to the equation.

So, 9 x 3 = 27

6. 10 and 11 times tricks

The trick to multiplying any number by 10 is to add a zero to the end of the number. For example, 62 x 10 = 620.

There is also an easy trick for multiplying any two-digit number by 11. Here it is:

Take the original two-digit number and put a space between the digits. In this example, that number is 25.

Now add those two numbers together and put the result in the center:

2_(2 + 5)_5

The answer to 11 x 25 is 275.

If the numbers in the center add up to a number with two digits, insert the second number and add 1 to the first one. Here is an example for the equation 11 x 88

(8 + 1)_6_8

There is the answer to 11 x 88: 968

7. Percentage

Finding a percentage of a number can be somewhat tricky, but thinking about it in the right terms makes it much easier to understand. For instance, to find out what 5% of 235 is, follow this method:

• Step 1: Move the decimal point over by one place, 235 becomes 23.5.
• Step 2: Divide 23.5 by the number 2, the answer is 11.75. That is also the answer to the original equation.

8. Quickly square a two-digit number that ends in 5

Let’s use the number 35 as an example.

• Step 1: Multiply the first digit by itself plus 1.
• Step 2: Put a 25 at the end.

35 squared = [3 x (3 + 1)] & 25

[3 x (3 + 1)] = 12

12 & 25 = 1225

35 squared = 1225

9. Tough multiplication

When multiplying large numbers, if one of the numbers is even, divide the first number in half, and then double the second number. This method will solve the problem quickly. For instance, consider

Step 1: Divide the 20 by 2, which equals 10. Double 120, which equals 240.

Then multiply your two answers together.

10 x 240 = 2400

The answer to 20 x 120 is 2,400.

10. Multiplying numbers that end in zero

Multiplying numbers that end in zero is actually quite simple. It involves multiplying the other numbers together and then adding the zeros at the end. For instance, consider:

Step 1: Multiply the 2 times the 4

Step 2: Put all four of the zeros after the 8

200 x 400= 80,000

Practicing these fast math tricks can help both students and teachers improve their math skills and become secure in their knowledge of mathematics—and unafraid to work with numbers in the future.

You may also like to read

• Research-Based Math Teaching Strategies
• Tips in Teaching a Hands-On Math Curriculum
• 5 Tips to Help Get Students Engaged in High School Math
• 3 Tips for Running an Elementary School Math Workshop
• Seven Everyday Online Math Resources for Teachers
• Three Tips for Developing Elementary Math Tests

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Maths Tricks

Maths tricks are the ways to solve complex mathematical problems easily and quickly.  Mathematics is not only limited to learning from textbooks, there are different learning styles that make mathematics easier. Simple Maths magic tricks help us with fast calculations and improve our mathematical skills.  For example, the multiplication tricks will help students to learn maths tables and quick multiplication.

Maths is not easy for some students. The Maths tricks are not only helpful for school-going kids but also supports you to manage time in final exams as well as in the comp etitive exam and solve the Maths questions with accuracy. These tricks are very helpful for Class 5, 6, 7, 8, 9 and 10 students.

Learning Quick Maths Tricks

Mathematical tricks are the best way to make Maths a fun subject. Therefore, learning simple Maths tricks will help the students to gain their confidence and enhance problem-solving skills. With these learning skills, they can achieve a big success in the competitive exams and upcoming future.    

16 Maths Tricks (With Examples)

Imagine how mathematics would be easy and interesting when you have the ability to calculate the problems in a matter of seconds using some Maths tricks. There are different kinds of arithmetic operations  like addition, subtraction, division, multiplication, squaring, roots, powers, logarithms, divisions, etc. Here are some of the best tricks, which will help students to perform arithmetic calculations easily.

1. Maths Tricks for Addition

With the help of basic principles of tens and unit places, the addition of two-digit numbers is performed by

• Take 43 + 34
• Split the second number into tens and unit places. 34 = 30 + 4
• Finish the ten’s addition. 43 + 30 = 73
• Finally, add the remaining unit place digit. 73 + 4 = 77.

2. Maths Tricks for Subtraction

Here is an example that requires a lot of borrowing

• Consider two numbers say 1000 and 676
• Subtract 1 from both the numbers; we get 999 and 675
• Then subtract 675 from 999, we get 324
• So, 1000 – 676 = 324.

 3.  Quick Multiplication Tricks by Breaking Down Numbers

• Let’s try the numbers 24 and 16
• First split the number 24, which gives 4 x 6
• Then multiply 6 with 16, we get 96
• Finally multiply the number, 96 x 4 = 384
• So, the multiplication of two numbers 24 x 16 gives the solution 384.

4. Multiplied By 15

• Consider the multiplication of two numbers say 56 and 15
• Now add zero at the end of the first number, it becomes 560.
• Divide that number by 2; we get 560/2 = 280
• Add the resultant number with 560, so 560 + 280 = 840.
• So the answer for 56 and 15 is 840.

5. Multiplication of Two-Digit Numbers

If anyone of the given numbers is an even number, then follow the steps to solve

• Consider an example, 18 x 37
• Here 18 is an even number, then divide the first number in half, so that 18/2 = 9
• Then double the second number. 37 x 2 =74
• Finally, multiply the resultant numbers. It becomes 74 x 9 = 666

6. Maths Division Tricks

The numbers that can be evenly divided by certain numbers are:

• If a number is an even number and ends in 0, 2, 4, 6 or 8, it is divided by 2.
• A number is divisible by 3 if the sum of the digits is divisible by 3. Consider the number 12 = 1 + 3 and 3 is divisible by 3.
• A number is divisible by 4 if the last two digits are divisible by 4. Example: 9312. Here the last two digits are 12, and 12 is divisible by 4.
• If the last digit is 0 or 5, it is divisible by 5
• If a number is divisible by 2 and 3, then it is divisible by 6, since 6 is the product of 2 and 3.
• If the number is divisible by 8, the last three digits of the numbers are divisible by 8.
• If a number is divisible by 9, the sum of the digits is divided by 9. Let us consider the example, 4518 = 4 + 5 + 1 + 8 = 18, which is divisible by 9.
• If the final digit of the number is 0, it is divisible by 10.

7. Maths Trick to Find Percentage

Let us take; we have to find the percentage of the number 5% of 475, follow the steps.

• For the given number, move the decimal point over by one place. 475 becomes 47.5
• Then divide the number 47.5 by 2, we get 23.75.
• 23.75 is the solution to the given problem.

8. Maths Magic tricks to Calculate Squares ending with digit 5

• Let’s consider the number 75 to find its square.
• Start writing the answer of last two digits number that is 25 because any number that ends with 5 is 25
• Take the first digit of the number 75. That is 7 and take the number that follows 7 is 8.
• Now, multiply 7 and 8, we get the number 56.
• Finally, write the number 56 in the prefix and combine with 25 what we already wrote.
• So, the answer is 5625.
• Squares Ending in 5:  n5 = n(n + 1)5 2 = n(n + 1)25 , where n is the first digit.
• Example: Let’s consider the number 75 to find its square. Here n = 7,

So, 75 = 7(7 + 1)25 = (7 x 8) 25  = 5625.

9. Tricks to Multiply by 2 and 4

When a number is multiplied by 2 or 4, then the last digit of the resulting value will be an even number always.

• 19 x 2 = 38
• 19 x 4 = 76

10. Multiplication by 5

When a number is multiplied by 5, then the resulting value will either end with 0 or 5.

• 11 x 5 = 55
• 121 x 5 = 605

11. Multiplication by 10

When a number is multiplied by 10, then the resulting value ends with 0 always.

• 5 x 10 = 50
• 10 x 10 = 100
• 11 x 10 = 110
• 17 x 10 = 170
• 211 x 10 = 2110

12. Tricks to Memorise Table of 9

It is easy to remember the table of 9. Just we need to focus on the pattern.

09, 18, 27, 36, 45, 54, 63, 72, 81, 90

We can see the numbers at the ten’s place are increasing by 1, and the numbers at the unit place are decreasing by 1.

13. Trick for Squaring a Two-digit Number

Given number: 57 

Here, Tens digit = 5 Unit digit = 7

7 + 57 = 64

64 × 5 = 320

3200 + 49 = 3249

14. Trick for Unit Digit of a Cube Number

Below are the numbers which have the same unit digit for their cubes.

This will help you in finding the cube root of a number (which is a definite cube). 

15. Trick for Cube Root of a Number

Let’s consider a number: 117649

To find the cube root of this number, we need to divide the number into parts.

Take the three rightmost digits as one part and the remaining digits as another part.

Part 1: 649

Here, the unit digit is 9, and we know the cube root of this number will contain 9 as its unit digit.

Part 2: 117

Identify the cube of a number which is smaller than 117.

4 3 = 64 and 5 3 = 125

Thus, the cube root of 117649 is 49.

Note: This trick will work for finding the cube root of definite cube numbers.

16. Trick for Multiplication Table of 15

Here is the trick for writing the 15 times table quickly.

Maths is a fun subject. Adding tricks to this subject will make it more interesting. Students will be able to solve all the complex problems using these Maths magic tricks. These tricks also help students to improve their problem-solving skills and boost their confidence.

Maths Tricks Practice Questions

Find more maths tricks questions for practice here.

• A merchant can place 8 large boxes or 10 small boxes into a carton for shipping. In one shipment, he sent a total of 96 boxes. If there are more large boxes than small boxes, how many cartons did he ship? (A) 11 (B) 10  (C) 12   (D) 15 (E) 17
• 339% of 803 + 77.8% of 1107 = ? (A) 3175 (B) 3320 (C) 3580 (D) 3710 (E) 3950
• 78.54 ÷ 0.03 + 22.8 ÷ 0.8 – 1470 × 1.25 = ? (A) 809 (B) 807.5 (C) 805 (D) 802.5 (E) 801
• The cost of 8 dozen eggs is Rs. 256. Which calculation is needed to find the cost of 9 eggs? (A) (9 × 256) × (8 ÷ 12) (B) (12 × 256) ÷ (8 × 9) (C) (8 × 256) ÷ (9 × 12) (D) (9 × 256) × (8 × 12) (E) (9 × 256) ÷ (8 × 12)

Maths Tricks Related Articles

Frequently asked questions on maths tricks, what is the use of maths tricks, what are multiplication tricks, how to add in a faster way, how to multiply quickly, how to find if a number is divisible by 9, how to find if a number is divisible by 13, how to multiply 5 with an even number.

Assume that, 5 should be multiplied by 48. Follow the below steps to find the product value. Step 1: Divide the given even number by 2. Hence, 48 divided by 2 is 24. Step 2: Add zero to the result, obtained in step 1 to get the product value. So, we get 240. Therefore, the product of 5 and 48 is 240.

How to multiply 5 with an odd number?

Let us consider an example, 5 × 25. Here, 25 is an odd number. Go through the below steps to find the product. Step 1: Subtract 1 from the odd number (i.e. the number that is being multiplied by 5). So, 25 – 1 = 24. Step 2: Half the number that is obtained in step 1. Hence, 24/2 = 12. Step 3: Keep the number 5 in unit digits and append the result obtained in step 2. Hence, we get 125. So, we get 5 × 25 is 125.

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Easy Finger Math Tricks to Help Kids Solve Problems

While using your fingers isn't the fastest way to recall a multiplication fact while doing a problem, finger math tricks can help kids figure out how to answer the problem at hand — and as they work on their math, they will eventually learn all the facts by repetition.

Note that before your child can understand other finger tricks, they must be able to count by 2s, 5s, and 10s and multiply by 2s, 3s, and 4s.

Quick Finger Math Tricks for Threes and Fours

The tricks for multiplying by threes and fours are really a matter of counting out the answer on your fingers. As your children count out the answer repeatedly, they'll memorize it and then be able to move on to larger numbers.

Multiplying by Three

Did you realize that all of your fingers have three segments? Therefore, you can figure out anything from 3 x 1 to 3 x 10 by counting the segments on each finger. To start:

• Hold up the number of fingers you're going to multiply by 3. For example, if the problem is 3 x 4 — hold up four fingers.
• Count each segment on each finger you're holding up, and you should come up with 12 — which is the correct answer.

Multiplying by Four

Multiplying by four is the same as multiplying by two — twice. To start:

• Hold up the number of fingers to correspond with the number you are multiplying by four. For example, if you are multiplying 4 x 6 — hold up six fingers.
• Count each finger by two, moving from left to right. Then count each finger again, continuing to count by twos, until you've counted every finger twice.

Helpful Hack To keep track of the fingers you've counted twice, sometimes it's easier to put your finger down as you count the first time, and back up as you count the second time.

Finger Math Tricks for Multiplying by 6, 7, 8, and 9

While numbers one through five are easy for most kids to remember, six and up often pose a problem. This handy trick will make it a little easier to work those problems out.

Multiplying 6, 7, 8, and 9 by Hand

To begin, assign each finger a number. For example, your thumbs represent 6, your index fingers each represent 7, etc. This will remain the same throughout the finger math hack.

Your left hand will represent the first number that you are multiplying and your right hand will represent the second number you are multiplying. In this example, we are multiplying 7 x 8. 

To Determine the Part of Your Answer:

• On your left hand, put down the finger that represents the number you are multiplying as well as any fingers whose number value is less than this figure. In this example, you are multiplying 7 x 8, so the left hand will represent 7. You will drop your index finger (number 7) and your thumb (number 6).
• Similarly, the right hand will represent eight, so you will drop down your middle finger (number 8), your index finger (number 7), and your thumb (number 6).
• Now, just multiply the fingers that are still pointed upwards. In this case, you will have three fingers on your left hand and two on your right, so you will multiply 3 x 2 to get 6. This is the first part of your answer!

To Determine the Second Part of Your Answer:

• Keeping your fingers in the same positions, count how many fingers are folded down. In the 7 x 8 example, you should have five fingers folded. 
• You will count each of these in quantities of ten. So, 10, 20, 30, 40, 50.
• 50 is your answer.

To Determine Your Final Answer:

• Add your two numbers together. In this example, you would add 6 + 50, which gives you 56!

Another Finger Math Trick Just for Nine

There is a trick that works separately, just for multiplying by the number nine.

• To start, hold up all ten fingers, with your palms facing you.
• Assign each finger a number, starting with your left-hand thumb and ending with your right-hand thumb. The left-hand thumb will be one, the left-hand index finger will be two, and so on until you reach the number 10 for your right-hand thumb.
• To tackle a problem, put down the corresponding finger of the number you're multiplying by nine. For example, if you are multiplying 9 x 8, you'd put down the eighth finger (which will be on your right hand).
• Count all the fingers to the left of the finger you have folded down. This will give you 7. This is the first digit of your answer.
• Count all the fingers to the right of the finger you have folded down. This will give you 2. This is the second digit of your answer.
• Put the numbers together! Your answer is 72.

Finger Multiplication Tricks Can Make Math Easy and Fun

While the hope is that your kids will eventually memorize their multiplication charts , using some quick hand tricks for multiplication and letting them count things out on their fingers is not a bad way to learn. It keeps frustration at bay since the answer is always a fingertip away, and the repetition of having to figure it out will help cement those facts into their brains.

Watch CBS News

How two high school students solved a 2,000-year-old math puzzle

By Bill Whitaker , Aliza Chasan , Sara Kuzmarov, Mariah Campbell

May 5, 2024 / 7:00 PM EDT / CBS News

A high school math teacher at St. Mary's Academy in New Orleans, Michelle Blouin Williams, was looking for ingenuity when she and her colleagues set a school-wide math contest with a challenging bonus question. That bonus question asked students to create a new proof for the Pythagorean Theorem, a fundamental principle of geometry, using trigonometry. The teachers weren't necessarily expecting anyone to solve it, as proofs of the Pythagorean Theorem using trigonometry were believed to be impossible for nearly 2,000 years.

But then, in December 2022, Calcea Johnson and Ne'Kiya Jackson, seniors at St. Mary's Academy, stepped up to the challenge. The $500 prize money was a motivating factor.

After months of work, they submitted their innovative proofs to their teachers. With the contest behind them, their teachers encouraged the students to present at a mathematics conference, and then to seek to publish their work. And even today, they're not done. Now in college, they've been working on further proofs of the Pythagorean Theorem and believe they have found five more proofs. Amazingly, despite their impressive achievements, they insist they're not math geniuses.

"I think that's a stretch," Calcea said.

The St. Mary's math contest

When the pair started working on the math contest they were familiar with the Pythagorean Theorem's equation: A² + B² = C², which explains that by knowing the length of two sides of a right triangle, it's possible to figure out the length of the third side.

When Calcea and Ne'Kiya set out to create a new Pythagorean Theorem proof, they didn't know that for thousands of years, one using trigonometry was thought to be impossible.  In 2009, mathematician Jason Zimba submitted one, and now Calcea and Ne'Kiya are adding to the canon.

Calcea and Ne'Kiya had studied geometry and some trigonometry when they started working on their proofs, but said they didn't feel math was easy. As the contest went on, they spent almost all their free time developing their ideas.

"The garbage can was full of papers, which she would, you know, work out the problems and if that didn't work, she would ball it up, throw it in the trash," Cal Johnson, Calcea's dad, said.

Neliska Jackson, Ne'Kiya's mother, says lightheartedly, that most of the time, her daughter's work was beyond her. 

To document Calcea and Ne'Kiya's work, math teachers at St. Mary's submitted their proofs to an American Mathematical Society conference in Atlanta in March 2023.

"Well, our teacher approached us and was like, 'Hey, you might be able to actually present this,'" Ne'Kiya said. "I was like, 'Are you joking?' But she wasn't. So we went. I got up there. We presented and it went well, and it blew up."

Why Calcea' and Ne'kiya's work "blew up"

The reaction was insane and unexpected, Calcea said. News of their accomplishment spread around the world. The pair got a write-up in South Korea and a shoutout from former first lady Michelle Obama. They got a commendation from the governor and keys to the city of New Orleans. 

Calcea and Ne'Kiya said they think there's several reasons why people found their work so impressive. 

"Probably because we're African American, one," Ne'Kiya said. "And we're also women. So I think-- oh, and our age. Of course our ages probably played a big part."

Ne'Kiya said she'd like their accomplishment to be celebrated for what it is: "a great mathematical achievement."

In spite of the community's celebration of the students' work, St. Mary's Academy president and interim principal Pamela Rogers said that with recognition came racist calls and comments. 

"[People said] 'they could not have done it. African Americans don't have the brains to do it.' Of course, we sheltered our girls from that," Rogers said. "But we absolutely did not expect it to come in the volume that it came."

Rogers said too often society has a vision of who can be successful.

"To some people, it is not always an African American female," Rogers said. "And to us, it's always an African American female."

Success at St. Marys 

St. Mary's, a private Catholic elementary and high school, was started for young Black women just after the Civil War. Ne'Kiya and Calcea follow a long line of barrier-breaking graduates. Leah Chase , the late queen of Creole cuisine, was an alum. So was Michelle Woodfork, the first African American female New Orleans police chief, and Dana Douglas, a judge for the Fifth Circuit Court of Appeals. 

Math teacher Michelle Blouin Williams, who initiated the math contest, said Calcea and Ne'Kiya are typical St. Mary's students. She said if they're "unicorns," then every student who's matriculated through the school is a "beautiful, Black unicorn."

Students hear that message from the moment they walk in the door, Rogers said. 

"We believe all students can succeed, all students can learn," the principal said. "It does not matter the environment that you live in."

About half the students at St. Mary's get scholarships, subsidized by fundraising to defray the $8,000 a year tuition. There's no test to get in, but expectations are high and rules are strict: cellphones are not allowed and modest skirts and hair in its natural color are required. 

Students said they appreciate the rules and rigor.

"Especially the standards that they set for us," junior Rayah Siddiq said. "They're very high. And I don't think that's ever going to change." 

What's next for Ne'Kiya and Calcea

Last year when Ne'Kiya and Calcea graduated, all their classmates were accepted into college and received scholarship offers. The school has had a 100% graduation rate and a 100% college acceptance rate for 17 years, according to Rogers.

Ne'Kiya got a full ride in the pharmacy department at Xavier University in New Orleans. Calcea, the class valedictorian, is studying environmental engineering at Louisiana State University. Neither one is pursuing a career in math, though Calcea said she may minor in math.

"People might expect too much out of me if I become a mathematician," Ne'Kiya said wryly. 

Bill Whitaker is an award-winning journalist and 60 Minutes correspondent who has covered major news stories, domestically and across the globe, for more than four decades with CBS News.

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Teens who discovered new way to prove Pythagoras’s theorem uncover even more proofs

As high school students, Calcea Johnson and Ne’Kiya Jackson worked to find valid new proof for the 2,000-year-old theorem

Two college freshmen who, during their final year of high school, found a new way to prove Pythagoras’s theorem by using trigonometry – which mathematicians for generations thought was impossible – have since uncovered multiple more such proofs, they revealed in a national interview on Sunday.

“We found five, and then we found a general format that could potentially produce at least five additional proofs,” Calcea Johnson said on CBS’s 60 Minutes, a little more than a year after she and Ne’Kiya Jackson collaborated on an accomplishment that earned them international recognition.

Nonetheless, in comments that stunned their interviewer, Bill Whitaker, the two graduates of St Mary’s Academy in New Orleans denied seeing themselves as math geniuses and dismissed any interest in pursuing careers in mathematics.

“People might expect too much out of me if I become a mathematician,” Jackson said, shaking her head. Johnson, for her part, added: “I may take up a minor in math, but I don’t want that to be my job job.”

Sunday’s conversation on CBS’s popular Sunday evening news magazine were perhaps their most extensive, widely broadcast remarks to date on the new ground that they broke with respect to the Pythagorean theorem.

The 2,000-year-old theorem established that the sum of the squares of a right triangle’s two shorter sides equals the square of the hypotenuse – the third, longest side opposite the shape’s right angle. Countless schoolchildren taking geometry have memorized the notation summarizing the theorem: a 2 + b 2 = c 2 .

For 2,000 years, mathematicians maintained that any alleged proof of the Pythagorean theorem that was based in trigonometry would constitute a logical fallacy known as circular reason – in essence, trying to validate an idea with the idea itself.

But the bonus question on a math contest that Johnson and Jackson took home to complete during the Christmas break of their final year at St Mary’s served as the impetus for them to plot out a new way to demonstrate that one could indeed use trigonometry to prove Pythagoras’s theorem.

Their work was so compelling that the pair went to a regional meeting of the American Mathematical Society in Atlanta in March 2023 to outline their findings. At the organization’s recommendation, Jackson and Johnson have submitted their discoveries for final peer review and publication – as well as working on additional proofs while that process is pending, as 60 Minutes noted.

The 60 Minutes interview gave Johnson and Jackson occasion to reflect on the intense reaction caused by initial media reports on their innovative work at St Mary’s, a Catholic high school that has been dedicated to educating Black girls since its founding shortly after the US civil war.

Some of it was negative. Some in the math community smarted at claims in a press release issued by St Mary’s that asserted Jackson and Johnson’s research was “unprecedented”. And they flocked to social media demanding that a 2009 trigonometry-based proof for Pythagoras’s theorem get its due.

Yet a lot of the reaction to Johnson and Jackson was positive, especially as mathematicians who picked apart their work confirmed that – by all indications – they had arrived at a valid new proof, a celebration-worthy accomplishment.

Michelle Obama wrote a post on social media that linked to a story about Johnson and Jackson, adding the text: “I just love this story. … Way to go, Ne’Kiya and Calcea! I’m rooting for you and can’t wait to see what you all do next.”

They also received a commendation from Louisiana’s then governor as well as symbolic keys to the city of New Orleans .

Asked on 60 Minutes why they thought people were so impressed with what they had done, Jackson said she thought the public was surprised young Black women could author such a feat.

“I’d like to be celebrated for what it is,” Jackson said. “Like – it’s a great mathematical achievement.”

Jackson is now attending New Orleans’ Xavier University and enrolled in its pharmacy department. Meanwhile, Johnson – who graduated from St Mary’s as its valedictorian – is now an environmental engineering student at Louisiana State University in Baton Rouge.

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