problem solving examples for grade 7

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28 7th Grade Math Problems With Answers And Worked Examples

Linda Bortnick

7th grade math problems build on students’ reasoning skills and understanding of key mathematical concepts.

We’ve created 28 7th grade math problems for your classroom, suitable for use as additional practice before an assessment, as a Do Now activity, or for an Exit Slip to check for understanding after a lesson. Math teachers designed these problems to build on students’ math skills and help to prepare them for the rigor expected ahead in 8th grade and in high school courses like Algebra, Geometry, and Statistics.

What are 7th grade math problems?

7th grade math problems combine 7th grade math concepts with grade-level appropriate reasoning and problem solving skills.

7th grade math content includes computing with all rational numbers in all four operations, using the correct order of operations  to simplify expressions, solving problems involving percentages, using proportional reasoning, solving 1 and 2 step equations, multi-step inequalities and extending knowledge of geometry.

Why focus on problem solving in 7th grade?

Excelling at computation is great; excelling at computation but not understanding application is pointless. The purpose of mathematics is to learn to reason, find patterns, and use logic, so solving math problems  that are rich, deep, and complex is necessary to meet those purposes.

Mathematics today has advanced beyond just computation and evolved into a marriage of numbers and words – not only do students need to get to a numerical answer they also need to be able to communicate their thinking and processes along the way. They need repeated practice in reasoning, perseverance, and rational thinking to solve problems throughout their lives.

7th grade is a perfect time to extend problem solving activities in the classroom. The curriculum moves away from much of the memorization required in elementary school of basic addition facts and times tables , and builds on 6th grade content to focus on making connections between mathematical reasoning , logic, and numbers.

28 7th Grade Math Problems

28 7th grade math questions targeting the skills and knowledge needed for the four operations, order of operations, percentages, proportional relationships, one-step and two-step equations, geometry and word problems. Includes answers and explanations.

The 7th grade math curriculum

This article focuses on problems for 7th grade students, especially concepts taught in pre-algebra. Pre-algebra includes concepts of proportional relationships, algebraic expressions, equivalent expressions, equations, rational numbers, graphs, and extension of knowledge of the number system, including whole numbers and mixed numbers. The problems align with common core standards.

Seventh grade math can be considered a turning point in mathematics. It bridges the concepts taught in elementary school, including the foundations of arithmetic and number sense, with the rigor expected ahead in high school courses like Algebra, Geometry, and Statistics.

7th grade math problems

We’ve designed 28 7th grade math problems to use as a whole class or in a Tier 2  or Tier 3 intervention  to help develop and secure students’ mathematical knowledge.  

7th grade math problems: Four operations

• Solve -4 + 10. Use the number line.

2. Solve: -8 – 12. Use the number line.

Solution :        6

Solution :         -20

3. Solve: 4(-3)(-2)

Solution :          24

4. Bill said the answer to -3 – 12 is 9. What mistake did he make? What is the correct answer?

Solution : Bill knows that the opposite of subtraction is addition, but he forgot to take the opposite of 12, so he re-wrote the problem as -3 + 12. Since we are subtracting 12 from -3, the answer is the same as -3+-12, which is -15.

7th grade math problems: Order of operations

5. Solve: 2(10-8) ÷ 2 + 4

    Solution : 2(10-8) ÷ 2 + 4

                    2(2) ÷ 2 + 4

                    4  ÷ 2 + 4

                    2 + 4

                    6

6. Solve: (3 + 10 ÷ 2 – 6) x 6

     Solution : (3 + 10 ÷ 2 – 6) x 6

                    (3 + 5 – 6) x 6

                    (8 – 6) x 6

                     2 x 6

                     12

7. Solve: -5(8) ÷ 2 + 6

    Solution : -5(8) ÷ 2 + 6

                    -40 ÷ 8

                     -5

8. Solve: (-2)3 – 2 + 6 ÷ 3

    Solution : (-2)3 – 2 + 6 ÷ 3

                     -8  – 2 + 6 ÷ 3

                     -8 – 2 + 2

                     -10 + 2

                     -8

7th grade math problems: Percentages

9. Isabella got 16 out of 40 questions wrong on her quiz. What percent did she get correct?

Solution : \frac{16}{40} can be simplified to \frac{2}{5} , which is equivalent to \frac{40}{100} or 40%. If Isabella got 40% incorrect, she got 60% correct (100-40=60).

10. Without doing any computation, explain whether \frac{38}{72} is greater than or less than 50%.

Solution: \frac{38}{72} is greater than 50%. \frac{36}{72} is equivalent to \frac{1}{2} , which is equivalent to 50%. Since 38 is a little greater than 36, \frac{38}{72} is a little greater than 50%.

11. Put the following in order from least to greatest: \frac{3}{4} , 76%, 0.68, \frac{3}{5} , \frac{35}{50} , 0.702

Solution: \frac{3}{5} , 0.68, \frac{35}{50} , 0.702, \frac{3}{4} , 76%

12. A store marked all shoes on sale for 30% off. What percent will Sam pay for shoes? Explain your answer.

Solution: Sam will pay 70% for shoes. The full price is 100%, so if 30% is saved, the remaining 70% will be the sales price.

7th grade math problems: Proportional relationships

13. \frac{5}{6}  = \frac{x+2}{15}

      Solution: 6( x +2) = 5(15)

                     6 x + 12 = 75

                  -12   -12

                    \frac{6x}{6} = \frac{63}{6}

      x =   10.5

14. Three out of every five students are wearing jeans. If there are 20 students in total, how many are wearing jeans?

Solution: \frac{3}{5}  = \frac{x}{20}

                3 (20) = 5 x

                60 = 5 x

                12 = x

15. Three out of every five students are wearing jeans. If there are 20 students in all, how many are not wearing jeans?

Solution: From the last problem, we saw that \frac{3}{5} is the same as the 12 students wearing jeans. If there are 20 students total, we can subtract the 12 wearing jeans from the 20 total to find that 8 are not wearing jeans. We could also set up this proportion and solve to get 8.

                \frac{2}{5}  = \frac{x}{20}

16. A museum requires 12 chaperones for the 60 students attending the field trip. How many students are assigned to each chaperone?

         \frac{12}{60} = \frac{1}{x}

        12 x = 60(1)

        12 x = 60

          x = 5

Each chaperone will have a group of 5 students.

7th grade math problems: One-step equations and two-step equations

17. Solve: x + 7.1 = 15.9

         Solution:   x + 7.1 = 15.9

                           -7.1      -7.1

                         x = 8.8

18. Solve: x – 63 = 106.75

         Solution: x – 63 = 106.75

                       +63     +63

                         x  = 169.75

 19. Solve: 6( x + 3) = -6

         Solution: 6 x + 18 = -6

                     – 18    -18

                          \frac{6x}{6} = \frac{-24}{6}

                         x = -24

20. Solve: 0.5 x + 10 = 36

         Solution: 0.5 x + 10 = 36

                           -10    -10

                        0.5 x = 26

                        x = 52

7th grade math problems: Geometry

21. Madison measured this angle with her protractor and said “It is 60°.”

Without measuring the angle, Bella said she could tell Madison’s answer was incorrect. How did Bella know this?

Solution : Bella knew this angle could not be 60° because this angle is obtuse

but a 60° angle is acute.

22. Find the circumference of the circle.

Solution: C = πd

            C =  15π

23. Use the figure to fill in the blanks:

Angles A and B are _________________ angles so their measures are _______________________________.

Solution: Angles A and B are vertical  angles so their measures are equal.

24. Find the value of x .

         Solution: 5 x + 4 x =90

                      9 x = 90

                        x = 10

7th grade math problems: Math word problems

25. The 7th Graders at Marxville Middle School voted for their student council representatives. There were 200 votes cast in all. How many votes did the winner get?

Solution: Alexandra won the election with 30% of the votes. To find 30% of the 200 total votes, we can multiply 0.3 (200) to discover that she got 60 votes in all.

26. Brian runs every 12 days and Stella every 8 days. Both Brian and Stella ran today. How many days will it be before they both run on the same day again?

Solution: This is a Least Common Multiple problem. Brian runs on days 12, 24, 36, 48… and Stella runs on days 8, 16, 24, 32…, so they will both run again on Day 24.

27. Mr. Orlando is planting his vegetable garden this summer. He plants \frac{3}{4} of the garden with peppers and \frac{1}{4} with tomatoes. Of the peppers, \frac{1}{3} are red peppers. What fraction of the entire garden will be red peppers?

Solution: Red peppers will make up \frac{1}{3} of \frac{3}{4} (the pepper section) of the garden. \frac{1}{3} x \frac{3}{4} = \frac{1}{4} , so \frac{1}{4} of the entire garden will consist of red peppers.

28. Will the product of -45(96) be positive or negative? Without solving, how do you know?

Solution: The answer will be negative. Multiplying a negative number by a positive one always leads to a negative product.

Challenges in teaching 7th grade mathematics

Let’s face it, most 7th graders don’t wake up every morning excited to get to math class that day to expand on their mathematical knowledge and thinking! Often, there are much more pressing matters in the mind of a 12 year old. However, with the right mindset and classroom climate, we can support our students to help them put forth their best effort every day.

• Praise effort:  Developing a growth mindset in the classroom  is key. Even if a student struggles to come to an answer, it is critical to praise the effort. We must make the thinking part of mathematics something to treasure rather than solely focus on correct numerical answers.
• Include low threshold high ceiling activities: These go a long way in piquing the interest level of most students.
• Have fun:  Don’t be afraid to have some fun math activities  and silliness in your lessons, whether that’s through funny estimation activities  or videos to get students interested, engaged and willing to try.

How can Third Space Learning help with 7th grade math?

STEM-specialist tutors help close learning gaps and address misconceptions for struggling 7th grade math students. One-on-one online math tutoring sessions help students deepen their understanding of the math curriculum and keep up with difficult math concepts. Each student works with a private tutor who adapts instruction and math lesson content in real-time according to the student’s needs to accelerate learning.

4 top tips for teaching 7th grade math problems

Here are some teaching tips to overcome common challenges and support problem solving in your classroom:

• Focus on effort, not accuracy:  Students come into 7th grade with many ideas about math. Many of these ideas are likely negative and may result in some math anxiety . They’ve heard math is hard, you’re never going to use it, it’s confusing. Be excited about what you’re teaching and learning! Praise kids for trying their best, not for always being correct. Consider establishing effort-based reward systems, such as publicly nominating students for Mathematician of the Month.
• Check for understanding:  Be sure students understand questions before they attempt to answer them. Have them rephrase the question or explain what they’re looking for to solve the problem.
• Ask for wrong answers: Students may feel a lack of confidence in math but if you ask them for a wrong answer, they may feel more inclined to answer. It can also prompt students to engage with and make sense of the information given in the problem in a less pressured environment.

Alternatively, as teachers, you can give some suggestions. For example, “3/10 of students got an A on the last test. There are 20 students in the class. How many got an A?” Could 20 of the students have gotten an A? No? Why not? Could 8.5 of the students have gotten an A? Why not? Then get more specific. Did more or less than 10 of the students get an A? How do you know?

• Use relatable, real-world problems:  Sometimes, problem solving uses questions about gas mileage or building fences that kids either don’t relate to or don’t care about. Find out what your students’ interests are and incorporate that into your math instruction. Alternatively, use seasonal and relatable contexts such as Thanksgiving math activities , summer math  or mardi gras math .

7th grade math worksheets

Looking for more resources? Please see our selection of seventh grade math worksheets covering 7th grade key math topics and more. Each includes printable resources and step-by-step answer keys:

• Surface Area Of Rectangular Prisms Worksheet
• Fraction to Decimal Worksheet
• Adding and Subtracting Integers Worksheet
• Adding and Subtracting Scientific Notation Worksheet
• Distributing Exponents Worksheet
• Simplifying Expressions Worksheet

READ MORE :

• 25 Fun Math Problems For Elementary School
• Math Questions For 5th Graders
• 4th Grade Math Problems
• 6th Grade Math Problems
• 8th grade math problems

7th grade math problems FAQ

What math should a 7th grader know?

A 7th grader should be able to compute using all four operations with positive and negative rational numbers. They should be able to simplify expressions, solve one- and two- step equations, and solve proportions. Additionally, 7th graders can work with probability of simple and compound events, as well as study concepts of geometry, including finding angle measurements and solving for area and perimeter of regular and irregular polygons and circles.

What does 7th grade math focus on?

Seventh grade math focuses on working with positive and negative rational numbers in simplifying expressions and equations as well as solving one- and two-step equations. Additionally, extensive work is done in solving proportions and ratios.

Is 7th grade math hard?

Seventh grade math is always achievable! Students who find it hard usually come into the grade without having mastered basic concepts and facts. Be sure students are automatic with facts across addition, subtraction, multiplication, and division and have the ability to work with fractions and decimals. It will make the new concepts in 7th grade easier to learn.

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Solving Inequalities Questions [FREE]

Downloadable skills and applied questions about solving inequalities.

Includes 10 skills questions, 5 applied questions and an answer key. Print and share with your classes to support their learning.

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7th Grade Math Word Problems

Related Pages More Math Word Problems Algebra Word Problems More Singapore Math Word Problems 7th Grade Math Word Problems 2

In these lessons, we will look at some examples of 7th grade word problems with answers, solved using the Singapore Math block diagram method as well as using Algebra.

The following are some examples of 7th Grade Math Word Problems that deals with ratio and proportions.

These are Grade 7 word problems from a Singapore text. The problems are solved both using algebra (the way it is generally done in the US) and using block diagrams (the way it was shown in the Singapore text). You can then decide which one you prefer.

Singapore Math 2 (A Grade 7 Algebra Word Problem from a Singapore text)

Example: Mr. Rozario bought some apples and oranges. The ratio of the number of apples to the number of oranges are 2:5. He gave 3/4 of the apples to his sister and 34 oranges to his brother. The ratio of the number of apples to the number of oranges that he has now is 2:3. How many apples did Mr. Rozario buy? How many oranges did Mr. Rozario buy?

Example: At first, the ratio of Sam’s savings to Ray’s savings was 5:4. After each of them donated $45 to charity, the ratio of Sam’s savings to Ray’s savings became 13:10. What was Sam’s savings at first?

Example: Dan had 50% fewer stickers than Jeff. After Jeff gave 20 of his stickers to Dan, Dan had 40% fewer stickers than Jeff (had after he gave away 20 of his stickers). How many stickers did Dan have at first?

Example: There are 8 more girls than boys in a particular class. 3/5 of the boys and 1/3 of the girls were born in Georgia. If the number of boys that were born in Georgia is equal to the number of girls that were born in Georgia, how many students (boys and girls together) in that class were born in Georgia?

Example: Ray and Omar collect stamps. Originally, 1/5 of Omar’s stamps were equivalent to 1/3 of Ray’s stamps. If Ray gave Omar 24 stamps, Omar would have 3 times as many stamps as Ray. Find the number of stamps each of them had in the beginning.

Example: Jim had 103 red and blue marbles. After giving 2/5 of his blue marbles and 15 of his red marbles to Samantha, Jim had 3/7 as many red marbles as blue marbles. How many blue marbles did he have originally?

Example: The ratio of the amount of turkey to the amount of chicken at the grocery store was 8:3 in the morning. By the end of the day, 14 pounds of turkey had been sold. The ratio of the amount of turkey to the amount of chicken was now 3:2. a. How many pounds of turkey did the grocery store have in the morning? b. How many pounds of chicken did the grocery store have in the morning?

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• Students learn about ratios, mixed properties, statistics and other seventh grade skills.
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7th Grade Math Problems

In 7th grade math problems you will get all types of examples on different topics along with the solutions.

Keeping in mind the mental level of child in Grade 7, every efforts has been made to introduce new concepts in a simple language, so that the child understands them easily. The difficulty level of the problems has been reduced and mathematical concepts have been explained in the simplest possible way. Each topic contains a large number of examples to understand the applications of concepts. If student follow math-only-math they can improve their knowledge by practicing the solutions step by step which will help you to score in your exam.

●  Set Theory

Sets : An introduction to sets, methods for defining sets, element of set and use of set notations.

Objects Form a Set : State, whether the following objects form a set or not by giving reasons.

Elements of a Set : Learn how to find the elements of a set with the help of various types of problems on the basic concepts of sets.

Properties of Sets : Using the basic properties to represent a set learn to solve various basic types of problems on sets.

Representation of a Set : Definition with examples of statement form, roster form or tabular form, set builder form cardinal number of a set and the standard sets of numbers.

Different Notations in Sets : Some of the familiar notations used in sets which are generally required to solve various types of problems on sets.

Standard Sets of Numbers : Learn to represent the standard sets of numbers using the three methods i.e. statement form, roster form and set builder form.

Types of Sets : Definition with examples of empty set or null set, singleton set, finite set, infinite set, cardinal number of a set, equivalent set and equal sets.

Pairs of Sets : Definition with examples of equal set, equivalent set, disjoint sets and overlapping set.

Subset : Definition with examples of subset and its types, super set, proper subset, power set and universal set.

Subsets of a Given Set : How to find the number of subsets of a given set and number of proper subsets of a given set.

Operations on Sets : Learn the meaning. What are the four basic operations on sets? How the operations are carried out in union of sets and intersection of sets?

Union of Sets : Definition of union of sets with examples. Learn how to find the union of two sets and worked-out examples.

Intersection of Sets : Definition of intersection of sets with examples. Learn how to find the intersection of two sets and worked-out examples.

Difference of two Sets : Learn how to find the difference between the two sets and worked-out examples.

Complement of a Set : Definition of complement of a set and their properties with some worked-out examples.

Cardinal number of a set : Definition of a cardinal number of a set, the symbol used for showing the cardinal number, worked-out examples.

Cardinal Properties of Sets : Learn how to solve the real-life word problems on set using the cardinal properties.

Venn Diagrams : Learn to represent the basic concepts of sets using Venn-diagram in different situations.

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Ordered Pair

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Decimal Numbers : Learn reading decimal numbers, writing decimal numbers etc,.

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Decimal Places : Learn reading and writing the place values of a decimal number in words.

Decimal and Fractional Expansion : Learn how to expand the decimal numbers and the decimal numbers in fractions

Like and Unlike Decimals : Learn to identify like decimals and unlike decimals.

Conversion of Unlike Decimals to Like Decimals : Learn to convert unlike decimals in like decimals

Comparing Decimals : Learn to order and compare decimal numbers, arranging decimals in ascending order and arranging decimals in descending order

Adding Decimals : Learn to add decimals in the correct order, word problems.

Subtracting Decimals : Learn to subtract decimals in the correct order, word problems

Simplify Decimals Involving Addition and Subtraction Decimals : Examples for simplifying decimals for adding and subtracting.

Multiplying Decimal by a Whole Number : Learn to multiply decimals by a whole number in the correct order, word problems

Multiplying Decimal by a Decimal Number : Learn to multiply decimals by a decimal number in the correct order, word problems

Dividing Decimal by a Whole Number : Learn to divide decimals by a whole number in the correct order, word problems

Dividing Decimal by a Decimal Number : Learn to divide decimals by a decimal number in the correct order, word problems

Simplification of Decimal : Examples for simplifying decimals, operation on decimals.

Converting Decimals to Fractions : Learn to express decimal number into fraction

Converting Fractions to Decimals : Learn to express fraction into decimal number

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Rounding Decimals to the Nearest Whole Number : Learn to round off decimals to the nearest whole number

Rounding Decimals to the Nearest Tenths : Learn to round off decimals to the nearest tenths

Rounding Decimals to the Nearest Hundredths : Learn to round off decimals to the nearest hundredths

Round a Decimal : Problems for rounding off decimals

H.C.F. and L.C.M. of Decimals : Learn to find the highest common factor, lowest common multiple of more than two decimal numbers

Terminating Decimal : Definition of terminating decimals, examples to identify terminating decimals

Non-Terminating Decimal : Definition of non-terminating decimal, examples to identify non-terminating decimals

Repeating or Recurring Decimal : Definition of recurring decimals, examples to identify recurring decimals

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Conversion of Pure Recurring Decimal into Vulgar Fraction : Learn to express pure recurring decimals to vulgar fractions with examples.

Conversion of Mixed Recurring Decimals into Vulgar Fractions : Learn to express mixed recurring decimals to vulgar fractions with examples.

BODMAS/PEMDAS rule:

BODMAS Rule : Learn how to follow the BODMAS rules to simply the order of operations.

BODMAS Rules - Involving Integers : Learn how to follow the BODMAS rules to simply the order of operations involving integers.

BODMAS/PEMDAS Rules - Involving Decimals : Learn how to follow the BODMAS rules to simply the order of operations involving decimals

PEMDAS Rule : Learn how to follow the PEMDAS rules to simply the order of operations.

PEMDAS Rules - Involving Integers : Learn how to follow the PEMDAS rules to simply the order of operations involving integers.

PEMDAS Rules - Involving Decimals : Learn how to follow the PEMDAS rules to simply the order of operations involving decimals.

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Worksheet on Work Done in a Given Period of Time

Worksheet on Time Required to Complete a Piece of Work

Worksheet on Pipes and Water Tank

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Problems on Unitary Method using Direct Variation

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● Unitary Method - Worksheets

Worksheet on Direct Variation using Unitary Method

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Worksheet on Word Problems on Unitary Method

Worksheet on Inverse Variation Using Unitary Method

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What is Simple Interest?

Calculate Simple Interest

Practice Test on Simple Interest

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Simple Interest Worksheet

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● Algebraic Expressions - Worksheets

Worksheet on Dividing Polynomial by Monomial

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Practice Test on Framing the Formula

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Worksheet on Framing the Formula

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Worksheet on Changing the Subject in an Equation or Formula

● Algebraic Identities

Square of The Sum of Two Binomials : Using the formula of (a + b) \(^{2}\) = a \(^{2}\) + b \(^{2}\) + 2ab, learn to evaluate the square of the sum of two terms.

Square of The Difference of Two Binomials : Using the formula of (a - b)\(^{2}\)  = a \(^{2}\) + b \(^{2}\) - 2ab, learn to evaluate the square of the difference of two terms.

Product of Sum and Difference of Two Binomials : Using the formula of a \(^{2}\) - b \(^{2}\) = (a + b) (a - b), learn to evaluate the product of sum and difference of two terms.

Product of Two Binomials whose First Terms are Same and Second Terms are Different : Learn to use the given formulas to evaluate the product of the two terms whose 1 \(^{st}\) terms are same and 2 \(^{nd}\) terms are different.

• (x + a) (x + b) = x \(^{2}\) + x(a + b) + ab

• (x + a) (x - b) = x \(^{2}\) + x (a – b) – ab

• (x - a) (x - b) = x \(^{2}\) – x (a + b) + ab

• (x - a) (x + b) = x \(^{2}\) + x (b – a) – ab

Square of a Trinomial : Learn to use the formula for the expansion of the square of a trinomial.

• (a + b + c) \(^{2}\) = a \(^{2}\) + b \(^{2}\) + c \(^{2}\) + 2ab + 2bc + 2ca

• (a + b - c) \(^{2}\) = a \(^{2}\) + b \(^{2}\) + c \(^{2}\) + 2ab – 2bc - 2ca

• (a - b + c) \(^{2}\) = a \(^{2}\) + b \(^{2}\) + c \(^{2}\) – 2ab – 2bc + 2ca

• (a - b - c) \(^{2}\) = a \(^{2}\) + b \(^{2}\) + c \(^{2}\) – 2ab + 2bc – 2ca

Cube of The Sum of Two Binomials : Learn the formula to determine the cube of the sum of two terms.

(a + b) \(^{3}\) = a \(^{3}\)   + 3a \(^{2}\) b + 3ab \(^{2}\)   + b \(^{3}\)

             = a \(^{3}\) + b \(^{3}\) + 3ab (a + b)

Cube of The Difference of Two Binomials : Learn the formula to determine the cube of the difference of two terms.

(a - b) \(^{3}\) = a \(^{3}\)   – 3a \(^{2}\) b + 3ab \(^{2}\)   – b \(^{3}\)

           = a \(^{3}\) – b \(^{3}\) – 3ab (a - b)

Cube of a Binomial :

Square of a Binomial :

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Congruent Shapes :

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Side Angle Side Congruence : 

Angle Side Angle Congruence :

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Right Angle Hypotenuse Side congruence :

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Word problems on Pythagorean Theorem :

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Number of Triangles Contained in a Polygon

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Sum of the Exterior Angles of a Polygon

● Polygons - Worksheets

Worksheet on Polygon and its Classification

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● Quadrilateral

Quadrilateral : Perimeter of Quadrilateral : Angle Sum Property of a Quadrilateral : Missing angle of a Quadrilateral :

Angles of a Quadrilateral are in Ratio :

●   Symmetrical Figure

Linear Symmetry :  Identify symmetric and non-symmetric figures and also identify shapes having horizontal line of symmetry, vertical line of symmetry, both horizontal and vertical lines of symmetry, infinite lines of symmetry and no line of symmetry.

Lines of Symmetry :  Identify the geometrical shapes having 1, 2, 3, 4, 0 and so on or infinite lines of symmetry.

Point Symmetry :  How to find the point symmetry of letters of the English alphabet and the different geometrical figures. Learn the important points to find the conditions that satisfy centre of symmetry.

Reflection Symmetry :

Reflection and Symmetry :

Nets of a Solids :

Rotational Symmetry :  Learn what is rotational symmetry and types of rotation i.e. clockwise rotation and anticlockwise rotation. 

Order of Rotational Symmetry :  Learn the different orders of rotation of the shapes through 360° in clockwise direction and anticlockwise direction.

Types of Symmetry :  Learn the various symmetries i.e. linear symmetry, point symmetry and rotational symmetry of the geometrical shapes. 

Reflection :  Learn how reflection is related to math, define reflection using an image and worked-out examples on math reflection. 

Reflection of a Point in x-axis :  Learn how to draw the image on the graph paper to find the reflection of a point in x-axis.

Reflection of a Point in y-axis :  Learn how to draw the image on the graph paper to find the reflection of a point in y-axis.

Reflection of a point in origin :  Learn how to draw the image on the graph paper to find the reflection of a point in origin. 

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90 Degree Clockwise Rotation :  Learn with the help of solved examples to rotate a figure 90 degrees clockwise direction around the origin on a graph paper.

90 Degree Anticlockwise Rotation :  Learn with the help of solved examples to rotate a figure 90 degrees anticlockwise direction around the origin on a graph paper.

180 Degree Rotation :  Learn with the help of solved examples to rotate a figure 180 degrees clockwise direction and anticlockwise direction around the origin on a graph paper.

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• How to Construct a Line Graph?

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Building Problem-solving Skills for 7th-Grade Math

Mathematics is a subject that requires problem-solving skills to excel. In 7th grade, students begin to encounter more complex math concepts, and the ability to analyze and solve problems becomes increasingly important. Building problem-solving skills for math not only helps students to master math concepts but also prepares them for success in higher-level math courses and in life beyond academics. 

In this article, we will several key skills that are needed for success in 7th-grade math, and also explore how they can benefit students both academically and personally. We will also provide tips and strategies to help students develop and improve their problem-solving skills. Let’s dive in!

Building Analytical Skills

The first of seven important skills to build is that of analytical skills. These allow students to analyze a problem and break it down into smaller parts. From there, they’re able to identify the key components that need to be addressed. Analytical skills also hone students’ abilities to identify patterns. Students should be able to identify patterns in mathematical data, such as in number sequences, geometric shapes, and graphs. Importantly, students should not just be able to recognize the pattens, but they should be able to describe them (more on that in communication) and use them to make predictions and solve problems.

We alluded to this earlier, but breaking down problems is an essential component of analytical skills. Students with strong analytical skills can break problems down into smaller and more manageable parts. They are then able to identify key components of a problem and use this information to develop a strategy for solving it. 

Along with identifying patterns comes identifying relationships. Students with good mathematical analytical skills can identify relationships between different mathematical concepts, such as the relationship between addition and subtraction, or the relationship between angles and shapes. Through strengthening this skill, students will be able to describe these relationships and use them to solve problems. 

An important part of analytical skills is the ability to analyze data. Students should be able to analyzeand interpret data presented in a variety of formats, such as graphs, charts, and tables. They should be able to use this data to make predictions, draw conclusions, and solve problems.

Speaking of conclusions, reaching sound conclusions based on mathematical data is a fundamental skill needed for making predictions based on trends in a graph, or drawing inferences from a set of data.

Another skill students should master is the ability to compare and contrast mathematical concepts, such as the properties of different shapes or the strategies for solving different types of problems. Through this, they’ll be able to use the information they gather to solve problems. 

With all these skills at play comes arguably the most important: Critical thinking. This is an indicator that a student really grasps the concepts and it’s just repeating them back to you on command. Critical thinking is the ability to evaluate information and arguments, and make judgements and decisions based on evidence, and apply logic and reasoning to solve problems.

Building Creative Thinking

This is the ability for students (or anyone, really) to think outside the box and come up with innovative solutions to problems. This involves being able to approach problems from different angles and to consider multiple perspectives. For a 7th-grader, this skill can be exercised through the following:

• Thinking Outside the Box: Students should be encouraged to think creatively and come up with innovative solutions to problems. This involves thinking outside the box and considering multiple perspectives.
• Finding Multiple Solutions: Students should be able to come up with multiple solutions to a problem and evaluate each one to determine which is the most effective.
• Developing Original Ideas: Students should be able to develop original ideas and approaches to solving problems. This involves being able to come up with unique and innovative solutions that may not have been tried before.
• Making Connections: Students should be able to make connections between different mathematical concepts and apply these connections to solve problems. This involves looking for similarities and differences between concepts and using this information to make new connections.
• Visualizing Solutions : Students should be able to visualize solutions to problems and use diagrams, charts, and other visual aids to help them solve problems.
• Using Metaphors and Analogies: Students should be able to use metaphors and analogies to help them understand complex mathematical concepts. This involves using familiar concepts to explain unfamiliar ones and making connections between different ideas.

Building Problem-Solving Strategies

It may sound like the same thing, but building problem-solving strategies is not the same as building problem-solving skills. Building strategies for problem-solving lends itself to actual problem-solving. Let’s expand on this: Say your student is presented a problem that they’re struggling with, these are some of the problem-solving strategies they may use in order to solve the puzzle.

• Identify the problem: The first step in problem-solving is to identify the problem and understand what is being asked. Students should carefully read the problem and make sure they understand the question before attempting to solve it.
• Draw a diagram: Students can draw a diagram to help visualize the problem and better understand the relationships between different parts of the problem.
• Use logic: Students can use logic to identify patterns and relationships in the problem. They can use this information to develop a plan to solve the problem.
• Break the problem down: Students can break a complex problem down into smaller, more manageable parts. They can then solve each part of the problem individually before combining the solutions to get the final answer.
• Guess and check: Students can guess and check different solutions to the problem until they find the correct answer. This method involves trying different solutions and evaluating the results until the correct answer is found.
• Use algebra: Algebraic equations can be used to solve a variety of mathematical problems. Students can use algebraic equations to represent the problem and solve for the unknown variable.
• Work backward: Students can work backward from the final answer to determine the steps required to solve the problem. This method involves starting with the end goal and working backward to determine the steps needed to get there.

Building Persistence and Perseverance

In an increasingly instant-gratification world with apps, searches and AI chatbots just a click away, this is an important skill not just in the math classroom, but for life in general. Problem-solving, whether that’s a math problem or a life challenge, often requires persistence and perseverance. Student need to learn to be able to stick with a problem even when it seems challenging, difficult, or seemingly impossible. Here are ways you can encourage your students to stick it out when working on problems:

• Trying multiple approaches: When faced with a challenging problem, students can demonstrate persistence by trying multiple approaches until they find one that works. They don’t give up after one attempt but keep trying until they find a solution.
• Reframing the problem: If a problem seems particularly difficult, students can demonstrate perseverance by reframing the problem in a different way. This can help them see the problem from a new perspective and come up with a different approach to solve it.
• Asking for help: Sometimes, even with persistence, a problem may still be difficult to solve. In these cases, students can demonstrate perseverance by asking for help from their teacher or classmates. This shows that they are willing to put in the effort to find a solution, even if it means seeking assistance.
• Learning from mistakes: Making mistakes is a natural part of the problem-solving process, but students can demonstrate persistence by learning from their mistakes and using them to improve their problem-solving skills. They don’t get discouraged by their mistakes, but instead, they use them as an opportunity to learn and grow.
• Staying focused: In order to solve complex math problems, it’s important for students to stay focused and avoid distractions. Students can demonstrate perseverance by staying focused on the problem at hand and not getting distracted by other things.

Building Communication Skills

We alluded to this earlier, but a central part of building problem-solving skills is building the ability to articulate a problem or a solution. This isn’t just for the sake of personal understanding, but critical for collaboration. Students need to be able to explain their thinking, ask questions, and work with others to solve problems. Here are some examples of communication skills that can be used to build problem-solving skills:

• Clarifying understanding: Students can ask questions to clarify their understanding of the problem. They can seek clarification from their teacher or classmates to ensure they are interpreting the problem correctly.
• Explaining their reasoning: When solving a math problem, students can explain their reasoning to show how they arrived at a particular solution. This can help others understand their thought process and can also help students identify errors in their own work.
• Collaborating with peers: Problem-solving can be a collaborative effort. Students can work together in groups to solve problems and communicate their ideas and solutions with each other. This can lead to a better understanding of the problem and can also help students learn from each other.
• Writing clear explanations: When presenting their solutions to a math problem, students can write clear and concise explanations that are easy to understand. This can help others follow their thought process and can also help them communicate their ideas more effectively.
• Using math vocabulary: Math has its own language and using math vocabulary correctly is essential for effective communication. Students can demonstrate their understanding of math concepts by using correct mathematical terms and symbols when explaining their solutions.

Building Mathematical Knowledge

This would seem like a no-brainer, since you’re a math educator clicking on an article about building math problem-solving skills. However, it’s worth being explicit that problem-solving in math requires a solid understanding of mathematical concepts, including arithmetic, algebra, geometry, and data analysis. Students need to be able to apply these concepts to solve problems in real-world contexts.

7th-grade math covers a wide range of mathematical concepts and skills. Here are some examples of mathematical knowledge that 7th-grade math students should have:

• Algebraic expressions and equations: Students should be able to write and simplify algebraic expressions and solve one-step and two-step equations.
• Proportional relationships: Students should be able to understand and apply proportional relationships, including identifying proportional relationships in tables, graphs, and equations.
• Geometry: Students should have a solid understanding of geometry concepts such as angles, triangles, quadrilaterals, circles, and transformations.
• Statistics and probability: Students should be able to analyze and interpret data using measures of central tendency and variability, and understand basic probability concepts.
• Rational numbers: Students should have a solid understanding of rational numbers, including ordering, adding, subtracting, multiplying, and dividing fractions and decimals.
• Integers: Students should be able to perform operations with integers, including adding, subtracting, multiplying, and dividing.
• Ratios and proportions: Students should be able to understand and use ratios and proportions in a variety of contexts, including scale drawings and maps.

In conclusion, problem-solving skills are essential for success in 7th grade math. Analytical skills, critical and creative thinking, problem-solving strategies, persistence, communication skills, and mathematical knowledge are all important components of effective problem-solving. By developing these skills, students can approach math problems with confidence and achieve their full potential.

If you enjoyed this read, be sure to browse more of our articles . More importantly, if you want to save yourself hours of preparation time by having full math curriculums, review guides and tests available at the click of a button, be sure to sign up to our 7th Grade Newsletter . You’ll receive loads of free lesson resources, tips and advice and exclusive subscription offers!

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Build Confidence with Math Problem Solving

Daily Problem Solving will help your 7th grade students master the skills they need to be successful with challenging word problems...and have fun doing it .

What you'll get with this download:

Your download includes a full week of Daily Problem Solving for Grade 7 to try out in your own classroom. Developed with the brain in mind, these multi-step word problems will challenge your learners without overwhelming them. Best of all, you'll be able to watch their skills and confidence grow as they begin to internalize strategies for conquering this difficult math skill.

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Themed-problems and a weekly fun fact make it easy to keep learners engaged

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Formatted for quick & easy implementation that won't add stress to your planning

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Includes both print and digital student options for added flexibility

I've been using this with my extremely anxious learner and it works wonders to build confidence because we see the improvement.

These have given me such an insight into my students' abilities, and I've watched them improve their skills and successfully solve multi-step word problems.

Jilliana D.

STUDENT PRINTABLES

You'll receive a full week of printable Daily Problem Solving practice for your students. This format includes:

• Paper-saving format that fits a full week on one page
• Space for student work, feedback, and self-reflection
• Themed problems & weekly fun fact to engage

DIGITAL SLIDES

Daily digital slides offer your remote or online learners the opportunity to build problem-solving skills in a structured format designed for success.

• Single slide per day prevents overwhelm or distraction
• Organized with clear space to solve & answer
• Engaging graphics, themed problems, & fun facts

Don’t wait, get started helping your learners master word problems today!

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Seventh Grade (Grade 7) Problem Solving Strategies Questions

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SAMPLE MATH PROBLEMS FOR GRADE 7

Question 1 :

What is the square root of 196?

(A) 14                (B) 19                 (C) 12

Question 2 :

A play ground is 100 m long and 70 m wide. How much distance does a girl run when she runs 5 times around the playground?

(A) 2500 m      (B) 2300 m        (C) 1700 m

Question 3 :

From a rectangular sheet of tin, of size 100 cm by 80 cm, are cut four squares of side 10 cm from each corner. Find the area of the remaining sheet.

(A) 2400 cm 2       (B) 3600 cm 2         (C) 7600 cm 2

Question 4 :

If x = 2, y = 3 and z =-2 then find the value of  6x-7y + 4z

(A) 12                  (B) -17             (C) -10

Question 5 :

Find the value of x in 4 4x-7  = 4  x-1

(A) 2            (B) 5            (C) 3

Question 6 :

Expand (x  + 5y) 2

(A) x 2  + 10 xy + 25y 2      (B) x 2  - 10 xy - 25y 2   

(C) x 2  + 10 xy - 25y 2

Question 7 :

On selling a clock for $240, a trader loses 4%. In order to gain 10%, he must sell the clock for

(A) $275      (B) $295      (C) $365

Question 8 :

Find the remainder when 8x 3  - 12x 2  + 6x +15 divided by 2x-3. 

(A) 20         (B) 24         (C) 10

Question 9 :

The difference of two numbers is 72 and quotient obtained by dividing the one by other is 3. Find the numbers.

(A) 25, 97             (B) 15, 87          (C) 36, 108

Question 10 :

In triangle ABC, DE parallel to EC find AC when AB = 15 cm, AD  =  10 cm and AE  =  8 cm 

(A) 8 cm      (B) 4 cm      (C) 2 cm

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Fractions and mixed numbers grade 7 math questions and problems with solutions and explanations.

Detailed solutions and full explanations to fractions and mixed numbers grade 7 problems are presented.

• Find fraction F with denominator less than 8 such that 2 / 8 + F = 1 Solution Solve for F F = 1 - 2 / 8 = 8 / 8 - 2 / 8 , common denominator = 6 / 8 , subtract numerator = 3 / 4 , reduce fraction
• Find two fractions F1 and F2 with same denominator equal to 6 such that F1 + F2 = 1 and F1 - F2 = 2/3 Solution Let us write F1 + F2 and F1 - F2 as follows F1 + F2 = 1 = 6 / 6 F1 - F2 = 2 / 3 = 4 / 6 F1 and F2 are fractions with denominator 6. Their numerator must add up to 6 and their difference is 4. Hence F1 and F2 are equal to F1 = 5 / 6 and F2 = 1 / 6
• Which fraction is equivalent to 16%? Solution 16% is written as a fraction and reduced 16% = 16 / 100 = 4 / 25
• 1/2 + 1/5 + 1/6 = Solution The LCM of 2, 5 and 6 is first calculated 2 = 2 5 = 5 6 = 2 × 3 LCM = 2 × 5 × 3 = 30 We use the LCM as the lowest common denominator for all 3 fractions 1 / 2 + 1 / 5 + 1 / 6 = (1×15) / (2×15) + (1×6) / (5×6) + (1×5) / (6×5) Simplify = 15 / 30 + 6 / 30 + 5 / 30 Add and reduce = 26 / 30 = 13 / 15
• 3 3/5 + 5 1/2 = Solution Add whole parts together and fractional parts together 3 3/5 + 5 1/2 = (3 + 5) + (3/5 + 1/2) Common denominator for fractions 3/5 and 1/2 = 8 + (6/10 + 5/10) = 8 + 11/10 Change improper fraction 11/10 into mixed number and add = 8 + 10/10 + 1/10 = 8 + 1 + 1/10 = 9 1/10
• 1/7 × 2 2/5 = Solution Change the mixed number 2 2/5 into fraction and multiply 1/7 × 2 2/5 = 1/7 × 12/5 = 12 / 35
• 1/12 × 0.2 = Solution Change decimal number 0.2 into fraction 0.2 = 2/10 = 1/5 Multipliply the two fractions 1/12 × 0.2 = 1/12 × 1/5 = 1 / 60
• 2/5 ÷ 6 = . Solution Use division rule 2/5 ÷ 6 = 2/5 ÷ 6/1 = 2/5 × 1/6 = 2/30 Reduce fraction = 1/15
• 9/7 + 2 = . Solution Change 2 into fraction 2/1 and set common denominator. 9/7 + 2 = 9/7 + 2/1 = 9/7 + 14/7 = 23/7 = 3 2/7
• 2 1/3 + 4/2 = . Solution Simplify 4/2 and add. 2 1/3 + 4/2 = 2 1/3 + 2 = 4 1/3
• 3 1/5 ÷ 5 = Solution Change mixed number 3 1/5 into frcation and rewrite 5 as fraction 5/1. 3 1/5 ÷ 5 = 16/5 ÷ 5/1 = Apply rule of division of fractions and simplify. = 16/5 × 1/5 = 16/25
• 1/2 + 4 1/3 - 3 2/5 = Solution Add/subtract whole parts and fractional parts separately. 1/2 + 4 1/3 - 3 2/5 = 4 - 3 + 1/2 + 1/3 - 2/5 Find LCM of 2, 3 and 5 . 2 = 2 3 = 3 5 = 5 LCM(2,3,5) = 2×3×5 = 30 Use LCM as common denominator. = 1 + (15/30 + 10/30 - 12/30) = 1 + 13/30 = 1 13/30
• 5/2 ÷ 7/2 - 1/5 = Solution Order of operation division first 5 / 2 ÷ 7 / 2 - 1 / 5 = 5 / 2 × 2 / 7 - 1 / 5 Simplify = 5 / 7 - 1 / 5 Common denominator and subtract = 25 / 35 - 7 / 35 = 18 / 35
• (0.2 + 1/5) × 2/7 = Solution Change decimal number 0.2 into fraction 1/5 (0.2 + 1/5) × 2/7 = (1/5 + 1/5) × 2/7 Use order of opeartions = 2/5 × 2/7 = 4 / 35
• (3 1/2 + 3/5) × 1/7 = Solution Change mixed number 3 1/2 into fraction (3 1/2 + 3/5) × 1/7 = (7/2 + 3/5) × 1/7 Find common denominator to 7/2 and 3/5 and add = (35/10 + 6/10) × 1/7 = 41/10 × 1/7 Multiply fractions = 41 / 70
• (1/2 + 2/3) ÷ 0.2 = Solution Add the fractions within the brackets and write 0.2 as a fraction 1/5 (1/2 + 2/3) ÷ 0.2 = (3/6 + 4/6) ÷ 1/5 = 7/6 ÷ 1/5 Use rule of division of fractions = 7/6 × 5/1 = 35 / 6 = 5 5/6
• Order from least to greatest: 3 4/7 , 3 3/5 , 3 1/2 , 3 11/20 . Solution The given mixed numbers all have same whole parts but different fractions. It is easier to compare fractions when they have common denominator. The common denominator of the fractional parts is the LCM of 7,5,2,and 20. 7 = 7 5 = 5 2 = 2 20 = 2 2 × 5 LCM(7,5,2,20) = 7 × 5 × 2 2 = 140 We now rewrite all mixed with fractional parts with same denominator 3 4/7 = 3 80/140 3 3/5 = 3 84/140 3 1/2 = 3 70/140 3 11/20 = 3 77/140 We now order mixed numbers from least to greatest 3 1/2 , 3 11/20 , 3 4/7 , 3 3/5
• Order from least to greatest: 2 7/8 , 2.66 , 262% , 25/8 . Solution We need to rewrite the given numbers in a single form. Let us write them in decimal form 2 7/8 = 2 + 7/8 = 2 + 0.875 = 2.875 2.66 = 2.66 262% = 262 / 100 = 2.62 25/8 = 24/8 + 1/8 = 3 + 0.125 = 3.125 We now use the decimal form of the given numbers to order them from least to greatest. 262% , 2.66 , 2 7/8 , 25/8

Links and References

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Unit 9: Geometry

About this unit.

Geometric shapes are all around us. The world is built with them. In this series of tutorials and exercises you'll become familiar with Euclidean geometry and terms like scale drawings, parts of a circle, area, angles, and geometric figures.

Area and circumference of circles

• Geometry FAQ (Opens a modal)
• Radius, diameter, circumference & π (Opens a modal)
• Labeling parts of a circle (Opens a modal)
• Radius, diameter, & circumference (Opens a modal)
• Radius & diameter from circumference (Opens a modal)
• Relating circumference and area (Opens a modal)
• Area of a circle (Opens a modal)
• Partial circle area and arc length (Opens a modal)
• Circumference review (Opens a modal)
• Area of circles review (Opens a modal)
• Radius and diameter Get 5 of 7 questions to level up!
• Circumference of a circle Get 3 of 4 questions to level up!
• Area of a circle Get 5 of 7 questions to level up!
• Circumference of parts of circles Get 3 of 4 questions to level up!
• Area of parts of circles Get 3 of 4 questions to level up!

Area and circumference challenge problems

• Finding circumference of a circle when given the area (Opens a modal)
• Area of a shaded region (Opens a modal)
• Impact of increasing the radius (Opens a modal)
• Circumference and rotations Get 3 of 4 questions to level up!
• Area and circumference of circles challenge Get 3 of 4 questions to level up!
• Shaded areas Get 3 of 4 questions to level up!

Vertical, complementary, and supplementary angles

• Angles: introduction (Opens a modal)
• Complementary & supplementary angles (Opens a modal)
• Vertical angles (Opens a modal)
• Complementary and supplementary angles review (Opens a modal)
• Vertical angles are congruent proof (Opens a modal)
• Name angles Get 3 of 4 questions to level up!
• Identifying supplementary, complementary, and vertical angles Get 5 of 7 questions to level up!
• Complementary and supplementary angles (visual) Get 3 of 4 questions to level up!
• Complementary and supplementary angles (no visual) Get 5 of 7 questions to level up!
• Vertical angles Get 3 of 4 questions to level up!
• Finding angle measures between intersecting lines Get 3 of 4 questions to level up!

Missing angle problems

• Find measure of vertical angles (Opens a modal)
• Find measure of angles word problem (Opens a modal)
• Equation practice with complementary angles (Opens a modal)
• Equation practice with supplementary angles (Opens a modal)
• Equation practice with vertical angles (Opens a modal)
• Finding missing angles Get 5 of 7 questions to level up!
• Create equations to solve for missing angles Get 5 of 7 questions to level up!
• Unknown angle problems (with algebra) Get 5 of 7 questions to level up!

Constructing triangles

• Construct a right isosceles triangle (Opens a modal)
• Construct a triangle with constraints (Opens a modal)
• Triangle inequality theorem (Opens a modal)
• Triangle side length rules Get 3 of 4 questions to level up!

Slicing geometric shapes

• Slicing a rectangular pyramid (Opens a modal)
• Cross sections of 3D objects (basic) Get 3 of 4 questions to level up!

Volume and surface area word problems

• Volume word problem: gold ring (Opens a modal)
• Volume of triangular prism & cube (Opens a modal)
• Volume of rectangles inside rectangles (Opens a modal)
• Volume and surface area word problems Get 3 of 4 questions to level up!

Age Problems

Age word problems.

Every now and then, we encounter word problems that require us to find the relationship between the ages of different people. Age word problems typically involve comparing two people’s ages at different points in time, i.e. at present, in the past, or in the future.

This lesson is divided into two parts. Part I involves age word problems that can be solved using a single variable while Part II contains age word problems that need to be solved using two variables .

Let’s get familiar with age word problems by working through some examples.

PART I: Age Word Problems Solvable with One Variable

Example 1: Tanya is 28 years older than Marcus. In 6 years, Tanya will be three times as old as Marcus. How old is Tanya now?

In this problem, we are only asked to find Tanya’s current age. However, the problem also gave us a lot of other information which can be overwhelming. To help us organize the important details, let’s create a table to list what we know so far.

Since we are only given details about their current ages and what they will be 6 years from now, we’ll go ahead and gray out the Past column.

You may notice that Tanya’s current age is defined using the age of Marcus. However, Marcus’s present age is currently unknown. So let’s express Marcus’s age using the variable [latex]x[/latex]. Since Tanya is 28 years older than Marcus , then Tanya’s present age must be [latex]x+28[/latex].

Next, let’s fill in the Future column which will consist of their ages in 6 years. All we have to do is add 6 to Tanya and Marcus’s present or current ages. Therefore, we have:

• Tanya: [latex]\left( {x + 28} \right) {\color{red}+ 6} = x + 34[/latex]
• Marcus: [latex]x {\color{red}+ 6}[/latex]

Now that our table is filled out, we can go ahead and create our equation based on the information provided. The problem states the following:

In 6 years , Tanya will be three times as old as Marcus.

Here we are trying to find the relationship between their ages in the future. We can simply say that,

Tanya’s age in 6 years = 3( Marcus’s age in 6 years )

With that in mind, we can easily construct our equation.

Our next step now is to solve for [latex]x[/latex]. But before that, remember that our problem is asking us to find Tanya’s current age. Since Tanya’s age is defined using Marcus’s current age (which is [latex]x[/latex]), we have to find his age first in order to determine what Tanya’s present age is.

Now that we have the value for [latex]x[/latex], let’s find out what Tanya and Marcus’s current ages are. We can do this by simply replacing the [latex]x[/latex]’s with [latex]8[/latex].

CURRENT AGES (present)

• Marcus: [latex]x = {\textbf{8}}[/latex] years old
• Tanya: [latex]x + 28 = {\color{red}8} + 28 = {\textbf{36}}[/latex] years old

Going back to the problem’s question, how old is Tanya now?

Answer: Tanya is 36 years old.

Answer Check:

At this point, we are confident that our answer is correct. But, how can we be 100% sure? Well, it’s always a good idea especially in math, to check our answers so we’re certain that we got the correct values.

For this problem, we can simply verify if our answer makes our future statement true. Do you remember this statement?

In 6 years, Tanya will be three times as old as Marcus.

We know the present ages of Marcus and Tanya which are [latex]8[/latex] and [latex]36[/latex], respectively. Hence in 6 years, Marcus will be [latex]14[/latex] years old while Tanya will be [latex]42[/latex] years old.

So, will Tanya be three times as old as Marcus in 6 years? The answer is Yes .

Example 2: Bruce is 4 years younger than Hector. Twenty years ago, Hector’s age was 13 years more than half the age of Bruce. How old are they now?

By just reading the problem, we can already tell that there is a great deal of information that we have to sort through and that this problem includes a fraction. Most students easily get lost in all the given information, let alone solving equations that involve fractions. But, don’t fret! As long as you stick with the basic principles and steps on how to solve age word problems, you’ll be fine.

Right now, we don’t know Bruce or Hector’s current age. But since Bruce’s age is expressed in relation to Hector’s age, then our unknown variable will be based on Hector’s age. In other words,

• Let [latex]{\textbf{\textit{h}}} =[/latex] Hector’s age
• [latex]{\textbf{\textit{h} – 4}} =[/latex] Bruce’s age, since he is 4 years younger than Hector

Let’s organize all these important data into a table. We’re only given details about their present and past (20 years ago) ages so we’ll gray out the Future column.

Twenty years ago, both Bruce and Hector were 20 years younger so we’ll subtract 20 from each of their present ages.

• Bruce: [latex]\left( {h – 4} \right) {\color{red}- 20} = h – 24[/latex]
• Hector: [latex]h {\color{red}- 20}[/latex]

Our table is now ready so we can proceed to create our equation. As you can see under the Past column, we were able to create algebraic expressions for Bruce and Hector’s ages 20 years ago. But our problem also told us that,

Twenty years ago , Hector’s age was 13 years more than half the age of Bruce.

Since Hector’s age 20 years ago is also 13 years more than half of Bruce’s age, we can take these two algebraic expressions and set them equal to each other, to create an equation.

Hector’s age 20 years ago = [latex]\Large{1 \over 2}[/latex]( Bruce’s age 20 years ago )[latex]+ 13[/latex]

We’re now ready to solve for the unknown variable, [latex]h[/latex].

Therefore, Hector’s present age is [latex]{\textbf{42}}[/latex] years old.

On the other hand, you may recall that Bruce’s current age is: [latex]h – 4[/latex]. Since [latex]h = 42[/latex], then Bruce’s current age is [latex]42 – 4 = {\textbf{38}}[/latex].

So, how old are they now?

Answer: Hector is 42 years old and Bruce is 38 years old .

The final step is to check our answers by substituting the unknown values into our original equation to verify if each side of the equation equals the other.

Great! Our answer checks. This just showed us that if we take Bruce’s age twenty years ago, which is 18, and divide it in half, we get 9. Adding 13 to that ([latex]9 + 13[/latex]), we get 22 which was Hector’s age twenty years ago.

Therefore, we are able to confirm that twenty years ago when Hector was 22 years old and Bruce was 18 years old, Hector’s age was 13 years more than half the age of Bruce.

Example 3: Stella is 13 years younger than Kwame. Nine years from now, the sum of their ages will be 43. Find the present age of each.

This problem is a little different from our previous two examples as we are given the sum of their ages in 9 years. But right off the bat, we can see that Stella’s age is defined in terms of Kwame’s age. Therefore, we’ll select a variable to represent Kwame’s current age. In this instance, let’s use “[latex]k[/latex]”.

• Let [latex]{\textbf{\textit{k}}} =[/latex] Kwame’s age
• [latex]{\textbf{\textit{k} – 13}} =[/latex] Stella’s age, since she is 13 years younger than Kwame

Nine years from now, both Kwame and Stella will be 9 years older. So we’ll simply add 9 to their present ages above to show their future ages.

• Kwame: [latex]k {\color{red}+ 9}[/latex]
• Stella: [latex]\left( {k – 13} \right) {\color{red}+ 9} = k – 4[/latex]

Let’s complete our table.

Now that we have the algebraic expressions for both their ages in 9 years, we can add these expressions to create our equation. We were given the following details:

Nine years from now , the sum of their ages will be 43 .

So we have,

Checking back at our table, [latex]k[/latex] stands for Kwame’s age. But since our problem asked us to find the current ages for both, let’s do a little bit more solving.

• Kwame: [latex]k = {\textbf{19}}[/latex] years old
• Stella: [latex]k – 13 = {\color{red}19} – 13 = {\textbf{6}}[/latex] years old

Answer: Kwame is 19 years old and Stella is 6 years old .

Let’s now verify if indeed the sum of Kwame and Stella’s ages in 9 years will be 43.

• Kwame’s age in 9 years: [latex]k + 9 = {\color{red}19} + 9 = {\textbf{28}}[/latex]
• Stella’s age in 9 years: [latex]k – 4 = {\color{red}19} – 4 = {\textbf{15}}[/latex]

Perfect! The total of their ages nine years from now is 43 so our answers are correct.

Example 4: Mr. Cook is 34 years old. His son is 22 years younger than him. In how many years will Mr. Cook’s age be 24 years less than three times as old as his son?

We already know their current ages, so before we delve any further, let’s start filling in our table.

Note that since the son is 22 years younger than Mr. Cook, we subtracted 22 from 34 to get his son’s current age, [latex]34 – {\color{red}22} = 12[/latex].

This problem is unique because it’s not asking us for their ages at a certain point in time like usual. Instead, it asks us to find out the number of years when Mr. Cook’s age will meet a certain relationship with his son’s age in the future.

But at this point, we don’t know how long it will take for Mr. Cook to be 24 years less than three times as old as his son. So, let’s assign the unknown variable “[latex]x[/latex]” to stand for the number of years then add [latex]x[/latex] to both of their current ages to create algebraic expressions that will represent how old they will be after [latex]x[/latex] years.

Since Mr. Cook’s age after [latex]x[/latex] number of years ([latex]x + 34[/latex]) will also be 24 years less than three times as old as his son , we can set these two algebraic expressions equal to each other, thus creating our equation.

Now that we have our equation, let’s solve for [latex]x[/latex].

As you may recall, [latex]x[/latex] stands for the number of years from now that will take for Mr. Cook to be 24 years less than three times as old as his son. Therefore,

Answer: In 11 years , Mr. Cook’s age will be 24 years less than three times as old as his son.

To check if our answer is correct, we must first find out how old will Mr. Cook and his son be in 11 years. Substituting the value of [latex]x[/latex] which is 11 into our algebraic expressions, we get:

• Mr. Cooks’s age in 11 years: [latex]x + 34 = {\color{red}11} + 34 = {\textbf{45}}[/latex]
• Son’s age in 11 years: [latex]x + 12 = {\color{red}11} + 12 = {\textbf{23}}[/latex]

So in 11 years, Mr. Cook will be 45 years old while his son will be 23 years old.

This time, I’ll leave it up to you to verify if indeed during that time, his age of 45 years old will be 24 years less than three times as old as his son. If it meets the condition, then our answer is correct.

Example 5: The sum of one-fifth of Annika’s age four years ago and half of her age in six years is 33. How old is she now?

Compared to our previous exercises, this problem only involves one person. Also, instead of comparing the ages of two people at a certain point in time, we will be comparing Annika’s ages at different points in time, i.e. 4 years ago and in 6 years.

We don’t know Annika’s current age so let’s select the variable [latex]{\textbf{\textit{a}}}[/latex] to represent this unknown value. We’ll use this variable as well to create algebraic expressions that will stand for her past and future ages.

• Let [latex]{\textbf{\textit{a}}} =[/latex] Annika’s current age
• [latex]{\textbf{\textit{a} – 4}} =[/latex] Annika’s age 4 years ago
• [latex]{\textbf{\textit{a} + 6}} =[/latex] Annika’s age 6 years from now

Our problem also told us that if we add [latex]\Large{1 \over 5}[/latex] of Annika’s age 4 years ago and [latex]\Large{1 \over 2}[/latex] of her age 6 years from now , the sum is 33 .

With this information, it’s easy for us to write our equation.

Our next step is to solve for the unknown variable, [latex]a[/latex].

So, how old is Annika now?

Answer: Annika is currently 44 years old.

As I mentioned before, it’s always a good practice to verify if you got the correct answer. To start, let’s find out what Annika’s past and future ages are.

• Annika’s age 4 years ago : [latex]a – 4 = {\color{red}44} – 4 = {\textbf{40}}[/latex]
• Annika’s age 6 years from now : [latex]a + 6 = {\color{red}44} + 6 = {\textbf{50}}[/latex]

Now that we know how old she was 4 years ago and how old she’ll be in 6 years, we’ll plug in these values into our original equation to see if both sides of the equation equal each other.

And they did! We were able to prove that the sum of [latex]\Large{1 \over 5}[/latex] of Annika’s age 4 years ago and [latex]\Large{1 \over 2}[/latex] of her age 6 years from now is indeed 33.

PART II: Age Word Problems Solvable with Two Variables

Example 6: The sum of Aaliyah and Harald’s ages is 28. Four years from now, Aaliyah will be three times as old as Harald. Find their present ages.

Neither Aaliyah nor Harald’s age is expressed in terms of the other. So for this problem, we will be using more than one variable to represent the unknown values. To start,

• Let [latex]{\textbf{\textit{a}}}[/latex] be Aaliyah’s age
• Let [latex]{\textbf{\textit{h}}}[/latex] be Harald’s age

Since they will be 4 years older in the next 4 years, we simply have to add 4 to their current ages to represent their future ages.

Looking back at our problem, there are two significant statements that can help us find our answers.

1) The sum of Aaliyah and Harald’s ages is 28.

From this statement, we can create the equation below:

2) Four years from now, Aaliyah will be three times as old as Harald.

Meanwhile, the statement above can be translated into the following equation:

We now have two equations to solve.

• Equation 1: [latex]a + h = 28[/latex]
• Equation 2: [latex]a + 4 = 3(h + 4)[/latex]

First, we’ll use equation 1 to solve for [latex]a[/latex].

Next, we’ll replace [latex]a[/latex] with [latex]28 – h[/latex] in equation 2 .

Perfect! We are able to find the values for both our unknown variables, [latex]a[/latex] and [latex]h[/latex], which also stand for the present ages for Aaliyah and Harald. So we have,

• Aaliyah’s present age: [latex]a = 28 – h = 28 – {\color{red}5} = {\textbf{23}}[/latex]
• Harald’s present age: [latex]h = {\textbf{5}}[/latex]

Answer: Currently, Aaliyah is 23 years old while Harald is 5 years old.

I’ll leave it up to you to check if our answers are correct. But as you can see, even with just using mental computation, we can already tell that the sum of Aaliyah and Harald’s ages is 28 ([latex]23 + 5 = 28[/latex]) which makes our first statement true. You may further check our answers by plugging in the values of [latex]a[/latex] and [latex]h[/latex] into equation 2 to verify if the left side of the equation equals the right, thus making our second statement true as well.

Example 7: The sum of the ages of Jaya and Nadia is three times Nadia’s age. Seven years ago, Jaya was three less than four times as old as Nadia. How old are they now?

This problem is similar to our previous example. However, for this one, we are not given the exact number for the sum. We first have to find out each of their current ages so we can determine what the sum is.

• Let [latex]{\textbf{\textit{y}}}[/latex] be Jaya’s age
• Let [latex]{\textbf{\textit{n}}}[/latex] be Nadia’s age

We then need to subtract 7 from their current ages to represent how old they were seven years ago.

Now that we’ve organized our data, let’s go through the significant statements given in our problem and translate each into an equation.

1) The sum of the ages of Jaya and Nadia is three times Nadia’s age.

2) Seven years ago, Jaya was three less than four times as old as Nadia.

Therefore, our two equations are:

• Equation 1: [latex]y + n = 3n[/latex]
• Equation 2: [latex]y – 7 = 4(n – 7) – 3[/latex]

Let’s first focus on equation 1 and solve for [latex]y[/latex].

Now we’ll solve for [latex]n[/latex] using the value of [latex]y[/latex] from equation 1. We’ll do this by replacing [latex]y[/latex] with [latex]2n[/latex] in equation 2 .

Taking the values of [latex]y[/latex] and [latex]n[/latex], we have:

• Jaya’s present age: [latex]y = 2n = 2({\color{red}12}) = {\textbf{24}}[/latex]
• Nadia’s present age: [latex]n = {\textbf{12}}[/latex]

So, going back to our problem. How old are they now?

Answer: Jaya is 24 years old and Nadia is 12 years old.

To check our answers, we’ll replace the values of [latex]y[/latex] and [latex]n[/latex] in equation 1 and equation 2. Again, I’ll leave it up to you to solve both equations and verify if each side of the equation equals the other. Once you’re done with your solutions, you’ll see that we are able to prove that both statements from our problem are true.

Example 8: The difference between the ages of Penelope and her son, Zack, is 34. In six years, Penelope will be four times as old as Zack’s age two years ago. How old are they now?

It’s easy to get lost in all the information given so we’ll focus first on assigning variables that will stand for the unknown values.

• Let [latex]{\textbf{\textit{p}}}[/latex] be Penelope’s current age
• Let [latex]{\textbf{\textit{z}}}[/latex] be Zack’s current age

One thing that’s unique about this problem is that it involves three different points in time. We are given not only the relationship between Penelope and her son’s age in the present time but also how their ages in 6 years are related to their ages two years ago.

To show this, we’ll subtract 2 from their ages now for their ages 2 years ago then add 6 to their current ages for their ages 6 years later .

Great! We now have variables and algebraic expressions to represent Penelope and Zack’s current ages as well as their ages in the past and in the future. Moving forward, let’s go through the important details given in the problem and create an equation from each statement.

1) The difference between the ages of Penelope and her son, Zack, is 34 .

Remember that Penelope is Zack’s mother so she’s definitely older than him. Therefore, we are subtracting Zack’s age from Penelope’s age to find the difference.

2) In six years, Penelope will be four times as old as Zack’s age two years ago.

Here are our two equations:

• Equation 1: [latex]p – z = 34[/latex]
• Equation 2: [latex]p + 6 = 4(z – 2)[/latex]

Let’s now work on equation 1 to solve for [latex]p[/latex].

Next, we’ll replace [latex]p[/latex] with [latex]34 + z[/latex] in equation 2 then solve for [latex]z[/latex].

• Penelope’s current age: [latex]p = 34 + z = 34 + ({\color{red}16}) = {\textbf{50}}[/latex]
• Zack’s current age: [latex]z = {\textbf{16}}[/latex]

How about we replace the unknown values in our table and also find out what their past and future ages are?

Going back to our original question, how old are they now?

Answer: Penelope is currently 50 years old while her son, Zack, is 16 years old.

Problem Solving Activities: 7 Strategies

• Critical Thinking

Problem solving can be a daunting aspect of effective mathematics teaching, but it does not have to be! In this post, I share seven strategic ways to integrate problem solving into your everyday math program.

In the middle of our problem solving lesson, my district math coordinator stopped by for a surprise walkthrough. 

I was so excited!

We were in the middle of what I thought was the most brilliant math lesson– teaching my students how to solve problem solving tasks using specific problem solving strategies. 

It was a proud moment for me!

Each week, I presented a new problem solving strategy and the students completed problems that emphasized the strategy. 

Genius right? 

After observing my class, my district coordinator pulled me aside to chat. I was excited to talk to her about my brilliant plan, but she told me I should provide the tasks and let my students come up with ways to solve the problems. Then, as students shared their work, I could revoice the student’s strategies and give them an official name. 

What a crushing blow! Just when I thought I did something special, I find out I did it all wrong. 

I took some time to consider her advice. Once I acknowledged she was right, I was able to make BIG changes to the way I taught problem solving in the classroom. 

When I Finally Saw the Light

To give my students an opportunity to engage in more authentic problem solving which would lead them to use a larger variety of problem solving strategies, I decided to vary the activities and the way I approached problem solving with my students. 

Problem Solving Activities

Here are seven ways to strategically reinforce problem solving skills in your classroom. 

Seasonal Problem Solving

Many teachers use word problems as problem solving tasks. Instead, try engaging your students with non-routine tasks that look like word problems but require more than the use of addition, subtraction, multiplication, and division to complete. Seasonal problem solving tasks and daily challenges are a perfect way to celebrate the season and have a little fun too!

Cooperative Problem Solving Tasks

Go cooperative! If you’ve got a few extra minutes, have students work on problem solving tasks in small groups. After working through the task, students create a poster to help explain their solution process and then post their poster around the classroom. Students then complete a gallery walk of the posters in the classroom and provide feedback via sticky notes or during a math talk session.

Notice and Wonder

Before beginning a problem solving task, such as a seasonal problem solving task, conduct a Notice and Wonder session. To do this, ask students what they notice about the problem. Then, ask them what they wonder about the problem. This will give students an opportunity to highlight the unique characteristics and conditions of the problem as they try to make sense of it. 

Want a better experience? Remove the stimulus, or question, and allow students to wonder about the problem. Try it! You’ll gain some great insight into how your students think about a problem.

Math Starters

Start your math block with a math starter, critical thinking activities designed to get your students thinking about math and provide opportunities to “sneak” in grade-level content and skills in a fun and engaging way. These tasks are quick, designed to take no more than five minutes, and provide a great way to turn-on your students’ brains. Read more about math starters here ! 

Create your own puzzle box! The puzzle box is a set of puzzles and math challenges I use as fast finisher tasks for my students when they finish an assignment or need an extra challenge. The box can be a file box, file crate, or even a wall chart. It includes a variety of activities so all students can find a challenge that suits their interests and ability level.

Calculators

Use calculators! For some reason, this tool is not one many students get to use frequently; however, it’s important students have a chance to practice using it in the classroom. After all, almost everyone has access to a calculator on their cell phones. There are also some standardized tests that allow students to use them, so it’s important for us to practice using calculators in the classroom. Plus, calculators can be fun learning tools all by themselves!

Three-Act Math Tasks

Use a three-act math task to engage students with a content-focused, real-world problem! These math tasks were created with math modeling in mind– students are presented with a scenario and then given clues and hints to help them solve the problem. There are several sites where you can find these awesome math tasks, including Dan Meyer’s Three-Act Math Tasks and Graham Fletcher’s 3-Acts Lessons . 

Getting the Most from Each of the Problem Solving Activities

When students participate in problem solving activities, it is important to ask guiding, not leading, questions. This provides students with the support necessary to move forward in their thinking and it provides teachers with a more in-depth understanding of student thinking. Selecting an initial question and then analyzing a student’s response tells teachers where to go next. 

Ready to jump in? Grab a free set of problem solving challenges like the ones pictured using the form below. 

Which of the problem solving activities will you try first? Respond in the comments below.

Shametria Routt Banks

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2 Responses

This is a very cool site. I hope it takes off and is well received by teachers. I work in mathematical problem solving and help prepare pre-service teachers in mathematics.

Thank you, Scott! Best wishes to you and your pre-service teachers this year!

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