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6th grade (Illustrative Mathematics)

Unit 1: area and surface area, unit 2: introducing ratios, unit 3: unit rates and percentages, unit 4: dividing fractions, unit 5: arithmetic in base ten, unit 6: expressions and equations, unit 7: rational numbers, unit 8: data sets and distribution.

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problem solving for grade 6 with solution and answer pdf

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Expressions and Equations 6th Grade

Welcome to our Expressions and Equations 6th Grade Worksheets.

Here you will find a range of algebra worksheets to help you learn about basic algebra, including generating and calculating algebraic expressions and solving simple equations.

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Generate the expression worksheets, calculate the expression worksheets.

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Want to gain a basic understanding of algebra?

Looking for some simple algebra worksheets?

Do you need a bank of useful algebra resources?

Look no further! The pages you need are below!

Here is our selection of basic algebra sheets to try.

We have split the worksheets up into 3 different sections:

  • Generate the algebra - and write your own algebraic expressions;
  • Calculate the algebra - work out the value of different expressions;
  • Solve the algebra - find the value of the term in the equation.
  • Mixed questions involving all 3 of the above.

By splitting the algebra up into sections, you only need to concentrate on one aspect at a time!

Each question sheet comes with its own separate answer sheet.

Want to test yourself to see how well you have understood this skill?.

  • Try our NEW quick quiz at the bottom of this page.

What is an algebraic expression?

An expression is a mathematical statement where variables and operations are combined.

  • 2a + 5 is an expression involving the variable a
  • 5(y 2 - 6) is another expression

What is an algebraic equation?

An equation is where an algebraic expression is equal to something, which might be a number, or another algebraic expression.

  • 2a + 5 = 7 is an equation
  • 5(y 2 - 6) = 3y + 8 is another equation

How to Generate an Expression

When we are generating an expression, we are taking a rule and turning it into algebra.

  • Subtract 6 from n could be written as n - 6.
  • Multiply d by 4 could be written as d x 4 or 4d.
  • Add 5 to p and then double the result is written as (p + 5) x 2 or 2(p + 5)

How to Calculate an Expression

When we are calculating the value of an expression, we work out the value of the expression when we give a value to the variable.

  • p + 5 has a value of 11 when p = 6 because 6 + 5 = 11
  • 4 - q has a value of 1 ½ when q = 2 ½ because 4 - 2 ½ = 1 ½
  • 4(n - 2) has a value of 32 when n = 10 because 4 x (10 - 2) = 4 x 8 = 32
  • 4(n - 2) has a value of -8 when n = 0 because 4 x (0 - 2) = 4 x (-2) = -8

How to Solve a Simple Equation

When we are solving an equation, we are finding out the value(s) of the variable in the equation.

  • then p = 4 because 4 + 5 = 9
  • Answer: p = 4
  • then n = 56 ÷ 7 = 8
  • Answer: n = 8
  • means that f = 4 because 12 - 4 = 8
  • Answer: f = 4
  • then t = 5 because 0.6 x 5 = 3
  • Answer: t = 5
  • then r = 32 because ½ x 32 = 16
  • Answer: r = 32

Expressions and Equations 6th Grade Worksheets

  • Generate the Expressions Sheet 6:1
  • PDF version
  • Generate the Expressions Sheet 6:2
  • Calculate the Expression Sheet 6:1
  • Calculate the Expression Sheet 6:2

Calculate the Expressions Walkthrough Video

This short video walkthrough shows several problems from our Calculate the Expression Worksheet 6:1 being solved and has been produced by the West Explains Best math channel.

If you would like some support in solving the problems on these sheets, please check out the video below!

Simplifying Expressions Worksheets

  • Simplifying Expressions Sheet 6:1
  • Simplifying Expressions Sheet 6:2

Simplifying Expressions Walkthrough Video

This short video walkthrough shows several problems from our Simplifying Expressions Worksheet 6:1 being solved and has been produced by the West Explains Best math channel.

Solve the Equation Worksheets

  • Solving Equations Sheet 6:1
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Expressions and Equations 6th Grade Mixed Worksheets

These questions involve solving equations, working out the value of expressions and also generating expressions.

They are a combination of questions from all the above categories.

  • Expressions and Equations Mixed Questions Sheet 6:1
  • Expressions and Equations Mixed Questions Sheet 6:2

Expressions and Equations Walkthrough Video

This short video walkthrough shows several problems from our Expressions and Equations Mixed Questions Worksheet 6:1 being solved and has been produced by the West Explains Best math channel.

More Recommended Math Worksheets

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The sheets on this page have been designed to factorize and expand a range of simple expressions using the distributive property..

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We have a selection of basic algebra worksheets which are aimed at 6th and 7th graders and involve similar skills to the sheets here.

Free Algebra Problem Solver

The Mathway Calculator is a great way to solve algebra problems that you can type into a calculator.

Try using this online calculator tool to solve one of your problems and watch it work!

There are a range of calculators to choose from to meet your needs.

The Mathway problem solver will answer your problem instantly and also give you a link to view each of the steps needed.

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Are you looking for some worksheets on factorising quadratic equations to print out?

Take a look at our support pages on quadratic equations where you will hopefully find what you are looking for.

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If you are looking for a fun printable algebra game to play then try out our algebra game page.

You will find a range of algebra games that make learning algebra fun and non-threatening.

The only equipment you need is a scientific calculator, some dice, and a few counters!

PEMDAS Worksheets

The sheets in this section involve using parentheses and exponents in simple calculations.

There are also lots of worksheets designed to practice and learn about PEMDAS.

Using these worksheets will help your child to:

  • know and understand how parentheses works;
  • understand how exponents work in simple calculations.
  • understand and use PEMDAS to solve a range of problems.
  • PEMDAS Problems Worksheets 5th Grade
  • 6th Grade Order of Operations

Interactive Equality Explorer

This interactive equality explorer has been produced by PhET Interactive Simulations at the University of Colorado.

It is a useful tool for exploring different ideas including negative numbers and algebra equations and equality.

Probably the most useful part of the app is to use the 'Solve It' section once you are confident how it works.

You can then select your level of difficulty and start solving some algebraic equations by getting your variables onto one side of the equation and the numerical values on the other, and then multiplying or dividing the equation until you find the value of the required variable.

interactive equality explorer by PhET

  • Interactive Equality Explorer by PhET

Expressions and Equations 6th Grade Quiz

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This quick expressions and equations 6th grade quiz tests your knowledge and skill at generating and calculating expressions, as well as solving equations.

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Maths word Problems With Solutions for Grade 6

Detailed solutions and full explanations to grade 6 maths word problems are presented.

Solutions to Word Problems

  • Two numbers N and 16 have LCM = 48 and GCF = 8. Find N. Solution The product of two integers is equal to the product of their LCM and GCF. Hence. 16 × N = 48 × 8 N = 48 × 8 / 16 = 24
  • If the area of a circle is 81? square feet, find its circumference. Solution The area is given by ? × r × r. Hence ? × r × r = 81? r × r = 81 ; hence r = 81 feet The circumference is given by 2 × ? × r = 2 × pi × 9 = 18 ? feet
  • Find the greatest common factor (GFC) of 24, 40 and 60. Solution We first write the prime factorization of each given number 24 = 2 × 2 × 2 × 3 = 2 3 × 3 40 = 2 × 2 × 2 × 5 = 2 3 × 5 60 = 2 × 2 × 3 × 5 = 2 2 × 3 × 5 GFC = 2 2 = 4
  • In a given school, there 240 boys and 260 girls. a) What is the ratio of the number of girls to the number of boys? b) What is the ratio of the number of boys to the total number of pupils in the school? Solution a) ratio of girls to boys 260:240 or 13:12 b) ratio of boys to the total number of pupils 240:(240+260) or 240:500 or 12:25
  • If Tim had lunch at $50.50 and he gave 20% tip, how much did he spend? Solution The tip is 20% of what he paid for lunch. Hence tip = 20% of 50.50 = (20/100)*50.50 = 101/100 = $10.10 Total spent 50.50 + 10.10 = $60.60
  • Find k if 64 ÷ k = 4. Solution Since 64 ÷ k = 4 and 64 ÷ 16 = 4, then k = 16
  • Little John had $8.50. He spent $1.25 on sweets and gave to his two friends $1.20 each. How much money was left? Solution John spent and gave to his two friends a total of 1.25 + 1.20 + 1.20 = $3.65 Money left 8.50 - 3.65 = $4.85
  • What is x if x + 2y = 10 and y = 3? Solution Substitute y by 3 in x + 2y = 10 x + 2(3) = 10 x + 6 = 10 If we substitute x by 4 in x + 6 = 10, we have 4 + 6 = 10. Hence x = 4
  • A telephone company charges initially $0.50 and then $0.11 for every minute. Write an expression that gives the cost of a call that lasts N minutes. Solution Cost C for a call of 1 minute C = 0.50 + 0.11 Cost C for a call of 2 minutes C = 0.50 + 0.11 + 0.11 = 0.50 + 2 × 0.11 Cost C for a call of 3 minutes C = 0.50 + 0.11 + 0.11 + 0.11 = 0.50 + 3 × 0.11 We note that the cost C is equal to C = 0.50 + (number of minutes) × 0.11 If N is the number of minutes, the cost C is given by C = 0.50 + N × 0.11
  • A car gets 40 kilometers per gallon of gasoline. How many gallons of gasoline would the car need to travel 180 kilometers? Solution Each 40 kilometers, 1 gallon is needed. We need to know how many 40 kilometers are there in 180 kilometers? 180 � 40 = 4.5 × 1 gallon = 4.5 gallons
  • A machine fills 150 bottles of water every 8 minutes. How many minutes it takes this machine to fill 675 bottles? Solution 8 minutes are needed to fill 150 bottles. How many groups of 150 bottles are there in 675 bottles? 675 � 150 = 4.5 = 4 and 1/2 For each of these groups 8 minutes are needed. For 4 groups and 1/2 8 × 4 + 4 = 32 + 4 = 36 minutes. (4 is for 1/2 a group that needs half time) We can also find the final answer as follows 4.5 x 8 = 32 minutes
  • A car travels at a speed of 65 miles per hour. How far will it travel in 5 hours? Solution During each hour, the car travels 65 miles. For 5 hours it will travel 65 + 65 + 65 + 65 + 65 = 5 × 65 = 325 miles
  • A small square of side 2x is cut from the corner of a rectangle with a width of 10 centimeters and length of 20 centimeters. Write an expression in terms of x for the area of the remaining shape. Solution Let us first find the total area A of the rectangle before cutting the small is cut A = length × width = 20 × 10 = 200 A square of side 2x has an area B given by B = (2x) × (2x) = 4 × x × x = 4 x 2 The small square of area B is cut from the large rectangle of area A. Hence the area of the remaining shape is given by A - B = 200 - 4 x 2
  • A rectangle A with length 10 centimeters and width 5 centimeters is similar to another rectangle B whose length is 30 centimeters. Find the area of rectangle B. Solution Two rectangles A and B are similar if their lengths and widths are proportinal. Let L1 = 10 cm and W1 = 5 cm be the length and width of rectangle A. Let L2 = 30 cm and W2 be the length and width of rectangle B. Proportionality of the dimensions of the two rectangles is written as: L2 / L1 = W2 / W1 Substitute by the known quantities and find W2 30 / 10 = W2 / 5 For the above ratios to be equal, W2 must be equal to 15. Hence the area of rectamgle B is given by L2 × W2 = 30 cm × 15 cm = 450 cm 2
  • A school has 10 classes with the same number of students in each class. One day, the weather was bad and many students were absent. 5 classes were half full, 3 classes were 3/4 full and 1/8 of the students in the last two classes were absent. A total of 70 students were absent. How many students are registered in this school? Solution The given information is related to different classes and since we have 10 classes with the same number of students we need to find the number of registered students in each class. Let x be the number of registered students in each class. Hence the number of students absent in the 5 classes that were half full is given by: (1/2 of the students were absent in each class) 5 × ( 1/2 × x) = 5 x / 2 The number of students absent in the 3 classes that were 3/4 full is given by: ( a class that 3/4 full has 1/4 of the students absent) 3 × ( 1/4 × x) = 3 x / 4 The number of students absent in the 2 classes that were 1/8 full is given by: 1/8 × (2 × x) = 2 x / 8 The total number of students who were absent is equal to 70 (given). Hence 5 x / 2 + 3 x / 4 + 2 x / 8 = 70 rewrite fractions with same denominator (4 / 4) × 5 x / 2 + (2/ 2) × 3 x / 4 + 2 x / 8 = 70 Simplify 20 x / 8 + 6 x / 8 + 2 x / 8 = 70 add fractions 28 x / 8 = 70 Multiply both sides of the above equation by 8 / 28 (8 / 28) × 28 x / 8 = (8 / 28) × 70 Simplify to solve for x. x = 20 students per class total number of students for all 10 classes in the school: 10 × 20 = 200 students
  • The perimeter of square A is 3 times the perimeter of square B. What is the ratio of the area of square A to the area of square B. Solution Let x be the size of the side of square A and y be the size of the side of square B. The perimeters of the two squares are given by: Perimeter of A: 4 x and Perimeter of B: 4 y The expression "The perimeter of square A is 3 times the perimeter of square B" is written mathematically as: 4 x = 3(4 y) = 12 y Divide both sides of the above equation by 4 x = 3 y Square both sides of the above equation. (x) 2 = (3 y) 2 Simplify to obtain. x 2 = 9 y 2 The area of the two squares are given by Area of square A := x 2 and Area of square B = y 2 The ratio of the area of square A to the area of square B is given by x 2 / y 2 Divide both sides of the equation x 2 = 9 y 2 obtained above by y 2 x 2 / y 2 = 9 y 2 / y 2 Simplify to obtain the ratio of the two areas x 2 / y 2 = 9
  • John gave half of his stamps to Jim. Jim gave gave half of his stamps to Carla. Carla gave 1/4 of the stamps given to her to Thomas and kept the remaining 12. How many stamps did John start with? Solution Let x be the number of stamps that John started with. John gave half of his stamps to Jim: Jim got (1/2) x Jim gave gave half of his stamps to Carla: Carla got (1/2) ((1/2) x) Carle gave 1/4 and therefore kept 3/4 of the stamps given to her. Carla kept 3/4 of (1/2) ((1/2) x ) = (3/4) ((1/2) ((1/2) ) x) Simplify the expression (3/4) ((1/2) ((1/2) ) ) x . (3/4) ((1/2) ((1/2) ) x) = (3 × 1 × 1) / (4 × 2 × 2) x = 3 x / 16 The number of stamps kept by Carla is equal to 12. Hence the equation to solve 3 x / 16 = 12 Mutliply both side of the above equation by 16 / 3 (16/3) × (3 x / 16) = (16 / 3) ×12 Simplify and solve for x x = (16 / 3) ×12 = 64
  • Two balls A and B rotate along a circular track. Ball A makes 4 full rotations in 120 seconds. Ball B makes 3 full rotation in 60 seconds. If they start rotating now from the same, how long will take them to be at the same starting point again? Solution It will take 120 / 4 = 30 seconds for ball A to make one full rotation It will take 60 / 3 = 20 seconds for ball B to make one full rotation Let us calculate the time for whole rotations of ball A and B Ball A : 1 rot: 30 sec , 2 rot: 60 sec , 3 rot 90 sec, 4 rot 120 sec, .... Ball B: 1 rot: 20 sec , 2 rot: 40 sec , 3 rot 60 sec, 4 rot 80 sec, .... The first time that they have made a whole number of rotations and therefore be at the same starting point is after 60 seconds which is the lowest common (LCM) multiple of 30 and 20.
  • A segment is 3 units long. It is divided into 9 parts. What fraction of a unit are 2 parts of the segment? Solution To divide a 3 unit segment to make 9 parts, you need to divide each unit by 3. Hence 1 part = 1/3 of a unit and therefore 2 parts = 2/3 of a unit
  • A car is traveling 75 kilometers per hour. How many meters does the car travel in one minute? Solution Rewrite the rate 75 kilometers per hour converting kilometers in meters and hours in minutes 1 kilometer = 1000 meters 1 hour = 60 minutes Hence 75 kilometers per hour = 75 × 1000 meters per 60 minutes = 75 000 / 60 meters/minute = 1250 meters/minute
  • Carla is 5 years old and Jim is 13 years younger than Peter. One year ago, Peter's age was twice the sum of Carla's and Jim's age. Find the present age of each one of them. Solution Let x be Peter's age now. Hence the present age of Carla, Jim and Peter are given by Carla : 5 Peter : x Jim : x - 13 One year ago their ages were Carla : 5 - 1 = 4 Peter : x - 1 Jim : x - 13 - 1 = x - 14 One year ago, Peter's age was twice the sum of Carla's and Jim's age which is written mathematically as x - 1 = 2 (4 + x - 14) Simplify and expand the right side of the above equation x - 1 = 2 (x - 10) x - 1 = 2x - 20 Add + 20 to both sides and simplify x - 1 + 20 = 2x - 20 + 20 x + 19 = 2x Subtract x to both sides and simplify x + 19 - x = 2x - x 19 = x , Peter's age now Jim's age: x - 13 = 19 - 13 = 6 Carla's age: 5
  • Linda spent 3/4 of her savings on furniture. She then spent 1/2 of her remaining savings on a fridge. If the fridge cost her $150, what were her original savings? Solution Let x be Linda's savings. If she spent 3/4 of her savings on furniture then 1/4 of her savings are remaining and are written as (1/4) x She spent 1/2 of her remaining saving on a fridge that costs 150. Hence (1/2) × ((1/4) x ) = 150 Simplify the above and rewrite as x / 8= 150 Multiply both sides of the above equation by 8 and solve for x 8 × (x / 8) = 8 × 150 x = $1200
  • The distance between Harry and Kate is 2500 meters. Kate and Harry start walking toward one another and Kate' dog start running back and forth between Harry and Kate at a speed of 120 meters per minute. Harry walks at the speed of 40 meters per minute while Kate walks at the speed of 60 meters per minute. What distance will the dog have travelled when Harry and Kate meet each other? Solution The dog runs during all the period of time while Kate and Harry are walking. This period of time may be calculated as follows: 2500 meters / (40 + 60) meters / minutes = 25 minutes The dog runs during 25 minutes at the rate of 120 meters per minutes. Hence the total distance covered by the dog is given by 120 meters/minute × 25 minutes = 3000 meters

Free Printable Math Word Problems Worksheets for 6th Grade

Math Word Problems: Discover a vast collection of free printable worksheets for Grade 6 students, created by educators to enhance their mathematical skills and problem-solving abilities. Dive into the world of numbers with Quizizz!

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Math Word Problems worksheets for Grade 6 are an essential resource for teachers who want to help their students develop strong problem-solving skills and a deep understanding of mathematical concepts. These worksheets provide a variety of engaging and challenging problems that require students to apply their knowledge of math in real-world situations. With a wide range of topics covered, including fractions, decimals, percentages, and geometry, Grade 6 Math Word Problems worksheets are designed to align with the Common Core State Standards and support teachers in their efforts to create a well-rounded math curriculum. By incorporating these worksheets into their lesson plans, teachers can ensure that their students are receiving the practice and reinforcement they need to excel in math.

Quizizz is an excellent platform for teachers to access a vast library of Math Word Problems worksheets for Grade 6, along with other valuable resources to enhance their students' learning experience. This interactive platform offers a variety of features, such as customizable quizzes, real-time feedback, and gamification elements, to keep students engaged and motivated. In addition to Grade 6 Math worksheets, Quizizz also provides resources for other subjects and grade levels, making it a one-stop-shop for teachers looking to diversify their instructional materials. By utilizing Quizizz in conjunction with Math Word Problems worksheets for Grade 6, teachers can create a dynamic and effective learning environment that fosters a love for math and sets their students up for success.

Best of Operations on Whole Numbers Worksheets for Grade 6 Pdf with Answers and Explanations

Best of Operations on Whole Numbers Worksheets for Grade 6 Pdf with Answers and Explanations

H i there, welcome to the mathSkills4Kids.com blog , where I share my passion for math and education. Today, you'll discover some original  operations on whole numbers worksheets for grade 6 pdf with answers and explanations  in addition, subtraction, multiplication, and division.

These worksheets are perfect for practicing and mastering the skills of adding and subtracting large numbers, as well as solving word problems involving multiplying and dividing whole numbers.

You can download them for free from the links below and print them out for your use or your students/children.

Multiple Choice Questions (MCQs) on operations on whole numbers worksheets for grade 6

Here are ten multiple-choice questions (MCQs) on operations on whole numbers for grade 6 for your kiddos. These questions will test their skills in addition, subtraction, multiplication, and division of large numbers, using estimation, and applying the order of operations.

What is the value of 567 + 432 - 99? A) 900 B) 810 C) 800 D) 700

What is the value of 1234 - (567 + 89)? A) 578 B) 668 C) 778 D) 888

What is the quotient of 144 ÷ 12? A) 10 B) 11 C) 12 D) 13

What is the value of 999 - 888 + 777? A) 888 B) 898 C) 998 D) 988

What is the missing factor in this equation: ___ x 9 = 81? A) 7 B) 8 C) 9 D) 10

What is the product of 12 x 15? A) 120 B) 150 C) 180 D) 200

What is the value of 456 + 789 - 123? A) 1122 B) 1112 C) 1222 D) 1212

What is the value of 9999 + 1111 - (2222 + 3333)? A) 5555 B) 6666 C) 7777 D) 8888

What is the missing divisor in this equation: 72 ÷ ___ = 9? A) 6 B) 7 C) 8 D) 9

What is the simplified value of this expression: (6 x 5) ÷ (3 x 2)? A) (30 ÷6) B) (18 ÷10) C) (5 ÷3) D) (10 ÷3)

Click the links below to download operations on whole numbers for grade 6:

Best of Operations on Whole Numbers Worksheets for Grade 6 Pdf with Answers and Explanations - addition and subtraction word problems

Multiple Choice Questions (MCQs) on operations on whole numbers for grade 6 answers and explanations

Here are multiple-choice questions (MCQs) on operations on whole numbers for grade 6 answers and explanations to help students/children to verify their work and get a better understanding of the given exercises.

Question 1:  

The correct answer is B) 810.

To solve this problem, you can use the associative property of addition and subtraction, which states that you can group the numbers in any way without changing the result.

For example, you can rewrite the expression as (567 - 99) + 432, which gives 468 + 432 = 810.

Question 2:  

The correct answer is A) 578.

To solve this problem, you need to apply the order of operations, which states that you must perform the operations inside the parentheses before doing anything else.

So, you have to calculate 567 + 89 first and get 656. Then, you subtract that from 1234, which gives you 578.

Question 3:  

Answer: C) 12.

To find the quotient of two numbers, you divide them. So 144 ÷ 12 = 12.

Question 4:  

The correct answer is A) 888. To solve this problem, we also need to follow the order of operations. First, we subtract 888 from 999, which gives us 111. Then, we add 777 to 111, which gives us 888. 

Question 5:  

Answer: C) 9.

To find the missing factor, you can use inverse operations. That means you can divide both sides of the equation by 9 to get the answer.

So ___ x 9 = 81 becomes ___ = 81 ÷ 9, which is ___ = 9.

Question 6:  

Answer: C) 180.

To find the product of two numbers, you multiply them together. So 12 x 15 = 180.

Question 7:  

The correct answer is B) 1112.

To solve this problem, we need to follow the order of operations.

First, we add 456 and 789, which gives us 1245. Then, we subtract 123 from 1245, which gives us 1112. 

Question 8:  

The correct answer is A) 5555.

To solve this problem, you can use a similar strategy as in question 2. So, you can first add (9999 + 1111) and (2222 + 3333), which makes it easier to solve.

Therefore the equation sentence becomes (11110 - 5555), which is equal to 5555.

Question 9:  

Answer: C) 8.

To find the missing divisor, you can also use inverse operations. That means you can multiply both sides of the equation by ___ to get the answer.

So 72 ÷ ___ = 9 becomes 72 = ___ x 9, which means ___ = 72 ÷ 9. So, ___ = 8.

Question 10: 

Answer: D) (10 ÷3).

To evaluate this expression, you need to follow the order of operations. That means you do what's inside the parentheses first, then multiply or divide from left to right.

So (6 x 5) ÷ (3 x2 ) becomes (30) ÷ (6), which is equal to (10 ÷3).

Writing Whole Numbers in Words Complete with the Missing Word Exercises for Grade 6

These Class 6 operations on whole numbers word problems with answers and explanations contain multiple real-life situations which involve addition, subtraction, multiplication, and division of whole numbers for grade 6. These problems will help your students/children to practice their arithmetic skills and also test their logical thinking.

Alice has 12 books in her backpack and 8 books on her desk. She wants to donate some of her books to the school library. She decides to donate half of the books in her backpack and a quarter of the books on her desk. How many books will Alice have left after donating?

To find out how many books Alice will have left, we must subtract the number of books she donated from the total number of books she had.

First, let's find out how many books she donated from her backpack and desk.

Half of 12 books is 12 ÷ 2 = 6 books.

A quarter of 8 books is 8 ÷ 4 = 2 books.

So, Alice donated 6 + 2 = 8 books in total.

Secondly, let's find out how many books Alice had before donating. We add the number of books in her backpack and on her desk to get it.

12 + 8 = 20 books.

Finally, let's find the remaining number of books after the donation. To get it, we subtract the number of books donated from the total number of books she had.

20 - 8 = 12 books.

Therefore, Alice will have 12 books left after donating.

Ben and his friends are playing a game of bowling. Ben scores 135 points in the first round and 120 points in the second round. His friend Sam scores 145 points in the first round and 110 points in the second round. How many more points does Ben score than Sam in total?

To find out how many more points Ben scores than Sam in total, we need to add up their scores for both rounds and then subtract them.

Ben's total score = 135 + 120 = 255

Sam's total score = 145 + 110 = 255

Ben's total score - Sam's total score = 255 - 255 = 0

Therefore, Ben and Sam score the same number of points in total.

Leona is making cookies for a bake sale. She has a recipe that makes 24 cookies using 3 cups of flour, 2 cups of sugar, and 1 cup of butter. She wants to make 72 cookies for the bake sale. How many cups of each ingredient does she need?

To find out how many cups of each ingredient Leona needs, we need to multiply the quantity of each ingredient in the recipe by a factor that will give us 72 cookies. We know that the recipe makes 24 cookies, which means we need to multiply by 3, since 24 × 3 = 72.

Flour: 3 × 3 = 9 cups

Sugar: 2 × 3 = 6 cups

Butter: 1 × 3 = 3 cups

Therefore, Leona needs 9 cups of flour, 6 cups of sugar, and 3 cups of butter to make 72 cookies.

Andy collects postcards. Currently, he has 360 postcards that he wants to share with 11 siblings. How many postcards will each person get?

To find out how many postcards each person will get, we need to divide the total number of postcards by the number of people.

360 ÷ 12 = 30

Therefore, each person will get 30 postcards.

Sixth-grade operations on whole numbers multi-step word problems with explanations

All that we have done up there are very interesting, but what if we go deep through our topic with this sixth-grade operations on whole numbers multi-step word problems with explanations ?

Because we need to practice addition, subtraction, multiplication, and division, we will solve this long word problem which resumes all arithmetic operations correctly.

Bob wants to buy a new bike that costs $240. He has saved up $80 so far. He earns $10 for every hour he works at his part-time job.

  • How long does Bob need to work to buy the bike?
  • Knowing that he can only get paid if he finishes a full hour, how many hours does Bob have to work to buy the bike and accessories worth $148?
  • If, besides what he had saved, Bob works 4 hours a day for 9 days to buy the bike and its accessories, how much will he have left after his purchases?

Here is the explanation of how to solve the sixth-grade operations on whole numbers multi-step word problems.

Question a) answer:

To solve this first question, we must find out how much more money Bob needs to save and divide that amount by his hourly wage.

We know that Bob saved $80. So, he is already own that amount.

To find out how much more money Bob needs, we must subtract the saved amount ($80) from the bike price ($240).

Therefore, Bob needs 240 - 80, which is equal to 160 more to buy the bike. So, $160 is what he needs more for the purchase of the bike.

Now, we need to know the number of work hours required for him to earn $160. To get it, we must divide $160 by $10, which is his hourly wage.

160 ÷ 10 = 16.

It means that Bob needs 16 hours of work to buy the bike.

Question b) answer:

For Bob to buy accessories worth $148 and the bike, we must find how many hours he needs to work to earn money for accessories. Then, we will add that number of hours to the precedent one to find the total hours to work.

We begin with the number of work hours required for him to earn $148.

To find it, we divide $148, which is the cost of the accessories by 10, which represents the amount earned per hour.  The answer is: 148 ÷ 10 = 14r8

So, Bob needs to work 14 hours and one-eighth of an hour to earn $148.

Since the work is paid per hour, he is supposed to work an overtime hour instead of one-eighth of an hour. Thus, he must work 15 hours to be able to buy the additional accessories.

Therefore, the time he needs to work to buy the bike and the accessories is 16 hours, which is the time he needs to work for the bike plus 15 hours, which is the time needed for the accessories.

16 + 15 = 31 hours.

So, Bob must work for 31 hours to buy the bike and accessories.

Question c) answer:

To solve this question, we need to find out how much money Bob earns from working, how much money he needs to buy the bike and its accessories, and how much money he has left after his purchases.

First, let's calculate how much money Bob earns from working. He works 4 hours a day for 9 days, so he works a total of 4 x 9 = 36 hours. He earns $10 for every hour he works, so he earns a total of 10 x 36 = $360 from working.

Next, let's calculate how much money Bob needs to buy the bike and its accessories. The bike costs $240, and the accessories cost $148. Thus, the total cost of the bike and its accessories is 240 + 148 = $388.

Finally, let's calculate how much money Bob has left after his purchases. He has saved up $80 and he earns $360 from working, so he has a total of 80 + 360 = $440 before his purchases. He spends $388 on the bike and its accessories, so he has a remaining amount of 440 - 388 = $52 after his purchases.

Therefore, Bob will have $52 left after buying the bike and its accessories.

Advantages for using our fun operations on whole numbers worksheets for grade 6 pdf

Why do I love these worksheets so much? Using our  operations on whole numbers worksheets for grade 6 pdf with answers and explanations  will bring many advantages to your students/children.

Well, first of all, they are fun and engaging. They have colorful graphics and exciting scenarios that make math more enjoyable and relevant.

For example, one of the worksheets asks to add up the number of books in a library, while another asks to subtract the number of people who left a concert from the total number of tickets sold. 

These are real-life situations that require math skills and logical thinking.

Secondly, these worksheets are challenging and rewarding. They cover a range of difficulty levels, from easy to difficult, so you can choose the ones that suit your needs and goals. 

Some worksheets have missing addends or subtrahends, meaning you have to find the unknown number that makes the equation true. Here is a great way to practice algebraic thinking and problem-solving skills. 

You'll also find worksheets with word problems that require students/children to apply their addition, subtraction, multiplication, or division knowledge to different contexts and situations. So, here is a great way to  develop students'/children's reading comprehension and critical thinking skills .

Thirdly, these grade 6 worksheets are comprehensive and helpful. They include answers and explanations at the end, so you can check students'/children's work and see where they made mistakes or needed more practice. It means that they get fantastic tips and tricks on adding, subtracting, multiplying, or dividing large numbers quickly and accurately, such as using place value, regrouping, borrowing, or estimating. These valuable strategies will help them improve their mental math skills and confidence.

So, what are you waiting for?  Download these excellent operations on whole numbers worksheets for grade 6 pdf with answers and explanations , and take advantage of our comprehensive educational resources. 

You'll be amazed at how much your students or kids can learn and improve their knowledge on this topic.

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    Solutions to Word Problems. Two numbers N and 16 have LCM = 48 and GCF = 8. Find N. Solution The product of two integers is equal to the product of their LCM and GCF. Hence. 16 × N = 48 × 8 N = 48 × 8 / 16 = 24. If the area of a circle is 81π square feet, find its circumference. Solution The area is given by π × r × r.

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    2. Use the expression 6t e 3f to find Seattle's final score in the 2006 Super Bowl. 3. GEOMETRY The expression 6s2 can be used to find the surface area of a cube, where s is the length of an edge of the cube. Find the surface area of a cube with an edge of length 10 centimeters. 10 cm.

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    The correct answer is A) 5555. To solve this problem, you can use a similar strategy as in question 2. So, you can first add (9999 + 1111) and (2222 + 3333), which makes it easier to solve. Therefore the equation sentence becomes (11110 - 5555), which is equal to 5555.

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    PERIMETER The perimeter of a rectangle can be w found using the formula 2 2w, where represents the length and w represents the width. Find the perimeter if 6 units and w 3 units. 4. PERIMETER Another formula for perimeter is 2( w). Find the perimeter of the rectangle in Exercise 3 using this formula.

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