8th maths assignment answer unit 2

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

8th grade (Illustrative Mathematics)

Unit 1: rigid transformations and congruence, unit 2: dilations, similarity, and introducing slope, unit 3: linear relationships, unit 4: linear equations and linear systems, unit 5: functions and volume, unit 6: associations in data, unit 7: exponents and scientific notation, unit 8: pythagorean theorem and irrational numbers.

Test and Worksheet Generators for Math Teachers

Test and worksheet generators, learning management system, how it works, you choose the topic..

You choose the mathematical properties of the questions.

Every topic has different options., it creates as many questions as you would like., distribute assignments to your students., print hard copies to give to students..

Post assignments to Kuta Works.

Students complete online., unlimited questions.

Once you have created an assignment, you can regenerate all of its questions with a single click. The new questions will conform to the same parameters as the original questions, but they will be completely new. This feature is at the heart of our software and is what makes it so powerful: you choose the properties of the questions, not the questions themselves. When a question is replaced, you get a new one that is similar to the original question. How it works. You can regenerate entire assignments, particular question groups, or individual questions.

Easy Spacing

Respace the entire assignment to the desired length with one click. Easily give your students enough room to show their work by increasing the spacing. Or you can save paper by decreasing the spacing.

Spacing can also be controlled manually.

Presentation Mode

Very useful as a teaching aid when used in combination with an LCD projector or other display system. One to four questions at a time are shown on the screen.

Use this feature while you teach. Prepare your examples with the software, and then use a projector to display the questions on the board. This saves time during planning and during the lesson, and it makes it very easy to present long questions or questions with graphs and diagrams. With one question displayed, you can:

• Change the zoom level -- so students in the back can read it
• Draw lines beside the question to help you organize your work if you solve the question
• Jump to another question -- useful while reviewing homework
• Reveal the answer
• Show / hide the question number and the directions.

Multiple-Version Printing

Print multiple versions of an assignment. You control how each new version is created: scramble the choices, scramble the questions, or make completely new questions. You can also save each new version after it is created.

Scale Assignment

Proportionally increase or decrease the number of questions in the assignment. This is very useful when planning a lesson. You can create a few questions to use as examples, and then scale up the number of question to create a homework assignment. The questions on the homework will be completely new, yet follow precisely from the lesson--and you don't need to design the questions again.

Export Questions

Export questions as bitmap images and paste them into your favorite word processing software. Questions created with our products can be added to existing assignments you have created with other programs. Or you can freshen old assignments by replacing old questions with new ones.

All questions are available for export.

Good Multiple-Choice Questions

Every question you create can be toggled between free-response and multiple-choice format. Multiple-choice questions come with smart, potentially misleading choices. Some are based on common mistakes students make while others are just random but near the correct answer.

You control the number of choices each question has, from two to five.

Merge Assignments

Merge two or more assignments into one. Easily create quizzes, tests, and reviews by merging the assignments from the unit and then scaling the total to an appropriate length. The questions will be new while following exactly from what you taught.

Diagrams Drawn to Scale

Diagrams are all accurately drawn, except if the answer would be given away. If an angle is labeled as 30°, then it really is 30°. If a triangle's sides are labeled 3, 4, and 5, then its lengths truly are in a 3:4:5 ratio. Seeing accurate diagrams helps students gain an intuitive understanding of angles and measurements.

Answer Format

When you print an assignment, you choose how the answers are reported:

• On an answer sheet
• On an answer sheet with just the odds
• In context (next to or within the question)
• No answer sheet

Graphing and Graph Paper Utility

Supplement your lessons with high-quality graphs and graph paper of any size. Each graph can have zero to two functions graphed on it. Graphs can be of any logical and physical size. You can also tile graphs across the page to maximize your paper use.

Custom Directions and Custom Questions

Enter your own directions to create new types of problems. Shown on the left was a standard order of operations question that has been modified to be more analytical. You can alter the directions on any question type.

From time to time, you will need to enter your own question. That's what custom questions are for. They can be either free response or multiple-choice and can contain math formatted text (equations, expressions, etc).

Modify Automatically-Generated Questions

Most automatically-generated questions can be modified manually. If there is a choice you don't like, you can change it. If you wish a question was slightly different, you can change it.

Paper Size and Margins

Print assignments on any sized paper that your printer supports. If you decide to print an assignment on legal-sized paper, no problem. The questions will automatically be repositioned for you--no cutting and pasting the assignment back together just to use a different paper size. You also have control over the margins, page numbering, and paper orientation.

• NCERT Solutions
• NCERT Class 8
• NCERT 8 Maths
• Chapter 2: Linear Equation

NCERT Solutions for Class 8 Maths Chapter 2 - Linear Equations in One Variable

Ncert solutions class 8 maths chapter 2 – free pdf download.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable are provided here in PDF format, which can be downloaded for free. The NCERT Solutions for the Chapter Linear Equations in One Variable have been designed by Mathematics experts at BYJU’S accurately. All the solved questions of Linear Equations in One Variable are with respect to the latest NCERT syllabus and guidelines to help students solve each exercise question present in the book and prepare for the exam.

Download Exclusively Curated Chapter Notes for Class 8 Maths Chapter – 2 Linear Equations in One Variable

Download most important questions for class 8 maths chapter – 2 linear equations in one variable.

These solutions serve as a reference tool for the students to do their homework and assignments as well. The NCERT Solutions for Class 8 contains the exercise-wise answers for all the chapters, thus being a very useful study material for the students studying in Class 8.  Practising these solutions help the students to perform well in the final exams and ace the subject. These solutions are devised based on the most updated NCERT syllabus , covering all the crucial topics of the respective subjects.

NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable

carouselExampleControls111

Previous Next

Access Answers to NCERT Class 8 Maths Chapter 2 – Linear Equations in One Variable

Exercise 2.1 page: 23.

Solve the following equations.

1. x – 2 = 7

2. y + 3 = 10

3. 6 = z + 2

4. 3/7 + x = 17/7

3/7 + x = 17/7

x = 17/7 – 3/7

6. t/5 = 10

7. 2x/3 = 18

2x = 18 × 3

8. 1.6 = y/15

1.6 = y/1.5

y/1.5 = 1.6

y = 1.6 × 1.5

9. 7x – 9 = 16

7x – 9 = 16

10. 14y – 8 = 13

14y – 8 = 13

11. 17 + 6p = 9

17 + 6p = 9

6p = 9 – 17

12. x/3 + 1 = 7/15

x/3 + 1 = 7/15

x/3 = 7/15 – 1

x/3 = (7 -15)/15

x/3 = -8/15

x = -8/15 × 3

Exercise 2.2 Page: 28

1. If you subtract ½ from a number and multiply the result by ½, you get 1/8. What is the number?

Let the number be x.

According to the question,

(x – 1/2) × ½ = 1/8

x/2 – ¼ = 1/8

x/2 = 1/8 + ¼

x/2 = 1/8 + 2/8

x/2 = (1+ 2)/8

x = (3/8) × 2

2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m, more than twice its breadth. What are the length and breadth of the pool?

Given that,

The perimeter of the rectangular swimming pool = 154 m. Let the breadth of the rectangle be = x

Length of the rectangle = 2x + 2 We know that,

Perimeter = 2(length + breadth)

⇒ 2(2x + 2 + x) = 154 m

⇒ 2(3x + 2) = 154

⇒ 3x +2 = 154/2

⇒ 3x = 77 – 2

Therefore, Breadth = x = 25 cm

Length = 2x + 2

= (2 × 25) + 2

Base of isosceles triangle = 4/3 cm

Let the length of equal sides of the triangle be x.

4/3 + x + x = 62/15 cm

⇒ 2x = (62/15 – 4/3) cm

⇒ 2x = (62 – 20)/15 cm

⇒ 2x = 42/15 cm

⇒ x = (42/30) × (½)

⇒ x = 42/30 cm

⇒ x = 7/5 cm

The length of either of the remaining equal sides is 7/5 cm.

4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Let one of the numbers be = x.

Then, the other number becomes x + 15. According to the question,

x + x + 15 = 95

⇒ 2x + 15 = 95

⇒ 2x = 95 – 15

First number = x = 40

And, other number = x + 15 = 40 + 15 = 55

5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Let the two numbers be 5x and 3x. According to the question,

5x – 3x = 18

The numbers are 5x = 5 × 9 = 45

And 3x = 3 × 9 = 27.

6. Three consecutive integers add up to 51. What are these integers?

Let the three consecutive integers be x, x+1 and x+2. According to the question,

x + (x+1) + (x+2) = 51

⇒ 3x + 3 = 51

⇒ 3x = 51 – 3

Thus, the integers are

7. The sum of three consecutive multiples of 8 is 888. Find the multiples.

Let the three consecutive multiples of 8 be 8x, 8(x+1) and 8(x+2). According to the question,

8x + 8(x+1) + 8(x+2) = 888

⇒ 8 (x + x+1 + x+2) = 888 (Taking 8 as common)

⇒ 8 (3x + 3) = 888

⇒ 3x + 3 = 888/8

⇒ 3x + 3 = 111

⇒ 3x = 111 – 3

⇒ x = 108/3

Thus, the three consecutive multiples of 8 are:

8x = 8 × 36 = 288

8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296

8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4, respectively, they add up to 74. Find these numbers.

2x + 3(x+1) + 4(x+2) = 74

⇒ 2x + 3x +3 + 4x + 8 = 74

⇒ 9x + 11 = 74

⇒ 9x = 74 – 11

Thus, the numbers are:

9. The ages of Rahul and Haroon are in the ratio 5:7. Four years later, the sum of their ages will be 56 years. What are their present ages?

Let the ages of Rahul and Haroon be 5x and 7x. Four years later,

The ages of Rahul and Haroon will be (5x + 4) and (7x + 4), respectively. According to the question,

(5x + 4) + (7x + 4) = 56

⇒ 5x + 4 + 7x + 4 = 56

⇒ 12x + 8 = 56

⇒ 12x = 56 – 8

⇒ x = 48/12

Therefore, Present age of Rahul = 5x = 5×4 = 20

And, present age of Haroon = 7x = 7×4 = 28

10. The number of boys and girls in a class is in the ratio of 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Let the number of boys be 7x, and girls be 5x.

7x = 5x + 8

⇒ 7x – 5x = 8

Therefore, number of boys = 7×4 = 28

And, number of girls = 5×4 = 20

Total number of students = 20+28 = 48

11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Let the age of Baichung’s father be x.

Then, the age of Baichung’s grandfather = (x+26)

and, the age of Baichung = (x-29). According to the question,

x + (x+26) + (x-29) = 135

⇒ 3x + 26 – 29 = 135

⇒ 3x – 3 = 135

⇒ 3x = 135 + 3

⇒ x = 138/3

Age of Baichung’s father = x = 46

Age of Baichung’s grandfather = (x+26) = 46 + 26 = 72

Age of Baichung = (x-29) = 46 – 29 = 17

12. Fifteen years from now, Ravi’s age will be four times his present age. What is Ravi’s present age?

Let the present age of Ravi be x.

Fifteen years later, Ravi’s age will be x+15 years. According to the question,

x + 15 = 4x

⇒ 4x – x = 15

Therefore, the present age of Ravi = 5 years.

13. A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?

Let the rational be x.

x × (5/2) + 2/3 = -7/12

⇒ 5x/2 + 2/3 = -7/12

⇒ 5x/2 = -7/12 – 2/3

⇒ 5x/2 = (-7- 8)/12

⇒ 5x/2 = -15/12

⇒ 5x/2 = -5/4

⇒ x = (-5/4) × (2/5)

⇒ x = – 10/20

Therefore, the rational number is -½.

14. Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, ₹50 and ₹10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹4,00,000. How many notes of each denomination does she have?

Let the numbers of notes of ₹100, ₹50 and ₹10 be 2x, 3x and 5x, respectively.

Value of ₹100 = 2x × 100 = 200x

Value of ₹50 = 3x × 50 = 150x

Value of ₹10 = 5x × 10 = 50x

200x + 150x + 50x = 4,00,000

⇒ 400x = 4,00,000

⇒ x = 400000/400

Numbers of ₹100 notes = 2x = 2000

Numbers of ₹50 notes = 3x = 3000

Numbers of ₹10 notes = 5x = 5000

15. I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of ₹2 coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Let the number of ₹5 coins be x.

Number ₹2 coins = 3x

And, number of ₹1 coins = (160 – 4x) Now,

Value of ₹5 coins = x × 5 = 5x

Value of ₹2 coins = 3x × 2 = 6x

Value of ₹1 coins = (160 – 4x) × 1 = (160 – 4x)

5x + 6x + (160 – 4x) = 300

⇒ 11x + 160 – 4x = 300

⇒ x = 140/7

Number of ₹5 coins = x = 20

Number of ₹2 coins = 3x = 60

Number of ₹1 coins = (160 – 4x) = 160 – 80 = 80

16. The organisers of an essay competition decide that a winner in the competition gets a prize of ₹100 and a participant who does not win gets a prize of ₹25. The total prize money distributed is ₹3,000. Find the number of winners, if the total number of participants is 63.

Let the number of winners be x.

Then, the number of participants who didn’t win = 63 – x

Total money given to the winner = x × 100 = 100x

Total money given to the participant who didn’t win = 25 × (63-x)

100x + 25 × (63-x) = 3,000

⇒ 100x + 1575 – 25x = 3,000

⇒ 75x = 3,000 – 1575

⇒ 75x = 1425

⇒ x = 1425/75

Therefore, the numbers of winners are 19.

Exercise 2.3 Page: 30

Solve the following equations and check your results.

1. 3x = 2x + 18

3x = 2x + 18

⇒ 3x – 2x = 18

Putting the value of x in RHS and LHS, we get, 3 × 18 = (2 × 18) +18

⇒ LHS = RHS

2. 5t – 3 = 3t – 5

5t – 3 = 3t – 5

⇒ 5t – 3t = -5 + 3

Putting the value of t in RHS and LHS, we get, 5× (-1) – 3 = 3× (-1) – 5

⇒ -5 – 3 = -3 – 5

3. 5x + 9 = 5 + 3x

5x + 9 = 5 + 3x

⇒ 5x – 3x = 5 – 9

Putting the value of x in RHS and LHS, we get, 5× (-2) + 9 = 5 + 3× (-2)

⇒ -10 + 9 = 5 + (-6)

4. 4z + 3 = 6 + 2z

4z + 3 = 6 + 2z

⇒ 4z – 2z = 6 – 3

Putting the value of z in RHS and LHS, we get,

(4 × 3/2) + 3 = 6 + (2 × 3/2)

⇒ 6 + 3 = 6 + 3

5. 2x – 1 = 14 – x

2x – 1 = 14 – x

⇒ 2x + x = 14 + 1

Putting the value of x in RHS and LHS, we get, (2×5) – 1 = 14 – 5

⇒ 10 – 1 = 9

6. 8x + 4 = 3 (x – 1) + 7

8x + 4 = 3 (x – 1) + 7

⇒ 8x + 4 = 3x – 3 + 7

⇒ 8x + 4 = 3x + 4

⇒ 8x – 3x = 4 – 4

Putting the value of x in RHS and LHS, we get, (8×0) + 4 = 3 (0 – 1) + 7

⇒ 0 + 4 = 0 – 3 + 7

7. x = 4/5 (x + 10)

x = 4/5 (x + 10)

⇒ x = 4x/5 + 40/5

⇒ x – (4x/5) = 8

⇒ (5x – 4x)/5 = 8

⇒ x = 8 × 5

Putting the value of x in RHS and LHS, we get,

40 = 4/5 (40 + 10)

⇒ 40 = 4/5 × 50

⇒ 40 = 200/5

8. 2x/3 + 1 = 7x/15 + 3

2x/3 + 1 = 7x/15 + 3

⇒ 2x/3 – 7x/15 = 3 – 1

⇒ (10x – 7x)/15 = 2

⇒ 3x = 2 × 15

9. 2y + 5/3 = 26/3 – y

2y + 5/3 = 26/3 – y

⇒ 2y + y = 26/3 – 5/3

⇒ 3y = (26 – 5)/3

⇒ 3y = 21/3

Putting the value of y in RHS and LHS, we get,

⇒ (2 × 7/3) + 5/3 = 26/3 – 7/3

⇒ 14/3 + 5/3 = 26/3 – 7/3

⇒ (14 + 5)/3 = (26 – 7)/3

⇒ 19/3 = 19/3

10. 3m = 5m – 8/5

3m = 5m – 8/5

⇒ 5m – 3m = 8/5

⇒ 2m × 5 = 8

Putting the value of m in RHS and LHS, we get,

⇒ 3 × (4/5) = (5 × 4/5) – 8/5

⇒ 12/5 = 4 – (8/5)

⇒ 12/5 = (20 – 8)/5

⇒ 12/5 = 12/5

Exercise 2.4 Page: 31

1. Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?

Let the number be x,

(x – 5/2) × 8 = 3x

⇒ 8x – 40/2 = 3x

⇒ 8x – 3x = 40/2

Thus, the number is 4.

2. A positive number is 5 times another number. If 21 is added to both numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Let one of the positive numbers be x, then the other number will be 5x. According to the question,

5x + 21 = 2(x + 21)

⇒ 5x + 21 = 2x + 42

⇒ 5x – 2x = 42 – 21

One number = x = 7

Other number = 5x = 5×7 = 35. The two numbers are 7 and 35.

3. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Let the digit at tens place be x, then the digit at ones place will be (9-x).

Original two-digit number = 10x + (9-x)

After interchanging the digits, the new number = 10(9-x) + x

10x + (9-x) + 27 = 10(9-x) + x

⇒ 10x + 9 – x + 27 = 90 – 10x + x

⇒ 9x + 36 = 90 – 9x

⇒ 9x + 9x = 90 – 36

Original number = 10x + (9-x) = (10×3) + (9-3) = 30 + 6 = 36

Thus, the number is 36.

4. One of the two digits of a two-digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Let the digit at tens place be x, then the digit at ones place will be 3x.

Original two-digit number = 10x + 3x

After interchanging the digits, the new number = 30x + x

(30x + x) + (10x + 3x) = 88

⇒ 31x + 13x = 88

Original number = 10x + 3x = 13x = 13×2 = 26

5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one-third of his mother’s present age. What are their present ages?

Let the present age of Shobo be x, then the age of her mother will be 6x.

Shobo’s age after 5 years = x + 5

(x + 5) = (1/3) × 6x

⇒ x + 5 = 2x

⇒ 2x – x = 5

Present age of Shobo = x = 5 years

The present age of Shobo’s mother = 6x = 30 years.

6. There is a narrow rectangular plot reserved for a school in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate ₹100 per metre, it will cost the village panchayat ₹75000 to fence the plot. What are the dimensions of the plot?

Let the length of the rectangular plot be 11x and the breadth be 4x.

Rate of fencing per metre = ₹100

Total cost of fencing = ₹75000

Perimeter of the plot = 2(l+b) = 2(11x + 4x) = 2×15x = 30x

Total amount of fencing = (30x × 100)

(30x × 100) = 75000

⇒ 3000x = 75000

⇒ x = 75000/3000

Length of the plot = 11x = 11 × 25 = 275m

Breadth of the plot = 4 × 25 = 100m.

7. Hasan buys two kinds of cloth materials for school uniforms; shirt material that costs him ₹50 per metre and trouser material that costs him ₹90 per metre. For every 3 meters of the shirt material, he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit, respectively. His total sale is ₹36,600. How much trouser material did he buy?

Let 2x m of trouser material and 3x m of shirt material be bought by him

Selling price of shirt material per meter = ₹ 50 + 50 ×(12/100) = ₹ 56

Selling price of trouser material per meter = ₹ 90 + 90 × (10/100) = ₹ 99

Total amount of sale = ₹36,600

(2x × 99) + (3x × 56) = 36600

⇒ 198x + 168x = 36600

⇒ 366x = 36600

⇒ x = 36600/366

Total trouser material he bought = 2x = 2 × 100 = 200 m.

8. Half of a herd of deer is grazing in the field, and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Let the total number of deer be x.

Deer grazing in the field = x/2

Deer playing nearby = x/2 × ¾ = 3x/8

Deer drinking water = 9

x/2 + 3x/8 + 9 = x

(4x + 3x)/8 + 9 = x

⇒ 7x/8 + 9 = x

⇒ x – 7x/8 = 9

⇒ (8x – 7x)/8 = 9

⇒ x = 9 × 8

9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Let the age of granddaughter be x and grandfather be 10x.

Also, he is 54 years older than her.

According to the question, 10x = x + 54

⇒ 10x – x = 54

Age of grandfather = 10x = 10×6 = 60 years.

Age of granddaughter = x = 6 years.

10. Aman’s age is three times his son’s age. Ten years ago, he was five times his son’s age. Find their present ages.

Let the age of Aman’s son be x, then the age of Aman will be 3x.

5(x – 10) = 3x – 10

⇒ 5x – 50 = 3x – 10

⇒ 5x – 3x = -10 + 50

Aman’s son age = x = 20 years

Aman age = 3x = 3×20 = 60 years

Exercise 2.5 Page: 33

Solve the following linear equations.

1. x/2 – 1/5 = x/3 + ¼

x/2 – 1/5 = x/3 + ¼

⇒ x/2 – x/3 = ¼+ 1/5

⇒ (3x – 2x)/6 = (5 + 4)/20

⇒ 3x – 2x = 9/20 × 6

⇒ x = 54/20

⇒ x = 27/10

2. n/2 – 3n/4 + 5n/6 = 21

n/2 – 3n/4 + 5n/6 = 21

⇒ (6n – 9n + 10n)/12 = 21

⇒ 7n/12 = 21

⇒ 7n = 21 × 12

⇒ n = 252/7

3. x + 7 – 8x/3 = 17/6 – 5x/2

x + 7 – 8x/3 = 17/6 – 5x/2

⇒ x – 8x/3 + 5x/2 = 17/6 – 7

⇒ (6x – 16x + 15x)/6 = (17 – 42)/6

⇒ 5x/6 = – 25/6

⇒ 5x = – 25

4. (x – 5)/3 = (x – 3)/5

(x – 5)/3 = (x – 3)/5

⇒ 5(x-5) = 3(x-3)

⇒ 5x-25 = 3x-9

⇒ 5x – 3x = -9+25

5. (3t – 2)/4 – (2t + 3)/3 = 2/3 – t

(3t – 2)/4 – (2t + 3)/3 = 2/3 – t

⇒ ((3t – 2)/4) × 12 – ((2t + 3)/3) × 12

⇒ (3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t

⇒ 9t – 6 – 8t – 12 = 8 – 12t

⇒ t – 18 = 8 – 12t

⇒ t + 12t = 8 + 18

6. m – (m – 1)/2 = 1 – (m – 2)/3

m – (m – 1)/2 = 1 – (m – 2)/3

⇒ m – m/2 – 1/2 = 1 – (m/3 – 2/3)

⇒ m – m/2 + ½ = 1 – m/3 + 2/3

⇒ m – m/2 + m/3 = 1 + 2/3 – ½

⇒ m/2 + m/3 = ½ + 2/3

⇒ (3m + 2m)/6 = (3 + 4)/6

⇒ 5m/6 = 7/6

⇒ m = 7/6 × 6/5

Simplify and solve the following linear equations.

7. 3 (t – 3) = 5(2t + 1)

3(t – 3) = 5(2t + 1)

⇒ 3t – 9 = 10t + 5

⇒ 3t – 10t = 5 + 9

⇒ t = 14/-7

8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0

15(y – 4) –2(y – 9) + 5(y + 6) = 0

⇒ 15y – 60 -2y + 18 + 5y + 30 = 0

⇒ 15y – 2y + 5y = 60 – 18 – 30

⇒ y = 12/18

9. 3 (5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17

⇒ 15z – 18z – 32z = -52 – 17 + 21 – 22

⇒ -35z = -70

⇒ z = -70/-35

10. 0.25(4f – 3) = 0.05(10f – 9)

0.25(4f – 3) = 0.05(10f – 9)

⇒ f – 0.75 = 0.5f – 0.45

⇒ f – 0.5f = -0.45 + 0.75

⇒ 0.5f = 0.30

⇒ f = 0.30/0.5

Exercise 2.6 Page: 35

1. (8x – 3)/3x = 2

(8x – 3)/3x = 2

⇒ 8x/3x – 3/3x = 2

⇒ 8/3 – 1/x = 2

⇒ 8/3 – 2 = 1/x

⇒ (8 – 6)/3 = 1/x

⇒ 2/3 = 1/x

2. 9x/(7 – 6x) = 15

9x/(7 – 6x) = 15

⇒ 9x = 15(7 – 6x)

⇒ 9x = 105 – 90x

⇒ 9x + 90x = 105

⇒ 99x = 105

⇒ x = 105/99 = 35/33

3. z/(z + 15) = 4/9

z/(z + 15) = 4/9

⇒ z = 4/9 (z + 15)

⇒ 9z = 4(z + 15)

⇒ 9z = 4z + 60

⇒ 9z – 4z = 60

4. (3y + 4)/(2 – 6y) = -2/5

(3y + 4)/(2 – 6y) = -2/5

⇒ 3y + 4 = -2/5 (2 – 6y)

⇒ 5(3y + 4) = -2(2 – 6y)

⇒ 15y + 20 = -4 + 12y

⇒ 15y – 12y = -4 – 20

5. (7y + 4)/(y + 2) = -4/3

(7y + 4)/(y + 2) = -4/3

⇒ 7y + 4 = -4/3 (y + 2)

⇒ 3(7y + 4) = -4(y + 2)

⇒ 21y + 12 = -4y – 8

⇒ 21y + 4y = -8 – 12

⇒ 25y = -20

⇒ y = -20/25 = -4/5

6. The ages of Hari and Harry are in the ratio of 5:7. Four years from now, the ratio of their ages will be 3:4. Find their present ages.

Let the age of Hari be 5x and Harry be 7x. 4 years later,

Age of Hari = 5x + 4

Age of Harry = 7x + 4

(5x + 4)/(7x + 4) = ¾

⇒ 4(5x + 4) = 3(7x + 4)

⇒ 20x + 16 = 21x + 12

⇒ 21x – 20x = 16 – 12

Hari’s age = 5x = 5 × 4 = 20 years

Harry’s age = 7x = 7 × 4 = 28 years

7. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.

Let the numerator be x, then the denominator will be (x + 8)

(x + 17)/(x + 8 – 1) = 3/2

⇒ (x + 17)/(x + 7) = 3/2

⇒ 2(x + 17) = 3(x + 7)

⇒ 2x + 34 = 3x + 21

⇒ 34 – 21 = 3x – 2x

The rational number is x/(x + 8) = 13/21

In order to find the value of some unknown quantities with minimal information about it given, a stronghold on the concept of Algebra is necessary. This chapter has 6 exercises that deal with the topic of Linear Equations in One Variable. The major concepts covered in this chapter include: 2.1 Introduction 2.2 Solving Equations which have Linear Expressions on 1 Side and Numbers on the other 2.3 Some Applications 2.4 Solving Equations having the Variable on Both Sides 2.5 Some More Applications 2.6 Reducing Equations to Simpler Form 2.7 Equations Reducible to the Linear Form. Exercise 2.1 Solutions 12 Questions (12 Short Answer Questions) Exercise 2.2 Solutions 16 Questions (6 Long Answer Questions, 10 Short Answer Questions) Exercise 2.3 Solutions 10 Questions (3 Long Answer Questions, 7 Short Answer Questions) Exercise 2.4 Solutions 10 Questions (4 Long Answer Questions, 6 Short Answer Questions) Exercise 2.5 Solutions 10 Questions (1 Long Answer Question, 9 Short Answer Questions) Exercise 2.6 Solutions 7 Questions (1 Long Answer Question, 6 Short Answer Questions)

Chapter 2 of NCERT Solutions for Class 8 Maths is the continuation of the concept of algebraic expressions and equations that the students learned in lower classes. In this chapter, the students learn with equations having one variable. Some of the main topics or concepts that are discussed in this chapter include:

• An algebraic equation is an equality involving variables. It says that the value of the expression on one side of the equality sign is equal to the value of the expression on the other side.
• A linear equation may have for its solution any rational number.
• An equation may have linear expressions on both sides.
• Just as numbers, variables can also be transposed from one side of the equation to the other.
• Occasionally, the expressions forming equations have to be simplified before we can solve them by usual methods. Some equations may not even be linear, to begin with, but they can be brought to a linear form by multiplying both sides of the equation by a suitable expression.
• The utility of linear equations is in their diverse applications; different problems on numbers, ages, perimeters, a combination of currency notes, and so on can be solved using linear equations.

Learning the chapter Linear Equations in One Variable enables the students to understand:

• Multiplication and division of algebraic exp. (Coefficient should be integers)
• Some common errors
• Factorisation
• The method of solving linear equations in one variable in contextual problems involving multiplication and division (word problems) (avoid complex coefficients in the equations).

Disclaimer:

Dropped Topics – 2.2 Solving Equations Which Have Linear Expressions on One Side and Numbers on the Other Side, 2.3 Some Applications, 2.5 Some More Applications and 2.7 Equations Reducible to the Linear Forms.

Frequently Asked Questions on NCERT Solutions for Class 8 Maths Chapter 2

How many exercises are present in ncert solutions for class 8 maths chapter 2, what are the main topics covered in ncert solutions for class 8 maths chapter 2, are ncert solutions for class 8 maths chapter 2 enough for board exam preparation.

BYJU’S is very nice learning app

Very nice app

Very nice app and very helpful

Byju’s is much most better than ever ❤

This app is wonderful best and very helpfull i loved this app

BYJU’S is very helpful app

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

Question and Answer forum for K12 Students

DAV Class 8 Maths Book Solutions Pdf – DAV Class 8 Maths Solutions

DAV Class 8 Maths Book Solutions Pdf: Many students feel difficulty finding the DAV Class 8 Maths Solutions. If you are also in the same situation then this is the right platform you came to. In this article, we provide complete solutions to DAV Class 8 Maths Book Pdf. With that, you can easily understand the concepts. Even DAV Maths Class 8 Solutions are available in pdf format, so you can just download them and practice.

Have a look at the complete article DAV Maths Book Class 8 Solutions Pdf, and practice all worksheets that are available for every chapter to score well in exams and get promoted to class 9 easily.

DAV Public School Class 8 Maths Book Solutions – DAV Maths Class 8 Solutions

In our DAV Class 8 Maths Solution, we have included all the chapters according to the latest DAV Maths Book Class 8 Pdf. And this PDF can be downloaded easily and follow all the chapter-wise worksheets that are provided with solutions to every problem by our math experts. With these DAV Class 8th Maths Solutions, you can prepare for exams very effectively and quickly.

Look into the below chapters and select the link of the worksheets that you like to check out the Secondary Mathematics Class 8th DAV Solutions for every problem.

Also, Check:

• DAV Books Solutions Class 8

DAV Class 8 Maths Book Solutions PDF Chapter 1 Squares and Square Roots

• DAV Class 8 Maths Chapter 1 Worksheet 1
• DAV Class 8 Maths Chapter 1 Worksheet 2
• DAV Class 8 Maths Chapter 1 Worksheet 3
• DAV Class 8 Maths Chapter 1 Worksheet 4
• DAV Class 8 Maths Chapter 1 Worksheet 5
• DAV Class 8 Maths Chapter 1 Brain Teasers

Class 8 DAV Maths Solutions Chapter 2 Cubes and Cube Roots

• DAV Class 8 Maths Chapter 2 Worksheet 1
• DAV Class 8 Maths Chapter 2 Worksheet 2
• DAV Class 8 Maths Chapter 2 Brain Teasers

DAV Class 8 Maths Solutions Chapter 3 Exponents and Radicals

• DAV Class 8 Maths Chapter 3 Worksheet 1
• DAV Class 8 Maths Chapter 3 Worksheet 2
• DAV Class 8 Maths Chapter 3 Brain Teasers

DAV Maths Class 8 Solutions Chapter 4 Direct and Inverse Variation

• DAV Class 8 Maths Chapter 4 Worksheet 1
• DAV Class 8 Maths Chapter 4 Worksheet 2
• DAV Class 8 Maths Chapter 4 Worksheet 3
• DAV Class 8 Maths Chapter 4 Brain Teasers

DAV Maths Book Class 8 Solutions Pdf Chapter 5 Profit, Loss and Discount

• DAV Class 8 Maths Chapter 5 Worksheet 1
• DAV Class 8 Maths Chapter 5 Worksheet 2
• DAV Class 8 Maths Chapter 5 Worksheet 3
• DAV Class 8 Maths Chapter 5 Brain Teasers

DAV Class 8 Maths Solution Pdf Chapter 6 Compound Interest

• DAV Class 8 Maths Chapter 6 Worksheet 1
• DAV Class 8 Maths Chapter 6 Worksheet 2
• DAV Class 8 Maths Chapter 6 Worksheet 3
• DAV Class 8 Maths Chapter 6 Worksheet 4
• DAV Class 8 Maths Chapter 6 Brain Teasers

DAV Class 8th Maths Solutions Chapter 7 Algebraic Identities

• DAV Class 8 Maths Chapter 7 Worksheet 1
• DAV Class 8 Maths Chapter 7 Worksheet 2
• DAV Class 8 Maths Chapter 7 Worksheet 3
• DAV Class 8 Maths Chapter 7 Worksheet 4
• DAV Class 8 Maths Chapter 7 Worksheet 5
• DAV Class 8 Maths Chapter 7 Worksheet 6
• DAV Class 8 Maths Chapter 7 Worksheet 7
• DAV Class 8 Maths Chapter 7 Brain Teasers

Class 8 DAV Maths Book Solution Chapter 8 Polynomials

• DAV Class 8 Maths Chapter 8 Worksheet 1
• DAV Class 8 Maths Chapter 8 Worksheet 2
• DAV Class 8 Maths Chapter 8 Worksheet 3
• DAV Class 8 Maths Chapter 8 Brain Teasers

DAV Maths Solution Class 8 Chapter 9 Linear Equations in One Variable

• DAV Class 8 Maths Chapter 9 Worksheet 1
• DAV Class 8 Maths Chapter 9 Worksheet 2
• DAV Class 8 Maths Chapter 9 Brain Teasers

DAV Solution Class 8 Maths Chapter 10 Parallel Lines

• DAV Class 8 Maths Chapter 10 Worksheet 1
• DAV Class 8 Maths Chapter 10 Worksheet 2
• DAV Class 8 Maths Chapter 10 Worksheet 3
• DAV Class 8 Maths Chapter 10 Brain Teasers

Class 8 DAV Maths Solution Chapter 11 Understanding Quadrilaterals

• DAV Class 8 Maths Chapter 11 Worksheet 1
• DAV Class 8 Maths Chapter 11 Worksheet 2
• DAV Class 8 Maths Chapter 11 Worksheet 3
• DAV Class 8 Maths Chapter 11 Brain Teasers

DAV School Class 8 Maths Solutions Chapter 12 Construction of Quadrilaterals

• DAV Class 8 Maths Chapter 12 Worksheet 1
• DAV Class 8 Maths Chapter 12 Worksheet 2
• DAV Class 8 Maths Chapter 12 Worksheet 3
• DAV Class 8 Maths Chapter 12 Worksheet 4
• DAV Class 8 Maths Chapter 12 Brain Teasers

Maths Class 8 DAV Solutions Chapter 13 Introduction to Graphs

• DAV Class 8 Maths Chapter 13 Worksheet 1
• DAV Class 8 Maths Chapter 13 Worksheet 2
• DAV Class 8 Maths Chapter 13 Brain Teasers

Class 8 Maths DAV Solution Chapter 14 Mensuration

• DAV Class 8 Maths Chapter 14 Worksheet 1
• DAV Class 8 Maths Chapter 14 Worksheet 2
• DAV Class 8 Maths Chapter 14 Worksheet 3
• DAV Class 8 Maths Chapter 14 Worksheet 4
• DAV Class 8 Maths Chapter 14 Worksheet 5
• DAV Class 8 Maths Chapter 14 Worksheet 6
• DAV Class 8 Maths Chapter 14 Worksheet 7
• DAV Class 8 Maths Chapter 14 Worksheet 8
• DAV Class 8 Maths Chapter 14 Worksheet 9
• DAV Class 8 Maths Chapter 14 Brain Teasers

DAV Class 8 Maths Book Solution Chapter 15 Statistics and Probability

• DAV Class 8 Maths Chapter 15 Worksheet 1
• DAV Class 8 Maths Chapter 15 Worksheet 2
• DAV Class 8 Maths Chapter 15 Worksheet 3
• DAV Class 8 Maths Chapter 15 Worksheet 4
• DAV Class 8 Maths Chapter 15 Brain Teasers

DAV Public School Class 8 Maths Book Solutions Chapter 16 Rotational Symmetry

• DAV Class 8 Maths Chapter 16 Worksheet 1
• DAV Class 8 Maths Chapter 16 Worksheet 2
• DAV Class 8 Maths Chapter 16 Brain Teasers

DAV Class 8 Maths Syllabus 2023-24

Here we are going to provide you with a complete syllabus of class 8 Maths with the names of the concepts in every chapter. And also provides you with the weightage of marks for every chapter, instructions, weightage to form of questions, guidelines for internal assessment, and many others.

General Instructions:

• The entire syllabus has to be covered in two terms i.e. Term I and Term II.
• Each term will comprise 100 marks, whereas in written exam to have 80% weightage and the internal assessment will have 20% weightage.
• Term II will cover the whole syllabus and the written examination will have 80% weightage and internal assessment will have 20% weightage.

Guidelines for Internal Assessment (20 Marks) It is suggested that, in each term, the internal assessment will be carried out as follows:

Weightage to Form of Questions

Class 8 Maths DAV Book Pdf Detailed Syllabus

The syllabus has been divided into two parts, one for the first term and the other for the second term.

Note: Final Term will cover the whole syllabus so time must be devoted to the revision.

Chapter 1 Square and Square Roots (12 Periods) Square of a number, triangular numbers, and numbers between two consecutive square numbers, finding the square root of a number by the repeated subtraction method, finding square roots of perfect squares by factorization, Using the division method, finding square roots of Positive integers which are perfect squares, Decimals which are perfect squares, Finding square roots of numbers which are not perfect squares by the division method up to three decimal places. Problems based on square roots (simple problems only).

Learning Outcomes: Students will be able to appreciate Squares of even numbers are even, Squares of odd numbers are odd, Perfect squares, and number ending in 2, 3, 7, or 8 is never perfect square, the Concept of Pythagorean triplet, find the square root of a number, By prime factorization, By long division method Students will be able to understand and apply the following rules:

• Rule 1: If a and b are perfect squares (b ≠ 0) then \(\sqrt{a \times b}=\sqrt{a} \times \sqrt{b}\), \(\sqrt{\frac{a}{b}} \times \frac{\sqrt{a}}{\sqrt{b}}\)
• Rule 2: The pairing of numbers in the division method starts from the decimal point. For the integral part, it goes from right to left, and for the decimal part, it goes from left to right.
• Rule 3: If p and q are not perfect squares, then to find \(\sqrt{\frac{p}{q}}\), we express \(\frac{p}{q}\) as a decimal and then apply division method.

Chapter 2 Cubes and Cube Roots (8 Periods) Cube of a number, Cube roots of perfect cubes by factorization (cube root should not exceed two digits).

Learning Outcomes: Students will be able to understand Cube and the cube root of a negative number is negative i.e. \(\sqrt[3]{-x}=-\sqrt[3]{x}\), Cube of an even natural number is even and cube of odd natural number is odd. Students will be able to apply the following rules: For any two integers a and b, we have \(\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}\), \(\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\), b ≠ 0

Chapter 3 Exponents and Radicals (8 Periods) Idea of rational exponents, Laws of exponents including rational numbers as exponents, and Idea of radicals and radicands.

Learning Outcomes: Students will be able to convert radical form to exponential form and vice versa. Students will be able to apply the following rules: If a is any rational number different from zero and x, y are any rational numbers, then

• \(a^x \times a^y=a^{x+y}\)
• \(a^x \div a^y=a^{x-y}\)
• \(\left(a^x\right)^y=a^{x y}\)
• \((a)^0=1\)

Chapter 4 Direct and Inverse Variations (10 Periods) Direct variation, Inverse variation, and examples. Problems with Time and Work and Time and Distance.

Learning Outcomes: Students will be able to distinguish between Direct Variation and Inverse Variation, solve the problems on time and work as well as time and distance using the concepts of direct and inverse variations.

Chapter 5 Profit and Loss and Discount (12 Periods) Problems with profit and loss including discount (rebate), marked price, selling price (only single discount to be discussed), VAT.

Learning Outcomes: The students will be able to understand the concept of profit and loss, calculate S.P./C.P., apply the concept of discount, and understand VAT and service tax and its calculation.

Chapter 6 Compound Interest (12 Periods) Meaning of Compound Interest. Calculation of amount and compound interest by unitary method. Calculation of amount and compound interest by formula up to three years. Interest compounded annually, half-yearly, or quarterly up to three conversion periods, Growth and Depreciation.

Learning Outcomes: Students will be able to distinguish between simple interest and compound interest, calculate compound interest from the amount, calculate compound interest when compounded annually, half-yearly, and quarterly, and analyze growth and depreciation applicable in various situations.

Chapter 7 Algebraic Identities (12 Periods) Study of the following identities:

• (a + b) 2 = a 2 + 2ab + b 2
• (a – b) 2 = a 2 – 2ab + b 2
• (a + b) (a – b) = a 2 – b 2

The above identities may be verified through cardboard models. Expansion of the square of a trinomial: (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca Product of two binomials: (x + a) (x + b) = x 2 + (a + b)x + ab Factorization of Algebraic Expressions based on the above identities.

Learning Outcomes: After the completion of this chapter, students will be able to distinguish between identity and equation, learn the application of identities, factorize algebraic expressions using the identities, and apply the identities in different practical situations.

Chapter 8 Polynomials (10 Periods) Idea of a polynomial in one variable and its terms, coefficients, and degree, Division of a polynomial in one variable by a monomial or binomial. (Restricted to polynomials in one variable of degree ‘4’). Verification of Dividend = Divisor × Quotient + Remainder. (Explain the cases of non-zero remainder and remainder equal to zero). Concept of a factor of a polynomial when the remainder is zero.

Learning Outcomes: The students will be able to identify coefficients and degree of a polynomial, divide a polynomial in one variable by a monomial as a binomial, verify the dividend by using Divisor × Quotient + Remainder, understand and appreciate the factor of a polynomial when the remainder is zero.

Chapter 9 Linear Equations in One Variable (10 Periods) Solving equations of the type \(\frac{a x+b}{c x+d}\) = k; cx + d ≠ 0 Word problems on linear equations in one variable, Simple problems from daily life situations like age, coins, number of students in a class, speed, distance, formation of ‘2’ digit numbers, etc. with special emphasis on the ability to translate word problems into mathematical statements.

Learning Outcomes: The students will be able to solve linear equations in one variable and convert the language into a linear equation based on different life situations.

Chapter 10 Parallel Lines (10 Periods) definition, Angle made by a transversal with two parallel lines and vice-versa. Verification of the following properties:

• Two lines parallel to the same line are parallel to each other.
• Two lines perpendicular to the same line are parallel to each other.
• To divide a line segment into a given number of equal segments.
• To divide a line segment in a given ratio internally (constructions should be done by using a ruler and compasses).

Learning Outcomes: After the completion of this unit, students will be able to appreciate different types of angles and their relation when a transversal intersects two parallel lines and vice-versa, divide a line segment into equal parts using parallel lines with the help of a ruler and compass, comprehend that two lines parallel/perpendicular to the same line are parallel to each other.

Chapter 11 Understanding Quadrilaterals (12 Periods) Introduction to curves, Polygons, squares, rectangles, rhombus, parallelograms, and trapezium (An example of a kite may be given as a special type of quadrilateral). Verification of the following Properties:

• Opposite sides of a parallelogram are equal.
• Opposite angles of a parallelogram are equal.
• Diagonals of a parallelogram bisect each other,
• Diagonals of a rectangle are equal and bisect each other.
• Diagonals of a square are equal, perpendicular to each other, and bisect each other.
• Diagonals of a rhombus bisect each other at right angles.

(Simple problems based on these properties involving one or two logical steps).

Learning Outcomes: After the completion of this chapter, students will be able to recognize different types of quadrilaterals i.e. trapezium, parallelogram, rectangle, rhombus, square, and kite, understand the properties of parallelogram, rectangle, rhombus, and square, distinguish between different type of quadrilaterals.

Chapter 12 Construction of Quadrilaterals (10 Periods) Construction of quadrilateral given:

• Four sides and one diagonal
• Three sides and both diagonals
• Two adjacent sides and three angles
• Three sides and two included angles.

(The sides should be in whole no. of cm or at least multiples of \(\frac{1}{2}\)a cm. Angles should be multiples of 15)

Learning Outcomes: After the completion of this chapter, students will be able to construct a quadrilateral with given conditions and comprehend whether the construction of a quadrilateral with given data is possible or not.

Chapter 13 Introduction to Graphs (5 Periods) Cartesian plane. Plotting a point on the Cartesian plane. Independent and dependent variables. Drawing of graphs and type of figure.

Learning Outcomes: After the completion of this chapter, students will be able to understand the Cartesian plane and its various elements, identify the coordinates of a point, evaluate the distance of a point from the x-axis and y-axis, plot the point on a Cartesian plane, join the points and identify the figure so formed, identify abscissa and ordinates of a point.

Chapter 14 Mensuration (5 + 10 Periods) Area of trapezium, general quadrilateral, and polygon. Surface area of cuboid, cube, and right Circular cylinder. Volume of cuboid, cube, and right circular cylinder. Visualizing solid shapes, and polyhedrons. Mapping space around us.

Learning Outcomes: The students will be able to find the area of the plane figure (trapezium and quadrilateral), find the area of a polygon by dividing into various quadrilaterals and triangles, calculate the surface area of rectilinear solid figures, calculate the surface area of rectilinear solids i.e. cube and cuboids, distinguish between S.A. of a right circular cylinder and cube/cuboid, calculate S.A. of a right circular cylinder, understand the formation of cubes, cuboid with the help of nets, locate side view, top view and front view of solid figures, verify Euler’s formula for polyhedrons, map the different routes.

Chapter 15 Statistics and Probability (12 Periods) Raw data, frequency, making frequency table from the given raw data. Ungrouped and grouped data. Range, class size, class limits, and class marks. Grouping the given data into classes. Drawing, reading, and interpretation of histogram. Circle graphs or pie charts and their drawing, Probability, Chance, Experiment, Outcome, Event, and Probability of an event. Simple cases.

Learning Outcomes: After studying this chapter, students will be able to understand the terms observation, raw data, range, class marks, frequency, frequency table, differentiate between raw data, ungrouped and grouped data, mark pictorial representation through histogram and pie charts, and interpret the same, define the term trial, outcome, probability, find probability under different given situations.

Chapter 16 Rotational Symmetry (4 Periods) Rotational symmetry and its order; Centre of Rotation, Angle of Rotation. Line symmetry and Rotational Symmetry. Rotational symmetry should be confined.

Learning Outcomes: The students will be able to understand symmetry, distinguish between line symmetry and rotational symmetry, understand rotational turns about a fixed point, know the order of rotation of symmetry i.e. four in a square and 3 in an equilateral triangle, and calculate the angle of rotation about a fixed point.

FAQs on DAV Class 8 Maths Solutions PDF Free Download

1. How many chapters are available in this DAV Public School Class 8 Maths Book Pdf?

In this Class 8 DAV Maths Book Pdf, there are 16 chapters available starting from squares and square roots to rotational symmetry.

2. Where can I download the Class 8 Maths DAV Solution for free?

You can download the Class 8 DAV Maths Solution for free of cost at our learncram.com website.

3. What are the most important chapters in DAV class 8 maths?

Some of the most important chapters of DAV Class 8 Maths are

• Chapter 5: Profit and Loss and Discount
• Chapter 7: Algebraic Identities
• Chapter 14: Mensuration
• Chapter 15: Statistics and Probability

These have the highest weightage of marks in exams.

4. How are these Maths Class 8 DAV Solutions useful for exams?

By this DAV Solution Class 8 Maths you can practice all the important questions that are available in worksheets along with answers and score good marks in the exam by attempting all the questions.

With this DAV Class 8 Maths Book Solution PDF we are hoping that it is very helpful for students of class 8 in DAV to practice and prepare for exams well. If you find this PDF useful even share this with your friends. Also, explore all other class solutions and worksheets like DAV Solution Class 8, at our DAV Solutions web page.

Even visit our site regularly to stay updated.

Test Grade Calculator

How to calculate test score, test grade calculator – how to use it, test grade calculator – advanced mode options.

This test grade calculator is a must if you're looking for a tool to help set a grading scale . Also known as test score calculator or teacher grader , this tool quickly finds the grade and percentage based on the number of points and wrong (or correct) answers. Moreover, you can change the default grading scale and set your own. Are you still wondering how to calculate test scores? Scroll down to find out – or simply experiment with this grading scale calculator.

If this test grade calculator is not the tool you're exactly looking for, check out our other grading calculators like the grade calculator .

Prefer watching rather than reading? We made a video for you! Check it out below:

To calculate the percentile test score, all you need to do is divide the earned points by the total points possible . In other words, you're simply finding the percentage of good answers:

percentage score = (#correct / #total) × 100

As #correct + #wrong = #total , we can write the equation also as:

percentage score = 100 × (#total - #wrong) / #total

Then, all you need to do is convert the percentage score into a letter grade . The default grading scale looks as in the table below:

If you don't like using the +/- grades, the scale may look like:

• An A is 90% to 100%;
• A B is 80% to 89%;
• A C is 70% to 79%;
• A D is 60% to 69%; and finally
• F is 59% and below – and it's not a passing grade

Above, you can find the standard grading system for US schools and universities. However, the grading may vary among schools, classes, and teachers. Always check beforehand which system is used in your case.

Sometimes the border of passing score is not 60%, but, e.g., 50 or 65%. What then? We've got you covered – you can change the ranges of each grade! Read more about it in the last section of this article: Advanced mode options .

🙋 You might also be interested in our semester grade calculator and the final grade calculator .

Our test score calculator is a straightforward and intuitive tool!

Enter the number of questions/points/problems in the student's work (test, quiz, exam – anything). Assume you've prepared the test with 18 questions.

Type in the number the student got wrong . Instead – if you prefer – you can enter the number of gained points. Let's say our exemplary student failed to answer three questions.

Here we go! Teacher grader tool shows the percentage and grade for that score. For our example, the student scored 83.33% on a test, which corresponds to a B grade.

Underneath you'll find a full grading scale table . So to check the score for the next students, you can type in the number of questions they've got wrong – or just use this neat table.

That was a basic version of the test grade calculator. But our teacher grader is a much more versatile and flexible tool!

You can choose more options to customize this test score calculator. Just hit the Advanced mode button below the tool, and two more options will appear:

Increment by box – Here, you can change the look of the table you get as a result. The default value is 1, meaning the student can get an integer number of points. But sometimes it's possible to get, e.g., half-points – then you can use this box to declare the increment between the next scores.

Percentage scale – In this set of boxes, you can change the grading scale from the default one. For example, assume that the test was challenging and you'd like to change the scale so that getting 50% is already a passing grade (usually, it's 60% or even 65%). Change the last box, Grade D- ≥ value, from default 60% to 50% to reach the goal. You can also change the other ranges if you want to.

And what if I don't need +/- grades ? Well, then just ignore the signs 😄

How do I calculate my test grade?

To calculate your test grade:

• Determine the total number of points available on the test.
• Add up the number of points you earned on the test.
• Divide the number of points you earned by the total number of points available.
• Multiply the result by 100 to get a percentage score.

That's it! If you want to make this easier, you can use Omni's test grade calculator.

Is 27 out of 40 a passing grade?

This depends mainly on the grading scale that your teacher is using. If a passing score is defined as 60% (or a D-), then 27 out of 40 would correspond to a 67.5% (or a D+), which would be a passing grade. However, depending on your teacher’s scale, the passing score could be higher or lower.

What grade is 7 wrong out of 40?

This is a B-, or 82.5% . To get this result:

Use the following percentage score formula: percentage score = 100 × (#total - #wrong) / #total

Here, #total represents the total possible points, and #wrong , the number of incorrect answers.

Substitute your values: percentage score = 100 × (40 - 7) / 40 percentage score = 82.5%

Convert this percentage into a letter grade. In the default grading scale, 82.5% corresponds to a B-. However, grading varies — make sure to clarify with teachers beforehand.

Is 75 out of 80 an A?

Yes , a score of 75 out of 80 is an A according to the default grading scale. This corresponds to a percentage score of 93.75%.

• Poker Odds Calculator
• RAID Calculator

Chilled drink

Christmas tree, coffee kick.

• Biology (100)
• Chemistry (100)
• Construction (144)
• Conversion (295)
• Ecology (30)
• Everyday life (262)
• Finance (571)
• Health (440)
• Physics (510)
• Sports (105)
• Statistics (184)
• Other (183)
• Discover Omni (40)

• CPA NEW SYLLABUS 2021
• KCSE MARKING SCHEMES
• ACTS OF PARLIAMENT
• UNIVERSITY RESOURCES PDF
• CPA Study Notes
• INTERNATIONAL STANDARDS IN AUDITING (ISA)
• Teach Yourself Computers

KNEC / TVET CDACC STUDY MATERIALS, REVISION KITS AND PAST PAPERS

Quality and Updated

KNEC, TVET CDACC NOTES AND PAST PAPERS

DIPLOMA MATERIALS

• KNEC NOTES –  Click to download
• TVET CDACC PAST PAPERS – Click to download

CERTIFICATE MATERIALS

• KNEC CERTIFICATE NOTES – Click to download

NCERT Class 5 English The Little Bully Worksheet With Answer

• Download file

In the realm of education, few things are as rewarding as witnessing young minds grow and learn. And in the NCERT Class 5 English curriculum, there's a captivating chapter that tackles an important topic - dealing with bullies. In Unit 8, Chapter 2, "The Little Bully," students are led on a journey of self-reflection and understanding.

This worksheet provides a comprehensive set of questions and exercises that delve into the intricacies of bullying. From analyzing the behavior of bullies to exploring strategies for standing up against them, students are encouraged to think critically and develop empathy. By incorporating engaging activities, this worksheet actively encourages participation and sparks lively discussions in the classroom.

Through the exploration of different scenarios, children are encouraged to consider the consequences of bullying and develop skills for conflict resolution. Incorporating the keywords "NCERT Class 5 English," "Unit 8," "Chapter 2," and "worksheet with answers," this resource not only aids in comprehension but also reinforces the importance of kindness and empathy. With the NCERT Class 5 English Unit 8 Chapter 2 "The Little Bully" Worksheet, educators can empower students to recognize the impact of their words and actions, fostering a more compassionate and inclusive school environment.

Diving into the heart of Class 5 English syllabus, Chapter 15, and specifically focusing on NCERT Class 5 English Unit 8 Chapter 2, we explore The Little Bully – a story that carries profound lessons on empathy, kindness, and the consequences of bullying. At witknowlearn, our educational materials are designed to not only cover the essentials of the curriculum but also to instill valuable life lessons in young minds. For teachers and parents looking for effective teaching resources, The Little Bully Class 5 worksheet with answer is a perfect tool. These worksheets are crafted to enhance comprehension, allowing students to engage deeply with the story’s themes and characters.

Furthermore, we provide The Little Bully Class 5 extra questions answers, which are excellent for extending learning beyond the basic curriculum. These additional questions aim to challenge students understanding and encourage them to think critically about the storys moral implications. Also available is a set of meticulously designed The Little Bully Class 5 MCQs (Multiple Choice Questions), which serve as an excellent way to test students recall and understanding of the story in a quick and interactive manner.

witknowlearn takes pride in offering comprehensive educational resources that cater to the needs of both educators and learners. Our goal is to ensure that every child grasps the essential messages within The Little Bully, promoting a classroom environment that is aware of the impacts of bullying and equipped with the knowledge to foster kindness and empathy among peers. By integrating these resources into your teaching or learning practice, youre not just covering the NCERT Class 5 English curriculum, but also contributing to the development of well-rounded, empathetic individuals.

• The little bully worksheet with answer
• All CBSE worksheet
• All ENGLISH worksheet
• All GRADE 5 worksheet

You may like these also

Same or Different Comparing Size Worksheets for Nursery

7 Pages Profit and Loss Worksheets for Class 5 with Answer Key

Best Measurement Worksheet for Class 5th with Answer Key

7 Engaging Fun with Numbers Class 3 Worksheets for Math Practice

Teaching resources, test generator, worksheet generator, elearning for students, practice question paper, mock test series, happy parenting, elearning for child, worksheet for child, mock test series for child, witknowlearn, privacy policy, terms and condition, refund/cancellation policy.

Or login with Google

Notification

Upgrade to better learning opportunities.