Important Questions Class 8 Maths Chapter 9

Important Questions Class 8 Mathematics Chapter 9 – Algebraic Expressions and Identities

Algebra is one of the wide areas of Mathematics that studies mathematical expressions involving variables, constants, and mathematical operations like addition, subtraction, multiplication, and division. Algebra is the basis for higher mathematics studies in many fields, including science, engineering, medicine, and economics.

Class 8 Mathematics Chapter 9 – Algebraic expressions and identities is the continuation of the earlier classes and chapters, where students learned about basic algebraic expressions like x + 3, 2y – 5, 3x², 4xy + 7 etc. Many more complex expressions can be formed using variables and constants. The topics covered in Chapter 9 are discussed in the points below.

• Expressions: The problem statement is formed by combining variables, constants and mathematical operators.
• Variables and constants: Consider the example 2x – 3. This expression is formed from the variable X and constants 2 and 3. With different values of the variable x, the result of the expression 2x – 3 changes.
• Terms, factors, and coefficients: Terms in the expression are the product of its factors. The expression 2x – 3 is formed by two terms, 2x and 3. The term 2x is the product of its factors 2 and x. The term 3 comprises just one factor, i.e., 3. 

A coefficient is the numerical factor of the term. In the expression 5xy + 3z, the coefficient of the term 5xy is 5, and the coefficient of the term 3z is 3.

• Types of expressions: The three main types of algebraic expressions are,

Monomials: Expression formed with only one term. For example, 4x², -5z, 9xz², etc.

Binomials: Expression formed with two terms. For example 5x – 2, 6y – 9x², a + b, etc

Polynomials: A polynomial is an expression containing one or more terms with a non-zero coefficient and variables having non-negative integers as exponents. For example 2xy, 6x³ + 4x³ + 3x – 1, etc.

• Like and unlike terms: Like terms have the same variables and exponents. However, the coefficients need not be the same. Unlike terms are two or more terms that do not have the same variables or exponents. The order of the variables does not matter unless there is power. To understand better, check the following examples of both types. 

Like terms: 4b + 10b, 2x² + 4x², 15w – 13.4w, etc.

Unlike terms: 2x + 9y, 10a² – 9b, x³ – x², etc.

• Subtraction and addition of algebraic expressions: While adding or subtracting polynomials, first add or subtract the like terms; and then handle the unlike terms.
• Multiplication of algebraic expressions: There are three ways to multiply expressions:

 (1) A monomial multiplied by a monomial,

 (2) A polynomial multiplied by a monomial and

 (3) A polynomial multiplied by a polynomial

• Identity: An algebra identity is an equality, which means that the left-hand side of the equation is equal to the right-hand side for all values of the variables. Identities are very useful for solving equations in simple and easy steps. The four standard identities are,

1. (a + b)² = a² + 2ab + b²
2. (a – b)² = a² – 2ab + b²
3. (a + b) (a – b) = a² – b²
4. (x + a) (x + b) = x² + (a + b)x + ab

Algebra is an overwhelming and difficult chapter for many students, requiring critical thinking and creative problem-solving skills. Extramarks is one of the most trusted online learning platforms to help students understand the concepts clearly and prepare well for the examinations. Our highly experienced and expert Mathematics teachers have prepared CBSE Mathematics preparation materials for Class 8 students. These study resources include NCERT solutions, solved question papers, sample papers, CBSE revision notes, and more, per the CBSE syllabus and NCERT guidelines.

Get Access to CBSE Class 8 Maths Important Questions 2022-23 with Chapter-Wise Solutions

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

Important Questions Class 8 Mathematics Chapter 9 – With Solutions

A complete list of Important Questions Class 8 Mathematics Chapter 9 provided on the Extramarks website consists of various questions for students to analyse their weak areas and practise until they thoroughly understand all the concepts and reduce mistakes. Chapter 9 Class 8 Mathematics Important Questions includes short answer questions, long answer questions, and MCQs backed with detailed and easy-to-understand solutions. These important questions are collated from the NCERT textbook and other genuine sources. By practising from solved question papers, sample papers, and the Mathematics Class 8 Chapter 9 Important Questions, students get familiar with the exam pattern and improve their time-management skills. Regular practise improves the calculating speed, which is helpful in the examination.

Register on the Extramarks website to refer to and practise the Important Questions Class 8 Mathematics Chapter 9 and other chapter-wise questions.

Below is a sample list of questions and their solutions from our Class 8 Mathematics Chapter 9 Important Questions:

Question 1.  Find the terms and their coefficients for each of the following expressions.

(i) 5xyz² – 3zy 

(ii) 1 + x+ x²

(iii) 4x²y² – 4x²y²z² + z²

(iv) 3- pq + qr – rp

(v) x/2 + y/2 – xy

(vi) 0.3a – 0.6ab + 0.5b

Answer 1. The terms and coefficients are given below,

Question 2. Classify the following polynomials as monomials, binomials, and trinomials. Also, state the 

polynomials do not fall in any of these three categories?

x + y, 1000, x + x² + x³, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd + da, pqr, 

p²q + pq², 2p + 2q, 

Answer 2. The classified terms are given below,

Monomials- 1000, pqr

Binomials- x + y, 2y – 3y², 4z – 15z², p²q + pq², 2p + 2q

Trinomials- x + x²+ x³, 7 + y + 5x, 2y – 3y² + 4y³, 5x – 4y + 3xy

Polynomials that do not fall in any of these categories are  x + y, x + x²+ x³,  ab + bc + cd + da

Question 3. Add the following.

(i) ab – bc, bc – ca, ca – ab

(ii) a – b + ab, b – c + bc, c – a + ac

(iii) 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²

(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl

Answer 3. (i)       (ab – bc) + (bc – ca) + (ca – ab)         

                            = ab – bc + bc – ca + ca – ab

                            = ab-ab-bc+bc-ca+ca

                            = 0

                    (ii)      (a – b + ab) + (b – c + bc) + (c – a + ac)

                            = a – b + ab + b – c + bc + c -a +ac

                            = a – a -b + b + ab – c + c + bc + ac

                            = ab + bc + ac

                    (iii)     ( 2p²q² – 3pq + 4) + ( 5 + 7pq – 3p²q²)

                            = 2p²q² – 3pq + 4 + 5 + 7pq – 3p²q²

                            = 2p²q² – 3p²q² + 7pq – 3pq + 4 + 5

                            = -1p²q² + 4pq + 9

                            = 4pq + 9 – p²q²

                    (iv)     ( l² + m²) + (m² + n²) + (n² + l²) + (2lm + 2mn + 2nl)

                            = l² + m² + m² + n² + n² + l² + 2lm + 2mn + 2nl

                            = l² +  l² + m² + m² + n² + n² + 2lm + 2mn + 2nl

                            = 2l² + 2m² + 2n² + 2lm + 2mn + 2nl

                            = 2( l² + m² + n² + lm + mn + nl)     

Question 4. Subtract the following.

(i) 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(ii 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

(iii) 4p²q – 3pq + 5pq²– 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2+ 5p²q

Answer 4.

                (i)      ( 12a – 9ab + 5b – 3 ) – ( 4a – 7ab + 3b + 12 )

                          = 12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12

                          = 12a – 4a – 9ab + 7ab + 5b –3b – 3 – 12

                          = 8a – 2ab + 2b – 15

                 (ii)     ( 5xy – 2yz – 2zx + 10xyz ) – ( 3xy + 5yz – 7zx )                        

                            = 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx                         

                            = 5xy – 3xy – 2yz – 5yz – 2zx + 7zx + 10xyz                        

                            = 2xy – 7yz + 5zx + 10xyz

                (iii)      ( 18 – 3p – 11q + 5pq – 2pq2+ 5p²q ) – ( 4p²q – 3pq + 5pq²– 8p + 7q – 10 )                         

                            = 18 – 3p – 11q + 5pq – 2pq2 + 5p²q – 4p²q + 3pq – 5pq² + 8p – 7q + 10                         

                            = 18 + 10 – 3p + 8p – 11q – 7q + 5pq + 3pq – 2pq2 – 5pq² + 5p²q – 4p²q                           

                            = 28 + 5p – 18q + 8pq – 7pq² + p²q

Question 5. Multiply the following. 

                      (i)  – 7pq²r³, – 13p³q²r

                      (ii) 3x²y²z², 17xyz

                     (iii) 15xy², 17yz²

                     (iv) –5a²bc, 11ab, 13abc²

                     (v) (pq – 2r), (pq – 2r)

                     (vi) (3/2p² + 2/3q²), (2p² –3q²)

Answer 5. (i)       ( – 7pq²r³ ) x ( – 13p³q²r )                           

                               = 91P4 q4 r4                  

                   (ii)      ( 3x²y²z² ) x ( 17xyz )

                                                  = 51x³y³z³

                  (iii)      ( 15xy² ) x ( 17yz² )

                              = 255xy³z²

                  (iv)      (  –5a²bc ) x ( 11ab ) x ( 13abc² )

                              = -715a⁴b³c³

                  (v)      ( pq – 2r ) x ( pq – 2r )

                             = pq ( pq – 2r) – 2r( pq – 2r )

                             = p²q² – 2pqr – 2rpq + 4r²

                             = p²q² – 4pqr + 4r²             

                  (vi)    (  3  p²  +  2 q² ) x ( 2p² – 3q² )

                               2            3                        

                            =   3p²  x 2p² –  3  p² x 3q² +  2q² x 2p² –  2q² x 3q²                      

                                 2                    2                     3                  3                           

                            =     6P4 –  9p²q² + 4q²p² – 6q4

                                   2        2             3          3                           

                            =    3P4 – 9p²q² + 4q²p² – 2q4

                                            2             3         

Question 6. Which term is the like term similar to 24a²bc?

(a) 13 × 8a × 2b × c × a 

(b) 8 × 3 × a × b × c

(c) 3 × 8 × a × b × c × c 

(d) 3 × 8 × a × b × b × c

Answer 6. Option (a)

Explanation: To find out the similar term as 24a²bc, let us find the product of each of the equations,

1. 13 × 8a × 2b × c × a =  208a²bc
2. 8 × 3 × a × b × c = 24abc
3. 3 × 8 × a × b × c × c = 24abc²
4. 3 × 8 × a × b × b × c = 24ab²c

Hence, we can get that option (a) is correct.

Question 7. Which of the following is an identity? 

(a) (p + q)²  = p² + q² 

(b) p² – q² = (p – q)² 

(c) p² – q² = p² + 2pq – q² 

(d) (p + q)² = p² + 2pq + q²

Answer 7. Option (d) 

Explanation: The equation  (p + q)² = p² + 2pq + q² follows the first standard algebraic identity

 ( a + b )² = a² + 2ab + b². The rest of the other options do not follow any of the standard identities. Hence option (d) is correct.

Question 8. Fill in the blanks.

(a) (x + a) (x + b) = x² + (a + b)x + ________.

(b) The product of two terms with like signs is a  ________ term.

(c) The product of two terms with unlike signs is a  ________ term.

(d) (a – b) _________ = a² – 2ab + b²

(e)  a² – b² = (a + b ) __________.

(f) (a – b)² + ____________ = a² – b²

(g) (a + b)² – 2ab = ___________ + ____________

(h) The product of two polynomials is a ________

(i) The coefficient in – 37abc is __________. 

(j) Number of terms in the expression a2 + bc × d is ________

Answer 8. 

(a) ab

As per the standard identity 4, (x + a) (x + b) =  x² + (a + b)x + ab

(b) Positive

(c) Negative

(d) ^2 or ( a – b)²

As per standard identity 2, (a – b)² = a² – 2ab + b²

(e) (a – b)

As per standard identity 3, (a + b ) ( a – b ) = a² – b²

(f) 2ab – 2b²

Let us solve the equation with x in the blank space. As per identity 2, (a – b)² = a² – 2ab + b². 

Hence, a² – 2ab + b² + x = a² – b²

x = a² – b² – a² + 2ab – b²

x = 2ab – 2b²  

(g) a² + b²

Using Identity 1 ( a + b )² = a² + 2ab + b²,

(a + b)² – 2ab = a² + 2ab + b² – 2ab = a² + b²

(h) Polynomial

(i) -37

(j) 2

Question 9. Solve the below using correct identities.

(a) (48)²

(b) 181² – 19²

(c) 497 × 505

(d) 2.07 × 1.93

Answer 9.

 (a) (48)²

      = (50 – 2)²

     As (a – b)² = a² – 2ab + b² , hence

     (50 – 2)² = (50)² – 2 × 50 × 2 + (2)²

     = 2500 – 200 + 4

     = 2300 + 4

     = 2304

(b) As a² – b² = (a – b) (a + b)

     181² – 19² = (181 – 19) (181 + 19)

     = 162 × 200

     = 32400

(c) By using the identity (x + a) (x + b) = x2 + (a + b) x + ab

     497 x 505 = ( 500 – 3 ) (500 + 5 )

     = 500² + (–3 + 5) × 500 + (–3) (5) 

     = 250000 + 1000 – 15 

     = 250985

(d) As (a + b) (a – b) = a² – b² 

      2.07 × 1.93 = (2 + 0.07) (2 – 0.07)

      = 2² – 0.07²

      = 3.9951

Question 10. The length of a rectangular box is  ( x + 9y) and the area is x² + 12xy + 27y². Find the breadth.

Answer 10.  Area of a rectangle =  length x breadth, hence breadth = area / length.

                      breadth = x² + 12xy + 27y²

                                         ( x + 9y )

                                     =  x² + 9xy + 3xy + 27y²

                                          ( x + 9y )

                                     =  x ( x + 9y ) + 3y (x + 9y)

                                          ( x + 9y )

                     breadth   =  x + 3y

Question 11. With the help of  identity (x + a) (x + b) = x² + (a + b) x + ab, find the following products.

(a) (x + 3) (x + 7) 

(b) (4x + 5) (4x + 1)

(c) (4x – 5) (4x – 1)

(d) (4x + 5) (4x – 1)

(e) (2x + 5y) (2x + 3y) 

(f) (2a²+ 9) (2a²+ 5)

(g) (xyz – 4) (xyz – 2)

Answer 11. 

1. ( x + 3 ) ( x + 7 )

= x² + ( 3 + 7 ) x + ( 3 x 7 )

= x² + 10x + 21

1. ( 4x + 5 ) ( 4x + 1 )

= 16x² + ( 5 + 1 ) 4x + ( 5 x 1 )

= 16x² + 24x + 5     

     (c) (4x – 5) (4x – 1)

           = 16x² + ( – 5 – 1 ) 4x + ( -5 x -1 )

           = 16x² – 24x + 5   

     (d) (4x + 5) (4x – 1)

          = 16x² + ( 5 + ( – 1 ) )4x + ( 5 x ( – 1 ) )

          = 16x² + 16x  – 5

    (e) (2x + 5y) (2x + 3y) 

           = 4x² + ( 5y + 3y )2x + ( 5y x 3y )

           = 4x² + 16xy + 15y²

     (f) (2a²+ 9) (2a²+ 5)

           = 4a4  + ( 9 + 5 )2a² + ( 9 x 5 )

           = 4a4  + 28a² + 45

    (g) (xyz – 4) (xyz – 2)

          = x²y²z² + ( -4 – 2)xyz + ( – 4 x – 2)

          =  x²y²z² – 6xyz + 8

Question 12. The exponents of the variables in a polynomial are always

 (a) Integers 

 (b) Positive integers 

 (c) Non-negative integers 

 (d) Non-positive integers

Answer 12. (c) Non-negative integers

Explanation: A polynomial will have a non-zero coefficient and variables having non-negative integers as exponents. For example : a + b + r + q, 3ab, 5xyz – 10, 2a + 3b + 7z, etc.

Question 13. The product of a binomial and monomial is a 

(a) Monomial 

(b) Binomial 

(c) Trinomial 

(d) None of these

Answer 13. (b) Binomial

Explanation: This can be demonstrated through an example  below,

 x ( y + z ) = xy + xz 

 This expression contains two terms, x is a monomial and ( y + z ) is a binomial.

 The product of multiplying these terms results in a binomial product xy + xz.

Question 14. Using identities, find products for the below.

(a) 71²

(b) 99²

(c) 102²

(d) 998²

(e) 5.2²

(f) 297 × 303

(g) 78 × 82 

(h) 8.9²

(i) 10.5 × 9.5

Answer 14. 

1. 71² =  ( 70 + 1 )²                                      Identity applied ( a + b )² = a² + 2ab + b²

= 70² + 2 ( 70 x 1 ) + 1²

= 4900 + 140 + 1

= 5041

1. 99² = ( 100 – 1 )²                                       Identity applied ( a – b )² = a² – 2ab + b² 

       = 100² – 2 ( 100 x 1 ) + 1² 

       = 10000 – 200 + 1

       = 9801

       (c) 102² = ( 100 + 2 )²                                      Identity applied ( a + b )² = a² + 2ab + b²

                      = 100² + 2 ( 100 x 2 ) + 2²

                      = 10000 + 400 + 4

                      = 10404      

       (d) 998² = ( 1000 – 2 )²                                      Identity applied ( a – b )² = a² – 2ab + b² 

                      = 1000² – 2 ( 1000 x 2 ) + 2²

                      = 1000000 – 4000 + 4

                      = 996004

       (e) 5.2² = ( 5 + 0.2 )²                                           Identity applied ( a + b )² = a² + 2ab + b²

                     = 5² + 2 ( 5 x 0.2 ) + 0.2²

                     = 25 + 2 +  0.04

                     = 27.04

      (f) 297 × 303 = ( 300 – 3 ) ( 300 + 3 )                 Identity applied  ( a + b ) ( a – b ) = a² – b²

                              = 300² – 3²

                              = 90000 – 9

                              = 89991

      (g) 78 × 82 = ( 80 – 2 ) ( 80 + 2 )                         Identity applied ( a + b ) ( a – b ) = a² – b²

                          =  80² – 2²

                          = 6400 – 4

                          = 6396     

      (h) 8.9² = ( 9.0 – 0.1 )²                                           Identity applied ( a – b )² = a² – 2ab + b²

                    = 9.0² – 2 ( 9.0 x 0.1 ) + 0.1²

                    = 81 – 1.8 + 0.01

                    =  79.21

     (i) 10.5 x 9.5 = ( 10 + 0.5 ) ( 10 – 0.5 )                 Identity applied ( a + b ) ( a – b ) = a² – b²

                            =  10² – 0.5²

                            = 100 – 0.25

                            = 99.75

Question 15. The Coefficient of y in the term −y is

                                                                                    3

(a) – 1 

(b) – 3 

(c) –  1

          3   

(d) 1 

      3 

Answer 15. (c)  –  1

                                 3

Explanation: Coefficient is defined as the numerical factor of a term. 

Hence, the numerical factor/ coefficient of the term  −y is  −1 

                                                                                                 3       3

Question 16.Obtain the volume of rectangular boxes with the following length, breadth and height

respectively.

(a) 5a, 3a², 7a4

(b) 2p, 4q, 8r

(c) xy, 2x²y, 2xy²

(d) a, 2b, 3c

Answer 16. The volume of a rectangular box is the product of its length,

breadth and height, i.e Volume = length x breadth x height.

Volumes are calculated as below,

(a) length = 5a, breadth = 3a², height = 7a4

 Volume = 5a x 3a² x 7a4

= 105a7

(b) length = 2p, breadth = 4q, height = 8r

 Volume = 2p x 4q x 8r

= 64pqr

(c) length = xy, breadth = 2x²y, height = 2xy²

Volume = xy x 2x²y x 2xy²

= 4x4y4

(d) length = a, breadth = 2b, height = 3c

Volume = a x 2b x 3c

= 6abc

Question 17. State whether the following  are True (T) or False (F):

(a) (a + b)² = a² + b² 

(b) (a – b)² = a² – b²

(c) (a + b) (a – b) = a² – b²

(d) The product of two negative terms is a negative term.

(e) The product of one negative and one positive term is a negative term. 

(f)  The coefficient of the term – 6x²y² is – 6. 

(g) p²q + q²r + r²q is a binomial

(h) An equation is true for all values of its variables.

Answer 17. 

(a) False. ( a + b )² = a² + 2ab + b²

(b) False. ( a – b )² = a² – 2ab + b²

(c) True.

(d) False. The product of two negative terms is positive.

(e) True

(f) True

(g) False. The equation p²q + q²r + r²q consists of three terms; hence it is a trinomial.

(h) False. An equation is not true for all values of its variables. For example :  4x + 2 = 10 is true, only for x = 2.

Question 18. Show that LHS = RHS for the below equations.

(a) ( 3x + 7 )² – 84x = ( 3x – 7 )²

(b) ( 9p – 5q )² + 180pq = ( 9p + 5q )²

(c) ( 4pq + 3q )² – ( 4pq – 3q )² = 48pq²

(d) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

Answer 18. 

(a) LHS =  ( 3x + 7 )² – 84x 

      = (3x)² + 2 ( 3x x 7 ) + 7² – 84x

      =  9x² + 42x + 49 – 84x

      = 9x² – 42x + 49

     RHS = ( 3x – 7 )² 

      = (3x)² – 2 ( 3x x 7 ) + 7² 

      = 9x² – 42x + 49

      Hence LHS = RHS

(b) LHS = ( 9p – 5q )² + 180pq 

             = (9p)² – 2 ( 9p x 5q ) + (5q)² + 180pq

             = 81p² + 90pq + 25q²

      RHS = ( 9p + 5q )²

              = (9p)² + 2 ( 9p x 5q ) + (5q)²

              = 81p² + 90pq + 25q²

      Hence LHS = RHS

(c) LHS = ( 4pq + 3q )² – ( 4pq – 3q )²

             = (4pq)² + 2 ( 4pq x 3q ) + (3q)² – ( (4pq)² – 2 ( 4pq x 3q ) + (3q)²)

             = 16p²q² + 24pq² + 9q² – 16p²q² + 24pq² – 9q²

             = 48pq²

     RHS = 48pq²

     Hence LHS = RHS

(d) LHS =  (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

              = a² + ab – ba – b² + b² + bc – cb – c² + c² + ca – ac – a²

              = 0

      RHS = 0

      Hence LHS = RHS

Question 19. Expand the following, using suitable identities.

(a) ( xy + yz )²

(b) ( x²y – xy² )²

(c) ( 7x + 5 )² 

(d) ( 0.9p – 0.5q )²

(e) ( x² + y² ) ( x² – y² )

Answer 19. 

(a) (xy + yz)²

 = (xy)² + 2 (xy x yz ) + (yz)²

 = x²y² + 2xy²z + y²z²

(b)  (x²y – xy²)²

= (x²y)² – 2 (x²y x xy²) + (xy²)²

= x4y² – 2x³y³ + x²y4

(c) (7x + 5)² 

= (7x)² + 2 (7x x 5) + (5)²

= 49x² + 70x + 25

(d) (0.9p – 0.5q)²

= (0.9p)² – 2 ( 0.9p x 0.5q) + (0.5q)²

= 0.81p² – 0.9pq + 0.25q²

(e) ( x² + y² ) ( x² – y² )

= x4 – x²y² + y²x² – y4

= x4 – y4

Question 20. Select the correct option of volume of a rectangular box  with length = 2ab, breadth = 3ac and height = 2ac 

(a) 12a³bc² 

(b) 12a³bc

(c) 12a²bc

(d) 2ab +3ac + 2ac

Answer 20. Option (a)

Explanation:  The formula for calculating the volume of a rectangular box is 

Volume = length x breadth x height 

With the length of the input = 2ab, breadth = 3ac and height = 2ac 

Volume = 2ab x 3ac x 2ac

                = 12a³bc²

Benefits of Solving Important Questions Class 8 Mathematics Chapter 9

Practising the questions from Important Questions Class 8 Mathematics Chapter 9 available on the Extramarks website will benefit the students in grasping the chapter concepts like variables, terms, and standard identities easily and solving the equations with better understanding and avoiding mistakes. Solving the important questions of the chapter rigorously will help you score well in the examination.

A few of the benefits of referring to Important Questions Class 8 Mathematics Chapter 9 are:

• The study and practice materials are based on the latest versions of the CBSE syllabus and NCERT guidelines. Students can rely upon these materials to get well versed in the examination’s basic question/answer format.
• Important Questions Class 8 Mathematics Chapter 9 contains various questions, including MCQs, long answer questions, etc., collated carefully by referring to some of the most trusted sources like NCERT textbooks, NCERT exemplar, past years’ question papers and other sources.
• By thoroughly practising these tricky questions from the Important Questions Class 8 Mathematics Chapter 9, students can analyse their weak areas of Chapter 9 Algebra and overcome them before facing their final examinations.
• All the Important Questions Class 8 Science Chapter 9 has step-by-step solutions to all the questions in the textbook. It will benefit you to read through the problem carefully and figure out what it’s about.

To access the answers given in Important Questions Class 8 Mathematics Chapter 9, you can register on the Extramarks website. Furthermore, students can access the other study materials from Classes 1 to 12 by clicking on the below links:

CBSE Revision Notes

CBSE syllabus

CBSE sample papers

CBSE previous years’ question papers

Important formulas 

CBSE extra questions