# 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:

 CBSE Class 8 Maths Important Questions Sr No. Chapters Chapters Name 1 Chapter 1 Rational Numbers 2 Chapter 2 Linear Equations in One Variable 3 Chapter 3 Understanding Quadrilaterals 4 Chapter 4 Practical Geometry 5 Chapter 5 Data Handling 6 Chapter 6 Squares and Square Roots 7 Chapter 7 Cubes and Cube Roots 8 Chapter 8 Comparing Quantities 9 Chapter 9 Algebraic Expressions and Identities 10 Chapter 10 Visualising Solid Shapes 11 Chapter 11 Mensuration 12 Chapter 12 Exponents and Powers 13 Chapter 13 Direct and Inverse Proportions 14 Chapter 14 Factorisation 15 Chapter 15 Introduction to Graphs 16 Chapter 16 Playing with Numbers

### 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,

 Terms Coefficient (i) 5xyz², -3zy 5, -3 (ii) 1, x, x² 1, 1, 1 (iii) 4x²y², -4x²y²z², z² 4, -4, 1 (iv) 3, -pq, qr, -rp 3, -1, 1, -1 (v) x/2, y/2, -xy 1/2, 1/2, -1 (vi) 0.3a, -0.6ab, 0.5b 0.3, -0.6, 0.5

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

(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

(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 + 4q²p²6q4

2        2             3          3

=    3P49p²q² + 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

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

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 ________

(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

(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.

breadth = x² + 12xy + 27y²

( x + 9y )

=  x² + 9xy + 3xy + 27y²

( x + 9y )

=  x ( x + 9y ) + 3y (x + 9y)

( x + 9y )

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)

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

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

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

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

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,

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.

(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

(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 )

(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

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.
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Important formulas

CBSE extra questions

Q.1 Using identity, find the value of (4.7)2.

Marks:2
Ans

Use the identity as (a b)2 = a2 2ab + b2.

$\begin{array}{l}{4.7}^{2}={\left(5\text{}\text{}0.3\right)}^{2}\\ \\ \text{}={\left(5\right)}^{2}\text{}\text{}2\text{}—\text{}5\text{}—\text{}0.3\text{}+\text{}{\left(0.3\right)}^{2}\\ \\ \text{}=\text{25}\text{3.0 + 0.09}\\ \\ \text{}=\text{}22.09\end{array}$

Hence, the required value is 22.09.

Q.2 Simplify: (a + 2) (c 4) + (a 2) (c + 2) + 2 (ac + 17)

Marks:2
Ans

(a + 2) (c 4) + (a 2) (c + 2) + 2 (ac + 17)
= ac 4a + 2c 8 + ac + 2a 2c 4 + 2ac + 34
= 4ac 2a + 22

Q.3 The value of 1001 × 1004 is

A. 1006004

B. 1005004

C. 1004004

D. 1003004

Marks:1
Ans [ 2049078 ]

Q.4 If the length of a rectangle is 2x2 + 3 units and its breadth is 2x2 7 units. What is the area and perimeter of the rectangle

Marks:4
Ans

(x + a) (x + b) = x2 + x (a + b) + ab
= (2x2 + 3) × (2x2 7)
= (2x2)2 + 2x2(3 7) + 3(7)
= 4x4 8x2 21 square units
= 2(2x2 + 3 + 2x2 7)
= 2(4x2 4)
= 8x2 8 units

Q.5

$\mathrm{If}{\text{x}}^{2}\text{+}\frac{1}{{\mathrm{x}}^{2}}=4,\mathrm{what}\text{is the value of x +}\frac{1}{\mathrm{x}}$

Marks:2
Ans

$\begin{array}{l}{\text{Given: x}}^{2}\text{+}\frac{1}{{\mathrm{x}}^{2}}=4\\ {\text{using identities: (a + b)}}^{2}{\text{= a}}^{2}{\text{+ b}}^{2}\text{+ 2ab}\\ {\left(\text{x +}\frac{1}{\mathrm{x}}\right)}^{2}={\text{x}}^{2}\text{+}\frac{1}{{\mathrm{x}}^{2}}+2\mathrm{x}×\frac{1}{\mathrm{x}}\\ \text{= 4 + 2}\\ \text{= 6}\end{array}$