NCERT Exemplar Solutions Class 8 Maths Solutions Chapter 7 Algebraic Expression, Identities & Factorisation

# NCERT Exemplar Solutions Class 8 Maths Solutions Chapter 7 Algebraic Expression, Identities & Factorisation

## NCERT Exemplar Solutions Class 8 Maths Solutions Chapter 7 Algebraic Expression, Identities & Factorization

Subject specialists have designed NCERT Exemplar Solutions for Maths Class 8 Chapter 7 which includes thorough solutions for reference. These solutions are updated according to the latest CBSE syllabus for 2024-25 and are provided in easy language for understanding. Tips and tricks are also provided. These solutions are provided so a student can clear his doubts and get help with a deep understanding of the concept. Also, you can refer them to make the chapter notes and revisions notes. PDF of this can also be downloaded from the website.

Fill Out the Form for Expert Academic Guidance!

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

1. The product of a monomial and a binomial is a

(a) monomial (b) binomial

(c) trinomial (d) none of these

Solution:-

(b) binomial

The product of a monomial and a binomial results in a binomial. Let’s consider a monomial as 2x and a binomial as x + y. The product of a monomial and a binomial is calculated as (2x) × (x + y) This simplifies to 2x^2 + 2xy.

1. In a polynomial, the exponents of the variables are always

(a) integers (b) positive integers

(c) non-negative integers (d) non-positive integers

Solution:-

(b) positive integers

In a polynomial, the exponents of the variables are always positive integers.

1. Which of the following is correct?

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

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

Solution:-

(b) (a – b)2 = a2 – 2ab + b2

We have, = (a – b) × (a – b)

= a × (a – b) – b × (a – b)

= a2 – ab – ba + b2

= a2 – 2ab + b2

The correct expansion for (a – b)^2 is a^2 – 2ab + b^2.

1. The sum of –7pq and 2pq is

(a) –9pq (b) 9pq (c) 5pq (d) – 5pq

Solution:-

(d) – 5pq

The sum of -7pq and 2pq is -5pq as they are like terms.

1. If we subtract –3x2y2 from x2y2, then we get

(a) – 4x2y2 (b) – 2x2y2 (c) 2x2y2 (d) 4x2y2

Solution:-

(d) 4x2y2

Subtracting -3x^2y^2 from x^2y^2 results in 4x^2y^2.

1. Like term as 4m3n2 is

(a) 4m2n2 (b) – 6m3n2 (c) 6pm3n2 (d) 4m3n

Solution:-

(b) – 6m3n2

A like term to 4m^3n^2 is -6m^3n^2.

1. Which of the following is a binomial?

(a) 7 × a + a (b) 6a2 + 7b + 2c

(c) 4a × 3b × 2c (d) 6 (a2 + b)

Solution:-

(d) 6 (a2 + b)

A binomial among the options is 6(a^2 + b).

1. Sum of a – b + ab, b + c – bc and c – a – ac is

(a) 2c + ab – ac – bc (b) 2c – ab – ac – bc

(c) 2c + ab + ac + bc (d) 2c – ab + ac + bc

Solution:-

(a) 2c + ab – ac – bc

The sum of a – b + ab, b + c – bc, and c – a – ac simplifies to 2c + ab – ac – bc.

1. Product of the following monomials 4p, – 7q3, –7pq is

(a) 196 p2q4 (b) 196 pq4 (c) – 196 p2q4 (d) 196 p2q3

Solution:-

(a) 196 p2q4

The product of 4p, -7q^3, and -7pq is 196p^2q^4.

1. Area of a rectangle with length 4ab and breadth 6b2 is

(a) 24a2b2 (b) 24ab3 (c) 24ab2 (d) 24ab

Solution:-

(b) 24ab3

The area of a rectangle with length 4ab and breadth 6b^2 is 24ab^3.

1. Volume of a rectangular box (cuboid) with length = 2ab, breadth = 3ac and height = 2ac is

(a) 12a3bc2 (b) 12a3bc (c) 12a2bc (d) 2ab +3ac + 2ac

Solution:-

(a) 12a3bc2

The volume of a rectangular box with dimensions 2ab, 3ac, and 2ac is 12a^3bc^2.

1. Product of 6a2 – 7b + 5ab and 2ab is

(a) 12a3b – 14ab2 + 10ab (b) 12a3b – 14ab2 + 10a2b2

(c) 6a2 – 7b + 7ab (d) 12a2b – 7ab2 + 10ab

Solution:-

(b) 12a3b – 14ab2 + 10a2b2

The product of 6a^2 – 7b + 5ab and 2ab simplifies to 12a^3b – 14ab^2 + 10a^2b^2.

1. Square of 3x – 4y is

(a) 9x2 – 16y2 (b) 6x2 – 8y2

(c) 9x2 + 16y2 + 24xy (d) 9x2 + 16y2 – 24xy

Solution:-

(d) 9x2 + 16y2 – 24xy

The square of 3x – 4y is 9x^2 + 16y^2 – 24xy.

1. Which of the following are like terms?

(a) 5xyz2, – 3xy2z (b) – 5xyz2, 7xyz2

(c) 5xyz2, 5x2yz (d) 5xyz2, x2y2z2

Solution:-

(b) – 5xyz2, 7xyz2

Like terms among the options are -5xyz^2, 7xyz^2.

1. Coefficient of y in the term –y/3 is

(a) – 1 (b) – 3 (c) -1/3 (d) 1/3

Solution:-

(c) -1/3

The coefficient of y in the term -y/3 is -1/3.

1. a2 – b2 is equal to

(a) (a – b)2 (b) (a – b) (a – b)

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

Solution:-

(c) (a + b) (a – b)

The expression a^2 – b^2 is equal to (a + b)(a – b).

1. Common factor of 17abc, 34ab2, 51a2b is

(a) 17abc (b) 17ab (c) 17ac (d) 17a2b2c

Solution:-

(b) 17ab

The common factor of 17abc, 34ab^2, and 51a^2b is 17ab.

1. Square of 9x – 7xy is

(a) 81x2 + 49x2y2 (b) 81x2 – 49x2y2

(c) 81x2 + 49x2y2 –126x2y (d) 81x2 + 49x2y2 – 63x2y

Solution:-

(c) 81x2 + 49x2y2 –126x2y

The square of 9x – 7xy simplifies to 81x^2 + 49x^2y^2 – 126x^2y.

1. Factorised form of 23xy – 46x + 54y – 108 is

(a) (23x + 54) (y – 2) (b) (23x + 54y) (y – 2)

(c) (23xy + 54y) (– 46x – 108) (d) (23x + 54) (y + 2)

Solution:-

(a) (23x + 54) (y – 2)

The factorised form of 23xy – 46x + 54y – 108 is (23x + 54)(y – 2)

1. Factorised form of r2 – 10r + 21 is

(a) (r – 1) (r – 4) (b) (r – 7) (r – 3)

(c) (r – 7) (r + 3) (d) (r + 7) (r + 3)

Solution:-

(b) (r – 7) (r – 3)

The factorised form of r^2 – 10r + 21 is (r – 7)(r – 3).

1. Factorised form of p2 – 17p – 38 is

(a) (p – 19) (p + 2) (b) (p – 19) (p – 2)

(c) (p + 19) (p + 2) (d) (p + 19) (p – 2)

Solution:-

(a) (p – 19) (p + 2)

The factorised form of p^2 – 17p – 38 is (p – 19)(p + 2).

1. On dividing 57p2qr by 114pq, we get

(a) ¼pr (b) ¾pr (c) ½pr (d) 2pr

Solution:-

(c) ½pr

On dividing 57p2qr by 114pq,

It can be expanded as = (57 × p × p × q × r)/(114 × p × q)

= 57pr/114 … [divide both numerator and denominator by 57]

= ½pr

1. On dividing p (4p2 – 16) by 4p (p – 2), we get

(a) 2p + 4 (b) 2p – 4 (c) p + 2 (d) p – 2

Solution:-

(c) p + 2

On dividing p (4p2 – 16) by 4p (p – 2)

= (p((2p)2 – (4)2))/ (4p(p – 2))

= ((2p – 4) × (2p + 4))/(4(p – 2))

Take out the common factors

= ((2(p – 2)) × (2 (p + 4)))/(4(p -2))

= (4(p – 2)(p + 2))/ (4(p – 2))

= p + 2

1. The common factor of 3ab and 2cd is

(a) 1 (b) – 1 (c) a (d) c

Solution:-

(a) 1

The common factor of 3ab and 2cd is 1, as there is no common factor except 1 between them.

1. An irreducible factor of 24x2y2 is

(a) x2 (b) y2 (c) x (d) 24x

Solution:-

(c) x

The irreducible factor of 24x²y² is x, as it cannot be further factored into simpler components.

24x2y2 = 2 × 2 × 2 × 3 × x × x × y × y

Therefore an irreducible factor is x.

1. Number of factors of (a + b)2 is

(a) 4 (b) 3 (c) 2 (d) 1

Solution:-

(c) 2

The number of factors of (a + b)² is 2, as further factorization is not possible beyond (a + b)(a + b).

1. The factorised form of 3x – 24 is

(a) 3x × 24 (b) 3 (x – 8) (c) 24 (x – 3) (d) 3(x – 12)

Solution:-

(b) 3 (x – 8)

The factorised form of 3x – 24 is 3(x – 8), where 3 is taken out as a common factor.

1. The factors of x2 – 4 are

(a) (x – 2), (x – 2) (b) (x + 2), (x – 2)

(c) (x + 2), (x + 2) (d) (x – 4), (x – 4)

Solution:-

(b) (x + 2), (x – 2)

The factors of x² – 4 are (x + 2) and (x – 2), as x² – 4 can be expressed as (x + 2)(x – 2).

1. The value of (– 27x2y) ÷ (– 9xy) is

(a) 3xy (b) – 3xy (c) – 3x (d) 3x

Solution:-

(d) 3x

The value of (-27x²y) ÷ (-9xy) is 3x, obtained by simplifying the division.

1. The value of (2x2 + 4) ÷ 2 is

(a) 2x2 + 2 (b) x2 + 2 (c) x2 + 4 (d) 2x2 + 4

Solution:-

(b) x2 + 2

The value of (2x2 + 4) ÷ 2 = (2x2 + 4)/2

= (2(x2 + 2))/2

= x2 + 2

1. The value of (3x3 +9x2 + 27x) ÷ 3x is

(a) x2 +9 + 27x (b) 3x3 +3x2 + 27x

(c) 3x3 +9x2 + 9 (d) x2 +3x + 9

Solution:-

(d) x2 +3x + 9

The value of (3x3 +9x2 + 27x) ÷ 3x = (3x3 + 9x2 + 27x)/3x

Takeout 3x as common,

= 3x (x2 + 3x + 9)/3x

= x2 + 3x + 9

1. The value of (a + b)2 + (a – b)2 is

(a) 2a + 2b (b) 2a – 2b (c) 2a2 + 2b2 (d) 2a2 – 2b2

Solution:-

(c) 2a2 + 2b2

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

= (a2 + a2) + (b2 + b2) + (2ab – 2ab)

= 2a2 + 2b2

1. The value of (a + b)2 – (a – b)2 is

(a) 4ab (b) – 4ab (c) 2a2 + 2b2 (d) 2a2 – 2b2

Solution:-

(a) 4ab

The value of (a + b)2 – (a – b)2 = (a2 + b2 + 2ab) – (a2 + b2 – 2ab)

= a2 – a2 + b2 – b2 + 2ab + 2ab

= 4ab

In questions 34 to 58, fill in the blanks to make the statements true:

1. The product of two terms with like signs is a term.

Solution:-

The product of two terms with like signs is a positive term.

Let us assume two like terms are, 3p and 2q

= 3p × 2q

= 6pq

1. The product of two terms with unlike signs is a term.

Solution:-

The product of two terms with unlike signs is a negative term.

Let us assume two unlike terms are, – 3p and 2q

= -3p × 2q

= – 6pq

1. a (b + c) = a × ____ + a × _____.

Solution:-

a (b + c) = a × b + a × c. … [by using left distributive law]

= ab + ac

1. (a – b) _________ = a2 – 2ab + b2

Solution:-

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

(a – b) (a – b)= a × (a – b) – b × (a – b)

= a2 – ab – ba + b2

= a2 – 2ab + b2

1. a2 – b2 = (a + b ) __________.

Solution:-

a2 – b2 = (a + b) (a – b) … [from the standard identities]

1. (a – b)2 + ____________ = a2 – b2

Solution:-

(a – b)2 + (2ab – 2b2) = a2 – b2

= (a – b)2 + (2ab – 2b2)

= a2 + b2 – 2ab + 2ab – 2b2

= a2 – b2

1. (a + b)2 – 2ab = ___________ + ____________

Solution:-

(a + b)2 – 2ab = a2 + b2

= (a + b)2 – 2ab

= a2 + 2ab + b2 – 2ab

= a2 + b2

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

Solution:-

(x + a) (x + b) = x2 + (a + b) x + ab

= (x + a) (x + b)

= x × (x + b) + a × (x + b)

= x2 + xb + xa + ab

= x2 + x (b + a) + ab

1. The product of two polynomials is a ________.

Solution:-

The product of two polynomials is a polynomial.

1. Common factor of ax2 + bx is __________.

Solution:-

Common factor of ax2 + bx is x (ax + b)

1. Factorised form of 18mn + 10mnp is ________.

Solution:-

Factorised form of 18mn + 10mnp is 2mn (9 + 5p)

= (2 × 9 × m × n) + (2 × 5 × m × n × p)

= 2mn (9 + 5p)

## Class 8 Maths Chapter 7 FAQs

##### What kind of questions are there in these NCERT Exemplar Solutions?

This chapter deals with multiple choice questions, descriptive type of questions, long answer type questions, short answer type questions, fill in the blanks, and daily life examples. In end, students can increase their problem-solving skills also with time management skills. This will help you to score good marks

##### Is NCERT Exemplar Solutions of this Chapter enough to attend all the questions that come in the board exam?

Yes, these NCERT Exemplar Solutions deal with solutions for all questions given in NCERT Textbook Maths for Class 8. Most of the questions coming in the exams are from these exercises. By studying these concepts, you can achieve good grades.

##### Is it necessary to learn all the topics provided in NCERT Exemplar Solutions for Class 8 Maths Chapter 7?

Yes. As these questions seem to be important for exams. These questions are solved by subject matter experts for helping students to crack these exercises easily. These solutions give students knowledge about data handling. Solutions can be downloaded in PDF format on the Infinity Learn website.

## Related content

 NCERT Exemplar for Class 6 Maths Solutions CBSE Notes for Class 8 Maths CBSE Notes for Class 7 Science CBSE Notes for Class 8 Science Lines and Angles Class 9 Extra Questions Maths Chapter 6 AMU Class 11 Entrance Exam Sample Papers Harappan Civilization Latitude and Longitude Sahara Desert Home Rule Movement