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Important Questions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities

Important Questions for Class 8 Maths Chapter 8 has been provided here. Students can refer to these questions before their examinations for better preparation.
authorImageNeha Tanna27 Dec, 2024
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Important Questions for Class 8 Maths Chapter 8

Important Questions for Class 8 Maths Chapter 8: Chapter 8 of Class 8 Maths, Algebraic Expressions, and Identities, introduces key concepts such as algebraic expressions, terms, factors, coefficients, and the classification of expressions (monomials, binomials, and polynomials). It covers the addition, subtraction, and multiplication of expressions.

Important questions focus on simplifying expressions, verifying identities, and applying them to solve problems. Practice is crucial for mastering operations on expressions and using identities in real-life problem-solving contexts. This chapter lays a foundation for advanced algebra, making conceptual clarity essential for higher classes.

Important Questions for Class 8 Maths Chapter 8 Overview

Chapter 8 of Class 8 Maths, Algebraic Expressions and Identities, is crucial as it forms the foundation for advanced algebra. This chapter helps students understand the structure and manipulation of algebraic expressions, including terms, coefficients, and their classifications (monomials, binomials, and polynomials). It emphasizes essential algebraic identities, which simplify complex expressions and solve practical problems efficiently. Important questions focus on simplifying, verifying, and applying these identities. Mastering this chapter enhances logical reasoning and problem-solving skills, preparing students for competitive exams and higher mathematical studies.

Important Questions for Class 8 Maths Chapter 8 PDF

Below, we have provided a comprehensive PDF of Important Questions for Class 8 Maths Chapter 8: Algebraic Expressions and Identities. This curated set includes key questions to strengthen your understanding of terms, coefficients, and identities. Download the PDF to practice simplifying expressions, verifying identities, and mastering algebraic concepts effectively for exams.

Important Questions for Class 8 Maths Chapter 8 PDF

Important Questions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities

Below is the Important Questions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities - 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 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

Benefits of Using Important Questions for Class 8 Maths Chapter 8

Using Important Questions for Class 8 Maths Chapter 8: Algebraic Expressions and Identities offers several benefits:

Concept Clarity : Focused questions help reinforce fundamental concepts like terms, coefficients, and identities, ensuring better understanding.

Practice with Identities : Questions on identities will improve fluency in using these formulas to simplify expressions.

Problem-Solving Skills : Regular practice enhances analytical thinking and the ability to tackle algebraic problems systematically.

Exam Preparation : Covers common and challenging patterns, helping students score better in exams.

Confidence Boost : Familiarity with diverse problems builds confidence in applying algebraic concepts effectively.

Foundation for Future Studies : Prepares students for advanced algebra in higher classes and competitive exams.

Important Questions for Class 8 Maths Chapter 8 FAQs

What is the use of algebraic identities in real life?

The two types of algebraic identities are Binomial and trinomial Algebraic Identities. These identities are used in real life. From calculating how many boxes, tons of raw material will fit into calculating the area of your room. These are used as algebraic identities all the time.

What are the basic concepts of algebraic expressions and identities?

An algebraic expression is one that has at least one variable operation and one constant.

What is the objective of algebraic identities?

The algebraic equations which are valid for all values of variables in them are called algebraic identities. They are also used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and solving different polynomials.

What is the purpose of algebra in real life?

It helps individuals to analyze complex situations, break them down into simpler parts, and solve problems step by step. Moreover, basic algebra can also be a gateway to other advanced math topics like calculus, which has even more applications in science, engineering, economics, and statistics.
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