ML Aggarwal Algebraic Expression and Identities Exe-10.1 Class 8 ICSE Ch-10 Maths Solutions. We Provide Step by Step Answer of Exe-10.1 Questions for Algebraic Expression and Identities as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.
ML Aggarwal Algebraic Expression and Identities Exe-10.1 Class 8 ICSE Maths Solutions
| Board | ICSE |
| Publications | Avichal Publishig Company (APC) |
| Subject | Maths |
| Class | 8th |
| Chapter-10 | Algebraic Expression and Identities |
| Writer | ML Aggarwal |
| Book Name | Understanding |
| Topics | Solution of Exe-10.1 Questions |
| Edition | 2023-2024 |
Algebraic Expression and Identities Exe-10.1
ML Aggarwal Class 8 ICSE Maths Solutions
Page-173
Question 1. Identify monomials, binomials, and trinomials from the following algebraic expressions :
(i) 5p × q × r2
(ii) 3x2 + y ÷ 2z
(iii) -3 + 7x2
(iv) (5a2−3b2+𝑐2 ) /2
(v) 7x5 – 3𝑥/𝑦
(vi) 5p ÷ 3q – 3p2 × q2
Answer:
(i) 5p × q × r2 = 5pqr2
As this algebraic expression has only one term, its therefore a monomial.
(ii) 3x2 + y ÷ 2z = 3x2/2z + y/2z
As this algebraic expression has two terms, its therefore a binomial.
(iii) -3 + 7x2
As this algebraic expression has two terms, its therefore a binomial.
(iv) (5a2−3b2+𝑐2 ) /2
= 5a2/2 = 3b2/2 + c/2
As this algebraic expression has three terms, its therefore a trinomial.
(v) 7x5 – 3x/y
As this algebraic expression has two terms, its therefore a binomial.
(vi) 5p ÷ 3q – 3p2 × q2 = 5p/3q – 3p2q2
As this algebraic expression has two terms, its therefore a binomial.
Question 2. Identify which of the following expressions are polynomials. If so, write their degrees.
(i) 25x4 – √3x2 + 5x – 1
(ii) 7x3 – 3𝑥2 + √5
(iii) 4a3b2 – 3ab4 + 5ab + 23
(iv) 2x2y – 3/𝑥𝑦 + 5y3 +√ 3
Answer:
(i) It is a polynomial and the degree of this expression is 4.
(ii) It is not a polynomial.
(iii) It is a polynomial and the degree of this expression is 5.
(iv) It is not a polynomial.
Question 3. Add the following expressions:
(i) ab – bv, bv – ca, ca – ab
(ii) 5p2q2 + 4pq + 7,3 + 9pq – 2p2q
(iii) l2 + m2 + n2, lm + mn, mn + nl, nl + lm
(iv) 4x3 – 7x2 + 9, 3x2 – 5x + 4, 7x3 – 11x + 1, 6x2 – 13x
Answer:
(i) ab – bc, bc – ca, ca – ab
⇒ ab – bc + bc – ca + ca – ab = 0
(ii) 5p2q2 + 4pq + 7,3 + 9pq – 2p2q2
= 5p2q2 + 4pq + 7 + 3 + 9pq – 2p2q2
= 5p2q2 – 2p2q2 + 4pq + 9pq + 7 + 3
= 3p2q2 + 13pq + 10
(iii) l2 + m2 + n2, lm + mn, mn + nl, nl + lm
= l2 + m2 + n2 + lm + mn + mn + nl + nl + lm
= l2 + m2 + n2 + 2lm + 2mn + 2nl
(iv) 4x3 – 7x2 + 9, 3x2 – 5x + 4, 7x3 – 11x + 1, 6x2 – 13x
= 4x3 – 7x2 + 9 + 3x2 – 5x + 4 + 7x3 – 112 + 1 + 6x2 – 13x
= 4x2 + 7x3 – 7x2 + 3x2 + 6x2 – 5x – 11x – 13x + 9 + 4 + 1
= 11x3 – 2x2 – 29x + 14
Question 4. Subtract:
(i) 8a + 3ab – 2b + 7 from 14a – 5ab + 7b – 5
(ii) 8xy + 4yz + 5zx from 12xy – 3yz – 4zx + 5xyz
(iii) 4p2q – 3pq + 5pq2 – 8p + 7q -10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q
Answer:
(i) Subtracting 8a + 3ab – 2b + 7 from 14a – 5ab + 7b – 5, we have
= (14a – 5ab + 7b – 5) – (8a + 3ab – 2b + 7)
= 14a – 5ab + 7b – 5 – 8a – 3ab + 2b – 7
= 6a – 8ab + 9ab – 12
(ii) Subtracting 8xy + 4yz + 5zx from 12xy – 3yz – 4zx + 5xyz, we have
= (12xy – 3yz – 4zx + 5xyz) – (8xy + 4yz + 5zx)
= 12xy – 3yz – 4zx + 5xyz – 8xy – 4yz – 5zx
= 4xy – 7yz – 9zx + 5xyz
(iii) Subtracting 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q, we have
= (18 – 3p – 11q + 5pq – 2pq2 + 5p2q) – (4p2q – 3pq + 5pq2 – 8p + 7q – 10)
= 18 – 3p – 11q + 5pq – 2pq2 + 5p2q – 7p2q + 3pq – 5pq2 + 8p – 7q + 10
= 28 + 5p – 78q + 8pq – 7pq2 + p2q
Question 5. Subtract the sum of 3x2 + 5xy + 7y2 + 3 and 2x2 – 4xy – 3y2 + 7 from 9x2 – 8xy + 11y2
Answer:
First, adding 3x2 + 5xy + 7y2 + 3 and 2x2 – 4xy – 3y2 + 7, we have
= 3x2 + 5xy + 7y2 + 3 + 2x2 – 4xy – 3y2 + 7
= 5x2 + xy + 4y2 + 10
Subtracting 5x2 + xy + 4y2 + 10 from 9x2 – 8xy + 11y2
= (9x2 – 8xy + 11y2) – (5x2 + xy + 4y2 + 10)
= 9x2 – 8xy + 11y2 – 5x2 – xy – 4y2 – 10
= 4x2 – 9xy + 7y2 – 10
Question 6. What must be subtracted from 3a2 – 5ab – 2b2 – 3 to get 5a2 – 7ab – 3b2 + 3a ?
Answer:
From the question, its understood that we have to subtract 5a2 – 7ab – 3b2 + 3a from 3a2 – 5ab – 2b2 – 3
= 3a2 – 5ab – 2b2 – 3 – (5a2 – 7ab – 3b2 + 3a)
= 3a2 – 5ab – 2b2 – 3 – 5a2 + 7ab + 3b2 – 3a
= -2a2 + 2ab + b2 – 3a – 3
Question 7. The perimeter of a triangle is 7p2 – 5p + 11 and two of its sides are p2 + 2p – 1 and 3p2 – 6p + 3. Find the third side of the triangle.
Answer:
Perimeter of a triangle = 7p2 – 5p + 11
And, two of its sides are p2 + 2p – 1 and 3p2 – 6p + 3
Perimeter of a triangle = Sum of three sides of triangle
⇒ 7p2 – 5p + 11 = (p2 + 2p – 1) + (3p2 – 6p + 3) + (Third side of triangle)
7p2 – 5p + 11 = (4p2 – 4p + 2) + (Third side of triangle)
⇒ Third side of triangle = (7p2 – 5p + 11) – (4p2 – 4p + 2)
= (7p2 – 4p2) + (- 5p + 4p) + (11 – 2)
= 3p2 – p + 9
Hence, the third side of the triangle is 3p2 – p + 9.
— End of Algebraic Expression and Identities Exe-10.1 Class 8 ICSE Maths Solutions :–
Return to : – ML Aggarwal Maths Solutions for ICSE Class -8
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