{"id":44453,"date":"2022-05-30T16:00:44","date_gmt":"2022-05-30T10:30:44","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44453"},"modified":"2023-01-25T11:36:54","modified_gmt":"2023-01-25T06:06:54","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-10-ex-10-2","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-10-ex-10-2\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.2"},"content":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.2<\/h2>\n

Question 1.
\nFind the product of:
\n(i) 4x3<\/sup> and -3xy
\n(ii) 2xyz and 0
\n(iii) –\\(\\frac{2}{3}\\)p2<\/sup>q, \\(\\frac{3}{4}\\)pq2<\/sup> and 5pqr
\n(iv) -7ab,-3a3<\/sup> and –\\(\\frac{2}{7}\\)ab2<\/sup>
\n(v) –\\(\\frac{1}{2}\\)x2<\/sup> – \\(\\frac{3}{5}\\)xy, \\(\\frac{2}{3}\\)yz and \\(\\frac{5}{7}\\)xyz
\nSolution:
\nProduct of
\n(i) 4x3<\/sup> and -3xy = 4x3<\/sup> \u00d7 (-3xy) = -12x3+1<\/sup> y = -12x4<\/sup>y
\n(ii) 2xyz and 0 = 2xyz \u00d7 0 = 0
\n\"ML
\n\"ML<\/p>\n

Question 2.
\nMultiply:
\n(i) (3x – 5y + 7z) by – 3xyz
\n(ii) (2p2<\/sup> – 3pq + 5q2<\/sup> + 5) by – 2pq
\n(iii) (\\(\\frac{2}{3}\\)a2<\/sup>b – \\(\\frac{4}{5}\\)ab2<\/sup> + \\(\\frac{2}{7}\\)ab + 3) by 35ab
\n(iv) (4x2<\/sup> – 10xy + 7y2<\/sup> – 8x + 4y + 3) by 3xy
\nSolution:
\n(i) – 3xyz (3x – 5y + 7z)
\n= (- 3xyz) \u00d7 3x + (- 3xyz) \u00d7 (- 5y) + (- 3xyz) \u00d7 (7z)
\n= – 9x2<\/sup>yz + 15xyz2<\/sup> – 21xyz2<\/sup><\/p>\n

(ii) -2pq (2p2<\/sup> – 3pq + 5q2<\/sup> + 5)
\n= (-2pq) \u00d7 2p2<\/sup> + (-2pq) \u00d7 (-3pq) + (- 2pq) \u00d7 (5q2<\/sup>) + (-2pq) \u00d7 5
\n= -4p3<\/sup>q + 6p2<\/sup>q2<\/sup> – 10pq3<\/sup> – 10pq<\/p>\n

(iii) \\(\\left(\\frac{2}{3} a^{2} b-\\frac{4}{5} a b^{2}+\\frac{2}{7} a b+3\\right)\\) by 35ab
\n= \\(\\frac{2}{3}\\)a2<\/sup>b \u00d7 35ab – \\(\\frac{4}{5}\\)ab2<\/sup> \u00d7 35ab = \\(\\frac{2}{7}\\)ab \u00d7 35ab + 3 \u00d7 35ab
\n= \\(\\frac{70}{3}\\)a3<\/sup>b2<\/sup> – 28a2<\/sup>b3<\/sup> + 10a2<\/sup>b2<\/sup> + 105ab<\/p>\n

(iv) (4x2<\/sup> – 10xy + 7y2<\/sup> – 8x + 4y + 3) by 3xy
\n4x2<\/sup> \u00d7 3xy – 10xy \u00d7 3xy + 7y2<\/sup> \u00d7 3xy – 8x \u00d7 3xy + 4y \u00d7 3xy + 3 \u00d7 3xy
\n= 12x3<\/sup>y – 30x2<\/sup>y2<\/sup> + 21xy3<\/sup> – 24x2<\/sup>y + 12xy2<\/sup> + 9xy<\/p>\n

Question 3.
\nFind the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively:
\n(i) (p2<\/sup>q, pq2<\/sup>)
\n(ii) (5xy, 7xy2<\/sup>)
\nSolution:
\n(i) Sides of a rectangle are p2<\/sup>q and pq2<\/sup>
\nArea = p2<\/sup>q \u00d7 pq2<\/sup> = p2+1<\/sup>q2+1<\/sup> = p3<\/sup>q3<\/sup><\/p>\n

(ii) Sides are 5xy and 7xy2<\/sup>
\nArea = 5xy \u00d7 7xy2<\/sup> = 35x1+1<\/sup> \u00d7 y1+2\u00a0<\/sup>= 35x2<\/sup>y3<\/sup><\/p>\n

Question 4.
\nFind the volume of rectangular boxes with the following length, breadth and height respectively:
\n(i) 5ab, 3a2<\/sup>b, 7a4<\/sup>b2<\/sup>
\n(ii) 2pq, 4q2<\/sup>, 8rp
\nSolution:
\nLength, breadth and height of a rectangular box are
\n(i) 5ab, 3a2<\/sup>b, 7a4<\/sup>b2<\/sup>
\n\u2234 Volume = Length \u00d7 breadth \u00d7 height
\n= 5ab \u00d7 3a2<\/sup>b \u00d7 7a4<\/sup>b2<\/sup>
\n= 5 \u00d7 3 \u00d7 7 \u00d7 a1+2+4<\/sup> \u00d7 b1+1+2<\/sup>
\n= 105a7<\/sup>b4<\/sup><\/p>\n

(ii) 2pq, 4q2<\/sup>, 8rp
\n\u2234 Volume = 2pq \u00d7 4q2<\/sup> \u00d7 8rp
\n= 2 \u00d7 4 \u00d7 8 \u00d7 p1+1<\/sup> \u00d7 q1+2<\/sup> \u00d7 r
\n= 64p2<\/sup>q3<\/sup>r<\/p>\n

Question 5.
\nSimplify the following expressions and evaluate them as directed:
\n(i) x2<\/sup>(3 – 2x + x2<\/sup>) for x = 1; x = -1; x = \\(\\frac{2}{3}\\) and x = –\\(\\frac{1}{2}\\)
\n(ii) 5xy(3x + 4y – 7) – 3y(xy – x2<\/sup> + 9) – 8 for x = 2, y = -1
\nSolution:
\n(i) x2<\/sup>(3 – 2x + x2<\/sup>)
\nfor x = 1; x = -1; x = – 1 x = \\(\\frac{2}{3}\\) and x = –\\(\\frac{1}{2}\\)
\nx2<\/sup>(3 – 2x + x2<\/sup>) = 3x2<\/sup> – 2x3<\/sup> + x4<\/sup>
\n(a) x = 1, then
\n3x2<\/sup> – 2x3<\/sup> + x4<\/sup> = 3(1)2<\/sup> – 2(1 )3<\/sup> + (1)4<\/sup>
\n= 3 \u00d7 1 – 2 \u00d7 1 + l
\n=3 – 2 + 1 = 2<\/p>\n

(b) x = -1
\n3x2<\/sup> – 2x3<\/sup> + x4<\/sup> = 3(-1)2<\/sup> – 2(-1)3<\/sup> + (-1)4<\/sup>
\n= 3 \u00d7 1 – 2 \u00d7 (-1) + 1 = 3 + 2 + 1 = 6<\/p>\n

(c) x = \\(\\frac{2}{3}\\)
\n\"ML
\n\"ML<\/p>\n

(ii) 5xy(3x + 4y – 7) – 3y(xy – x2<\/sup> + 9) – 8
\n= 15x2<\/sup>y + 20xy2<\/sup> – 35xy – 3xy2<\/sup> + 3 x2<\/sup>y – 21y – 8
\n= 18x2<\/sup>y + 17xy2<\/sup> – 35xy – 27y – 8
\nWhen x = 2, y = -1
\n= 18(2)2<\/sup> \u00d7 (-1) + 17(2) (-1)2<\/sup> – 35(2) (-1) – 27(-1) – 8
\n= 18 \u00d7 4 \u00d7 (-1) + 17 \u00d7 2 \u00d7 1 – 35 \u00d7 2 \u00d7 (-1) – 27 \u00d7 (-1) – 8
\n= -74 + 34 + 70 + 27 – 8
\n= 131 – 80 = 51<\/p>\n

Question 6.
\nAdd the following:
\n(i) 4p(2 – p2<\/sup>) and 8p3<\/sup> – 3p
\n(ii) 7xy(8x + 2y – 3) and 4xy2<\/sup>(3y – 7x + 8)
\nSolution:
\nAdd
\n(i) 4p(2 – p2<\/sup>) and 8p3<\/sup> – 3p
\n= 8p – 4p3<\/sup> + 8p3<\/sup> – 3p
\n= 5p + 4p3<\/sup>
\n= 4p3<\/sup> + 5p<\/p>\n

(ii) 7xy(8x + 2y – 3) and 4xy2<\/sup>(3y – 7x + 8)
\n= 56x2<\/sup>y + 14xy2<\/sup> – 21xy + 12xy3<\/sup> – 28x2<\/sup>y2<\/sup> + 32xy2<\/sup>
\n= 12xy3<\/sup> – 28x2<\/sup>y2<\/sup> + 56x2<\/sup>y +46xy2<\/sup> = 21xy<\/p>\n

Question 7.
\nSubtract:
\n(i) 6x(x – y + z)- 3y(x + y – z) from 2z(-x + y + z)
\n(ii) 7xy(x2<\/sup> -2xy + 3y2<\/sup>) – 8x(x2<\/sup>y – 4xy + 7xy2<\/sup>) from 3y(4x2<\/sup>y – 5xy + 8xy2<\/sup>)
\nSolution:
\n(i) 6x(x – y + z) – 3y(x + y – z) from 2z(-x + y + z)
\n6x2<\/sup> – 6xy + 6xz – 3xy – 3y2<\/sup> + 3yz from – 2xz + 2yz + 2z2<\/sup>
\n= (-2xz + 2yz + 2z2<\/sup>) – (6x2<\/sup> – 6xy + 6xz – 3xy – 3y2<\/sup> + 3yz)
\n= – 2xz + 2yz + 2z2<\/sup> – 6x2<\/sup> + 6xy – 6xz + 3xy + 3y2<\/sup> – 3yz
\n= 9xy – yz – 8zx – 6x2<\/sup> + 3y2<\/sup> + 2z2<\/sup><\/p>\n

(ii) 7xy(x2<\/sup> – 2xy + 3y2<\/sup>) – 8x(x2<\/sup>y – 4xy + 7xy2<\/sup>) from 3y(4x2<\/sup>y – 5xy + 8xy2<\/sup>)
\n7x3<\/sup>y – 14x2<\/sup>y2<\/sup> + 21xy3<\/sup> – 8x3<\/sup>y + 32x2<\/sup>y – 56x2<\/sup>y2<\/sup> from 12x2<\/sup>y2<\/sup> – 15xy2<\/sup> + 24xy3<\/sup>
\n= (12x2<\/sup>y2<\/sup> – 15xy2<\/sup> + 24xy3<\/sup>) – (7x3<\/sup>y – 14x2<\/sup>y2<\/sup> + 21xy3<\/sup> – 8x3<\/sup>y + 32x2<\/sup>y – 56x2<\/sup>y2<\/sup>
\n= 12x2<\/sup>y2<\/sup> – 15xy2<\/sup> + 24xy3<\/sup> – 7x3<\/sup>y + 14x2<\/sup>y2<\/sup> – 12xy3<\/sup> + 8x3<\/sup>y – 32x2<\/sup>y + 56x2<\/sup>y2<\/sup>
\n= 82x2<\/sup>y2<\/sup> + 3xy3<\/sup> + x3<\/sup>y – 15xy2<\/sup> – 32x2<\/sup>y<\/p>\n

ML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.2 Question 1. Find the product of: (i) 4×3 and -3xy (ii) 2xyz and 0 (iii) -p2q, pq2 and 5pqr (iv) -7ab,-3a3 and -ab2 (v) -x2 – xy, yz and xyz Solution: Product of (i) 4×3 and -3xy = 4×3 … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.2 - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-10-ex-10-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.2\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.2 Question 1. Find the product of: (i) 4x3 and -3xy (ii) 2xyz and 0 (iii) -p2q, pq2 and 5pqr (iv) -7ab,-3a3 and -ab2 (v) -x2 – xy, yz and xyz Solution: Product of (i) 4x3 and -3xy = 4x3 ... 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Find the product of: (i) 4x3 and -3xy (ii) 2xyz and 0 (iii) -p2q, pq2 and 5pqr (iv) -7ab,-3a3 and -ab2 (v) -x2 – xy, yz and xyz Solution: Product of (i) 4x3 and -3xy = 4x3 ... 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