{"id":43491,"date":"2022-05-28T17:00:03","date_gmt":"2022-05-28T11:30:03","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=43491"},"modified":"2023-11-09T16:09:29","modified_gmt":"2023-11-09T10:39:29","slug":"ml-aggarwal-class-6-solutions-for-icse-maths-chapter-9-ex-9-4","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-6-solutions-for-icse-maths-chapter-9-ex-9-4\/","title":{"rendered":"ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 9 Algebra Ex 9.4"},"content":{"rendered":"
Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. (ii) (x – y)2<\/sup> = x2<\/sup> – 2xy + y2<\/sup> (iii) x2<\/sup> – y2<\/sup> = (x + y)(x – y) (iv) (x + y)2<\/sup> = (x – y)2<\/sup> + 4xy (v) (x + y)3<\/sup> = x3<\/sup> + y2<\/sup> + 3x2<\/sup>y + 3xy2<\/sup> ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 9 Algebra Ex 9.4 Question 1. Find the value of the following: (i) 43 (ii) (-6)4 (iii) (iv) (-2)3 \u00d7 52 Solution: (i) 43 = 4 \u00d7 4 \u00d7 4 = 64 (ii) (-6)4 = (-6) \u00d7 (-6) \u00d7 (-6) \u00d7 (-6) = 1296 (iii) (iv) … 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":"\n
\nFind the value of the following:
\n(i) 43<\/sup>
\n(ii) (-6)4<\/sup>
\n(iii) \\(\\left(\\frac{2}{3}\\right)^{4}\\)
\n(iv) (-2)3<\/sup> \u00d7 52<\/sup>
\nSolution:
\n(i) 43<\/sup> = 4 \u00d7 4 \u00d7 4 = 64
\n(ii) (-6)4<\/sup> = (-6) \u00d7 (-6) \u00d7 (-6) \u00d7 (-6) = 1296
\n(iii) \\(\\left(\\frac{2}{3}\\right)^{4}=\\frac{2}{3} \\times \\frac{2}{3} \\times \\frac{2}{3} \\times \\frac{2}{3}\\)
\n\\(=\\frac{2 \\times 2 \\times 2 \\times 2}{3 \\times 3 \\times 3 \\times 3}=\\frac{16}{81}\\)
\n(iv) (-2)3<\/sup> \u00d7 52<\/sup>
\n= (-2) \u00d7 (-2) \u00d7 (-2) \u00d7 5 \u00d7 5
\n= (-8) \u00d7 25 = -200<\/p>\n
\nFind the value of:
\n(i) 3x + 2y when x = 3 and y – 2
\n(ii) 5x – 3y when x = 2 and y = -5
\n(iii) a + 2b – 5c when a = 2, b = -3 and c = 1
\n(iv) 2p + 3q + 4r + pqr when p = -1, q = 2 and r = 3
\n(v) 3ab + 4bc – 5ca when a = 4, 6 = 5 and c = -2.
\nSolution:
\n(i) 3x + 2y, x = 3,y = 2
\n(3 \u00d7 3) + (2 \u00d7 2) = 9 + 4 = 13
\n(ii) 5x – 3y, x = 2, y = -5
\n(5 \u00d7 2) – (3 \u00d7 -5) = 10 + 15 = 25
\n(iii) a + 2b – 5c, a =2,b = -3, c = 1
\n2 + (2 \u00d7 -3) -5 \u00d7 (1)
\n= 2 – 6 – 5 = -9
\n(iv) 2p + 3q + 4r + pqr, p = -1, q = 2, r = 3
\n= (2 \u00d7 -1) + (3 \u00d7 2) + (4 \u00d7 3) + (-1) \u00d7 2 \u00d7 3
\n= -2 + 6 + 12 – 6= 10
\n(v) 3ab + 4bc – 5ca, a = 4, b = 5, c = -2 (3 \u00d7 4 \u00d7 5) + (4 \u00d7 5 \u00d7 -2) -5 \u00d7 -2 \u00d7 4
\n= 60 – 40 + 40 = 60<\/p>\n
\nFind the value of:
\n(i) 2x2<\/sup> – 3x + 4 when \u00d7 = 2
\n(ii) 4x3<\/sup> – 5x2<\/sup> – 6x + 7 when x = 3
\n(iii) 3x3<\/sup> + 9x2<\/sup> – x + 8 when x = -2
\n(iv) 2x4<\/sup> – 5x3<\/sup> + 7x – 3 when x = -3
\nSolution:
\n(i) 2x2<\/sup> – 3x + 4, x = 2
\n= 2 \u00d7 (2)2<\/sup> -3x2<\/sup> + 4
\n= 8 – 6 + 4 = 6
\n(ii) 4x3<\/sup> – 5x2<\/sup> – 6x + 7, x = 3
\n= 4(3)3<\/sup> – 5(3)2<\/sup> – 6(3) + 7
\n= 108 – 45 – 18 + 7 = 52
\n(iii) 3x3<\/sup> + 9x2<\/sup> – x + 8, x = -2
\n= 3(-2)3<\/sup> + 9(-2)2<\/sup> – (-2) + 8
\n= -24 + 36 + 2 + 8 = 22
\n(iv) 2x4<\/sup> – 5x3<\/sup> + 7x – 3, x = -3
\n= 2(-3)4<\/sup> – 5(-3)3<\/sup> + 7(-3) – 3
\n= 162 + 135 – 21 – 3 = 273<\/p>\n
\nIf x = 5, find the value of:
\n(i) 6 – 7x2<\/sup>
\n(ii) 3x2<\/sup> + 8x – 10
\n(iii) 2x3<\/sup> – 4x2<\/sup> – 6x + 25
\nSolution:
\n(i) 6 – 7x2<\/sup> = 6 – 7(5)2<\/sup> = 6 – 7(25)
\n= 6 – 175 = -169
\n(ii) 3(5)2<\/sup> + 8(5) – 10
\n= 3(25) + 40 – 10
\n= 75 + 40 – 10 = 75 + 30 = 105
\n(iii) 2(5)3<\/sup> – 4(5)2<\/sup> – 6(5) + 25
\n= 2(125) – 4(25) – 30 + 25
\n= 250 – 100 – 30 + 25= 145<\/p>\n
\nIf x = 2, y = 3 and z = -1, find the values of:
\n(i) x + y
\n(ii) \\(\\frac{x y}{z}\\)
\n(iii) \\(\\frac{2 x+3 y-4 z}{3 x-z}\\)
\nSolution:
\n<\/p>\n
\nIf a = 2, b = 3 and c = -2, find the value of a2<\/sup> + b2<\/sup> + c2<\/sup> – 2ab – 2bc – 2ca + 3abc.
\nSolution:
\na = 2,b = 3,c = -2
\na2<\/sup> + b2<\/sup> + c2<\/sup> – 2ab – 2bc – 2ca + 3abc
\n= (2)2<\/sup> + (3)2<\/sup> + (-2)2<\/sup> – 2 \u00d7 2 \u00d7 3 – 2 \u00d7 3 \u00d7 – 2 – 2 \u00d7 -2 \u00d7 2 + 3 \u00d7 2 \u00d7 3 \u00d7 -2
\n= 4 + 9 + 4 – 12 + 12 + 8 – 36
\n= 25 – 36 = -11<\/p>\n
\nIf p = 4, q = -3 and r = 2, find the value of: p3<\/sup> + q3<\/sup> – r3<\/sup> – 3pqr.
\nSolution:
\np = 4, q = -3, r = 2
\np3<\/sup> + q3<\/sup> – r3<\/sup> – 3pqr
\n= (4)3<\/sup> + (-3)3<\/sup> – (2)3<\/sup> – 3 \u00d7 4 \u00d7 -3 \u00d7 2
\n= 64 – 27 – 8 + 72
\n= 136 – 35 = 101<\/p>\n
\nIf m = 1, n = 2 and p = -3, find the value of 2mn4<\/sup> – 15m2<\/sup>n + p.
\nSolution:
\nm = 1, n = 2, p = -3
\n2mn4<\/sup> – 15m2<\/sup>n + p
\n= 2(1 )(2)4<\/sup> – 15(1 )2<\/sup>(2) + (-3)
\n= 32 – 30 – 3 = \u20141<\/p>\n
\nState true or false:
\n(i) The value of 3x – 2 is 1 when x = 0.
\n(ii) The value of 2x2<\/sup> – x – 3 is 0 when x = -1.
\n(iii) p2<\/sup> + q2<\/sup> – r2<\/sup> when p = 5, q = 12 and r = 13.
\n(iv) 16 – 3x = 5x when x = 2.
\nSolution:
\n(i) The value of 3x – 2 is 1 when x = 0. False
\nCorrect :
\n\u2235 3 \u00d7 0 – 2 = -2
\n(ii) The value of 2x2<\/sup> – x – 3 is 0 when x = -1. True
\n2(-1 )2<\/sup> – (-1) – 3
\n= 2 + 1 – 3 = 0
\n(iii) p2<\/sup> + q2<\/sup> = r2<\/sup> when p = 5, q = 12 and r = 13. True
\n(5)2<\/sup> + (12)2<\/sup> = (13)2<\/sup>
\n= 25 + 144 = 169
\n\u21d2 169= 169
\n(iv) 16 – 3x = 5x when x = 2. True
\n16 – 3x2<\/sup> = 5x2<\/sup>
\n16 – 6 = 10
\n\u21d2 10 = 10<\/p>\n
\nFor x = 2 and y = -3, verify the following:
\n(i) (x + y)2<\/sup> = x2<\/sup> + 2xy + y2<\/sup>
\n(ii) (x – y)2<\/sup> = x2<\/sup> – 2xy + y2<\/sup>
\n(iii) x2<\/sup> – y2<\/sup> = (x + y) (x – y)
\n(iv) (x + y)2<\/sup> = (x – y)2<\/sup> + 4xy
\n(v) (x + y)3<\/sup> = x3<\/sup> + y3<\/sup> + 3x2<\/sup>y + 3xy2<\/sup>
\nSolution:
\nx = 2, y = -3
\n(i) (x + y)2<\/sup> = x2<\/sup> + 2xy + y2<\/sup>
\nL.H.S. = (x + y)2<\/sup> = (2 – 3)2<\/sup> = (-1)2<\/sup> = 1
\nR.H.S. = x2<\/sup> + 2xy + y2<\/sup>
\n= (2)2<\/sup> + 2 \u00d7 2 (-3) + (-3)2<\/sup>
\n= 4 – 12 + 9 = 13 – 12 = 1
\nL.H.S. = R.H.S.<\/p>\n
\nL.H.S. = (x – y)2<\/sup> = [2 – (-3)]2<\/sup>
\n= (2 + 3)2<\/sup> = (5)2<\/sup> = 25
\nR.H.S. = x2<\/sup> – 2xy + y2<\/sup>
\n= (2)2<\/sup> – 2 \u00d7 2 \u00d7 (-3) + (-3)2<\/sup>
\n= 4 + 12 + 9 = 25
\n\u2234 L.H.S. = R.H.S.<\/p>\n
\nL.H.S. = (x)2<\/sup> – (y)2<\/sup> = (2)2<\/sup> – (-3)2<\/sup>
\n= 4 – 9 = -5
\nR.H.S. = (x + y)(x – y)
\n= (2 – 3) [2 – (-3)]
\n= (2 – 3) (2 + 3) = -1 \u00d7 5 = -5
\n\u2234 L.H.S. = R.H.S.<\/p>\n
\nL.H.S. = (x + y)2<\/sup> = [2 + (-3)]2<\/sup>
\n= (2 – 3)2<\/sup> = (-1)2<\/sup> = 1
\nR.H.S. = (x – y)2<\/sup> + 4xy
\n= [2 – (-3)]2<\/sup> + 4 \u00d7 2 \u00d7 (-3)
\n= (2 + 3)2<\/sup> + 4 \u00d7 2 \u00d7 (-3)
\n= (5)2<\/sup> – 24 = 25 – 24 = 1
\n\u2234 L.H.S. = R.H.S.<\/p>\n
\nL.H.S. = (x + y)3<\/sup> = [2 + (-3)]3<\/sup> = (2 – 3)3<\/sup>
\n= (-1)3<\/sup> = (-1) \u00d7 (-1) \u00d7 (-1) = -1
\nR.H.S. = x3<\/sup> + y3<\/sup> + 3x2<\/sup>y + 3xy2<\/sup>
\n= (2)3<\/sup> + (-3)3<\/sup> + 3 (2)2<\/sup> (-3) + 3 \u00d7 2 (-3)2<\/sup>
\n=8 – 27 + 3 \u00d7 4 \u00d7 (-3) + 6 (9)
\n= 8 – 27 – 36 + 54 = 62 – 63 = -1
\n\u2234 L.H.S. = R.H.S.<\/p>\nML Aggarwal Class 6 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"