{"id":95407,"date":"2019-10-04T11:06:21","date_gmt":"2019-10-04T05:36:21","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=95407"},"modified":"2021-09-18T15:16:26","modified_gmt":"2021-09-18T09:46:26","slug":"ncert-solutions-for-class-8-maths-algebraic-expressions-and-identities-ex-9-5","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-8-maths-algebraic-expressions-and-identities-ex-9-5\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5"},"content":{"rendered":"
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5<\/strong><\/p>\n Ex 9.5 Class 8 Maths Question 1. <\/p>\n <\/p>\n <\/p>\n Ex 9.5 Class 8 Maths\u00a0Question 2. Ex 9.5 Class 8 Maths\u00a0Question 3. <\/p>\n Ex 9.5 Class 8 Maths\u00a0Question 4. <\/p>\n <\/p>\n Ex 9.5 Class 8 Maths\u00a0Question 5. <\/p>\n <\/p>\n Ex 9.5 Class 8 Maths\u00a0Question 6. <\/p>\n <\/p>\n Ex 9.5 Class 8 Maths\u00a0Question 7. Ex 9.5 Class 8 Maths\u00a0Question 8. <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Class 8 Maths Algebraic Expressions and Identities Exercise 9.1 Class 8 Maths Algebraic Expressions and Identities Exercise 9.2 Class 8 Maths Algebraic Expressions and Identities Exercise 9.3 Class 8 Maths Algebraic Expressions and Identities Exercise 9.4 Class 8 Maths Algebraic Expressions …<\/p>\n","protected":false},"author":27,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\n
\nUse a suitable identity to get each of the following products:
\n(i) (x + 3) (x + 3)
\n(ii) (2y + 5) (2y + 5)
\n(iii) (2a – 7) (2a – 7)
\n(iv) (3a – \\(\\frac { 1 }{ 2 }\\)) (3a – \\(\\frac { 1 }{ 2 }\\))
\n(v) (1.1m – 0.4) (1.1m + 0.4)
\n(vi) (a2<\/sup> + b2<\/sup>) (-a2<\/sup> + b2<\/sup>)
\n(vii) (6x – 7) (6x + 7)
\n(viii) (-a + c) (-a + c)
\n(ix) (\\(\\frac { x }{ 2 }\\) + \\(\\frac { 3y }{ 4 }\\)) (\\(\\frac { x }{ 2 }\\) + \\(\\frac { 3y }{ 4 }\\))
\n(x) (7a – 9b) (7a – 9b)
\nSolution:
\n<\/p>\n
\nUse the identity (x + a)(x + b) = x2<\/sup> + (a + b)x + ab to find the following products.
\n(i) (x + 3) (x + 7)
\n(ii) (4x + 5)(4x + 1)
\n(iii) (4x – 5) (4x – 1)
\n(iv) (4x + 5) (4x – 1)
\n(v) (2x + 5y) (2x + 3y)
\n(vi) (2a2<\/sup> + 9) (2a2<\/sup> + 5)
\n(vii) (xyz – 4) (xyz – 2)
\nSolution:
\n<\/p>\n
\nFind the following squares by using the identities.
\n(i) (b – 7)2<\/sup>
\n(ii) (xy + 3z)2<\/sup>
\n(iii) (6x2<\/sup> – 5y)2<\/sup>
\n(iv) (\\(\\frac { 2 }{ 3 }\\) m + \\(\\frac { 3 }{ 2 }\\) n)2<\/sup>
\n(v) (0.4p – 0.5q)2<\/sup>
\n(vi) (2xy + 5y)2<\/sup>
\nSolution:
\n<\/p>\n
\nSimplify:
\n(i) (a2<\/sup> – b2<\/sup>)2<\/sup>
\n(ii) (2x + 5)2<\/sup> – (2x – 5)2<\/sup>
\n(iii) (7m – 8n)2<\/sup> + (7m + 8n)2<\/sup>
\n(iv) (4m + 5n)2<\/sup> + (5m + 4n)2<\/sup>
\n(v) (2.5p – 1.5q)2<\/sup> – (1.5p – 2.5q)2<\/sup>
\n(vi) (ab + bc)2<\/sup> – 2ab2<\/sup>c
\n(vii) (m2<\/sup> – n2<\/sup>m)2<\/sup> + 2m3<\/sup>n2<\/sup>
\nSolution:
\n<\/p>\n
\nShow that:
\n(i) (3x + 7)2<\/sup> – 84x = (3x – 7)2<\/sup>
\n(ii) (9p – 5q)2<\/sup> + 180pq = (9p + 5q)2<\/sup>
\n(iii) (\\(\\frac { 4 }{ 3 }\\) m – \\(\\frac { 3 }{ 4 }\\) n)2<\/sup> + 2mn = \\(\\frac { 16 }{ 9 }\\) m2<\/sup> + \\(\\frac { 9 }{ 16 }\\) n2<\/sup>
\n(iv) (4pq + 3q)2<\/sup> – (4pq – 3q)2<\/sup> = 48pq2<\/sup>
\n(v) (a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
\nSolution:
\n<\/p>\n
\nUsing identities, evaluate:
\n(i) 712<\/sup>
\n(ii) 992<\/sup>
\n(iii) 1022<\/sup>
\n(iv) 9982<\/sup>
\n(v) 5.22<\/sup>
\n(vi) 297 \u00d7 303
\n(vii) 78 \u00d7 82
\n(viii) 8.92<\/sup>
\n(ix) 1.05 \u00d7 9.5
\nSolution:
\n<\/p>\n
\nUsing a2<\/sup> – b2<\/sup> = (a + b) (a – b), find
\n(i) 512<\/sup> – 492<\/sup>
\n(ii) (1.02)2<\/sup> – (0.98)2<\/sup>
\n(iii) 1532<\/sup> – 1472<\/sup>
\n(iv) 12.12<\/sup> – 7.92<\/sup>
\nSolution:
\n(i) 512<\/sup> – 492<\/sup> = (51 + 49) (51 – 49) = 100 \u00d7 2 = 200
\n(ii) (1.02)2<\/sup> – (0.98)2<\/sup> = (1.02 + 0.98) (1.02 – 0.98) = 2.00 \u00d7 0.04 = 0.08
\n(iii) 1532<\/sup> – 1472<\/sup> = (153 + 147) (153 – 147) = 300 \u00d7 6 = 1800
\n(iv) 12.12<\/sup> – 7.92<\/sup> = (12.1 + 7.9) (12.1 – 7.9) = 20.0 \u00d7 4.2 = 84<\/p>\n
\nUsing (x + a) (x + b) = x2<\/sup> + (a + b)x + ab, find
\n(i) 103 \u00d7 104
\n(ii) 5.1 \u00d7 5.2
\n(iii) 103 \u00d7 98
\n(iv) 9.7 \u00d7 9.8
\nSolution:
\n(i) 103 \u00d7 104 = (100 + 3)(100 + 4) = (100)2<\/sup> + (3 + 4) (100) + 3 \u00d7 4 = 10000 + 700 + 12 = 10712
\n(ii) 5.1 \u00d7 5.2 = (5 + 0.1) (5 + 0.2) = (5)2<\/sup> + (0.1 + 0.2) (5) + 0.1 \u00d7 0.2 = 25 + 1.5 + 0.02 = 26.5 + 0.02 = 26.52
\n(iii) 103 \u00d7 98 = (100 + 3) (100 – 2) = (100)2<\/sup> + (3 – 2) (100) + 3 \u00d7 (-2) = 10000 + 100 – 6 = 10100 – 6 = 10094
\n(iv) 9.7 \u00d7 9.8 = (10 – 0.3) (10 – 0.2) = (10)2<\/sup> – (0.3 + 0.2) (10) + (-0.3) (-0.2) = 100 – 5 + 0.06 = 95 + 0.06 = 95.06<\/p>\nMore CBSE Class 8 Study Material<\/h4>\n
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