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      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

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      • NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

       

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1

      Ex 9.1 Class 8 Maths Question 1.
      Identify the terms, 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) x2 + y2 – xy
      (vi) 0.3a – 0.6ab + 0.5b
      Solution:
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 Q1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 Q1.1

      Ex 9.1 Class 8 Maths Question 2.
      Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
      x + y, 1000, x + x2 + x3 + x4, 7 + y + 5x, 2y – 3y2, 2y – 3y2 + 4y3, 5x – 4y + 3xy, 4z – 15z2, ab + bc + cd + da, pqr, p2q + pq2, 2p + 2q
      Solution:
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 Q2

      Ex 9.1 Class 8 Maths 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
      Solution:
      (i) Given: ab – bc, bc – ca, ca – ab
      We have
      (ab – bc) + (bc – ca) + (ca – ab) (Adding all the terms)
      = ab – bc + bc – ca + ca – ab
      = (ab – ab) + (bc – bc) + (ca – ca) (Collecting the like terms together)
      = 0 + 0 + 0
      = 0

      (ii) Given:
      a – b + ab, b – c + bc, c – a + ac
      We have (a – b + ab) + (b – c + bc) + (c – a + ac) (Adding all the terms)
      = a – b + ab + b – c + bc + c – a + ac
      = (a – a) + (b – b) + (c – c) + ab + bc + ac (Collecting all the like terms together)
      = 0 + 0 + 0 + ab + bc + ac
      = ab + bc + ac

      (iii) Given:
      2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
      By arranging the like terms in the same column, we have
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 Q3
      (Adding columnwise)

      (iv) Given: l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + nl
      By arranging the like terms in the same column, we have
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 Q3.1
      Thus, the sum of the given expressions is 2(l2 + m2 + n2 + lm + mn + nl)

      Ex 9.1 Class 8 Maths Question 4.
      (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
      (6) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
      (c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q
      Solution:
      (a) Arranging the like terms column-wise, we have
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 Q4
      [Change the signs of all the terms of lower expressions and then add]
      (b) Arranging the like terms column-wise, we have
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 Q4.1
      [Change the signs of all the terms of lower expressions and then add]
      (c) Arranging the like terms column-wise, we have
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 Q4.2
      [Change the signs of all the terms of lower expressions and then add]
      The terms are p2q – 7pq2 + 8pq – 18q + 5p + 20

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 q-1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 q-1.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 q-2

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 q-3

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 q-4

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.2

      Ex 9.2 Class 8 Maths Question 1.
      Find the product of the following pairs of monomials.
      (i) 4, 7p
      (ii) -4p, 7p
      (iii) -4p, 7pq
      (iv) 4p3, -3p
      (v) 4p, 0
      Solution:
      (i) 4 × 7p = (4 × 7) × p = 28p
      (ii) -4p × 7p = (-4 × 7) × p × p = -28p2
      (iii) -4p × 7pq = (-4 × 7) × p × pq = -28p2q
      (iv) 4p3 × -3p = (4 × -3) × p3 × p = -12p4
      (v) 4p x 0 = (4 × 0) × p = 0 × p = 0

      Ex 9.2 Class 8 Maths Question 2.
      Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
      (p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)
      Solution:
      (i) Length = p units and breadth = q units
      Area of the rectangle = length × breadth = p × q = pq sq units
      (ii) Length = 10 m units, breadth = 5n units
      Area of the rectangle = length × breadth = 10 m × 5 n = (10 × 5) × m × n = 50 mn sq units
      (iii) Length = 20x2 units, breadth = 5y2 units
      Area of the rectangle = length × breadth = 20x2 × 5y2 = (20 × 5) × x2 × y2 = 100x2y2 sq units
      (iv) Length = 4x units, breadth = 3x2 units
      Area of the rectangle = length × breadth = 4x × 3x2 = (4 × 3) × x × x2 = 12x3 sq units
      (v) Length = 3mn units, breadth = 4np units
      Area of the rectangle = length × breadth = 3mn × 4np = (3 × 4) × mn × np = 12mn2p sq units

      Ex 9.2 Class 8 Maths Question 3.
      Complete the table of Products.
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Q3
      Solution:
      Completed Table
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Q3.1

      Ex 9.2 Class 8 Maths Question 4.
      Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
      (i) 5a, 3a2, 7a4
      (ii) 2p, 4q, 8r
      (iii) xy, 2x2y, 2xy2
      (iv) a, 2b, 3c
      Solution:
      (i) Here, length = 5a, breadth = 3a2, height = 7a4
      Volume of the box = l × b × h = 5a × 3a2 × 7a4 = 105 a7 cu. units
      (ii) Here, length = 2p, breadth = 4q, height = 8r
      Volume of the box = l × b × h = 2p × 4q × 8r = 64pqr cu. units
      (iii) Here, length = xy, breadth = 2x2y, height = 2xy2
      Volume of the box = l × b × h = xy × 2x2y × 2xy2 = (1 × 2 × 2) × xy × x2y × xy2 = 4x4y4 cu. units
      (iv) Here, length = a, breadth = 2b, height = 3c
      Volume of the box = length × breadth × height = a × 2b × 3c = (1 × 2 × 3)abc = 6 abc cu. units

      Ex 9.2 Class 8 Maths Question 5.
      Obtain the product of
      (i) xy, yz, zx
      (ii) a, -a2, a3
      (iii) 2, 4y, 8y2, 16y3
      (iv) a, 2b, 3c, 6abc
      (v) m, -mn, mnp
      Solution:
      (i) xy × yz × zx = x2y2z2
      (ii) a × (-a2) × a3 = -a6
      (iii) 2 × 4y × 8y2 × 16y3 = (2 × 4 × 8 × 16) × y × y2 × y3 = 1024y6
      (iv) a × 2b × 3c × 6abc = (1 × 2 × 3 × 6) × a × b × c × abc = 36 a2b2c2
      (v) m × (-mn) × mnp = [1 × (-1) × 1 ]m × mn × mnp = -m3n2p

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 q-1, q-2

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 q-3

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 q-4

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3

      Ex 9.3 Class 8 Maths Question 1.
      Carry out the multiplication of the expressions in each of the following pairs:
      (i) 4p, q + r
      (ii) ab, a – b
      (iii) a + b, 7a2b2
      (iv) a2 – 9, 4a
      (v) pq + qr + rp, 0
      Solution:
      (i) 4p × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr
      (ii) ab, a – b = ab × (a – b) = (ab × a) – (ab × b) = a2b – ab2
      (iii) (a + b) × 7a2b2 = (a × 7a2b2) + (b × 7a2b2) = 7a3b2 + 7a2b3
      (iv) (a2 – 9) × 4a = (a2 × 4a) – (9 × 4a) = 4a3 – 36a
      (v) (pq + qr + rp) × 0 = 0
      [∵ Any number multiplied by 0 is = 0]

      Ex 9.3 Class 8 Maths Question 2.
      Complete the table.

      S.No.

      First Expression Second
      Expression
      Product
      (i) a b + c + d –
      (ii) x + y – 5 5xy –
      (iii) p 6p2 – 7p + 5 –
      (iv) 4p2q2 p2 – q2 –
      (v) a + b + c abc –

      Solution:
      (i) a × (b + c + d) = (a × b) + (a × c) + (a × d) = ab + ac + ad
      (ii) (x + y – 5) (5xy) = (x × 5xy) + (y × 5xy) – (5 × 5xy) = 5x2y + 5xy2 – 25xy
      (iii) p × (6p2 – 7p + 5) = (p × 6p2) – (p × 7p) + (p × 5) = 6p3 – 7p2 + 5p
      (iv) 4p2q2 × (p2 – q2) = 4p2q2 × p2 – 4p2q2 × q2 = 4p4q2 – 4p2q4
      (v) (a + b + c) × (abc) = (a × abc) + (b × abc) + (c × abc) = a2bc + ab2c + abc2

      Completed Table:

      S.No. First Expression Second
      Expression
      Product
      (i) a b + c + d ab + ac + ad
      (ii) x + y – 5 5xy 5x2y + 5xy2 – 25xy
      (iii) p 6p2 – 7p + 5 6p3 – 7p2 + 5p
      (iv) 4p2q2 p2 – q2 4p4q2 – 4p2q4
      (v) a + b + c abc a2bc + ab2c + abc2

      Ex 9.3 Class 8 Maths Question 3.
      Find the products.
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Q3
      Solution:
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Q3.1

      Ex 9.3 Class 8 Maths Question 4.
      (a) Simplify: 3x(4x – 5) + 3 and find its values for (i) x = 3 (ii) x = 12.
      (b) Simplify: a(a2 + a + 1) + 5 and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1
      Solution:
      (a) We have 3x(4x – 5) + 3 = 4x × 3x – 5 × 3x + 3 = 12x2 – 15x + 3
      (i) For x = 3, we have
      12 × (3)2 – 15 × 3 + 3 = 12 × 9 – 45 + 3 = 108 – 42 = 66
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Q4
      (b) We have a(a2 + a + 1) + 5
      = (a2 × a) + (a × a) + (1 × a) + 5
      = a3 + a2 + a + 5
      (i) For a = 0, we have
      = (0)3 + (0)2 + (0) + 5 = 5
      (ii) For a = 1, we have
      = (1)3 + (1)2 + (1) + 5 = 1 + 1 + 1 + 5 = 8
      (iii) For a = -1, we have
      = (-1)3 + (-1)2 + (-1) + 5 = -1 + 1 – 1 + 5 = 4

      Ex 9.3 Class 8 Maths Question 5.
      (a) Add: p(p – q), q(q – r) and r(r – p)
      (b) Add: 2x(z – x – y) and 2y(z – y – x)
      (c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
      (d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)
      Solution:
      (a) p(p – q) + q(q – r) + r(r – p)
      = (p × p) – (p × q) + (q × q) – (q × r) + (r × r) – (r × p)
      = p2 – pq + q2 – qr + r2 – rp
      = p2 + q2 + r2 – pq – qr – rp
      (b) 2x(z – x – y) + 2y(z – y – x)
      = (2x × z) – (2x × x) – (2x × y) + (2y × z) – (2y × y) – (2y × x)
      = 2xz – 2x2 – 2xy + 2yz – 2y2 – 2xy
      = -2x2 – 2y2 + 2xz + 2yz – 4xy
      = -2x2 – 2y2 – 4xy + 2yz + 2xz
      (c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)
      = (4l × 10n) – (4l × 3m) + (4l × 2l) – (3l × l) – (3l × -4m) – (3l × 5n)
      = 40ln – 12lm + 8l2 – 3l2 + 12lm – 15ln
      = (40ln – 15ln) + (-12lm + 12lm) + (8l2 – 3l2)
      = 25ln + 0 + 5l2
      = 25ln + 5l2
      = 5l2 + 25ln
      (d) [4c(-a + b + c)] – [3a(a + b + c) – 2b(a – b + c)]
      = (-4ac + 4bc + 4c2) – (3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc)
      = -4ac + 4bc + 4c2 – 3a2 – 3ab – 3ac + 2ab – 2b2 + 2bc
      = -3a2 – 2b2 + 4c2 – ab + 6bc – 7ac

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 q-1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 q-2, q-3

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 q-4

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 q-5

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 q-5.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4

      Ex 9.4 Class 8 Maths Question 1.
      Multiply the binomials:
      (i) (2x + 5) and (4x – 3)
      (ii) (y – 8) and (3y – 4)
      (iii) (2.5l – 0.5m) and (2.5l + 0.5m)
      (iv) (a + 3b) and (x + 5)
      (v) (2pq + 3q2) and (3pq – 2q2)
      (vi) (34a2 + 3b2) and 4(a2 – 23 b2)
      Solution:
      (i) (2x + 5) × (4x – 3)
      = 2x × (4x – 3) + 5 × (4x – 3)
      = (2x × 4x) – (3 × 2x) + (5 × 4x) – (5 × 3)
      = 8x2 – 6x + 20x – 15
      = 8x2 + 14x – 15

      (ii) (y – 8) × (3y – 4)
      = y × (3y – 4) – 8 × (3y – 4)
      = (y × 3y) – (y × 4) – (8 × 3y) + (-8 × -4)
      = 3y2 – 4y – 24y + 32
      = 3y2 – 28y + 32

      (iii) (2.5l – 0.5m) × (2.5l + 0.5m)
      = (2.5l × 2.5l) + (2.5l × 0.5m) – (0.5m × 2.5l) – (0.5m × 0.5m)
      = 6.25l2 + 1.25ml – 1.25ml – 0.25m2
      = 6.25l2 + 0 – 0.25m2
      = 6.25l2 – 0.25m2

      (iv) (a + 3b) × (x + 5)
      = a × (x + 5) + 36 × (x + 5)
      = (a × x) + (a × 5) + (36 × x) + (36 × 5)
      = ax + 5a + 3bx + 15b

      (v) (2pq + 3q2) × (3pq – 2q2)
      = 2pq × (3pq – 2q2) + 3q2 (3pq – 2q2)
      = (2pq × 3pq) – (2pq × 2q2) + (3q2 × 3pq) – (3q2 × 2q2)
      = 6p2q2 – 4pq3 + 9pq3 – 6q4
      = 6p2q2 + 5pq3 – 6q4

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Q1

      Ex 9.4 Class 8 Maths Question 2.
      Find the product:
      (i) (5 – 2x) (3 + x)
      (ii) (x + 7y) (7x – y)
      (iii) (a2 + b) (a + b2)
      (iv) (p2 – q2)(2p + q)
      Solution:
      (i) (5 – 2x) (3 + x)
      = 5(3 + x) – 2x(3 + x)
      = (5 × 3) + (5 × x) – (2x × 3) – (2x × x)
      = 15 + 5x – 6x – 2x2

      (ii) (x + 7y) (7x – y)
      = x(7x – y) + 7y(7x – y)
      = (x × 7x) – (x × y) + (7y × 7x) – (7y × y)
      = 7x2 – xy + 49xy – 7y2
      = 7x2 + 48xy – 7y2

      (iii) (a2 + b) (a + b2)
      = a2 (a + b2) + b(a + b2)
      = (a2 × a) + (a2 × b2) + (b × a) + (b × b2)
      = a3 + a2b2 + ab + b3

      (iv) (p2 – q2)(2p + q)
      = p2(2p + q) – q2(2p + q)
      = (p2 × 2p) + (p2 × q) – (q2 × 2p) – (q2 × q)
      = 2p3 + p2q – 2pq2 – q3

      Ex 9.4 Class 8 Maths Question 3.
      Simplify:
      (i) (x2 – 5) (x + 5) + 25
      (ii) (a2 + 5)(b3 + 3) + 5
      (iii) (t + s2) (t2 – s)
      (iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
      (v) (x + y) (2x + y) + (x + 2y) (x – y)
      (vi) (x + y)(x2 – xy + y2)
      (vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
      (viii) (a + b + c) (a + b – c)
      Solution:
      (i) (x2 – 5) (x + 5) + 25
      = x2(x + 5) + 5(x + 5) + 25
      = x3 + 5x2 – 5x – 25 + 25
      = x3 + 5x2 – 5x + 0
      = x3 + 5x2 – 5x

      (ii) (a2 + 5)(b3 + 3) + 5
      = a2(b3 + 3) + 5(b3 + 3) + 5
      = a2b3 + 3a2 + 5b3 + 15 + 5
      = a2b3 + 3a2 + 5b3 + 20

      (iii) (t + s2) (t2 – s)
      = t(t2 – s) + s2(t2 – s)
      = t3 – st + s2t2 – s3
      = t3 + s2t2 – st – s3

      (iv) (a + b)(c – d) + (a – b) (c + d) + 2(ac + bd)
      = a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2ac + 2bd
      = ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
      = ac + ac + 2ac + bc – bc – ad + ad – bd – bd + 2bd
      = 4ac + 0 + 0 + 0
      = 4ac

      (v) (x + y) (2x + y) + (x + 2y) (x – y)
      = x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)
      = 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2
      = 2x2 + x2 + xy + 2xy – xy + 2xy + y2 – 2y2
      = 3x2 + 4xy – y2

      (vi) (x + y)(x2 – xy + y2)
      = x(x2 – xy + y2) + y(x2 – xy + y2)
      = x3 – x2y + x2y + xy2 – xy2 + y3
      = x3 – 0 + 0 + y3
      = x3 + y3

      (vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x.+ 12y
      = 1.5x (1.5x + 4y + 3) – 4y(1.5x + 4y + 3) – 4.5x + 12y
      = 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y
      = 2.25x2 + 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y2
      = 2.25x2 + 0 + 0 + 0 – 16y2
      = 2.25x2 – 16y2

      (viii) (a + b + c) (a + b – c)
      = a(a + b – c) + b(a + b – c) + c(a + b – c)
      = a2 + ab – ac + ab + b2 – bc + ac + bc – c2
      = a2 + ab + ab – bc + bc – ac + ac + b2 – c2
      = a2 + 2ab + b2 – c2 + 0 + 0
      = a2 + 2ab + b2 – c2

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 q-1
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 q-1.1
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 q-2
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 q-3
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 q-3.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5

      Ex 9.5 Class 8 Maths Question 1.
      Use a suitable identity to get each of the following products:
      (i) (x + 3) (x + 3)
      (ii) (2y + 5) (2y + 5)
      (iii) (2a – 7) (2a – 7)
      (iv) (3a – 12) (3a – 12)
      (v) (1.1m – 0.4) (1.1m + 0.4)
      (vi) (a2 + b2) (-a2 + b2)
      (vii) (6x – 7) (6x + 7)
      (viii) (-a + c) (-a + c)
      (ix) (x2 + 3y4) (x2 + 3y4)
      (x) (7a – 9b) (7a – 9b)
      Solution:
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q1.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q1.2

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q1.3

      Ex 9.5 Class 8 Maths Question 2.
      Use the identity (x + a)(x + b) = x2 + (a + b)x + ab to find the following products.
      (i) (x + 3) (x + 7)
      (ii) (4x + 5)(4x + 1)
      (iii) (4x – 5) (4x – 1)
      (iv) (4x + 5) (4x – 1)
      (v) (2x + 5y) (2x + 3y)
      (vi) (2a2 + 9) (2a2 + 5)
      (vii) (xyz – 4) (xyz – 2)
      Solution:
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q2

      Ex 9.5 Class 8 Maths Question 3.
      Find the following squares by using the identities.
      (i) (b – 7)2
      (ii) (xy + 3z)2
      (iii) (6x2 – 5y)2
      (iv) (23 m + 32 n)2
      (v) (0.4p – 0.5q)2
      (vi) (2xy + 5y)2
      Solution:
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q3

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q3.1

      Ex 9.5 Class 8 Maths Question 4.
      Simplify:
      (i) (a2 – b2)2
      (ii) (2x + 5)2 – (2x – 5)2
      (iii) (7m – 8n)2 + (7m + 8n)2
      (iv) (4m + 5n)2 + (5m + 4n)2
      (v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
      (vi) (ab + bc)2 – 2ab2c
      (vii) (m2 – n2m)2 + 2m3n2
      Solution:
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q4

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q4.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q4.2

      Ex 9.5 Class 8 Maths Question 5.
      Show that:
      (i) (3x + 7)2 – 84x = (3x – 7)2
      (ii) (9p – 5q)2 + 180pq = (9p + 5q)2
      (iii) (43 m – 34 n)2 + 2mn = 169 m2 + 916 n2
      (iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
      (v) (a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
      Solution:
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q5

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q5.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q5.2

      Ex 9.5 Class 8 Maths Question 6.
      Using identities, evaluate:
      (i) 712
      (ii) 992
      (iii) 1022
      (iv) 9982
      (v) 5.22
      (vi) 297 × 303
      (vii) 78 × 82
      (viii) 8.92
      (ix) 1.05 × 9.5
      Solution:
      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q6

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q6.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Q6.2

      Ex 9.5 Class 8 Maths Question 7.
      Using a2 – b2 = (a + b) (a – b), find
      (i) 512 – 492
      (ii) (1.02)2 – (0.98)2
      (iii) 1532 – 1472
      (iv) 12.12 – 7.92
      Solution:
      (i) 512 – 492 = (51 + 49) (51 – 49) = 100 × 2 = 200
      (ii) (1.02)2 – (0.98)2 = (1.02 + 0.98) (1.02 – 0.98) = 2.00 × 0.04 = 0.08
      (iii) 1532 – 1472 = (153 + 147) (153 – 147) = 300 × 6 = 1800
      (iv) 12.12 – 7.92 = (12.1 + 7.9) (12.1 – 7.9) = 20.0 × 4.2 = 84

      Ex 9.5 Class 8 Maths Question 8.
      Using (x + a) (x + b) = x2 + (a + b)x + ab, find
      (i) 103 × 104
      (ii) 5.1 × 5.2
      (iii) 103 × 98
      (iv) 9.7 × 9.8
      Solution:
      (i) 103 × 104 = (100 + 3)(100 + 4) = (100)2 + (3 + 4) (100) + 3 × 4 = 10000 + 700 + 12 = 10712
      (ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2) = (5)2 + (0.1 + 0.2) (5) + 0.1 × 0.2 = 25 + 1.5 + 0.02 = 26.5 + 0.02 = 26.52
      (iii) 103 × 98 = (100 + 3) (100 – 2) = (100)2 + (3 – 2) (100) + 3 × (-2) = 10000 + 100 – 6 = 10100 – 6 = 10094
      (iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2) = (10)2 – (0.3 + 0.2) (10) + (-0.3) (-0.2) = 100 – 5 + 0.06 = 95 + 0.06 = 95.06

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-1.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-2

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-3

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-4

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-4.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-5

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-5.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-6

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-6.1

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-7

      NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 q-8

      Algebraic Expressions and Identities Class 8 Extra Questions Maths Chapter 9

      Extra Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

      Algebraic Expressions and Identities Class 8 Extra Questions Very Short Answer Type

      Question 1.
      Write two examples of each of
      (i) Monomials
      (ii) Binomials
      (iii) Trinomials
      Solution:
      (i) Monomials:
      (a) 3x
      (b) 5xy2
      (ii) Binomials:
      (a) p + q
      (b) -5a + 2b
      (iii) Trinomials:
      (a) a + b + c
      (b) x2 + x + 2

      Question 2.
      Identify the like expressions.
      5x, -14x, 3x2 + 1, x2, -9x2, xy, -3xy
      Solution:
      Like terms: 5x and -14x, x2 and -9x2, xy and -3xy

      Question 3.
      Identify the terms and their coefficients for each of the following expressions:
      (i) 3x2y – 5x
      (ii) xyz – 2y
      (iii) -x – x2
      Solution:
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q3

      Question 4.
      Add: -3a2b2, –52 a2b2, 4a2b2, 23 a2b2
      Solution:
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q4
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q4.1

      Question 5.
      Add: 8x2 + 7xy – 6y2, 4x2 – 3xy + 2y2 and -4x2 + xy – y2
      Solution:
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q5

      Question 6.
      Subtract: (4x + 5) from (-3x + 7)
      Solution:
      (-3x + 7) – (4x + 5) = -3x + 7 – 4x – 5 = -3x – 4x + 7 – 5 = -7x + 2

      Question 7.
      Subtract: 3x2 – 5x + 7 from 5x2 – 7x + 9
      Solution:
      (5x2 – 7x + 9) – (3x2 – 5x + 7)
      = 5x2 – 7x + 9 – 3x2 + 5x – 7
      = 5x2 – 3x2 + 5x – 7x + 9 – 7
      = 2x2 – 2x + 2

      Question 8.
      Multiply the following expressions:
      (a) 3xy2 × (-5x2y)
      (b) 12 x2yz × 23 xy2z × 15 x2yz
      Solution:
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q8

      Question 9.
      Find the area of the rectangle whose length and breadths are 3x2y m and 5xy2 m respectively.
      Solution:
      Length = 3x2y m, breadth = 5xy2 m
      Area of rectangle = Length × Breadth = (3x2y × 5xy2) sq m = (3 × 5) × x2y × xy2 sq m = 15x3y3 sq m

      Question 10.
      Multiply x2 + 7x – 8 by -2y.
      Solution:
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q10

      Algebraic Expressions and Identities Class 8 Extra Questions Short Answer Type

      Question 11.
      Simplify the following:
      (i) a2 (b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)
      (ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)
      Solution:
      (i) a2 (b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)
      = a2b2 – a2c2) + b2c2 – b2a2) + c2a2 – c2b2)
      = 0
      (ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)
      = x3 – 3x2y2 – xy3 + 2x2y2 – xy3 + 5x3
      = x3 + 5x3 – 3x2y2 + 2x2y2 – xy3 – xy3
      = 6x3 – x2y2 – 2xy3

      Question 12.
      Multiply (3x2 + 5y2) by (5x2 – 3y2)
      Solution:
      (3x2 + 5y2) × (5x2 – 3y2)
      = 3x2(5x2 – 3y2) + 5y2(5x2 – 3y2)
      = 15x4 – 9x2y2 + 25x2y2 – 15y4
      = 15x4 + 16x2y2 – 15y4

      Question 13.
      Multiply (6x2 – 5x + 3) by (3x2 + 7x – 3)
      Solution:
      (6x2 – 5x + 3) × (3x2 + 7x – 3)
      = 6x2(3x2 + 7x – 3) – 5x(3x2 + 7x – 3) + 3(3x2 + 7x – 3)
      = 18x4 + 42x3 – 18x2 – 15x3 – 35x2 + 15x + 9x2 + 21x – 9
      = 18x4 + 42x3 – 15x3 – 18x2 – 35x2 + 9x2 + 15x + 21x – 9
      = 18x4 + 27x3 – 44x2 + 36x – 9

      Question 14.
      Simplify:
      2x2(x + 2) – 3x (x2 – 3) – 5x(x + 5)
      Solution:
      2x2(x + 2) – 3x(x2 – 3) – 5x(x + 5)
      = 2x3 + 4x2 – 3x3 + 9x – 5x2 – 25x
      = 2x3 – 3x3 – 5x2 + 4x2 + 9x – 25x
      = -x3 – x2 – 16x

      Question 15.
      Multiply x2 + 2y by x3 – 2xy + y3 and find the value of the product for x = 1 and y = -1.
      Solution:
      (x2 + 2y) × (x3 – 2xy + y3)
      = x2(x3 – 2xy + y3) + 2y(x3 – 2xy + y3)
      = x5 – 2x3y + x2y3 + 2x3y – 4xy2 + 2y4
      = x5 + x2y3 – 4xy2 + 2y4
      Put x = 1 and y = -1
      = (1)5 + (1)2 (-1)3 – 4(1)(-1)2 + 2(-1)4
      = 1 + (1) (-1) – 4(1)(1) + 2(1)
      = 1 – 1 – 4 + 2
      = -2

      Question 16.
      Using suitable identity find:
      (i) 482 (NCERT Exemplar)
      (ii) 962
      (iii) 2312 – 1312
      (iv) 97 × 103
      (v) 1812 – 192 = 162 × 200 (NCERT Exemplar)
      Solution:
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q16

      Question 17.
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q17
      Solution:
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q17.1
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q17.2

      Question 18.
      Verify that (11pq + 4q)2 – (11pq – 4q)2 = 176pq2 (NCERT Exemplar)
      Solution:
      LHS = (11pq + 4q)2 – (11pq – 4q)2 = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q)
      [using a2 -b2 = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q]
      = (22pq) (8q)
      = 176 pq2
      = RHS.
      Hence Verified.

      Question 19.
      Find the value of 382−22216, using a suitable identity. (NCERT Exemplar)
      Solution:
      Algebraic Expressions and Identities NCERT Extra Questions for Class 8 Maths Q19

      Question 20.
      Find the value of x, if 10000x = (9982)2 – (18)2 (NCERT Exemplar)
      Solution:
      RHS = (9982)2 – (18)2 = (9982 + 18)(9982 – 18)
      [Since a2 -b2 = (a + b) (a – b)]
      = (10000) × (9964)
      LHS = (10000) × x
      Comparing L.H.S. and RHS, we get
      10000x = 10000 × 9964
      x = 9964

      algebraic-expressions-and-identities-ncert-extra-questions-for-class-8-maths-chapter-9-01
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