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Ncert Solutions for Algebraic Expressions and Identities Class 8 CBSE Part 2


In this page we have NCERT book Solutions for Class 8th Maths:Algebraic Expressions and Identities for EXERCISE 2 . Hope you like them and do not forget to like , social share and comment at the end of the page.

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

Answer: 

  1. 4 x 7 p = 28p
  2. - 4p x 7p = -28p2
  3.  - 4p x 7pq = -28p2q
  4. 4p3 x - 3p = -12p4
  5. 4p x 0 = 0

     

Question 2

 Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

  1. (p, q)
  2. (10m, 5n)
  3. (20x2 5y2)
  4. (4x, 3x2)\
  5. (3mn, 4np)

Answer: Now we know that

Area of rectangle = Length x breadth

So it is multiplication of monomials

(i) p x q = pq

(ii)10m x  5n = 50mn

(iii) 20x2 x  5y2 =  100x2y2

(iv) 4x x 3x2 = 12x3

(v)3mn x  4np = 12mn2p

Question 3.

Complete the following table of products:

Answer

Question 4

 Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

  1. 5a, 3a2, 7a8
  2.  (2p, 4q, 8r)
  3. xy, 2x2y, 2xy2
  4. (a, 2b, 3c

Answer: We know that volume of rectangular box is given by  

Volume = length x  breadth x  height

(i) 5a x 3a2 x 7a8 = 105a11

(ii) 2p x 4q x 8r = 64pqr

(iii) xy x 2x2y x 2xy2 = 4x4y4

(iv) a x  2b x 3c = 6abc

Question 5

 Obtain the product of

  1. xy,  yz, zx
  2. a, – a2 a3
  3. 2, 4y, 8y2 16y3
  4. a, 2b, 3c, 6abc
  5. m, – mn, mnp

Answer:

 (i) x2y2z2

(ii) –a5

(iii) 1024y6

(iv) 36a2b2c2

(v) –m3n2p

 


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