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
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**Question 1**

Find the product of the following pairs of monomials.

(i) 4, 7p

(ii) – 4p, 7p

(iii) – 4p, 7pq

(iv) 4p^{3}, – 3p

(v) 4p, 0

**Answer:**

- 4 x 7 p = 28p
- - 4p x 7p = -28p
^{2} - - 4p x 7pq = -28p
^{2}q - 4p
^{3}x - 3p = -12p^{4} - 4p x 0 = 0

**Question 2**

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

- (p, q)
- (10m, 5n)
- (20x
^{2}5y^{2}) - (4x, 3x
^{2})\ - (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) 20x ^{2} x *

*(iv) 4x x 3x ^{2} = 12x^{3}*

*(v)3mn x * *4np** = 12mn ^{2}p*

**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.

*5a, 3a*^{2}, 7a^{8}- (
*2p, 4q, 8r*) *xy, 2x*^{2}y, 2xy^{2}- (
*a, 2b, 3c*

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

Volume = length x breadth x height

(i) *5a* x *3a*^{2} x *7a*^{8} = *105a ^{11}*

(ii) *2p* x *4q* x *8r* = *64pqr*

(iii) *xy* x *2x ^{2}*y x

(iv) a x 2b x 3c = 6abc

**Question 5**

Obtain the product of

- xy, yz, zx
- a, – a
^{2}a^{3} - 2, 4y, 8y
^{2}16y^{3} - a, 2b, 3c, 6abc
- m, – mn, mnp

**Answer:**

(i) *x ^{2}*y

(ii) –a^{5}

(iii) 1024y^{6}

(iv) 36a^{2}b^{2}c^{2}

(v) –m^{3}n^{2}p

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