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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.5

Miscellaneous Practice Problems

Question 1.

Subtract: -2(xy)^{2} (y^{3} + 7x^{2}y + 5) from 5y^{2} (x^{2}y^{3} – 2x^{4}y + 10x^{2})

Solution:

5y^{2} (x^{2}y^{3} – 2x^{4}y + 10x^{2}) – [(-2)(xy)^{2} (y^{3} + 7x^{2}y + 5)]

= [5y^{2} (x^{2}y^{3}) – 5y^{2} (2x^{4}y) + 5y^{2} (10x^{2})] – [(-2)x^{2}y^{2} (y^{3} + 7x^{2}y + 5)]

= (5y^{5}5x^{2} – 10x^{4}y^{3} + 50x^{2}y^{2})

= 5x^{2}y^{5} – 10x^{4}y^{3} + 50x^{2}y^{2} – [(-2x^{2}y^{5}) – 14x^{4}y^{3} – 10x^{2}y^{2}]

= 5x^{2}y^{5} – 10x^{4}y^{3} + 50x^{2})y^{2} + 2x^{2} y^{5} + 14x^{4})y^{3} + 10x^{2})y^{2}

= (5 + 2)x^{2}y^{5} + (-10 + 14)x^{4}y^{3} + (+50 + 10)x^{2}y^{2}

= 7x^{2}y^{5} + 4x^{4}y^{3} + 60x^{2}y^{2}

Question 2.

Multiply (4x^{2} + 9) and (3x – 2).

Solution:

(4x^{2} + 9) (3x – 2) = 4x^{2}(3x – 2) + 9(3x – 2)

= (4x^{2})(3x) – (4x^{2})(2) + 9(3x) – 9(2) = (4 × 3 × x × x^{2}) – (4 × 2 × x^{2}) + (9 × 3 × x) – 18

= 12x^{3} – 8x^{2} + 27x – 18 (4x^{3} + 9) (3x – 2) = 12x^{3} – 8x^{2} + 27x – 18

Question 3.

Find the simple interest on Rs. 5a^{2}b^{2} for 4ab years at 7b% per annum.

Solution:

Question 4.

The cost of a note book is Rs. 10ab. If Babu has Rs. (5a^{2}b + 20ab^{2} + 40ab). Then how many note hooks can he buy?

Solution:

For ₹ 10 ab the number of note books can buy = 1.

Question 5.

Factorise : (7y^{2} – 19y – 6)

Solution:

7y^{2} – 19y – 6 is of the form ax^{2} + bx + c where a = 7; b = -19; c = – 6

The product a × c = 7 × -6 = -42

sum b = – 19

The middle term – 19y can be written as – 21y + 2y

7y^{2} – 19y – 6 = 7y^{2} – 21y + 2y – 6

= 7y(y – 3) + 2(y – 3) = (y – 3)(7y + 2)

7y^{2} – 19y – 6 = (y – 3)(7y + 2)

Challenging Problems

Question 6.

A contractor uses the expression 4x^{2} + 11x + 6 to determine the amount of wire to order when wiring a house. If the expression comes from multiplying the number of rooms times the number of outlets and he knows the number of rooms to be (x + 2), find the number of outlets in terms of ‘x’ [Hint: factorise 4x^{2} + 11x + 6]

Solution:

Given Number of rooms = x + 2

Number of rooms × Number of outlets = amount of wire.

Now factorising 4x^{2} + 11x + 6 which is of the form ax^{2} + bx + c with a = 4 b = 11 c = 6.

The product a × c = 4 × 6 = 24

sum b = 11

The middle term 11x can be written as 8x + 3x

∴ 4x^{2} + 11x + 6 = 4x^{2} + 8x + 3x + 6 = 4x (x + 2) + 3 (x + 2)

4x^{2} + 11x + 6 = (x + 2) (4x + 3)

Now from (1) the number of outlets

∴ Number of outlets = 4x + 3

Question 7.

A mason uses the expression x^{2} + 6x + 8 to represent the area of the floor of a room. If the decides that the length of the room will be represented by (x + 4), what will the width of the room be in terms of x ?

Solution:

Given length of the room = x + 4

Area of the room = x^{2} + 6x + 8

Length × breadth = x^{2} + 6x + 8

Question 8.

Find the missing term: y^{2} + (-) x + 56 = (y + 7)(y + -)

Solution:

We have (x + a) (x + b) = x^{2} + (a + b)x + ab

56 = 7 × 8 .

∴ y^{2} + (7 + 8)x + 56 = (y + 7)(y + 8)

Question 9.

Factorise: 16p^{4} – 1

Solution:

16p^{4} – 1 = 2^{4}p^{4} – 1 = (2^{2})^{2}(p^{2})^{2} – 1^{2}

Comparing with a^{2} – b^{2} = (a + b)(a – b) where a = 2^{2}p^{2} and b = 1

∴ (2^{2}p^{2})^{2} – 1^{2} = (2^{2}p^{2} + 1) (2^{2}p^{2} – 1) = (4p^{2} + 1) (4p^{2} – 1)

∴ 16p^{4} – 1 = (4p^{2} + 1)(4p^{2} – 1)(4p^{2} + 1)(2^{2}p^{2} – 1^{2})

= (4p^{2} + 1) [(2p)^{2}– 1^{2}] = (4p^{2} + 1) (2p + 1)(2p – 1)

[∵ using a^{2} – b^{2} = (a + b) (a – b)]

∴ 16p^{4} – 1 = (4psup>2 + 1)(2p + 1)(2p – 1)

Question 10.

Factorise : x^{6} – 64y^{3}

Solution:

x^{6} – 64y^{3} = (x^{2})^{3} – 4^{3}y^{3} = (x^{2})^{3} – (4y)^{3}

This is of the form a^{3} – b^{3} with a = x^{2}, b = 4y

a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2})

(x^{2})^{3} – (4y)^{3} = (x^{2} – 4y) [(x^{2})^{2} + (x^{2})(4y) + (4y)^{2}]

= (x^{2} – 4y) [x^{4} + 4x^{2}y + 16y^{2}]

∴ x^{6} – 64y^{3} = (x^{2} – 4y) [x^{4} + 4x^{2}y + 16y^{2}]