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## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.3

Miscellaneous Practice problems

Question 1.

Using identity, find the value of

(i) (4.9)^{2}

(ii) (100.1)^{2}

(iii) (1.9) × (2.1)

Solution:

(i) (4.9)^{2}

(4.9)^{2} = (5 – 0.1)^{2}

Substituting a = 5 and b = 0.1 in

(a – b)^{2} = a^{2} – 2ab + b^{2}, we have

(5 – 0.1)^{2} = 5^{2} – 2(5) (0.1) + (0.1)^{2}

(4.9)^{2} = 25 – 1 + 0.01 = 24 + 0.01

(4.9)^{2} = 24.01

(ii) (100.1)^{2}

(100.1)^{2} = (100 + 0.1)^{2}

Substituting a = 100 and b = 0.1 in

(a + b)^{2} = a^{2} + 2ab + b^{2}, we have

(100 + 0.1)^{2} = (100)^{2} + 2(100) (0.1) + (0.1)^{2}

(100.1)^{2} = 10000 + 20 + 0.01

(100.1)^{2} = 10020.01

(iii) (1.9) × (2.1)

(1.9) × (2.1) = (2 – 0.1) × (2 + 0.1)

Substituting a = 100 and b = 0.1 in

(a – b) (a + b) = a^{2} – b^{2} we have

(2 – 0.1) (2 + 0.1) = 2^{2} – (0.1)^{2}

(1.9) × (2.1) = 4 – 0.01

(9.9) (2.1) = 3.99

Question 2.

Factorise: 4x^{2} – 9y^{2}

Solution:

4x^{2} – 9y^{2} = 22 x^{2} – 3^{2} y^{2} = (2x)^{2} – (3y)^{2}

Substituting a = 2x and b = 3y in

(a^{2} – b^{2}) = (a + b) (a – b), we have

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

∴ Factors of 4x^{2} – 9y^{2} are (2x + 3y) and (2x – 3y)

Question 3.

Simplify using identities

(i) (3p + q) (3p + r)

(ii) (3p + q) (3p – q)

Solution:

(i) (3p + q) (3p + r)

Substitute x = 3p,a = q and b = r in

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

(3p + q)(3p + r) = (3p)^{2} + 3p (q + r) + (q × r)

= 3^{2} p^{2} + 3p (q + r) + qr

(3p + q)(3p + r) = 9p^{2} + 3p(q + r) + qr

(ii) (3p + q) (3p – q)

Substitute a = 3p and b = q in

(a + b) (a – b) = a^{2} – b^{2}, we have

(3p + q) (3p – q) = (3p)^{2} – q^{2} = 32 p^{2} – q^{2}

(3P + q) (3p – q) = 9p^{2} – q^{2}

Question 4.

Show that (x + 2y)^{2} – (x – 2y)^{2} = 8xy.

Solution:

LHS = (x + 2y)^{2} – (x – 2y)^{2}

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

= x^{2} + 4xy + 4y^{2} – [x^{2} – 4xy + 2^{2}y^{2}]

= x^{2} + 4xy + 4y^{2} – x^{2} + 4xy – 4y^{2}

= x^{2} – x^{2} + 4xy + 4xy + 4y^{2} – 4y^{2}

= x^{2} (1 – 1) + xy (4 + 4) + y^{2} (4 – 4)

= 0x^{2} + 8xy + 0y^{2} = 8xy = RHS

∴ (x + 2y)^{2} – (x – 2y)^{2} = 8xy

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

Question 5

The pathway of a square paddy field has 5 m width and length of its side is 40 m. Find the total area of its pathway. (Note: Use suitable identity)

Solution:

Given side of the square = 40 m

Also width of the pathway = 5 m

∴ Side of the larger square = 40m + 2(5)m = 40m + 10m = 50m

Area of the path way = area of large square – area of smaller square

= 50^{2 }– 40^{2
}

Substituting a = 50 and b = 40 in

a^{2} – b^{2} = (a + b) (a – b) we have

50^{2} – 40^{2} = (50 + 40) (50 – 40)

Area of pathway = 90 × 10

Area of the pathway = 900 m^{2}

Challenge Problems

Question 1.

If X = a^{2} – 1 and Y = 1 – b^{2}, then find X + Y and factorize the same.

Solution:

Given X = a^{2} – 1

Y = I – b^{2}

X + Y = (a^{2} – 1) + (1 – b^{2})

= a^{2} – 1 + 1 – b^{2}

We know the identity that a^{2} – b^{2} = (a + b) (a – b)

∴ X + Y = (a + b) (a – b)

Question 2.

Find the value of (x – y) (x + y) (x^{2} + y^{2}).

Solution:

We know that (a – b) (a + b) = a^{2} – b^{2}

Put a = x and b = y in the identity (1) then

(x – y) (x + y) = x^{2} – y^{2}

Now (x – y) (x + y)(x^{2} + y^{2}) = (x^{2} – y^{2}) (x^{2} + y^{2})

Again put a = x^{2} and b = y^{2} in (1)

We have (x^{2} – y^{2}) (x^{2} + y^{2}) = (x^{2})^{2} – (y^{2})^{2} = x^{4} – y^{4}

So (x – y) (x + y) (x^{2} + y^{2}) = x^{4} – y^{4}

Question 3.

Simplify (5x – 3y)^{2} – (5x + 3y)^{2}.

Solution:

We have the identities

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

(a – b)^{2} = a^{2} – 2ab + b^{2}

So (5x – 3y)^{2} – (5x + 3y)^{2} = (5x)^{2} – (2 × 5x × 3y) + (3y)^{2}

= 5^{2}x^{2} – 30xy + 3^{2} y^{2} – [5^{2}x^{2} – 30xy + 3^{2} y^{2}]

= 25x^{2} – 30xy + 9y^{2} – [25x^{2} + 30xy + 9y^{2}]

= 25x^{2} – 30xy + 9y^{2} – 25x^{2} – 30xy – 9y^{2}

= x^{2} (25 – 25) – xy (30 + 30) + y^{2} (9 – 9)

= 0x^{2} – 60xy + 0y^{2} = – 60 xy

∴ (5x – 3y)^{2} – (5x + 3y)^{2} = -60xy

Question 4.

Simplify : (i) (a + b)^{2} – (a – b)^{2}

(ii) (a + b)^{2} + (a – b)^{2}

Solution:

Applying the identities

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

(a – b)^{2} = a^{2} – 2ab + b^{2}

(i) (a + b)^{2} – (a – b)^{2} = a^{2} + 2ab + b^{2} – [a^{2} – 2ab + b^{2}]

= a^{2} + 2ab + b^{2} – a^{2} + 2ab – b^{2}

= a^{2} (1 – 1) + ab (2 + 2) + b^{2} (1 – 1)

= 0a^{2} + 4 ab + 0b^{2} = 4ab

(a + b)^{2} – (a – b)^{2} = 4ab

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

= a^{2} + 2ab + b^{2} + a^{2} – 2ab + b^{2}

= a^{2} (1 + 1) + ab (2 – 2) + b^{2} (1 + 1)

= 2a^{2} + 0 ab + 2b^{2} = 2a^{2} + 2b^{2} = 2 (a^{2} + b^{2})

∴ (a + b)^{2} – (a – b)^{2} = 2 (a^{2} + b^{2})

Question 5.

A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m^{2}, find the area of lawn.

Solution:

Let the side of the lawn = a m

then side Of big square = (a + 2(2)) m

= (a + 4)m

Area of the path – Area Of large square – Area of smaller square

136 = (a + 4)2 – a2

136 = a^{2} + (2 × a × 4) + 4^{2} – a^{2}

136 = a^{2} + 8a + 16 – a^{2}

136 = 8a + 16

136 = 8 (a + 2)

Dividing by 8

17 = a + 2

Subtracting 2 on both sides

17 – 3 = a + 2 – 2

15 = a

∴ side of small square = 15 m

Area of square = (side × side) Sq. units

∴ Area of the lawn = (15 × 15)m^{2} = 225 m^{2}

∴ Area of the lawn = 225 m^{2}

Question 6.

Solve the following inequalities.

(i) 4n + 7 > 3n + 10, n is an integer

(ii) 6(x + 6) > 5 (x – 3), x is a whole number.

(iii) -13 < 5x + 2 < 32, x is an integer.

Solution:

(i) 4n + 7 > 3n + 10, n is an integer.

4n + 7 – 3n > 3n + 10 – 3n

n(4 – 3) + 7 > 3n + 10 – 3n

n (4 – 3) + 7 > n (3 – 3) + 10

n + 7 > 10

Subtracting 7 on both sides

n + 7 – 7 > 10 – 7

n > 3

Since the solution is an integer and is greater than or equal to 3, the solution will be 3,

4, 5, 6, 7, …..

n = 3, 4, 5, 6,7, ….

(ii) 6 (x + 6) > 5 (x – 3), x is a whole number.

6x + 36 > 5x – 15

Subtracting 5x on both sides

6x + 36 – 5x > 5x – 15 – 5x

x (6 – 5) + 36 > x(5 – 5) – 15

x + 36 > -15

Subtracting 36 on both sides

x + 36 – 36 > -15 -36

x > -51

The solution is a whole number and which is greater than or equal to -51

∴ The solution is 0, 1, 2, 3, 4,…

x = 0,1,2, 3,4,…

(iii) -13 < 5x + 2 < 32, x is an integer.

Subtracting throughout by 2

-13 – 2 < 5x + 2 – 2 < 32 – 2

-15 < 5x < 30

Dividing throughout by 5

\(\frac { -15 }{ 5 } \) < \(\frac { 5x }{ 5 } \) < \(\frac { 30 }{ 5 } \)

– 3 < x < 6

∴ Since the solution is an integer between -3 and 6 both inclusive, we have the solution

as -3, -2, -1,0, 1,2, 3, 4, 5, 6.

i.e. x = -3, -2, 0, 1, 2, 3,4, 5 and 6.