Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.2 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.2

Question 1.

Fill in the blanks

(i)

Answer:

\(\frac{18 m^{4}\left(n^{8}\right)}{2 m^{(3)} n^{3}}\) = 9 mn^{5}

(ii)

Answer:

\(\frac{l^{4} m^{5} n^{(7)}}{2 l m^{(3)} n^{6}}=\frac{l^{3} m^{2} n}{2}\)

(iii)

Answer:

\(\frac{42 a^{4} b^{5}\left(c^{2}\right)}{6(a)^{4}(b)^{2}}\) = (7)b^{3}c^{2}

Question 2.

Say True or False

(i) 8x^{3}y ÷ 4x^{2} = 2xy

Answer:

True

(ii) 7ab^{3} ÷ 14 ab = 2b^{2}

Answer:

False

Question 3.

Divide

(i) 27 y^{3} by 3y

(ii) x^{3} y^{2} by x^{2}y

(iii) 45x^{3} y^{2} z^{4} by (-15 xyz)

(iv) (3xy)^{2} by 9xy

Answer:

(i) 27 y^{3} by 3y

\(\frac{27 y^{3}}{3 y}\) = \(\frac{27}{3} y^{3-1}\) = 9y^{2}

(ii) x^{3} y^{2} by x^{2}y

\(\frac{x^{3} y^{2}}{x^{2} y}\) = x^{3-2} y^{2-1} = x^{1} y^{1} = xy

(iii) 45x^{3} y^{2} z^{4} by (-15 xyz)

\(\frac{45 x^{3} y^{2} z^{4}}{-15 x y z}\) = \(\frac{45}{-15}\) x^{3-2} y^{2-1} y^{4-1} z^{4-1} = -3x^{2} yz^{3}

(iv) (3xy)^{2} by 9xy

\(\frac{(3 x y)^{2}}{3 \times(3 x y)}\) = \(\frac{(3 x y)^{2}}{3 \times(3 x y)}\) = \(\frac{1}{3}\) (3xy)^{2-1} = \(\frac{1}{3}\) 3xy = xy

Question 4.

Simplify

(i) \(\frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}}\)

(ii) \(\frac{14 p^{5} q^{3}}{2 p^{2} q} \frac{12 p^{3} q^{4}}{3 q^{2}}\)

Answer:

(i) \(\frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}}\)

\(\frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}}\) = 3m^{2-1} + 2m^{4-3}

= 3m + 2m

= (3 + 2) m

= 5m

(ii) \(\frac{14 p^{5} q^{3}}{2 p^{2} q} \frac{12 p^{3} q^{4}}{3 q^{2}}\)

\(\frac{14 p^{5} q^{3}}{2 p^{2} q} \frac{12 p^{3} q^{4}}{3 q^{2}}\) = \(\frac{14}{2}\)p^{5-2}q^{3-1} – \(\frac{12}{3}\) p^{3}q^{4-3}

= 7p^{3}q^{2} – 4p^{3}q

Question 5.

Divide:

(i) 32y^{2} – 8yz by 2y

(ii) (4m^{2}n^{3} + 16m^{4} n^{2} – mn) by 2mn

(iii) 5xy^{2} – 18x^{2}y^{3} + 6xy by 6xy

(iv) 81(p^{4} q^{2} r^{3} + 2p^{3}q^{3} r^{2} – 5p^{2}q^{2}r^{2}) by (3pqr)^{2}

Answer:

(i) 32y^{2} – 8yz by 2y

(ii) (4m^{2}n^{3} + 16m^{4} n^{2} – mn) by 2mn

(iii) 5xy^{2} – 18x^{2}y^{3} + 6xy by 6xy

(iv) 81(p^{4} q^{2} r^{3} + 2p^{3}q^{3} r^{2} – 5p^{2}q^{2}r^{2}) by (3pqr)^{2}

= \(\frac{81}{9}\)(p^{2}q^{2}r^{2})^{1-1} (p^{2}r + 2pq – 5)

= 9(p^{2}r + 2pq – 5) = 9 p^{2}r + 18pq – 45

Question 6.

Identify the errors and correct them.

(i) 7y^{2} – y^{2} + 3y^{2} = 10y^{2}

Answer:

7y^{2} – y^{2} + 3y^{2} = 10y^{2} = (7 – 1 + 3)y^{2}

= (6 + 3)y^{2}

= 9y^{2}

(ii) 6xy + 3xy = 9x^{2}y^{2}

Answer:

6xy + 3xy = (6 + 3) xy

= 9 xy

(iii) m(4m – 3) = 4m^{2} – 3

Answer:

m(4m – 3) = m(4m) + m(-3)

= 4m^{2} – 3m

(iv) (4n)^{2} – 2n + 3 = 4n^{2} – 2n + 3

Answer:

(4n)^{2} – 2n + 3 = 16n^{2} – 2n + 3

(v) (x – 2)(x + 3) = x^{2} – 6

Answer:

(x – 2)(x + 3) = x(x + 3) – 2 (x + 3)

= x(x) + (x) × 3 + (-2) (x) + (-2) (3)

= x^{2} + 3x – 2x – 6

= x^{2} + x – 6

(vi) -3p^{2} + 4p – 7 = -(3p^{2} + 4p – 7)

Answer:

-3p^{2} + 4p – 7 = -(3p^{2} – 4p + 7)

Question 7.

Statement A: If 24p^{2}q is divided by 3pq, then the quotient is 8p.

Statement B: Simplification of \(\frac{(5 x+5)}{5}\) is 5x.

(i) Both A and B are true

(ii) A is true but B is false

(iii) A is false but B is true

(iv) Both A and B are false

Answer:

(ii) A is true but B is false

Hint:

Question 8.

Statement A: 4x^{2} + 3x – 2 = 2(2x^{2} + \(\frac{3 x}{2}\) – 1)

Statement B: (2m – 5) – (5 – 2m) = (2m – 5) + (2m – 5)

(i) Both A and B are true

(ii) A is true but B is false

(iii) A is false but B is true

(iv) Both A and B are false

Answer:

(i) Both A and B are true

Hint:

(2m – 5) – (5 – 2m) = 2m – 5 – 5 + 2m = 4m – 10

(2m – 5) + (2m – 5) = 4m – 10