Students can Download Maths Chapter 3 Algebra Ex 3.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

**8th Maths Algebra Exercise 3.1 Question 1.**

Multiply a monomial by a monomial.

(i) 6x, 4

(ii) -3x, 7y

(iii) -2m^{2}, (-5m)^{3}

(iv) a^{3}, – 4a^{2}b

(v) 2p^{2}q^{3}, -9pq^{2}

Solution:

(i) 6x × 4 = (6 × 4) (x) = 24x

(ii) -3x × 7y = (-3 × 7) (x × y) = -21xy

(iii) (-2m^{2}) × (-5m)^{3} = -2m^{2} × (-)^{3} (5^{3} (m)^{3}) = -2m^{2} × (-125m^{3})

= (-) × (-)(2 × 125)(m^{2} × m^{3}) = + 250m5 = 250 m

(iv) a^{3} × (-4a^{2} b) = (-4) × (a^{3} × a^{2}) × (b) = -4a^{5}b

(v) (2p^{2}q^{3}) × (-9pq^{2}) = (+) × (-) × (2 × 9) (p^{2} × p(q^{3} × q^{2})) = -18p^{3}q^{5}

**Maths 8th Samacheer Kalvi Question 2.**

Complete the table

Solution:

**8th Standard Maths 3.1 Solution Question 3.**

Find the product of the terms.

(i) -2mn, (2m)^{2}, -3mn

(ii) 3x^{2}y, -3xy^{3}, x^{2}y^{2}

Solution:

(i) (-2mn) × (2m)^{2} × (-3mn) = (-2mn) × 2^{2} m^{2} × (-3mn) = (-2mn) × 4m^{2} × (-3mn)

= (-) (+)(-) (2 × 4 × 3) (m × m^{2} × m) (n × n)

= + 24 m^{4} n^{2}

(ii) (3^{2}y) × (-31xy^{3}) × (x^{2}y^{2}) = (+) × (-) × (+) × (3 × 3 × 1) (x^{2} × x × x^{2}) x (y × y^{3} × y^{2})

= -9x^{5}y^{6}

**Samacheer Kalvi Guru 8th Maths Question 4.**

If l = 4pq^{2}, b = -3p 2q, h = 2p^{3}q^{3} then, find the value of 1 × b × h.

Solution:

Given l = 4pq^{2}

b = -3p^{2}q

h = 2p^{3}q^{3}

l × b × h = (4pq^{2}) × (-3p^{2} q) × (2p^{3}q^{3})

= (+) (-) (+) (4 × 3 × 2) (p × p^{2} × p^{3}) (q^{2} × q × q^{3})

= -24p^{6}q^{6}

**Samacheer Kalvi.Guru 8th Maths Question 5.**

Expand

(i) 5x (2y – 3)

(ii) -2p (5p^{2} – 3p + 7)

(iii) 3mn (m^{3}n^{3} – 5m^{2}n + 7mn^{2})

(iv) x^{2} (x + y + z) y^{2} (x + y + z) + z^{2} (x – y – z)

Solution:

(i) 5x(2y – 3) = (5x) (2y) – (5x) (3)

= (5 × 2) (x × y) – (5 × 3) x

= 10xy – 15x

(ii) -2p (5p^{2} – 3p + 7) = (-2p) (5p^{2}) + (-2p) (-3p) + (-2p) (7)

= [(-) (+) (2 × 5) (p × p^{2})] + [(-) (+) (2 × 3) (p × q)] + (-) (+) (2 × 7) p

= -10p^{3} + 6p^{2} – 14p

(iii) 3mn(m^{3}n^{3} – 5m^{2}n + 7mn^{2})

= (3mn) (m^{3}n^{3}) + (3mn) (-5m^{2}n) + (3mn)(7mn^{2})

= (3) (m × m^{3}) (n × n^{3}) + (+) (-) (3 × 5) (m × m^{2}) (n × n) + (3 × 7) (m × m)(n × n^{2})

= 3m^{4}n^{4} – 15m^{3} n^{2} + 21m^{2}n^{3}

(iv) x^{2} (x + y + z) + y^{2} (x + y + z) + z^{2} (x – y – z)

= (x^{2} × x) + (x^{2} × y) + (x^{2} × z) + (y^{2} × x) + (y^{2} × y) + (y^{2} × z) + (z^{2} × x) + z^{2} (-y) + z^{2} (-z)

= x^{3} + x^{2}y + x^{2}z + xy^{2} + y^{3} + y^{2}z + xz^{2} – yz^{2} – z^{3}

= x^{3 }+ y^{3} – z^{2} + x^{2}y + x^{2}z + xy^{2} + zy^{2} + xz^{2} – yz^{2}

**Samacheer Kalvi Guru Maths 8th Question 6.**

Find the product of

(i) (2x + 3)(2x – 4)

(ii) (y^{2} – 4) (2y^{2} + 3y)

(iii) (m^{2} – m) (5m^{2}n^{2} – n^{2})

(iv) 3(x – 5) × 2(x – 1)

Solution:

(2x) (2x – 4) + 3 (2x – 4) = (2x) (2x – 4) + 3 (2x – 4)

= (2x × 2x) – 4 (2x) + 3(2x) – 3 (4)

= 4x^{2} – 8x + 6x – 12

= 4x^{2} + (- 8 + 6)x – 12

= 4x^{2} – 2x – 12

(ii) (y^{2} -4) (2y^{2} + 3y) = y^{2} (2y^{2} + 3y) – 4 (2y^{2} + 37)

= y^{2}(2y^{2}) + y^{2}(3y) – 4(2y^{2}) -4 (3y)

= 2y^{4} + 3y^{3} – 8y^{2} – 12y

(iii) (m^{2} – n) (5m^{2}n^{2} – n^{2}) = m^{2} (5m^{2}n^{2} – n^{2}) – n (5m^{2}n^{2} – n^{2})

= m^{2} (5m^{2}n^{2}) + m^{2} (-n^{2}) – n (5m^{2}n^{2}) + (-) (-) n (n^{2})

= 5m^{4}n^{2} – m^{2}n^{2} – 5m^{2}n^{3} + n^{3}

(iv) 3(x – 5) × 2(x – 1) = (3 × 2) (x – 5) (x – 1)

= 6 × [x (x – 1) – 5 (x- 1)]

= 6 [x.x – x . 1 – 5.x + (-1) (-) 5 1]

= 6 [x^{2} – x – 5x + 5] = 6 [x^{2} + (-1 – 5)x + 5]

= 6 [x^{2} – 6x + 5] = 6x^{2} – 36x + 30

**Samacheer Kalvi 8 Maths Question 7.**

Find the missing term.

(i) 6xy – × ______ = -12x^{3}y

(ii) ________ × (-15m^{2}n^{3}p) = 45m^{3}n^{3}p^{2}

(iii) 2y(5x^{2}y – ___ + 3 ___) = 10x^{2}y^{2} – 2xy + 6y^{3}

Solution:

(i) 6xy – × (-2x^{2}) = -12x^{3}y

(ii) -3mp × (-15m^{2}n^{3}p) = 45m^{3}n^{3}p^{2}

(iii) 2y(5x^{2}y – x + 3 y^{2}) = 10x^{2}y^{2} – 2xy + 6y^{3}

**Samacheer Kalvi 8th Maths Book Solutions Question 8.**

Match the following

(A) iv, v, ii, i, iii

(B) v, iv, iii, ii, i

(C) iv, v, ii, iii, i

(D) iv, v, ii, iii, i

Solution:

(a) iv

(b) v

(c) ii

(d) iii

(e) i

**Samacheer Kalvi 8 Maths Book Question 9.**

A car moves at a uniform speed of (x + 30) km/hr. Find the distance covered by the car in (y + 2)hours. (Hint: distance = speed × time).

Solution:

Sppeed of the car = (x + 30) km / hr.

Time = (y + 2) hours

Distance = Speed × time = (x + 30) (y + 2) = x(y + 2) + 30 (y + 2) = x (y + 2) + 30 (y + 2)

= (x) (y) + (x) (2) + (30) (y) + (30) (2)

= xy + 2x + 30y + 60

Distance covered = (xy + 2x + 30y + 60) km

Objective Type Questions

**Samacheer Kalvi Class 8 Maths Solutions Question 10.**

The product of 7p^{3} and (2p^{2})^{2} is

(A) 14p^{2}

(B) 28p^{7}

(C) 9p^{7}

(D) 11p^{12}

Solution:

(B) 28p^{7}

**Samacheerkalvi.Guru 8th Maths Question 11.**

The missing terms in the product -3m^{3} n × 9(- -) = ____ m^{4}n^{3} are

(A) mn^{2}, 27

(B) m^{2}n, 27

(C) m^{2}n^{2}, -27

(D) mn^{2}, -27

Solution

(A) mn^{2} ,27

**Samacheer Kalvi Guru 8th Maths Guide Question 12.**

If the area of a square is 36x^{4}y^{2} then, its side is ______ .

(A) 6x^{4}y^{2}

(B) 8x^{2}y^{2}

(C) 6x^{2}y

(D) -6x^{2}y

Solution:

(C) 6x^{2}y

**Samacheer Kalvi 8th Maths Question 13.**

If the area of a rectangle is 48m^{2}n^{3} and whose length is 8mn2 then, its breadth is ____ .

(A) 6 mn

(B) 8m^{2}n

(C) 7m^{2}n^{2}

(D) 6m^{2}n^{2}

Solution:

(A) 6mn

**Maths Class 8 Samacheer Kalvi Question 14.**

If the area of a rectangular land is (a^{2} – b^{2}) sq.units whose breadth is (a – b) then, its length is _____

(A) a – b

(B) a + b

(C) a^{2} – b

(D) (a + b)^{2}

Solution:

(B) a + b