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

Question 2.

Complete the table

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}

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}

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}

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

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}

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

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

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}

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

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

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

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