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

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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) -2m2, (-5m)3
(iv) a3, – 4a2b
(v) 2p2q3, -9pq2
Solution:
(i) 6x × 4 = (6 × 4) (x) = 24x
(ii) -3x × 7y = (-3 × 7) (x × y) = -21xy
(iii) (-2m2) × (-5m)3 = -2m2 × (-)3 (53 (m)3) = -2m2 × (-125m3)
= (-) × (-)(2 × 125)(m2 × m3) = + 250m5 = 250 m
(iv) a3 × (-4a2 b) = (-4) × (a3 × a2) × (b) = -4a5b
(v) (2p2q3) × (-9pq2) = (+) × (-) × (2 × 9) (p2 × p(q3 × q2)) = -18p3q5

Question 2.
Complete the table Solution: Question 3.
Find the product of the terms.
(i) -2mn, (2m)2, -3mn
(ii) 3x2y, -3xy3, x2y2
Solution:
(i) (-2mn) × (2m)2 × (-3mn) = (-2mn) × 22 m2 × (-3mn) = (-2mn) × 4m2 × (-3mn)
= (-) (+)(-) (2 × 4 × 3) (m × m2 × m) (n × n)
= + 24 m4 n2

(ii) (32y) × (-31xy3) × (x2y2) = (+) × (-) × (+) × (3 × 3 × 1) (x2 × x × x2) x (y × y3 × y2)
= -9x5y6

Question 4.
If l = 4pq2, b = -3p 2q, h = 2p3q3 then, find the value of 1 × b × h.
Solution:
Given l = 4pq2
b = -3p2q
h = 2p3q3
l × b × h = (4pq2) × (-3p2 q) × (2p3q3)
= (+) (-) (+) (4 × 3 × 2) (p × p2 × p3) (q2 × q × q3)
= -24p6q6 Question 5.
Expand
(i) 5x (2y – 3)
(ii) -2p (5p2 – 3p + 7)
(iii) 3mn (m3n3 – 5m2n + 7mn2)
(iv) x2 (x + y + z) y2 (x + y + z) + z2 (x – y – z)
Solution:
(i) 5x(2y – 3) = (5x) (2y) – (5x) (3)
= (5 × 2) (x × y) – (5 × 3) x
= 10xy – 15x

(ii) -2p (5p2 – 3p + 7) = (-2p) (5p2) + (-2p) (-3p) + (-2p) (7)
= [(-) (+) (2 × 5) (p × p2)] + [(-) (+) (2 × 3) (p × q)] + (-) (+) (2 × 7) p
= -10p3 + 6p2 – 14p

(iii) 3mn(m3n3 – 5m2n + 7mn2)
= (3mn) (m3n3) + (3mn) (-5m2n) + (3mn)(7mn2)
= (3) (m × m3) (n × n3) + (+) (-) (3 × 5) (m × m2) (n × n) + (3 × 7) (m × m)(n × n2)
= 3m4n4 – 15m3 n2 + 21m2n3

(iv) x2 (x + y + z) + y2 (x + y + z) + z2 (x – y – z)
= (x2 × x) + (x2 × y) + (x2 × z) + (y2 × x) + (y2 × y) + (y2 × z) + (z2 × x) + z2 (-y) + z2 (-z)
= x3 + x2y + x2z + xy2 + y3 + y2z + xz2 – yz2 – z3
= x3 + y3 – z2 + x2y + x2z + xy2 + zy2 + xz2 – yz2

Question 6.
Find the product of
(i) (2x + 3)(2x – 4)
(ii) (y2 – 4) (2y2 + 3y)
(iii) (m2 – m) (5m2n2 – n2)
(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)
= 4x2 – 8x + 6x – 12
= 4x2 + (- 8 + 6)x – 12
= 4x2 – 2x – 12

(ii) (y2 -4) (2y2 + 3y) = y2 (2y2 + 3y) – 4 (2y2 + 37)
= y2(2y2) + y2(3y) – 4(2y2) -4 (3y)
= 2y4 + 3y3 – 8y2 – 12y

(iii) (m2 – n) (5m2n2 – n2) = m2 (5m2n2 – n2) – n (5m2n2 – n2)
= m2 (5m2n2) + m2 (-n2) – n (5m2n2) + (-) (-) n (n2)
= 5m4n2 – m2n2 – 5m2n3 + n3

(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 [x2 – x – 5x + 5] = 6 [x2 + (-1 – 5)x + 5]
= 6 [x2 – 6x + 5] = 6x2 – 36x + 30 Question 7.
Find the missing term.
(i) 6xy – × ______ = -12x3y
(ii) ________ × (-15m2n3p) = 45m3n3p2
(iii) 2y(5x2y – ___ + 3 ___) = 10x2y2 – 2xy + 6y3
Solution:
(i) 6xy – × (-2x2) = -12x3y
(ii) -3mp × (-15m2n3p) = 45m3n3p2
(iii) 2y(5x2y – x + 3 y2) = 10x2y2 – 2xy + 6y3

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 7p3 and (2p2)2 is
(A) 14p2
(B) 28p7
(C) 9p7
(D) 11p12
Solution:
(B) 28p7

Question 11.
The missing terms in the product -3m3 n × 9(- -) = ____ m4n3 are
(A) mn2, 27
(B) m2n, 27
(C) m2n2, -27
(D) mn2, -27
Solution
(A) mn2 ,27 Question 12.
If the area of a square is 36x4y2 then, its side is ______ .
(A) 6x4y2
(B) 8x2y2
(C) 6x2y
(D) -6x2y
Solution:
(C) 6x2y

Question 13.
If the area of a rectangle is 48m2n3 and whose length is 8mn2 then, its breadth is ____ .
(A) 6 mn
(B) 8m2n
(C) 7m2n2
(D) 6m2n2
Solution:
(A) 6mn

Question 14.
If the area of a rectangular land is (a2 – b2) sq.units whose breadth is (a – b) then, its length is _____
(A) a – b
(B) a + b
(C) a2 – b
(D) (a + b)2
Solution:
(B) a + b