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