Samacheer Kalvi 8th Maths Book Solutions Term 1 Chapter 3 – Algebra is available here. The Samacheer Kalvi 8th Maths book answers of Term 1 Chapter 3, available at BYJU’S, contain step by step explanations designed by our Mathematics experts. All these important questions are based on the new pattern prescribed by the Tamil Nadu board. Students can also get the solutions of other chapters on Samacheer Kalvi 8th Maths solutions. Here in this article below, we have compiled the Samacheer Kalvi 8th Class Maths textbook Term 1 Chapter 3 Questions with Solutions for the students to solve and practice so that they can face the final exams more confidently.
Samacheer Kalvi Class 8 Maths Book Term 1 Chapter 3 Questions and Solutions
Term 1 Chapter 3 of the Samacheer Kalvi 8th Maths guide will help the students to solve problems related to the multiplication of algebraic expressions, division of algebraic expressions, algebraic identities, factorisation.
Samacheer Kalvi 8th Maths Term 1 Chapter 3: Algebra Book Exercise 3.1 Questions and Solutions
Question 1: Find the product of the terms:
[i] -2mn, (2m)2, -3mn
[ii] 3x2y, -3xy3, x2y2
Solution:
[i] -2mn, (2m)2, -3mn-2mn * (2m)2 * -3mn
= (-2mn) * (4m2) * (-3mn)
= (-) (+) (-) (2 * 4 * 3) (mn * m2 * mn)
= 24 m4n2
[ii] 3x2y, -3xy3, x2y23x2y * -3xy3 * x2y2
= (-) (3 * 3 * 1) (x2y * xy3 * x2y2)
= -9x5y6
Question 2: 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:
Speed of the car = (x + 30) km / hr
Time = (y + 2) hours
Distance = Speed * Time
= (x + 30) * (y + 2)
= xy + 2x + 30y + 60
Samacheer Kalvi 8th Maths Term 1 Chapter 3: Algebra Book Exercise 3.2 Questions and Solutions
Question 1: Divide
(i) (32y2 – 8yz) by 2y
(ii) 10 (4x – 8y) by 5 (x – 2y)
Solution:
(i) (32y2 – 8yz) by 2y
= [32y2 / 2y] – [8yz / 2y]
= 16y – 4z
(ii) 10 (4x – 8y) by 5 (x – 2y)
= 2 (4x – 8y) / (x – 2y)
= 8 (x – 2y) / (x – 2y)
= 8
Question 2: Find Adirai’s percentage of marks who scored 25m3n2p out of 100m2np.
Solution:
Total score = 100m2np
Adirai’s score = 25m3n2p
Adirai’s percentage = {[Adirai’s score] / [Total score]} * 100%
= {[25m3n2p] / [100m2np]} * 100%
= 25 mn%
Samacheer Kalvi 8th Maths Term 1 Chapter 3: Algebra Book Exercise 3.3 Questions and Solutions
Question 1: Expand (3m + 5)2.
Solution:
(3m + 5)2
On comparing (3m + 5)2 to (a + b)2, a = 3m, b = 5
(a + b)2 = a2 + b2 + 2ab
= (3m)2 + (5)2 + 2 * 3m * 5
= 9m2 + 25 + 30m
Question 2: Find the volume of the cube whose side is x + 1 cm.
Solution:
Side of the cube = (x + 1) cm
Volume of the cube = (side)3 cubic units
= (x + 1)3 cm3
(a + b)3 = a3 + 3a2b + 3ab2 + b3 cm3
(x + 1)3 = x3 + 3 * x2 * 1 + 3 * x * 12 + 13
= x3 + 3x2 + 3x + 1 cm3
Samacheer Kalvi 8th Maths Term 1 Chapter 3: Algebra Book Exercise 3.4 Questions and Solutions
Question 1: Factorise the following by taking out the common factor
(i) 18xy − 12yz
(ii) (ax + ay) (bx + by)
Solution:
(i) 18xy − 12yz
= 2 * 3 * y (3x – 2z)
= 6y (3x – 2z)
(ii) (ax + ay) (bx + by)
Taking a and b as common from each of them,
= a (x + y) b (x + y)
= (a + b) (x + y)
Question 2: Factorise the expression: x2 + 14x + 49.
Solution:
x2 + 14x + 49
= x2 + 7x + 7x + 49
= x (x + 7) + 7 (x + 7)
= (x + 7) (x + 7)
= (x + 7)2
Samacheer Kalvi 8th Maths Term 1 Chapter 3: Algebra Book Exercise 3.5 Questions and Solutions
Question 1: Find the missing term: y2 + ( _ ) x + 56 = (y + 7) (y + _ ).
Solution:
(x + a) (x + b) = x2 + (a + b) x + ab
56 = 7 * 8
= y2 + (8 + 7) x + 8 * 7
= y2 + 15x + 56
= (y + 8) (y + 7)
Question 2: Factorise x6 – 64y3
Solution:
(x6 – 64y3)
= (x2)3 – (4y)3
a3 – b3 = (a – b) (a2 + ab + b2)
= (x2 – 4y) [(x2)2 + 4x2y + 16y2]
= (x2 – 4y) (x4 + 4x2y + 16y2)
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