8th Standard Syllabus & Materials
8th Standard
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NEW8th Standard
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NEW8th Standard
Tamilnadu 8th Standard роХрогро┐родроорпН роОрогрпНроХро│рпН Important Questions And Answers Study Material - QB365 Set A
NEW8th Standard
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NEW8th Standard
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NEW8th Standard
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Published on: 15/06/2021
QB365 provides detailed and simple solution for every book back questions in class 8 Maths subject.It will helps to get more idea about question pattern in every book back questions with solution.
Download Tamil Nadu 8th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Maths Test1.
Expand (x + 4)3
2.
Find the value of (103)3
3.
Expand y2 - 16 using a2 − b2 identity
4.
Find the value of (3a + 4c)2 by using (a+b)2 identity.
5.
If Guru wants to multiply the expressions (2x+3y+50) and 3xy , what is the resultant expression?
1.
Comparing (x + 4)3 with (a + b)3, we get a = x, b = 4
We know(a + b)3 = a3 + 3a2b + 3ab2 + b3
(x + 4)3 = (x)3 + 3(x)2 + 3(x)(4)2 + (4)3
= (x)5 + 3x2(4)+ 3(x)(16) + 64
(x+4)3 = x3 + 12x2 + 48x + 64
2.
Now,(103)3 = (100 + 3)3
Comparing this with (a+ b)3,we get a = 100,b = 3
(ab)3 = a3 + 3a2b + 3ab2 + b3 replacing a, b values,
(100+3)3 = (100)3 +3(100)2(3) + 3(100)(3)2 + (3)3
= 1000000 + 3(10000)(3) + 3(100)(9) + 27
= 1000000 + 9000 + 2700 + 27
(103)3 = 1092727
3.
y2 - 16 can be written as y2 - 42
Comparing it with a2 - b2,we get ay, b = 4,
Now a2-b2 = (a + b)(a - b)
y2-42 = (y + 4)(y - 4)
y2-16 = (y + 4)(y - 4)
4.
\(\therefore\) (3a + 4c)2 = (3a)2 + 2(3a)(4c) + (4c)2 (replacing a and b values)
= 32 a2 + (2 x 3 x 4)(a x c) + 42c2
(3a + 4c)2 = 9a2 + 24ac + 16c2
5.
The resultant expression
=3xy × (2x+3y+50)
=3xy(2x)+3xy(3y)+3xy(50)
=6x2y+9xy2+150xy
8th Standard Syllabus & Materials
8th Standard
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NEW8th Standard
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NEW8th Standard
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Tamilnadu Stateboard 8th Standard Subjects
Tamilnadu Stateboard Standards
