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: 13/05/2022
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Questions + Answers key
Take MCQ Maths Test2 Marks
1.
Expand (3m + 5)2
2.
Factorise the following by taking out the common factor (ax + ay) + (bx + by)
3.
Factorise the following by taking out the common factor a3− 3a2 + a −3
4.
Factorise :(7y2- 19y- 6)
5.
Pick out the like terms from the following.
6.
The sides of a triangle are 2x - 5y + 9, 3y + 6x -7 and -4x + y +10. Find perimeter of the triangle.
7.
The length of a log is 3a + 4b - 2 and a piece (2a - b) is remove from it. What is the length of the remaining log?
8.
Multiply (m4+ n4) and (m - n)
9.
Are the following correct?
\(\cfrac { { 10m }^{ 4 } }{ 10{ m }^{ 4 } } =0\)
10.
4m2n + 9n2m + 3mn \(\div \)4mn
2 Marks
1.
(a + b)2 = a2 + 2ab + b2
(3m+5)2 = (3m)2 + 2(3m)(5) + 52
= 32m2 + 30m + 25 = 9m2 + 30m + 25
2.
(ax + ay) + (bx+ by)
= a (x + y) + b (x + y) = (x +y)(a + b)
3.
a3 − 3a2 + a −3 = a2(a - 3) + 1(a - 3)
= (a - 3)(a2 + 1)
4.
| Product | Sum |
| -42 | -19 |
| (-21) x 2 | -21 + 2 |
7y2 - 19y - 6 = 7y2-21y + 2y - 6
= 7y(y - 3) + 2(y - 3)
= (y - 3)(7y + 2)
5.
6.
Perimeter of the triangle = Sum of three sides
= (2x - 5y + 9) + (3y + 6x - 7) + (-4x + y + 10)
= 2x - 5y + 9 + 3y + 6x - 7 -4x + y +10
= 2x + 6x - 4x -:-5y + 3y + Y + 9 -7 + 10
= (2 + 6 - 4) x + (-5 + 3 + 1) y + (9 -7 + 10)
= 4x - y+ 12
\(\therefore\) Perimeter of the triangle = 4x -y + 12 units.
7.
Length of the log = 3a + 4b-2
Remaining length of the log = (3a + 4b - 2)-(2a - b)
= (3a - 2a) + [4b - (-b)]- 2
= (3 - 2) a + (4 + 1) b - 2
= a + 5b-2
8.
(m4 + n4) (m - n)
= m4(m - n) + n4 (m - n)
= m5-m4n + mn4-n5
= m5 - n5 - mn(m3 + n3)
9.
\(\cfrac { { 10m }^{ 4 } }{ 10{ m }^{ 4 } } =\cfrac { 10 }{ 10 } { m }^{ 4-4 }=1{ m }^{ 0 }=1\)
\(\therefore\) The given answer is not correct.
10.
\(=\frac{4 m^{2} n+9 n^{2} m+3 m n}{4 m n}\)
\(=m^{2-1}+\frac{9}{4} n^{2-1}+\frac{3}{4}=m+\frac{9}{4} n+\frac{3}{4}\)
8th Standard Syllabus & Materials
8th Standard
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NEW8th 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
