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
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.
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Questions + Answers key
Take MCQ Maths Test2 Marks
1.
Multiply a monomial by a monomial
−3x, 7y
2.
Multiply a monomial by a monomial
a3, -4a2b
3.
Expand 5x(2y− 3)
4.
Find the product of (2x + 3)(2x − 4)
5.
Find the missing term 6xy x _________= −12x3y
6.
Find the missing term 2y(5x2y −___+3___) = 10x2y2 − 2xy + 6y3
7.
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 x time).
8.
Divide (4m2n3 + 16m4n2 - mn) by 2mn
9.
Find Adirai’s percentage of marks who scored 25m3n2p out of 100m2np.
10.
Identify the errors and correct them. m(4m − 3) = 4m2 −3
2 Marks
1.
-3x \(\times\) 7y = (-3 \(\times\) 7)(x \(\times\) y) = - 21xy
2.
a3 \(\times\) (-42 b) = (-4) \(\times\) (a3 \(\times\) a2) \(\times\) (b) = -4a5b
3.
5x (2y - 3) = (5x) (2y) - (5x) (3)
= (5 x 2) (x x y) - (5 x 3)x
= 10xy - 15x
4.
(2x + 3) (2x - 4)
= 4x2 - 8x + 6x - 12
= 4x2 - 2x - 12
5.
6xy x a = −12x3y
\(a=-\frac{12 x y}{6 x y}\)
a = -2x2
6.
2y(5x2y − c + 3 d = 10x2y2 − 2xy + 6y3
\(10 x^{2} y^{2}-2 y c+6 y d=10 x^{2} y^{2}-2 x y+6 y^{3}\)
Equating the coefficients, we get
- 2yc = 2xy and 6yd = 6y3
\(c=\frac{-2 x y}{-2 y}\) and \(d=\frac{6 y^{3}}{6 y}\)
C = x and d = y2
7.
Speed of the car = (x + 30) km / hr.
Time = (y + 2) hours
Distance = Speed x time
= (x + 30) (y + 2) km
Distance covered = (xy + 2x + 30y + 60) km
8.
\(=\frac{4 m^{2} n^{3}+16 m^{4} n^{2}-m n}{2 m n}\)
\(
=\frac{4 m^{2} n^{3}}{2 m n}+\frac{16 m^{4} n^{2}}{2 m n}-\frac{m n}{2 m n}
\)
\(=2 m n^{2}+8 m^{3} n-\frac{1}{2}\)
9.
Total marks = 100 m2np
Adirai's score = 25 m3n2p
\(Adirai'spercentage=\cfrac { Adirai'sscore }{ Totalmarks } \times 100\%\)
= \(\cfrac { 25{ m }^{ 3 }{ n }^{ 2 }p }{ { 100m }^{ 2 }np } \times 100\%\)
= 25m3-2n2-1p1-1% = 25 mnp% = 25 mn%
10.
m(4m − 3) = 4m2 −3 (false)
m(4m − 3) = 4m2 −3 m
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
