8th Standard Syllabus & Materials
8th Standard
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NEW8th Standard
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TN 8th Tamil இயல் 2 - ஈடில்லா இயற்கை - இயற்கையை போற்றுவோம் Important Questions And Answers Study Material - QB365 Set A
NEW8th Standard
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Published on: 27/07/2019
Algebra
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.
Multiply (2x + 5y) and (3x − 4y)
2.
Multiply 3x2y and (2x3y3 − 5x2y + 9xy)
3.
If the side of a square carpet is 3x2 metre, then find its area
4.
Factorise :(7y2- 19y- 6)
5.
Multiply (4x2 + 9) and (3x-2)
6.
Factorise the following expressions using a3-b3 = (a-b) (a2+ab+b2) identity c3-27b3a3
7.
Factorise the following expressions m2 + m − 72
8.
Factorise the following by taking out the common factor 18xy - 12yz
9.
Expand (52)3
10.
Expand (3m + 5)2
11.
Divide 27y3 ÷ 3y
12.
Find the missing term 6xy x _________= −12x3y
13.
Find the product of (2x + 3)(2x − 4)
14.
Expand −2p(5p2−3p +7)
15.
Expand 5x(2y− 3)
16.
Multiply a monomial by a monomial
2p2q3, −9pq2
17.
Multiply a monomial by a monomial
a3, -4a2b
18.
Multiply a monomial by a monomial
6x,4
19.
If the area of a rectangular land is(a2 - b2 )− sq.units whose breadth is (a - b) then, its length is__________
a - b
a + b
a2- b
(a + b)2
20.
If the area of a rectangle is 48m2n3 and whose length is 82mn2 then, its breadth is______
6mm
8m2n
7m2n2
6m2n2
21.
If the area of a square is 36x4y2t hen, its side is______
6x4y2
8x2y2
6x2y2
6x2y
22.
The product of 7p3 and (2p2)2 is
14p12
28P7
9p7
11p12
23.
Factorise: a3 – 8
24.
Factorise : 49x2 - 64y2
25.
Divide (10m2 − 5m)by (2m−1)
26.
\(\cfrac { { 42a }^{ 4 }{ b }^{ 5 }\left( \_ \right) }{ { 6a }^{ 4 }{ b }^{ 2 } } =\left( \_ \right) { b }^{ 3 }{ c }^{ 2 }\)
27.
\(\frac{18 m^{4}(-)}{2 m^{3} n^{3}}\) ______mn5
1.

= 6x2 − 8xy + 15xy − 20y2
= 6x2 + 7xy− 20y2(simplify the like terms
2.

= 3x2y(2x3y3) − 3x2y(5x2y) + 3x2y(9xy)
multiplying each term of the polynomial by the monomial
= (3 × 2)(x2 × x3)(y × y3)−(3 × 5)(x2 × x2)(y × y) + (3 × 9)(x2 × x)(y × y)
= 6x5y4−15x4y2 + 27x3y2
3.
The area of the square carpet, A = (side × side) sq. units.
= 3x2×3x2
= 3×3×x2×x2
A = 9x4 sq.m
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.
(4x2 + 9)(3x - 2) = 12x3 - 8x2 + 27x - 18
6.
c3-27b3a3 = c3-33b3a3
= c3-(3ba)3
Comparing this with a3-b3 we have a = x and b = 3ba
a3 - b3 = (a-b)(a2+ab+b2)
\(\therefore\) c3-(3ba)3 = (c-3ba)(c2+(c)(3ba)+(3ba)2)
= (c-3ba) (c2+3bac+32b2a2)
c3-27b3a3 = (c-3ab)(c2+3bac+9a2b2)
7.
| Product | Sum |
| -72 | 1 |
| 9 x (-8) | +9 -8 |
m2 + m - 72 = m2 + 9m - 8m - 72
= m (m + 9) - 8 (m + 9)
= (m + 9)(m - 8)
8.
18xy - 12yz = 6y (3x - 2z)
9.
(52)3 = (50 + 2)3
Comparing (50 + 2)3 with (a + b)3we have a = 50 and b = 2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(50 + 2)3 = 503 + 3 (50)22 + 3 (50)(2)2 + 23
523 = 125000 + 6(2,500) + 150(4) + 8
= 1,25,000 + 15,000 + 600 + 8
523 = 1,40,608
10.
(a + b)2 = a2 + 2ab + b2
(3m+5)2 = (3m)2 + 2(3m)(5) + 52
= 32m2 + 30m + 25 = 9m2 + 30m + 25
11.
\(\cfrac { 27y^{ 3 } }{ \quad 3y } =\cfrac { 27 }{ 3 } { y }^{ 3-1 }=9y^{ 2 }\)
12.
6xy x a = −12x3y
\(a=-\frac{12 x y}{6 x y}\)
a = -2x2
13.
(2x + 3) (2x - 4)
= 4x2 - 8x + 6x - 12
= 4x2 - 2x - 12
14.
-10p3+6p2-14p
15.
5x (2y - 3) = (5x) (2y) - (5x) (3)
= (5 x 2) (x x y) - (5 x 3)x
= 10xy - 15x
16.
(2p2q3)\(\times\)(-9pq)2 = (+)(-) \(\times\) (2\(\times\)9) (p2\(\times\) p (q3\(\times\)q2)) = -18p3q5
17.
a3 \(\times\) (-42 b) = (-4) \(\times\) (a3 \(\times\) a2) \(\times\) (b) = -4a5b
18.
6x \(\times\) 4= (6 \(\times\) 4)(x) = 24x
19.
(b)
a + b
20.
(a)
6mm
21.
(d)
6x2y
22.
(b)
28P7
23.
Here a3 – 8 can be written as a3 – 23
Comparing this with a3−b3, we get a=a,b=2
\(\therefore\) a3-b3 = (a-b)(a2+ab+b2)
a3-23=(a-2)(a2+a(2)+22)
a3-8=(a-2)(a2+2a+4)
24.
Now, 49x2- 64y2 = 72x2 - 82y2
= (7x)2 - (8y)2
Comparing this with a2-b2 = (a + b)(a - b)
we get a = 7x, b = 8y
(7x)2- (8y)2 = (7x + 8y)(7x - 8y)
25.
We have = \(\cfrac { { 10m }^{ 2 }-5m }{ (2m-1) } \)
(
(taking common factor from the numerator)
= 5m
26.
( )
\(\cfrac { { 42a }^{ 4 }{ b }^{ 5 }\left( c^2 \right) }{ { 6a }^{ 4 }{ b }^{ 2 } } =\left( 7 \right) { b }^{ 3 }{ c }^{ 2 }\)
27.
( )
\(\cfrac { 18{ m }^{ 4 }(n^{ 98 }) }{ 2m^{ (3) }{ n }^{ 3 } } =9mn^{ 5 }\)
8th Standard Syllabus & Materials
8th Standard
TN 8th Tamil இயல் 3 - கல்வி கரையில - பாடறிந்து ஒழுகுதல் Important Questions And Answers Study Material - QB365 Set A
NEW8th Standard
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NEW8th Standard
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Tamilnadu Stateboard 8th Standard Subjects
Tamilnadu Stateboard Standards