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Published on: 03/12/2019
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.
Find the product of the following.
\(\cfrac { -8 }{ 5 } { x }^{ 2 }{ yz }^{ 2 }by-\cfrac { 3 }{ 4 } { xy }^{ 2 }z\)
2.
Find the product of the following
\({ 4x }^{ 2 }yzby\cfrac { 3 }{ 2 } { x }^{ 2 }{ yz }^{ 2 }\)
3.
Expand (x + 4)3
4.
Expand (y - 5)3
5.
Find the product of 2x2y2, 3y2z and –z2x3
6.
Divide 6x3y2z2 by 3x2yz
7.
Find the product of the following
(4a, 3a2)
8.
Find the missing term _________x (−15m2n3p) = 45m3n3p2
9.
Expand 5x(2y− 3)
10.
Multiply a monomial by a monomial
-2m2, (-5,)3
11.
Square of (3x - 4y) is _______.
9x2-16y2
6x2-8y2
9x2+ 16y2 + 24xy
9x2 + 16y2-24xy
12.
Product of 4p, -7q3, -7pq is _________.
196p2q4
196pq4
-196p2q4
196pq4
13.
In a polynomial, the exponents of the variables are always ___________.
integers
positive integers
non negative integers
non-positive integers
14.
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
15.
The missing terms in the product - 3m3n x 9(_) = ___________m4n3 are
mn2, 27
m2n, 27
m2n2, -27
mn2, -27
16.
Factorize
9-a6+2a3-b6
17.
Factoris x2 + 8x + 15
1.
\(\left( -\cfrac { 8 }{ 5 } { x }^{ 2 }{ yz }^{ 3 } \right) \times \left( -\cfrac { 3 }{ 4 } { xy }^{ 2 }z \right) =\left( -\cfrac { 8 }{ 5 } \times -\cfrac { 3 }{ 4 } \right) \times \left( { x }^{ 2 }\times x\times y\times z\times { z }^{ 2 } \right)\)
= \(\cfrac { 6 }{ 5 } { x }^{ 2+1 }{ y }^{ 3+1 }=\cfrac { 6 }{ 5 } { x }^{ 3 }{ y }^{ 2 }{ z }^{ 4 }\)
2.
\(\left( { 4x }^{ 2 }yz \right) \times \left( -\cfrac { 3 }{ 2 } { x }^{ 2 }{ yz }^{ 2 } \right) =\left( { x }^{ 2 }\times { x }^{ 2 }\times y\times y\times z\times { z }^{ 2 } \right)\)
= -6x2+2y1+1z1+2 = -6x4y2z3
3.
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
4.
Comparing (y-5)3 with (a-b)3,we get a = y,b = 5
(a-b)3 = a3- 3a2b + 3ab + 3ab2-b2
(y - 5)3 = (y)3 + 3(y)2(5) + 3(y)(5)2 - (-5)3
= (y)3+ 3y2(5) + 3(y)(25)-125
(y - 5)3 = y3-15y2 + 75y-125
5.
We have, (2x2y2) × (3y2z) × (−z2x3)
= (+) × (+) × (−)(2 × 3 × 1)(x2 × x3)(y2 × y2)(z × z2)
= − 6x5y4z3
6.
\(\cfrac { { 6x }^{ 3 }{ y }^{ 2 }{ z }^{ 2 } }{ { 3x }^{ 2 }yz } =\cfrac { 6 }{ 3 } { x }^{ 3-2 }{ y }^{ 2-1 }{ z }^{ 2-1 }=2xyz\)
7.
4a\(\times\)3a2 = (4\(\times\)3) (a\(\times\)a2) = 12a3
8.
b x (−15m2n3p) = 45m3n3p2
\(\mathrm{b}=\frac{45 m^{3} n^{3} p^{2}}{-15 m^{2} n^{3} p}\)
b = -3mp
9.
5x (2y - 3) = (5x) (2y) - (5x) (3)
= (5 x 2) (x x y) - (5 x 3)x
= 10xy - 15x
10.
(-2m2) \(\times\) (-5m)3 = -2m2 \(\times\) (-)3 (53 (m)3) = -2m2 \(\times\) (-125m3)
= (-), \(\times\) (-) (2 \(\times\) 125) (m2 \(\times\) m3) = + 250m5 = 250 m5
11.
(d)
9x2 + 16y2-24xy
12.
(a)
196p2q4
13.
(c)
non negative integers
14.
(b)
a + b
15.
(d)
mn2, -27
16.
9-a6+2a3-b6 = 9 - (a6 - 2a3b3 + b6),
= 32-{(a3)2-2Xa3Xb3+(b3)2}
= 32-(a3-b3)2
={3 +(a3 - b3)}{3 - (a3 - b3)}
= (3 + a3 - b3) (3 - a3 + b3)
= (a3 - b3 + 3)( - a3 + b3 + 3)
17.
This is in the form of ax2 + bx + c
We get a = 1, b = 8, c = 15
Now, the product = a x c and sum = b
= 1 x 15 b = 8
= x2+ 8x +15
= (x2 + 3x) + (5x + 15) (the middle term 8x can be written as 3x + 5x)
= x(x + 3) + 5(x + 3) (taking out the common factor x + 3 )
x2 + 8x + 15 = (x + 5) (x + 3)
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Tamilnadu Stateboard 8th Standard Subjects
Tamilnadu Stateboard Standards
