9th Standard CBSE Syllabus & Materials
9th Standard CBSE
CBSE 9th Science Is matter around us pure? - New Model Questions Papers Study Material - QB365 Set A
NEW9th Standard CBSE
CBSE 9th Science Matter in our surroundings - New Model Questions Papers Study Material - QB365 Set A
NEW9th Standard CBSE
CBSE 9th Mathematics Heron's Formula Sample Question Papers Study Material - QB365 Set A
NEW9th Standard CBSE
CBSE 9th Mathematics Circles Sample Question Papers Study Material - QB365 Set A
NEW9th Standard CBSE
CBSE 9th Mathematics Quadrilaterals Sample Question Papers Study Material - QB365 Set A
NEW9th Standard CBSE
CBSE 9th Mathematics Triangles Sample Question Papers Study Material - QB365 Set A
Published on: 06/07/2018
Dear Students,
Please refer the questions paper
Download CBSE Class 9th Standard CBSE Mathematics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 9th Standard CBSE Mathematics
Questions + Answers key
Take MCQ Mathematics Test
1.
Factorise each of the following: \(64m^3-343n^3\)
2.
Factorise each of the following: \(27y3+125z^3\)
3.
Write the following cubes in expanded from: \({ \left[ x-\frac { 2 }{ 3 } y \right] }^{ 3 }\)
4.
Write the following cubes in expanded from: \((2a-3b)^3\)
5.
Write the following cubes in expanded from: \((2x+1)^3\)
6.
Simplify: \((\frac{x}{3}+\frac{y}{5})^3-(\frac{x}{3}-\frac{y}{5})^3\)
7.
Expand \( (3x-\frac{1}{2}y+2z)^2\)
8.
Factorise: \(a^{12}y^4-a^4y^{12}.\)
9.
Factorise: \(216x^3-125y^3\)
10.
Factorise: \(x^4-125xy^3\)
11.
Factorise: \(16x^3-2y^3\)
12.
Factorise: \(x^4-y^4\)
13.
Factorise: \(216x^3+\frac{1}{125}\)
14.
Factorise:
(i) \(49a^2+70ab+25b^2\)
(ii) \(\frac{25}{4}x^2-\frac{y^2}{9}.\)
15.
Write the following cubes in the expanded form:
(i) \((3a+4b)^3\)
(ii) \((5p-3q)^3\)
16.
Expand \((4a-2b-3c)^2\)
17.
Write \((3a+4b+5c)^2\) in expanded form.
1.
\(64m^3-343n^3\)
\(64m^3-343n^3=(4m)^3-(7n)^3\)
\(=(4m-7n)\left\{ (4m^{ 2 })(7n)+(7n)^{ 2 } \right\} \)
\(=(4m-7n)(16m^2+28mn+49n^2)\)
2.
\(27y3+125z^3\)
\(27y3+125z^3=(3y)^3+(5z)^3\)
\(=(3y+5z)\left\{ (3y)^{ 2 }-(3y)(5z)+(5z)^{ 3 } \right\} \)
\(=(3y+5z)(9y^2-15yz+25^2)\)
3.
\({ \left[ x-\frac { 2 }{ 3 } y \right] }^{ 3 }\)
\(={ x }^{ 3 }-{ \left( \frac { 2 }{ 3 } y \right) }^{ 3 }-3(x)\left( \frac { 2 }{ 3 } y \right) \left( x-\frac { 2 }{ 3 } y \right) \) | Using Identity VII
\(={ x }^{ 3 }-\frac { 8 }{ 27 } { y }^{ 3 }-2xy\left( x-\frac { 2 }{ 3 } y \right) \)
\(={ x }^{ 3 }-\frac { 8 }{ 27 } { y }^{ 3 }-2{ x }^{ 2 }y+\frac { 4 }{ 3 } { xy }^{ 2 }\)
\(={ x }^{ 3 }-2{ x }^{ 2 }y+\frac { 4 }{ 3 } { xy }^{ 2 }-\frac { 8 }{ 27 } { y }^{ 3 }.\)
4.
\((2a-3b)^3\)
\({ (2a-3b) }^{ 3 }={ (2a) }^{ 3 }-{ (3b) }^{ 3 }-2(2a)(3b)(2a-3b)\) | Using Identity VII
\(=8{ a }^{ 3 }-27{ b }^{ 3 }-18ab(2a-3b)\)
\(=8{ a }^{ 3 }-27{ b }^{ 3 }-36{ a }^{ 2 }b+54{ ab }^{ 2 }\)
5.
\((2x+1)^3\)
\({ (2x+1) }^{ 2 }={ (2x) }^{ 3 }+{ (1) }^{ 3 }+3(2x)(1)(2x+1)\) | Using Identity VI
\(=8{ x }^{ 3 }+1+6(2x+1)\)
\(=8{ x }^{ 3 }+1+12{ x }^{ 2 }+6x\)
\(=8{ x }^{ 3 }+12{ x }^{ 2 }+6x+1\)
6.
\(\frac{2y^3}{125}+\frac{2x^2y}{15}\)
7.
\(9x^2+\frac{1}{4}y^2+4z^2-3xy-2yz+12zx\)
8.
\(a^4y^4(a^2+y^2)(a-y)(a+y)(a^2+y^2-\sqrt{2}ay(a^2+y^2+\sqrt{2}ay)\)
9.
\((6x-5y)(36x^2+30xy+25y^2)\)
10.
\(x(x-5y)(x^2+5xy+25y^2)\)
11.
\(2(2x-y)(4x^2+2xy+y^2)\)
12.
\((x-y)(x+y)(x^2+y^2)\)
13.
\((6x+\frac{1}{5})(36x^2-\frac{6x}{5}+\frac{1}{25})\)
14.
(i) \((7a+5b)^2\)
(ii) \((\frac{5}{2}x+\frac{y}{3})(\frac{5}{2}x-\frac{y}{3})\)
15.
(i) Comparing the given expression with (x + y)3, we find that
x = 3a and y = 4b.
So, using Identity VI, we have:
(3a + 4b)3 = (3a)3 + (4b)3 + 3(3a)(4b)(3a + 4b)
= 27a3 + 64b3 + 108a2b + 144ab2
(ii) Comparing the given expression with (x – y)3, we find that
x = 5p, y = 3q.
So, using Identity VII, we have:
(5p – 3q)3 = (5p)3 – (3q)3 – 3(5p)(3q)(5p – 3q)
= 125p3 – 27q3 – 225p2q + 135pq2
16.
Using Identity V, we have
(4a – 2b – 3c)2 = [4a + (–2b) + (–3c)]2
= (4a)2 + (–2b)2 + (–3c)2 + 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a)
= 16a2 + 4b2 + 9c2 – 16ab + 12bc – 24ac
17.
Comparing the given expression with (x + y + z)2, we find that
x = 3a, y = 4b and z = 5c.
Therefore, using Identity V, we have
(3a + 4b + 5c)2 = (3a)2 + (4b)2 + (5c)2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)
= 9a2 + 16b2 + 25c2 + 24ab + 40bc + 30ac
9th Standard CBSE Syllabus & Materials
9th Standard CBSE
CBSE 9th Mathematics Lines and Angles Sample Question Papers Study Material - QB365 Set A
NEW9th Standard CBSE
CBSE 9th Mathematics Introduction to Euclid's Geometry Sample Question Papers Study Material - QB365 Set A
NEW9th Standard CBSE
CBSE 9th Mathematics Linear Equations in Two Variables Sample Question Papers Study Material - QB365 Set A
NEW9th Standard CBSE
CBSE 9th Mathematics Coordinate Geometry Sample Question Papers Study Material - QB365 Set A
CBSE 9th Standard CBSE Subjects
CBSE Standards
