Class 9 Exam > Class 9 Notes > Class 9 Math: Solutions of Sample Practice Question Paper- 2
Class 9 Math: Solutions of Sample Practice Question Paper- 2 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
CBSE IX | Mathematics
Sample Paper 2 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 2 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
?
6 27 3 3 1 2 3
6 3 3 3 3 1 2 3
4
2. Given polynomial is x
2
– x – 1
Substituting x = –1 in x
2
– x – 1, we have
(–1)
2
– (–1) – 1 = 1 + 1 – 1 = 1
3. Two lines l and m in a plane are said to be parallel lines if they do not have common
point, i.e. they do not intersect.
4. Substituting x = 2 and y = 0 in x – 2y = 4, we get
L.H.S = 2 – 2 × 0 = 2 ? 4
i.e. L.H.S. ? R.H.S
Therefore, (2, 0) is not a solution of x – 2y = 4.
OR
Put y = 0 to find the coordinate of x-axis.
? 2x – y = 4
? 2x = 4
? x = 2
Hence, the graph of the linear equation 2x – y = 4 cut x-axis at (2, 0).
5. ?PSR = ?RQP = 125° (opposite angles will be equal since PQRS is a parallelogram)
?PQT = 180° (PQT is a straight line)
? ?PQR + ?RQT = 180°
? 125° + ?RQT = 180°
? ?RQT = 55°
6. Class size is the difference between two successive class marks.
? Class size = 10 – 6 = 4
OR
According to the given distribution,
Class size = 52 – 47 = 5.
Page 2
CBSE IX | Mathematics
Sample Paper 2 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 2 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
?
6 27 3 3 1 2 3
6 3 3 3 3 1 2 3
4
2. Given polynomial is x
2
– x – 1
Substituting x = –1 in x
2
– x – 1, we have
(–1)
2
– (–1) – 1 = 1 + 1 – 1 = 1
3. Two lines l and m in a plane are said to be parallel lines if they do not have common
point, i.e. they do not intersect.
4. Substituting x = 2 and y = 0 in x – 2y = 4, we get
L.H.S = 2 – 2 × 0 = 2 ? 4
i.e. L.H.S. ? R.H.S
Therefore, (2, 0) is not a solution of x – 2y = 4.
OR
Put y = 0 to find the coordinate of x-axis.
? 2x – y = 4
? 2x = 4
? x = 2
Hence, the graph of the linear equation 2x – y = 4 cut x-axis at (2, 0).
5. ?PSR = ?RQP = 125° (opposite angles will be equal since PQRS is a parallelogram)
?PQT = 180° (PQT is a straight line)
? ?PQR + ?RQT = 180°
? 125° + ?RQT = 180°
? ?RQT = 55°
6. Class size is the difference between two successive class marks.
? Class size = 10 – 6 = 4
OR
According to the given distribution,
Class size = 52 – 47 = 5.
CBSE IX | Mathematics
Sample Paper 2 – Solution
Section B
7. ky
2
– 6ky + 8k
= k(y
2
– 6y + 8)
= k(y
2
– 4y – 2y + 8)
= k(y – 4)(y – 2)
Thus, the dimensions of the cuboid are given by the expressions k, (y – 4) and (y – 2).
8.
9. Here ? ADC = y = ? ACD
Ext. ? ACD = ? ABC + ? BAC
? 2 ? BAC = ? ACD = y
? ? BAC =
y
2
?
y
2
+ (180° – 2y) = 180° – 75°
?
y
2
+ 180° – 2y = 180° – 75
?
y
2
– 2y = – 75°
? –
3y
2
= –75°
? y = 50°
10. Let ‘l’ be the length of the cube.
Now, T.S.A. of the cube = 294 cm
2
….(given)
? 6l
2
= 294
2
294
l 49
6
? ? ?
? Side (l) = 7 cm.
Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm
3
Page 3
CBSE IX | Mathematics
Sample Paper 2 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 2 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
?
6 27 3 3 1 2 3
6 3 3 3 3 1 2 3
4
2. Given polynomial is x
2
– x – 1
Substituting x = –1 in x
2
– x – 1, we have
(–1)
2
– (–1) – 1 = 1 + 1 – 1 = 1
3. Two lines l and m in a plane are said to be parallel lines if they do not have common
point, i.e. they do not intersect.
4. Substituting x = 2 and y = 0 in x – 2y = 4, we get
L.H.S = 2 – 2 × 0 = 2 ? 4
i.e. L.H.S. ? R.H.S
Therefore, (2, 0) is not a solution of x – 2y = 4.
OR
Put y = 0 to find the coordinate of x-axis.
? 2x – y = 4
? 2x = 4
? x = 2
Hence, the graph of the linear equation 2x – y = 4 cut x-axis at (2, 0).
5. ?PSR = ?RQP = 125° (opposite angles will be equal since PQRS is a parallelogram)
?PQT = 180° (PQT is a straight line)
? ?PQR + ?RQT = 180°
? 125° + ?RQT = 180°
? ?RQT = 55°
6. Class size is the difference between two successive class marks.
? Class size = 10 – 6 = 4
OR
According to the given distribution,
Class size = 52 – 47 = 5.
CBSE IX | Mathematics
Sample Paper 2 – Solution
Section B
7. ky
2
– 6ky + 8k
= k(y
2
– 6y + 8)
= k(y
2
– 4y – 2y + 8)
= k(y – 4)(y – 2)
Thus, the dimensions of the cuboid are given by the expressions k, (y – 4) and (y – 2).
8.
9. Here ? ADC = y = ? ACD
Ext. ? ACD = ? ABC + ? BAC
? 2 ? BAC = ? ACD = y
? ? BAC =
y
2
?
y
2
+ (180° – 2y) = 180° – 75°
?
y
2
+ 180° – 2y = 180° – 75
?
y
2
– 2y = – 75°
? –
3y
2
= –75°
? y = 50°
10. Let ‘l’ be the length of the cube.
Now, T.S.A. of the cube = 294 cm
2
….(given)
? 6l
2
= 294
2
294
l 49
6
? ? ?
? Side (l) = 7 cm.
Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm
3
CBSE IX | Mathematics
Sample Paper 2 – Solution
OR
Given that:
Side (S) of metal cube = 4 cm
Length (l) of the tank = 8 cm
Breadth (b) of the tank = 4 cm
Since the volume of cube immersed = volume of the cuboidal tank
We get,
S x S x S = l x b x h
4
3
= 8 x 4 x h
h =
64
32
? h = 2 cm
Thus, rise in water level = 2 cm
11. Number of students born in August = 6
Total number of students = 40
Number of students born in August 6 3
Required probability=
Total number of students 40 20
??
12. Let the angles of a quadrilateral be 2x, 5x, 8x and 9x respectively.
By the angle sum property of a quadrilateral, we have
2x + 5x + 8x + 9x = 360°
? 24x = 360°
? x = 15°
Now,
First angle = 2x = 2 × 15 = 30°,
Second angle = 5x = 5 × 15 = 75°,
Third angle = 8x = 8 × 15 = 120° and
Fourth angle = 9x = 9 × 15 = 135°.
Thus, the angles of a quadrilateral are 30°, 75°, 120° and 135°.
OR
Let x be the fourth angle of a quadrilateral.
According to the question,
The sum of the angles of a quadrilateral is 360°.
? 50° + 110° + 40° + x = 360°
?200° + x = 360°
? x = 160°
Hence, the fourth angle of a quadrilateral is 160°.
Page 4
CBSE IX | Mathematics
Sample Paper 2 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 2 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
?
6 27 3 3 1 2 3
6 3 3 3 3 1 2 3
4
2. Given polynomial is x
2
– x – 1
Substituting x = –1 in x
2
– x – 1, we have
(–1)
2
– (–1) – 1 = 1 + 1 – 1 = 1
3. Two lines l and m in a plane are said to be parallel lines if they do not have common
point, i.e. they do not intersect.
4. Substituting x = 2 and y = 0 in x – 2y = 4, we get
L.H.S = 2 – 2 × 0 = 2 ? 4
i.e. L.H.S. ? R.H.S
Therefore, (2, 0) is not a solution of x – 2y = 4.
OR
Put y = 0 to find the coordinate of x-axis.
? 2x – y = 4
? 2x = 4
? x = 2
Hence, the graph of the linear equation 2x – y = 4 cut x-axis at (2, 0).
5. ?PSR = ?RQP = 125° (opposite angles will be equal since PQRS is a parallelogram)
?PQT = 180° (PQT is a straight line)
? ?PQR + ?RQT = 180°
? 125° + ?RQT = 180°
? ?RQT = 55°
6. Class size is the difference between two successive class marks.
? Class size = 10 – 6 = 4
OR
According to the given distribution,
Class size = 52 – 47 = 5.
CBSE IX | Mathematics
Sample Paper 2 – Solution
Section B
7. ky
2
– 6ky + 8k
= k(y
2
– 6y + 8)
= k(y
2
– 4y – 2y + 8)
= k(y – 4)(y – 2)
Thus, the dimensions of the cuboid are given by the expressions k, (y – 4) and (y – 2).
8.
9. Here ? ADC = y = ? ACD
Ext. ? ACD = ? ABC + ? BAC
? 2 ? BAC = ? ACD = y
? ? BAC =
y
2
?
y
2
+ (180° – 2y) = 180° – 75°
?
y
2
+ 180° – 2y = 180° – 75
?
y
2
– 2y = – 75°
? –
3y
2
= –75°
? y = 50°
10. Let ‘l’ be the length of the cube.
Now, T.S.A. of the cube = 294 cm
2
….(given)
? 6l
2
= 294
2
294
l 49
6
? ? ?
? Side (l) = 7 cm.
Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm
3
CBSE IX | Mathematics
Sample Paper 2 – Solution
OR
Given that:
Side (S) of metal cube = 4 cm
Length (l) of the tank = 8 cm
Breadth (b) of the tank = 4 cm
Since the volume of cube immersed = volume of the cuboidal tank
We get,
S x S x S = l x b x h
4
3
= 8 x 4 x h
h =
64
32
? h = 2 cm
Thus, rise in water level = 2 cm
11. Number of students born in August = 6
Total number of students = 40
Number of students born in August 6 3
Required probability=
Total number of students 40 20
??
12. Let the angles of a quadrilateral be 2x, 5x, 8x and 9x respectively.
By the angle sum property of a quadrilateral, we have
2x + 5x + 8x + 9x = 360°
? 24x = 360°
? x = 15°
Now,
First angle = 2x = 2 × 15 = 30°,
Second angle = 5x = 5 × 15 = 75°,
Third angle = 8x = 8 × 15 = 120° and
Fourth angle = 9x = 9 × 15 = 135°.
Thus, the angles of a quadrilateral are 30°, 75°, 120° and 135°.
OR
Let x be the fourth angle of a quadrilateral.
According to the question,
The sum of the angles of a quadrilateral is 360°.
? 50° + 110° + 40° + x = 360°
?200° + x = 360°
? x = 160°
Hence, the fourth angle of a quadrilateral is 160°.
CBSE IX | Mathematics
Sample Paper 2 – Solution
Section C
13. Given: a = 3 + b
? a – b = 3
Applying the cubic identity on both the sides, we have
(a – b)
3
= 3
3
? a
3
– b
3
– 3(a)(b)(a – b) = 27
? a
3
– b
3
– 3ab(3) = 27 (? a – b = 3)
? a
3
– b
3
– 9ab = 27
OR
(3x + 2y)
2
= 12
2
9x
2
+ 12xy + 4y
2
= 144
9x
2
+ 4y
2
+ 12xy = 144
9x
2
+ 4y
2
+ 12 × 6 = 144 ? xy = 6
9x
2
+ 4y
2
= 144 – 72
9x
2
+ 4y
2
= 72
14. Since AB ? DC,
?x = 30° [Alternate angles]
In ?ABD,
80° + 30° + ?y = 180°
? ?y = 180° – 110° = 70°
In ?BDC,
30° + (70° – 30°) + ?z = 180°
? ?z = 110°
OR
Since, PQRS is a square. PS = SR and ? PSR = 90°.
In ?PSR,
PS = SR
? SRP = ?QRP angles opposite to equal sides
? SRP + ?QRP + ? PSR = 180°
2?SRP + 90° = 180°
?SRP = 45°
Page 5
CBSE IX | Mathematics
Sample Paper 2 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 2 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
?
6 27 3 3 1 2 3
6 3 3 3 3 1 2 3
4
2. Given polynomial is x
2
– x – 1
Substituting x = –1 in x
2
– x – 1, we have
(–1)
2
– (–1) – 1 = 1 + 1 – 1 = 1
3. Two lines l and m in a plane are said to be parallel lines if they do not have common
point, i.e. they do not intersect.
4. Substituting x = 2 and y = 0 in x – 2y = 4, we get
L.H.S = 2 – 2 × 0 = 2 ? 4
i.e. L.H.S. ? R.H.S
Therefore, (2, 0) is not a solution of x – 2y = 4.
OR
Put y = 0 to find the coordinate of x-axis.
? 2x – y = 4
? 2x = 4
? x = 2
Hence, the graph of the linear equation 2x – y = 4 cut x-axis at (2, 0).
5. ?PSR = ?RQP = 125° (opposite angles will be equal since PQRS is a parallelogram)
?PQT = 180° (PQT is a straight line)
? ?PQR + ?RQT = 180°
? 125° + ?RQT = 180°
? ?RQT = 55°
6. Class size is the difference between two successive class marks.
? Class size = 10 – 6 = 4
OR
According to the given distribution,
Class size = 52 – 47 = 5.
CBSE IX | Mathematics
Sample Paper 2 – Solution
Section B
7. ky
2
– 6ky + 8k
= k(y
2
– 6y + 8)
= k(y
2
– 4y – 2y + 8)
= k(y – 4)(y – 2)
Thus, the dimensions of the cuboid are given by the expressions k, (y – 4) and (y – 2).
8.
9. Here ? ADC = y = ? ACD
Ext. ? ACD = ? ABC + ? BAC
? 2 ? BAC = ? ACD = y
? ? BAC =
y
2
?
y
2
+ (180° – 2y) = 180° – 75°
?
y
2
+ 180° – 2y = 180° – 75
?
y
2
– 2y = – 75°
? –
3y
2
= –75°
? y = 50°
10. Let ‘l’ be the length of the cube.
Now, T.S.A. of the cube = 294 cm
2
….(given)
? 6l
2
= 294
2
294
l 49
6
? ? ?
? Side (l) = 7 cm.
Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm
3
CBSE IX | Mathematics
Sample Paper 2 – Solution
OR
Given that:
Side (S) of metal cube = 4 cm
Length (l) of the tank = 8 cm
Breadth (b) of the tank = 4 cm
Since the volume of cube immersed = volume of the cuboidal tank
We get,
S x S x S = l x b x h
4
3
= 8 x 4 x h
h =
64
32
? h = 2 cm
Thus, rise in water level = 2 cm
11. Number of students born in August = 6
Total number of students = 40
Number of students born in August 6 3
Required probability=
Total number of students 40 20
??
12. Let the angles of a quadrilateral be 2x, 5x, 8x and 9x respectively.
By the angle sum property of a quadrilateral, we have
2x + 5x + 8x + 9x = 360°
? 24x = 360°
? x = 15°
Now,
First angle = 2x = 2 × 15 = 30°,
Second angle = 5x = 5 × 15 = 75°,
Third angle = 8x = 8 × 15 = 120° and
Fourth angle = 9x = 9 × 15 = 135°.
Thus, the angles of a quadrilateral are 30°, 75°, 120° and 135°.
OR
Let x be the fourth angle of a quadrilateral.
According to the question,
The sum of the angles of a quadrilateral is 360°.
? 50° + 110° + 40° + x = 360°
?200° + x = 360°
? x = 160°
Hence, the fourth angle of a quadrilateral is 160°.
CBSE IX | Mathematics
Sample Paper 2 – Solution
Section C
13. Given: a = 3 + b
? a – b = 3
Applying the cubic identity on both the sides, we have
(a – b)
3
= 3
3
? a
3
– b
3
– 3(a)(b)(a – b) = 27
? a
3
– b
3
– 3ab(3) = 27 (? a – b = 3)
? a
3
– b
3
– 9ab = 27
OR
(3x + 2y)
2
= 12
2
9x
2
+ 12xy + 4y
2
= 144
9x
2
+ 4y
2
+ 12xy = 144
9x
2
+ 4y
2
+ 12 × 6 = 144 ? xy = 6
9x
2
+ 4y
2
= 144 – 72
9x
2
+ 4y
2
= 72
14. Since AB ? DC,
?x = 30° [Alternate angles]
In ?ABD,
80° + 30° + ?y = 180°
? ?y = 180° – 110° = 70°
In ?BDC,
30° + (70° – 30°) + ?z = 180°
? ?z = 110°
OR
Since, PQRS is a square. PS = SR and ? PSR = 90°.
In ?PSR,
PS = SR
? SRP = ?QRP angles opposite to equal sides
? SRP + ?QRP + ? PSR = 180°
2?SRP + 90° = 180°
?SRP = 45°
CBSE IX | Mathematics
Sample Paper 2 – Solution
15. 27p
3
+ 8q
3
+ 54p
2
q + 36pq
2
= (3p)
3
+ (2q)
3
+ 18pq(3p+2q)
= (3p)
3
+ (2q)
3
+ 3 × 3p × 2q (3p + 2q)
= (3p + 2q)
3
[(a + b)
3
= a
3
+ b
3
+ 3ab (a + b) [where a = 3p and b = 2q ]
= (3p + 2q) (3p + 2q) (3p + 2q)
16. b
2
+ c
2
+ 2(ab + bc + ca)
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ca – a
2
[Adding and subtracting a
2
]
= [a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ca] – a
2
= (a + b + c)
2
– (a)
2
[Using x
2
+ y
2
+ 2xy + 2yz + 2zx = (x + y + z)
2
]
= (a + b + c + a)(a + b + c – a) [Because a
2
– b
2
= (a+ b) (a – b)]
= (2a + b +c )(b + c)
17. 2x = z (Alternate angles, as l1 || l2)
y = z (Alternate angles, as a1 || a2)
So, 2x = y
Now, y + 4x – 15 = 180° (linear pair)
2x + 4x – 15 = 180°
? 6x = 195°
? x = 32.5
18. Number of white balls = x
Total number of balls = 12
If 6 white balls are added, we have
Total number of balls = 18
Number of white balls = x + 6
x6
Now, P(getting a white ball)
18
?
?
According to the given information,
? ??
?
??
??
?
??
? ? ?
??
??
x 6 x
2
18 12
x 6 x
18 6
6x 36 18x
12x 36
x3
x
P(white ball)
12
??
Read More
FAQs on Class 9 Math: Solutions of Sample Practice Question Paper- 2
| 1. What are the solutions of a quadratic equation? |  |
Ans. The solutions of a quadratic equation are the values of the variable that satisfy the equation and make it true. In other words, they are the values of the variable that make both sides of the equation equal.
| 2. How can I find the solutions of a quadratic equation? | |
Ans. To find the solutions of a quadratic equation, you can use various methods such as factoring, completing the square, or using the quadratic formula. These methods allow you to manipulate the equation and solve for the variable.
| 3. What is factoring in quadratic equations? | |
Ans. Factoring is a method used to break down a quadratic equation into its factors. By factoring, you can rewrite the equation as a product of two binomials, which makes it easier to find the solutions. Factoring involves finding two numbers that multiply to give the constant term and add up to give the coefficient of the linear term.
| 4. What is completing the square in quadratic equations? | |
Ans. Completing the square is a technique used to convert a quadratic equation into a perfect square trinomial. By adding and subtracting a constant term, you can rewrite the equation in a form that can be easily solved by taking the square root. Completing the square is useful when factoring is not possible.
| 5. What is the quadratic formula? | |
Ans. The quadratic formula is a formula that gives the solutions of any quadratic equation. It states that for a quadratic equation in the form ax^2 + bx + c = 0, the solutions can be found using the formula x = (-b ± √(b^2 - 4ac)) / 2a. By substituting the values of a, b, and c into the formula, you can calculate the solutions of the equation.
Related Exams
About this Document
|
4.73/5 Rating |
|
Oct 25, 2024 Last updated |
Document Description: Class 9 Math: Solutions of Sample Practice Question Paper- 2 for Class 9 2024 is part of Class 9 preparation. The notes and questions for Class 9 Math: Solutions of Sample Practice Question Paper- 2 have been prepared according to the Class 9 exam syllabus. Information about Class 9 Math: Solutions of Sample Practice Question Paper- 2 covers topics like and Class 9 Math: Solutions of Sample Practice Question Paper- 2 Example, for Class 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Class 9 Math: Solutions of Sample Practice Question Paper- 2.
Introduction of Class 9 Math: Solutions of Sample Practice Question Paper- 2 in English is available as part of
our Class 9 preparation & Class 9 Math: Solutions of Sample Practice Question Paper- 2 in Hindi for Class 9
courses. Download more important topics, notes, lectures and mock test series for Class 9
Exam by signing up for free. Class 9: Class 9 Math: Solutions of Sample Practice Question Paper- 2
Description
Full syllabus notes, lecture & questions for Class 9 Math: Solutions of Sample Practice Question Paper- 2 - Class 9 | Plus excerises question with solution to help you revise complete syllabus | Best notes, free PDF download
Information about Class 9 Math: Solutions of Sample Practice Question Paper- 2
In this doc you can find the meaning of Class 9 Math: Solutions of Sample Practice Question Paper- 2 defined & explained in the simplest way possible.
Besides explaining types of Class 9 Math: Solutions of Sample Practice Question Paper- 2 theory,
EduRev gives you an ample number of questions to practice Class 9 Math: Solutions of Sample Practice Question Paper- 2 tests, examples and also practice Class 9 tests.
Download as PDF
|
Explore Courses for Class 9 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Class 9 Math: Solutions of Sample Practice Question Paper- 2
,Previous Year Questions with Solutions
,Sample Paper
,practice quizzes
,Viva Questions
,past year papers
,Class 9 Math: Solutions of Sample Practice Question Paper- 2
,Extra Questions
,mock tests for examination
,Exam
,MCQs
,Important questions
,Semester Notes
,shortcuts and tricks
,Summary
,Class 9 Math: Solutions of Sample Practice Question Paper- 2
,ppt
,video lectures
,Objective type Questions
,study material
,Free
;
Additional Information about Class 9 Math: Solutions of Sample Practice Question Paper- 2 for Class 9 Preparation
Class 9 Math: Solutions of Sample Practice Question Paper- 2 Free PDF Download
The Class 9 Math: Solutions of Sample Practice Question Paper- 2 is an invaluable resource that delves deep into the core of the Class 9 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Class 9 Math: Solutions of Sample Practice Question Paper- 2 now and kickstart your journey towards success in the Class 9 exam.
Importance of Class 9 Math: Solutions of Sample Practice Question Paper- 2
The importance of Class 9 Math: Solutions of Sample Practice Question Paper- 2 cannot be overstated, especially for Class 9 aspirants.
This document holds the key to success in the Class 9 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Class 9 Math: Solutions of Sample Practice Question Paper- 2 Notes
Class 9 Math: Solutions of Sample Practice Question Paper- 2 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Class 9 Math: Solutions of Sample Practice Question Paper- 2.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Class 9 Math: Solutions of Sample Practice Question Paper- 2 Notes on EduRev are your ultimate resource for success.
Class 9 Math: Solutions of Sample Practice Question Paper- 2 Class 9 Questions
The "Class 9 Math: Solutions of Sample Practice Question Paper- 2 Class 9 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 9 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Class 9 Math: Solutions of Sample Practice Question Paper- 2 on the App
Students of Class 9 can study Class 9 Math: Solutions of Sample Practice Question Paper- 2 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Class 9 Math: Solutions of Sample Practice Question Paper- 2,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Class 9 Math: Solutions of Sample Practice Question Paper- 2 is prepared as per the latest Class 9 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup