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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.
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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.
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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.
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The content of Class 9 Math: Solutions of Sample Practice Question Paper- 2 is prepared as per the latest Class 9 syllabus.
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