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## Class 9 : Notes | EduRev

``` 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
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
??
```
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