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Sample Papers For Class 9

Created by: Indu Gupta

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
        [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
??
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