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Sample Papers For Class 9

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 Page 1


  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
CBSE Board 
Class IX Mathematics 
Sample Paper 3 – Solution 
Time: 3 hrs  
Total Marks: 80 
      
Section A 
1.  
? ? ? ? ? ?
? ?
2 2 2
2
22
L.H.S 5 6 5 2 5 6 5 a b a 2ab b
5 2 30 6
11 2 30
On compairing with R.H.S,
a + b 30  = 11 + 2 30
a 11 and b 2
? ? ? ? ? ? ? ? ?
? ? ?
??
? ? ?
 
 
2. Since, p(x) is a zero polynomial, 
? p(x) = 0 
? cx + d = 0 
? cx = –d 
? x   = 
d
c
?
 
 
3. In ?PTQ , base = PQ = 6 cm and height = 6 cm                                                           
? Area of ?PTQ = 
1
2
?  base ? height = 
1
2
 × 6 × 6 = 18 cm
2 
OR 
 
  ar ( ? APB) = 
1
2
 x ar(parallelogram ABCD) 
      (The area of a triangle is half that of a parallelogram on the same base and between         
the same parallels) 
 ar( ? APB) = 
1
2
 x 60 cm
2 
= 30 cm
2
 
Page 2


  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
CBSE Board 
Class IX Mathematics 
Sample Paper 3 – Solution 
Time: 3 hrs  
Total Marks: 80 
      
Section A 
1.  
? ? ? ? ? ?
? ?
2 2 2
2
22
L.H.S 5 6 5 2 5 6 5 a b a 2ab b
5 2 30 6
11 2 30
On compairing with R.H.S,
a + b 30  = 11 + 2 30
a 11 and b 2
? ? ? ? ? ? ? ? ?
? ? ?
??
? ? ?
 
 
2. Since, p(x) is a zero polynomial, 
? p(x) = 0 
? cx + d = 0 
? cx = –d 
? x   = 
d
c
?
 
 
3. In ?PTQ , base = PQ = 6 cm and height = 6 cm                                                           
? Area of ?PTQ = 
1
2
?  base ? height = 
1
2
 × 6 × 6 = 18 cm
2 
OR 
 
  ar ( ? APB) = 
1
2
 x ar(parallelogram ABCD) 
      (The area of a triangle is half that of a parallelogram on the same base and between         
the same parallels) 
 ar( ? APB) = 
1
2
 x 60 cm
2 
= 30 cm
2
 
  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
4. Substituting x = 
1
2
 and y = 0 in the equation 2x + y = 1 
? 2 ? 
1
2
+ 0 = 1 
? 1 = 1 
Since, L.H.S = R.H.S 
The values of x and y are satisfying the given equation. 
Therefore, (
1
2
, 0) is the solution of 2x + y = 1. 
OR 
(4, 19) is a solution of the equation y = ax + 3 then (4, 19) must satisfy the equation. 
? 19 = 4x + 3 
? 4a = 16 
? a = 4 
 
5. Median of a Triangle: A median of a triangle is the line segment that joins any vertex 
of the triangle with the midpoint of its opposite side. In the given Triangle, the Median 
from A meets the midpoint of the opposite side, BC at point D. 
 
 
 
6. In a parallelogram, the sum of consecutive angles are Supplementary. 
Here ABCD is a parallelogram,  
? ?A + ?B = 180°   
o
11
A B 90
22
? ? ? ? ? 
o
o
oo
o
In  AOB,
AOB OBA OAB 180
BA
AOB 180 ....( OA and OB are the angle bisectors of A and B)
22
AOB 90 180
AOB 90
?
? ? ? ? ? ?
??
? ? ? ? ? ? ?
? ? ? ?
? ? ?
 
 
 
 
 
 
Page 3


  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
CBSE Board 
Class IX Mathematics 
Sample Paper 3 – Solution 
Time: 3 hrs  
Total Marks: 80 
      
Section A 
1.  
? ? ? ? ? ?
? ?
2 2 2
2
22
L.H.S 5 6 5 2 5 6 5 a b a 2ab b
5 2 30 6
11 2 30
On compairing with R.H.S,
a + b 30  = 11 + 2 30
a 11 and b 2
? ? ? ? ? ? ? ? ?
? ? ?
??
? ? ?
 
 
2. Since, p(x) is a zero polynomial, 
? p(x) = 0 
? cx + d = 0 
? cx = –d 
? x   = 
d
c
?
 
 
3. In ?PTQ , base = PQ = 6 cm and height = 6 cm                                                           
? Area of ?PTQ = 
1
2
?  base ? height = 
1
2
 × 6 × 6 = 18 cm
2 
OR 
 
  ar ( ? APB) = 
1
2
 x ar(parallelogram ABCD) 
      (The area of a triangle is half that of a parallelogram on the same base and between         
the same parallels) 
 ar( ? APB) = 
1
2
 x 60 cm
2 
= 30 cm
2
 
  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
4. Substituting x = 
1
2
 and y = 0 in the equation 2x + y = 1 
? 2 ? 
1
2
+ 0 = 1 
? 1 = 1 
Since, L.H.S = R.H.S 
The values of x and y are satisfying the given equation. 
Therefore, (
1
2
, 0) is the solution of 2x + y = 1. 
OR 
(4, 19) is a solution of the equation y = ax + 3 then (4, 19) must satisfy the equation. 
? 19 = 4x + 3 
? 4a = 16 
? a = 4 
 
5. Median of a Triangle: A median of a triangle is the line segment that joins any vertex 
of the triangle with the midpoint of its opposite side. In the given Triangle, the Median 
from A meets the midpoint of the opposite side, BC at point D. 
 
 
 
6. In a parallelogram, the sum of consecutive angles are Supplementary. 
Here ABCD is a parallelogram,  
? ?A + ?B = 180°   
o
11
A B 90
22
? ? ? ? ? 
o
o
oo
o
In  AOB,
AOB OBA OAB 180
BA
AOB 180 ....( OA and OB are the angle bisectors of A and B)
22
AOB 90 180
AOB 90
?
? ? ? ? ? ?
??
? ? ? ? ? ? ?
? ? ? ?
? ? ?
 
 
 
 
 
 
  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
Section B 
 
7. Let x = 0.975 0.975975975 ....(1) ? 
 On multiplying both sides of equation (1) by 1000: 
 1000x = 975.975975    ....(2) 
 On subtracting equation (1) from equation (2), 
 999x = 975 
 ? ? ?
975 325
x
999 333
 
 
8.  
2
2
2
2
2
12
x 2 2x
xx
11
x 2 2 x
xx
11
x 2 x
xx
11
x x 2
xx
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 
 
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° 
Page 4


  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
CBSE Board 
Class IX Mathematics 
Sample Paper 3 – Solution 
Time: 3 hrs  
Total Marks: 80 
      
Section A 
1.  
? ? ? ? ? ?
? ?
2 2 2
2
22
L.H.S 5 6 5 2 5 6 5 a b a 2ab b
5 2 30 6
11 2 30
On compairing with R.H.S,
a + b 30  = 11 + 2 30
a 11 and b 2
? ? ? ? ? ? ? ? ?
? ? ?
??
? ? ?
 
 
2. Since, p(x) is a zero polynomial, 
? p(x) = 0 
? cx + d = 0 
? cx = –d 
? x   = 
d
c
?
 
 
3. In ?PTQ , base = PQ = 6 cm and height = 6 cm                                                           
? Area of ?PTQ = 
1
2
?  base ? height = 
1
2
 × 6 × 6 = 18 cm
2 
OR 
 
  ar ( ? APB) = 
1
2
 x ar(parallelogram ABCD) 
      (The area of a triangle is half that of a parallelogram on the same base and between         
the same parallels) 
 ar( ? APB) = 
1
2
 x 60 cm
2 
= 30 cm
2
 
  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
4. Substituting x = 
1
2
 and y = 0 in the equation 2x + y = 1 
? 2 ? 
1
2
+ 0 = 1 
? 1 = 1 
Since, L.H.S = R.H.S 
The values of x and y are satisfying the given equation. 
Therefore, (
1
2
, 0) is the solution of 2x + y = 1. 
OR 
(4, 19) is a solution of the equation y = ax + 3 then (4, 19) must satisfy the equation. 
? 19 = 4x + 3 
? 4a = 16 
? a = 4 
 
5. Median of a Triangle: A median of a triangle is the line segment that joins any vertex 
of the triangle with the midpoint of its opposite side. In the given Triangle, the Median 
from A meets the midpoint of the opposite side, BC at point D. 
 
 
 
6. In a parallelogram, the sum of consecutive angles are Supplementary. 
Here ABCD is a parallelogram,  
? ?A + ?B = 180°   
o
11
A B 90
22
? ? ? ? ? 
o
o
oo
o
In  AOB,
AOB OBA OAB 180
BA
AOB 180 ....( OA and OB are the angle bisectors of A and B)
22
AOB 90 180
AOB 90
?
? ? ? ? ? ?
??
? ? ? ? ? ? ?
? ? ? ?
? ? ?
 
 
 
 
 
 
  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
Section B 
 
7. Let x = 0.975 0.975975975 ....(1) ? 
 On multiplying both sides of equation (1) by 1000: 
 1000x = 975.975975    ....(2) 
 On subtracting equation (1) from equation (2), 
 999x = 975 
 ? ? ?
975 325
x
999 333
 
 
8.  
2
2
2
2
2
12
x 2 2x
xx
11
x 2 2 x
xx
11
x 2 x
xx
11
x x 2
xx
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 
 
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° 
  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
 
 
10. AD is the bisector of  ?A  
? ?BAD = ?CAD    
Exterior ?BDA > ?CAD  
? ?BDA > ?BAD  
? AB > BD         (side opposite the bigger angle is longer) 
 
OR 
In ?PQT, we have 
 PT = PQ  … (1) 
 In ?PQR, 
 PQ + QR > PR 
 PQ + QR > PT + TR 
 PQ + QR > PQ + TR [Using (1)] 
 QR > TR 
Hence, proved.  
 
11. 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 
OR 
  Given that: 
          Diagonal of a cube = 48 cm 
       
        i.e.,              3 x l = 48    [ Diagonal of cube = 3 x l ] 
                    l =  
48
3
 
                    l = 
48
3
 
                                      
                      = 16 
                      = 4 cm ?     Side (l)   =   4 cm 
Page 5


  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
CBSE Board 
Class IX Mathematics 
Sample Paper 3 – Solution 
Time: 3 hrs  
Total Marks: 80 
      
Section A 
1.  
? ? ? ? ? ?
? ?
2 2 2
2
22
L.H.S 5 6 5 2 5 6 5 a b a 2ab b
5 2 30 6
11 2 30
On compairing with R.H.S,
a + b 30  = 11 + 2 30
a 11 and b 2
? ? ? ? ? ? ? ? ?
? ? ?
??
? ? ?
 
 
2. Since, p(x) is a zero polynomial, 
? p(x) = 0 
? cx + d = 0 
? cx = –d 
? x   = 
d
c
?
 
 
3. In ?PTQ , base = PQ = 6 cm and height = 6 cm                                                           
? Area of ?PTQ = 
1
2
?  base ? height = 
1
2
 × 6 × 6 = 18 cm
2 
OR 
 
  ar ( ? APB) = 
1
2
 x ar(parallelogram ABCD) 
      (The area of a triangle is half that of a parallelogram on the same base and between         
the same parallels) 
 ar( ? APB) = 
1
2
 x 60 cm
2 
= 30 cm
2
 
  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
4. Substituting x = 
1
2
 and y = 0 in the equation 2x + y = 1 
? 2 ? 
1
2
+ 0 = 1 
? 1 = 1 
Since, L.H.S = R.H.S 
The values of x and y are satisfying the given equation. 
Therefore, (
1
2
, 0) is the solution of 2x + y = 1. 
OR 
(4, 19) is a solution of the equation y = ax + 3 then (4, 19) must satisfy the equation. 
? 19 = 4x + 3 
? 4a = 16 
? a = 4 
 
5. Median of a Triangle: A median of a triangle is the line segment that joins any vertex 
of the triangle with the midpoint of its opposite side. In the given Triangle, the Median 
from A meets the midpoint of the opposite side, BC at point D. 
 
 
 
6. In a parallelogram, the sum of consecutive angles are Supplementary. 
Here ABCD is a parallelogram,  
? ?A + ?B = 180°   
o
11
A B 90
22
? ? ? ? ? 
o
o
oo
o
In  AOB,
AOB OBA OAB 180
BA
AOB 180 ....( OA and OB are the angle bisectors of A and B)
22
AOB 90 180
AOB 90
?
? ? ? ? ? ?
??
? ? ? ? ? ? ?
? ? ? ?
? ? ?
 
 
 
 
 
 
  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
Section B 
 
7. Let x = 0.975 0.975975975 ....(1) ? 
 On multiplying both sides of equation (1) by 1000: 
 1000x = 975.975975    ....(2) 
 On subtracting equation (1) from equation (2), 
 999x = 975 
 ? ? ?
975 325
x
999 333
 
 
8.  
2
2
2
2
2
12
x 2 2x
xx
11
x 2 2 x
xx
11
x 2 x
xx
11
x x 2
xx
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 
 
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° 
  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
 
 
10. AD is the bisector of  ?A  
? ?BAD = ?CAD    
Exterior ?BDA > ?CAD  
? ?BDA > ?BAD  
? AB > BD         (side opposite the bigger angle is longer) 
 
OR 
In ?PQT, we have 
 PT = PQ  … (1) 
 In ?PQR, 
 PQ + QR > PR 
 PQ + QR > PT + TR 
 PQ + QR > PQ + TR [Using (1)] 
 QR > TR 
Hence, proved.  
 
11. 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 
OR 
  Given that: 
          Diagonal of a cube = 48 cm 
       
        i.e.,              3 x l = 48    [ Diagonal of cube = 3 x l ] 
                    l =  
48
3
 
                    l = 
48
3
 
                                      
                      = 16 
                      = 4 cm ?     Side (l)   =   4 cm 
  
 
CBSE IX | Mathematics 
Sample Paper 3 – Solution 
 
     
 
12. Given equation is 7x – 5y = –3 
i. (–1, –2) 
Putting x = –1 and y = –2 in the L.H.S. of the given equation, we get 
7x – 5y = 7(–1) – 5(–2) = –7 + 10 = 3 ? R.H.S. 
? L.H.S.? R.H.S. 
Hence, (–1, –2) is not a solution of this equation. 
 
ii. (–4, –5) 
Putting x = –4 and y = –5 in the L.H.S. of the given equation, we get 
7x – 5y = 7(–4) – 5(–5) = –28 + 25 = –3 = R.H.S. 
? L.H.S. = R.H.S. 
Hence, (–4, –5) is a solution of this equation. 
  
 
Section C 
  
13.   
? ?
?2
3
343 
= (343)
-2/3
 
= [(7)
3
}
-2/3
 
=  
=  = 
1
49
 
OR 
12
23
1
0.01 27
4
 
 = 
1
2
1 2
2
1
0.1 3
2
 
 = 
1 2 1
0.1 3
2
 
 = 
11
9
2 0.1
 
 = 
1 10
9
21
 
 = 
1
1
2
= 
3
2
 
 
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