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CBSE IX | Mathematics
Sample Paper 3 â€“ Solution

CBSE Board
Class IX Mathematics
Sample Paper 3 â€“ 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 = â€“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 â€“ Solution

CBSE Board
Class IX Mathematics
Sample Paper 3 â€“ 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 = â€“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 â€“ 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 â€“ Solution

CBSE Board
Class IX Mathematics
Sample Paper 3 â€“ 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 = â€“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 â€“ 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 â€“ 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° â€“ 2y) = 180° â€“ 75°
?
y
2
+ 180° â€“ 2y = 180° â€“ 75
?
y
2
â€“ 2y = â€“ 75°
? â€“
3y
2
= â€“75°
? y = 50°
Page 4

CBSE IX | Mathematics
Sample Paper 3 â€“ Solution

CBSE Board
Class IX Mathematics
Sample Paper 3 â€“ 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 = â€“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 â€“ 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 â€“ 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° â€“ 2y) = 180° â€“ 75°
?
y
2
+ 180° â€“ 2y = 180° â€“ 75
?
y
2
â€“ 2y = â€“ 75°
? â€“
3y
2
= â€“75°
? y = 50°

CBSE IX | Mathematics
Sample Paper 3 â€“ Solution

10. AD is the bisector of  ?A
? AB > BD         (side opposite the bigger angle is longer)

OR
In ?PQT, we have
PT = PQ  â€¦ (1)
In ?PQR,
PQ + QR > PR
PQ + QR > PT + TR
PQ + QR > PQ + TR [Using (1)]
QR > TR
Hence, proved.

11. Let â€˜lâ€™ be the length of the cube.
Now, T.S.A. of the cube = 294 cm
2
â€¦(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 â€“ Solution

CBSE Board
Class IX Mathematics
Sample Paper 3 â€“ 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 = â€“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 â€“ 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 â€“ 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° â€“ 2y) = 180° â€“ 75°
?
y
2
+ 180° â€“ 2y = 180° â€“ 75
?
y
2
â€“ 2y = â€“ 75°
? â€“
3y
2
= â€“75°
? y = 50°

CBSE IX | Mathematics
Sample Paper 3 â€“ Solution

10. AD is the bisector of  ?A
? AB > BD         (side opposite the bigger angle is longer)

OR
In ?PQT, we have
PT = PQ  â€¦ (1)
In ?PQR,
PQ + QR > PR
PQ + QR > PT + TR
PQ + QR > PQ + TR [Using (1)]
QR > TR
Hence, proved.

11. Let â€˜lâ€™ be the length of the cube.
Now, T.S.A. of the cube = 294 cm
2
â€¦(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 â€“ Solution

12. Given equation is 7x â€“ 5y = â€“3
i. (â€“1, â€“2)
Putting x = â€“1 and y = â€“2 in the L.H.S. of the given equation, we get
7x â€“ 5y = 7(â€“1) â€“ 5(â€“2) = â€“7 + 10 = 3 ? R.H.S.
? L.H.S.? R.H.S.
Hence, (â€“1, â€“2) is not a solution of this equation.

ii. (â€“4, â€“5)
Putting x = â€“4 and y = â€“5 in the L.H.S. of the given equation, we get
7x â€“ 5y = 7(â€“4) â€“ 5(â€“5) = â€“28 + 25 = â€“3 = R.H.S.
? L.H.S. = R.H.S.
Hence, (â€“4, â€“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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