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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
Read More
FAQs on Class 9 Math: Solutions of Sample Practice Question Paper- 3
| 1. What are the solutions of a quadratic equation? |  |
Ans. The solutions of a quadratic equation are the values of the variable that make the equation true. They can be found by solving the equation and determining the values of the variable that satisfy the equation.
| 2. How can we solve a quadratic equation graphically? | |
Ans. A quadratic equation can be solved graphically by plotting the equation on a graph and finding the points where the graph intersects the x-axis. These points represent the solutions of the equation.
| 3. What is the discriminant of a quadratic equation? | |
Ans. The discriminant of a quadratic equation is a value that is calculated from the coefficients of the equation. It helps determine the nature of the solutions. If the discriminant is positive, the equation has two distinct real solutions. If it is zero, the equation has one real solution. And if the discriminant is negative, the equation has no real solutions.
| 4. How can we find the roots of a quadratic equation using the quadratic formula? | |
Ans. The roots of a quadratic equation can be found using the quadratic formula. The formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the roots can be calculated using the formula x = (-b ± √(b^2 - 4ac))/(2a). By substituting the values of a, b, and c from the equation into the formula, we can find the roots.
| 5. Can a quadratic equation have imaginary solutions? | |
Ans. Yes, a quadratic equation can have imaginary solutions. If the discriminant of the equation is negative, it means that the equation has no real solutions but rather two complex solutions. These complex solutions involve the imaginary unit "i" and are of the form a + bi, where a and b are real numbers.
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