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CBSE IX | Mathematics
Sample Paper 4 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 4 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
56
0.056
1000
?
OR
4 4 3 4 3
3
3 3 3
2. Linear equation in two variables:
An equation in the form ax + by + c = 0, where a, b and c are real numbers, and a and b
are not both zero is called a linear equation in two variables x and y.
3. If a triangle and a parallelogram are on the same base and between the same parallels,
then the area of the triangle is equal to half the area of the Parallelogram.
? Area of triangle =
1
2
? Area of a Parallelogram
?
Area of triangle
Area of a Parallelogram
=
1
2
Hence, the ratio is 1 : 2.
4. p(x) = x
3
+ 10x
2
+ px
(x – 1) is a factor of p(x).
? x – 1 = 0
? x = 1
Sunstituting x = 1 in p(x) = 0
? 1 + 10 + p = 0
? p = –11
OR
g(x) = vx – 3 = x
1/2
- 3. Here, one term of exponent is ½ which is not an integer. Hence,
it is not a polynomial.
Page 2
CBSE IX | Mathematics
Sample Paper 4 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 4 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
56
0.056
1000
?
OR
4 4 3 4 3
3
3 3 3
2. Linear equation in two variables:
An equation in the form ax + by + c = 0, where a, b and c are real numbers, and a and b
are not both zero is called a linear equation in two variables x and y.
3. If a triangle and a parallelogram are on the same base and between the same parallels,
then the area of the triangle is equal to half the area of the Parallelogram.
? Area of triangle =
1
2
? Area of a Parallelogram
?
Area of triangle
Area of a Parallelogram
=
1
2
Hence, the ratio is 1 : 2.
4. p(x) = x
3
+ 10x
2
+ px
(x – 1) is a factor of p(x).
? x – 1 = 0
? x = 1
Sunstituting x = 1 in p(x) = 0
? 1 + 10 + p = 0
? p = –11
OR
g(x) = vx – 3 = x
1/2
- 3. Here, one term of exponent is ½ which is not an integer. Hence,
it is not a polynomial.
CBSE IX | Mathematics
Sample Paper 4 – Solution
5. Arranging the data in the ascending order: 35, 36, 39, 40, 41, 55, 61, 62, 65, 70, 71,
? Range = Maximum value – Minimum value = 75 – 35 = 40
6. Let the fourth angle of the quadrilateral be x.
Sum of all angles of a quadrilateral = 360°
? 60° + 110° + 86° + x = 360°
? 256° + x = 360°
? x = 104°
Section B
7.
1
2
1
2
12
27
?
?
?
?
1
2
11
22
1
2
34
39
2
3
8. Let f(z) = 3z
3
+ 8z
2
– 1
The possible integral zeros of f(z) are –1 and 1.
f(z) = 3z
3
+ 8z
2
– 1
f(–1) = 3(–1)
3
+ 8(–1)
2
– 1 0
–1 is not a zero of f(z)
f(1) = 3(1)
3
+ 8(1)
2
– 1 0
1 is not a zero of f(z)
Therefore, f(z) has no integral zero.
9. (i) Coordinates of A are (–7, 3)
(ii) Abscissa of point D is 4.
(iii) Point is B.
(iv) Coordinates of C are (–3, –2)
Page 3
CBSE IX | Mathematics
Sample Paper 4 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 4 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
56
0.056
1000
?
OR
4 4 3 4 3
3
3 3 3
2. Linear equation in two variables:
An equation in the form ax + by + c = 0, where a, b and c are real numbers, and a and b
are not both zero is called a linear equation in two variables x and y.
3. If a triangle and a parallelogram are on the same base and between the same parallels,
then the area of the triangle is equal to half the area of the Parallelogram.
? Area of triangle =
1
2
? Area of a Parallelogram
?
Area of triangle
Area of a Parallelogram
=
1
2
Hence, the ratio is 1 : 2.
4. p(x) = x
3
+ 10x
2
+ px
(x – 1) is a factor of p(x).
? x – 1 = 0
? x = 1
Sunstituting x = 1 in p(x) = 0
? 1 + 10 + p = 0
? p = –11
OR
g(x) = vx – 3 = x
1/2
- 3. Here, one term of exponent is ½ which is not an integer. Hence,
it is not a polynomial.
CBSE IX | Mathematics
Sample Paper 4 – Solution
5. Arranging the data in the ascending order: 35, 36, 39, 40, 41, 55, 61, 62, 65, 70, 71,
? Range = Maximum value – Minimum value = 75 – 35 = 40
6. Let the fourth angle of the quadrilateral be x.
Sum of all angles of a quadrilateral = 360°
? 60° + 110° + 86° + x = 360°
? 256° + x = 360°
? x = 104°
Section B
7.
1
2
1
2
12
27
?
?
?
?
1
2
11
22
1
2
34
39
2
3
8. Let f(z) = 3z
3
+ 8z
2
– 1
The possible integral zeros of f(z) are –1 and 1.
f(z) = 3z
3
+ 8z
2
– 1
f(–1) = 3(–1)
3
+ 8(–1)
2
– 1 0
–1 is not a zero of f(z)
f(1) = 3(1)
3
+ 8(1)
2
– 1 0
1 is not a zero of f(z)
Therefore, f(z) has no integral zero.
9. (i) Coordinates of A are (–7, 3)
(ii) Abscissa of point D is 4.
(iii) Point is B.
(iv) Coordinates of C are (–3, –2)
CBSE IX | Mathematics
Sample Paper 4 – Solution
10. In parallelogram ABCD, CD = AB = 16 cm
[Opposite sides of a parallelogram are equal]
We know that,
Area of parallelogram = Base × corresponding altitude
Area of parallelogram ABCD = CD × AE = AD × CF
16 cm × 8 cm = AD × 10 cm
Thus, the length of AD is 12.8 cm.
OR
Area of rhombus =
1
Product of diagonals
2
=
1
16 12
2
= 96 cm
2
11. Area of cross section of pipe = 5 cm
2
Speed of water flowing out of the pipe = 30 cm/sec
Volume of water that flows out in 1 sec = 5 × 30 = 150 cm
3
Volume of water that flows out in
1 minute = 150 × 60 = 9000 cm
3
= 9 litres.
OR
l = 15 m, b = 12 m and h = 4.5 m
Volume of cuboid = lbh = 15 × 12 × 4.5 = 810 m
3
Total surface area of cuboid = 2(lb + bh + lh)
= 2(15 × 12 + 12 × 4.5 + 15 × 4.5)
= 603 m
2
12. Let the measure of the smaller angle be x and that of the larger angle be y.
The larger angle is 3° less than twice the measure of the smaller angle, so
y = 2x - 3° ….(1)
Given, that the two angles are complementary,
x + y = 90°
? x + (2x - 3°) = 90°
? x + 2x - 3° = 90°
? 3x = 93°
? x = 31°
Substitute value of x in equation (1)
y = 2(31) - 3°
Page 4
CBSE IX | Mathematics
Sample Paper 4 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 4 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
56
0.056
1000
?
OR
4 4 3 4 3
3
3 3 3
2. Linear equation in two variables:
An equation in the form ax + by + c = 0, where a, b and c are real numbers, and a and b
are not both zero is called a linear equation in two variables x and y.
3. If a triangle and a parallelogram are on the same base and between the same parallels,
then the area of the triangle is equal to half the area of the Parallelogram.
? Area of triangle =
1
2
? Area of a Parallelogram
?
Area of triangle
Area of a Parallelogram
=
1
2
Hence, the ratio is 1 : 2.
4. p(x) = x
3
+ 10x
2
+ px
(x – 1) is a factor of p(x).
? x – 1 = 0
? x = 1
Sunstituting x = 1 in p(x) = 0
? 1 + 10 + p = 0
? p = –11
OR
g(x) = vx – 3 = x
1/2
- 3. Here, one term of exponent is ½ which is not an integer. Hence,
it is not a polynomial.
CBSE IX | Mathematics
Sample Paper 4 – Solution
5. Arranging the data in the ascending order: 35, 36, 39, 40, 41, 55, 61, 62, 65, 70, 71,
? Range = Maximum value – Minimum value = 75 – 35 = 40
6. Let the fourth angle of the quadrilateral be x.
Sum of all angles of a quadrilateral = 360°
? 60° + 110° + 86° + x = 360°
? 256° + x = 360°
? x = 104°
Section B
7.
1
2
1
2
12
27
?
?
?
?
1
2
11
22
1
2
34
39
2
3
8. Let f(z) = 3z
3
+ 8z
2
– 1
The possible integral zeros of f(z) are –1 and 1.
f(z) = 3z
3
+ 8z
2
– 1
f(–1) = 3(–1)
3
+ 8(–1)
2
– 1 0
–1 is not a zero of f(z)
f(1) = 3(1)
3
+ 8(1)
2
– 1 0
1 is not a zero of f(z)
Therefore, f(z) has no integral zero.
9. (i) Coordinates of A are (–7, 3)
(ii) Abscissa of point D is 4.
(iii) Point is B.
(iv) Coordinates of C are (–3, –2)
CBSE IX | Mathematics
Sample Paper 4 – Solution
10. In parallelogram ABCD, CD = AB = 16 cm
[Opposite sides of a parallelogram are equal]
We know that,
Area of parallelogram = Base × corresponding altitude
Area of parallelogram ABCD = CD × AE = AD × CF
16 cm × 8 cm = AD × 10 cm
Thus, the length of AD is 12.8 cm.
OR
Area of rhombus =
1
Product of diagonals
2
=
1
16 12
2
= 96 cm
2
11. Area of cross section of pipe = 5 cm
2
Speed of water flowing out of the pipe = 30 cm/sec
Volume of water that flows out in 1 sec = 5 × 30 = 150 cm
3
Volume of water that flows out in
1 minute = 150 × 60 = 9000 cm
3
= 9 litres.
OR
l = 15 m, b = 12 m and h = 4.5 m
Volume of cuboid = lbh = 15 × 12 × 4.5 = 810 m
3
Total surface area of cuboid = 2(lb + bh + lh)
= 2(15 × 12 + 12 × 4.5 + 15 × 4.5)
= 603 m
2
12. Let the measure of the smaller angle be x and that of the larger angle be y.
The larger angle is 3° less than twice the measure of the smaller angle, so
y = 2x - 3° ….(1)
Given, that the two angles are complementary,
x + y = 90°
? x + (2x - 3°) = 90°
? x + 2x - 3° = 90°
? 3x = 93°
? x = 31°
Substitute value of x in equation (1)
y = 2(31) - 3°
CBSE IX | Mathematics
Sample Paper 4 – Solution
? y = 59°
So, the measures of the two angles are 31° and 59°.
Section C
13. Let x = 0.001
Then, x = 0.001001001……….. (i)
Therefore, 1000x = 1.001001001…………… (ii)
Subtracting (i) from (ii), we get
999x = 1 ? x =
1
999
Hence, 0.001 =
1
999
OR
6 4 3 6 4 3 6 4 3
6 4 3 6 4 3 6 4 3
36 48 3 48
36 48
84 48 3
12
48 3 84
12
4 3 7
14.
i. No.
?
??
? ? ?
2
1
x 3x 2
x 3 2x
x
has negative power of
ii. Yes
? ?
?
?
22
42
t t 3
t 3t
iii. Yes
? ?
2
22
x 4x 5
x 4x 5 x 5
2x
2 2 2 2 2 2
??
? ? ? ? ? ?
Page 5
CBSE IX | Mathematics
Sample Paper 4 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 4 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
56
0.056
1000
?
OR
4 4 3 4 3
3
3 3 3
2. Linear equation in two variables:
An equation in the form ax + by + c = 0, where a, b and c are real numbers, and a and b
are not both zero is called a linear equation in two variables x and y.
3. If a triangle and a parallelogram are on the same base and between the same parallels,
then the area of the triangle is equal to half the area of the Parallelogram.
? Area of triangle =
1
2
? Area of a Parallelogram
?
Area of triangle
Area of a Parallelogram
=
1
2
Hence, the ratio is 1 : 2.
4. p(x) = x
3
+ 10x
2
+ px
(x – 1) is a factor of p(x).
? x – 1 = 0
? x = 1
Sunstituting x = 1 in p(x) = 0
? 1 + 10 + p = 0
? p = –11
OR
g(x) = vx – 3 = x
1/2
- 3. Here, one term of exponent is ½ which is not an integer. Hence,
it is not a polynomial.
CBSE IX | Mathematics
Sample Paper 4 – Solution
5. Arranging the data in the ascending order: 35, 36, 39, 40, 41, 55, 61, 62, 65, 70, 71,
? Range = Maximum value – Minimum value = 75 – 35 = 40
6. Let the fourth angle of the quadrilateral be x.
Sum of all angles of a quadrilateral = 360°
? 60° + 110° + 86° + x = 360°
? 256° + x = 360°
? x = 104°
Section B
7.
1
2
1
2
12
27
?
?
?
?
1
2
11
22
1
2
34
39
2
3
8. Let f(z) = 3z
3
+ 8z
2
– 1
The possible integral zeros of f(z) are –1 and 1.
f(z) = 3z
3
+ 8z
2
– 1
f(–1) = 3(–1)
3
+ 8(–1)
2
– 1 0
–1 is not a zero of f(z)
f(1) = 3(1)
3
+ 8(1)
2
– 1 0
1 is not a zero of f(z)
Therefore, f(z) has no integral zero.
9. (i) Coordinates of A are (–7, 3)
(ii) Abscissa of point D is 4.
(iii) Point is B.
(iv) Coordinates of C are (–3, –2)
CBSE IX | Mathematics
Sample Paper 4 – Solution
10. In parallelogram ABCD, CD = AB = 16 cm
[Opposite sides of a parallelogram are equal]
We know that,
Area of parallelogram = Base × corresponding altitude
Area of parallelogram ABCD = CD × AE = AD × CF
16 cm × 8 cm = AD × 10 cm
Thus, the length of AD is 12.8 cm.
OR
Area of rhombus =
1
Product of diagonals
2
=
1
16 12
2
= 96 cm
2
11. Area of cross section of pipe = 5 cm
2
Speed of water flowing out of the pipe = 30 cm/sec
Volume of water that flows out in 1 sec = 5 × 30 = 150 cm
3
Volume of water that flows out in
1 minute = 150 × 60 = 9000 cm
3
= 9 litres.
OR
l = 15 m, b = 12 m and h = 4.5 m
Volume of cuboid = lbh = 15 × 12 × 4.5 = 810 m
3
Total surface area of cuboid = 2(lb + bh + lh)
= 2(15 × 12 + 12 × 4.5 + 15 × 4.5)
= 603 m
2
12. Let the measure of the smaller angle be x and that of the larger angle be y.
The larger angle is 3° less than twice the measure of the smaller angle, so
y = 2x - 3° ….(1)
Given, that the two angles are complementary,
x + y = 90°
? x + (2x - 3°) = 90°
? x + 2x - 3° = 90°
? 3x = 93°
? x = 31°
Substitute value of x in equation (1)
y = 2(31) - 3°
CBSE IX | Mathematics
Sample Paper 4 – Solution
? y = 59°
So, the measures of the two angles are 31° and 59°.
Section C
13. Let x = 0.001
Then, x = 0.001001001……….. (i)
Therefore, 1000x = 1.001001001…………… (ii)
Subtracting (i) from (ii), we get
999x = 1 ? x =
1
999
Hence, 0.001 =
1
999
OR
6 4 3 6 4 3 6 4 3
6 4 3 6 4 3 6 4 3
36 48 3 48
36 48
84 48 3
12
48 3 84
12
4 3 7
14.
i. No.
?
??
? ? ?
2
1
x 3x 2
x 3 2x
x
has negative power of
ii. Yes
? ?
?
?
22
42
t t 3
t 3t
iii. Yes
? ?
2
22
x 4x 5
x 4x 5 x 5
2x
2 2 2 2 2 2
??
? ? ? ? ? ?
CBSE IX | Mathematics
Sample Paper 4 – Solution
iv. No
?
2
3x 6 x = ? ? ?
1/2
2
3x 6 x has fractional power of
v. No
?
1
z
z
i.e.
?
?
1
zz has negative power of
15. When p(x) = ax
3
+ 3x
2
– 3 is divided by (x – 4 ), the remainder is given by
R1 = a(4)
3
+ 3(4)
2
– 3 = 64a + 45
When q(x) = 2x
3
– 5x + a is divided by (x – 4), the remainder is given by
R2 = 2(4)
3
– 5(4) + a = 108 + a
Given: R1 + R2 = 0
? 65a + 153 = 0
? a =
?153
65
OR
p = 2 – a
a + p – 2 = 0
x + y + z = 0 where x = a, y = p and z = -2
x
3
+ y
3
+ z
3
= 3xyz ? x + y + z = 0
a
3
+ p
3
+ (-2)
3
= 3ap(-2)
a
3
+ p
3
– 8 = -6ap
a
3
+ 6ap + p
3
– 8 = 0
16. In ?DCB, ?DBC = ?DCB (given)
DC = DB [Side opp. To equal ?’s are equal]…..(i)
In ?ABD and ?ACD
AB = AC (given)
BD = CD [from (i) ]
AD = AD common
?ABD ? ? ACD [SSS Rule]
?BAD = ?CAD (CPCT)
Hence, AD is bisector of ?BAC.
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Nov 24, 2024 Last updated |
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