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CBSE IX | Mathematics
Sample Paper 5 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 5 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
11
0.011
1000
?
OR
Yes, because 0 can be written as
0
1
which is of the form
p
q
, where p and q are integers
and q ? 0.
2. x + y = 2 ……(i)
and x – y = 4 ……(ii)
Adding (i) and (ii), we get
x = 3
Putting x = 3 in the equation (i), we get
y = –1
? x = 3 and y = –1
3. Since AOB is a Straight Line,
? ?AOB = 180°
? ?AOD + ?COD + ?COB = 180°
? x + 10° + x + x + 20° = 180°
? 3x = 150°
? x = 50°
4. If for one of the solutions of the equation ax + by + c = 0, x is negative and y is positive,
then a portion of the above line definitely lies in the II Quadrant.
As in the II Quadrant the x-axis contains only negative numbers and y-axis contains
only positive numbers.
OR
Let the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y.
Hence, the linear equation is 5x = 2y
5x – 2y = 0.
Page 2
CBSE IX | Mathematics
Sample Paper 5 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 5 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
11
0.011
1000
?
OR
Yes, because 0 can be written as
0
1
which is of the form
p
q
, where p and q are integers
and q ? 0.
2. x + y = 2 ……(i)
and x – y = 4 ……(ii)
Adding (i) and (ii), we get
x = 3
Putting x = 3 in the equation (i), we get
y = –1
? x = 3 and y = –1
3. Since AOB is a Straight Line,
? ?AOB = 180°
? ?AOD + ?COD + ?COB = 180°
? x + 10° + x + x + 20° = 180°
? 3x = 150°
? x = 50°
4. If for one of the solutions of the equation ax + by + c = 0, x is negative and y is positive,
then a portion of the above line definitely lies in the II Quadrant.
As in the II Quadrant the x-axis contains only negative numbers and y-axis contains
only positive numbers.
OR
Let the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y.
Hence, the linear equation is 5x = 2y
5x – 2y = 0.
CBSE IX | Mathematics
Sample Paper 5 – Solution
5. Here the class marks are uniformly spaced,
Therefore,
Class size is the difference between any two consecutive class marks.
? Class Size = 52 – 47 = 5
6. ?PSR = ?RQP = 125° (since PQRS is a parallelogram, then the opposite angles will be
equal)
? ?PQT = 180°…………….. (PQT is a straight line)
? ?PQR + ?RQT = 180°
? 125° + ?RQT = 180°
? ?RQT = 55°
Section B
7. a = 2 + 3
?
11
a
23
?
?
=
1 2 3
2 3 2 3
?
?
??
=
? ?
2
2
23
23
?
?
= 2 – 3
So, a +
1
a
= (2 + 3 ) + (2 – 3 ) = 4
8. (i) I quadrant: (5, 2)
(ii) II quadrant: (–5, 2)
(iii) III quadrant: (–5, –2)
(iv) IV quadrant: (5, –2)
9. Let ‘l’ be the length of the cube.
Now, T.S.A. of the cube = 294 cm
2
…(given)
? 6l
2
= 294
? Side, l = 7 cm
Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm
3
OR
According to the question,
Volume of one matchbox = 4 cm × 2.5 cm × 1.5 cm = 15 cm
3
Volume of 12 such matchboxes = 15 × 12 = 180 cm
3
Page 3
CBSE IX | Mathematics
Sample Paper 5 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 5 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
11
0.011
1000
?
OR
Yes, because 0 can be written as
0
1
which is of the form
p
q
, where p and q are integers
and q ? 0.
2. x + y = 2 ……(i)
and x – y = 4 ……(ii)
Adding (i) and (ii), we get
x = 3
Putting x = 3 in the equation (i), we get
y = –1
? x = 3 and y = –1
3. Since AOB is a Straight Line,
? ?AOB = 180°
? ?AOD + ?COD + ?COB = 180°
? x + 10° + x + x + 20° = 180°
? 3x = 150°
? x = 50°
4. If for one of the solutions of the equation ax + by + c = 0, x is negative and y is positive,
then a portion of the above line definitely lies in the II Quadrant.
As in the II Quadrant the x-axis contains only negative numbers and y-axis contains
only positive numbers.
OR
Let the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y.
Hence, the linear equation is 5x = 2y
5x – 2y = 0.
CBSE IX | Mathematics
Sample Paper 5 – Solution
5. Here the class marks are uniformly spaced,
Therefore,
Class size is the difference between any two consecutive class marks.
? Class Size = 52 – 47 = 5
6. ?PSR = ?RQP = 125° (since PQRS is a parallelogram, then the opposite angles will be
equal)
? ?PQT = 180°…………….. (PQT is a straight line)
? ?PQR + ?RQT = 180°
? 125° + ?RQT = 180°
? ?RQT = 55°
Section B
7. a = 2 + 3
?
11
a
23
?
?
=
1 2 3
2 3 2 3
?
?
??
=
? ?
2
2
23
23
?
?
= 2 – 3
So, a +
1
a
= (2 + 3 ) + (2 – 3 ) = 4
8. (i) I quadrant: (5, 2)
(ii) II quadrant: (–5, 2)
(iii) III quadrant: (–5, –2)
(iv) IV quadrant: (5, –2)
9. Let ‘l’ be the length of the cube.
Now, T.S.A. of the cube = 294 cm
2
…(given)
? 6l
2
= 294
? Side, l = 7 cm
Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm
3
OR
According to the question,
Volume of one matchbox = 4 cm × 2.5 cm × 1.5 cm = 15 cm
3
Volume of 12 such matchboxes = 15 × 12 = 180 cm
3
CBSE IX | Mathematics
Sample Paper 5 – Solution
10. 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.
OR
An equation of the form ax + by + c = 0, where a, b, c are real numbers such that a ? 0
and b ? 0, is called a linear equation in two variables x and y.
11. Perimeter = 36 cm = 10 cm + 2(Length of an equal side)
Length of an equal side = 13 cm
Here,
36
s 18
2
?? , and the sides are 10, 13 and 13.
By Heron's formula,
Area = 18 8 5 5 60 ? ? ? ?
sq. cm
12. (–2x + 5y – 3z)
2
= (–2x)
2
+ (5y)
2
+ (–3z)
2
+ 2(–2x)(5y) + 2(5y)( –3z) + 2(–2x)( –3z)
= 4x
2
+ 25y
2
+ 9z
2
– 20xy – 30yz + 12zx
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
Page 4
CBSE IX | Mathematics
Sample Paper 5 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 5 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
11
0.011
1000
?
OR
Yes, because 0 can be written as
0
1
which is of the form
p
q
, where p and q are integers
and q ? 0.
2. x + y = 2 ……(i)
and x – y = 4 ……(ii)
Adding (i) and (ii), we get
x = 3
Putting x = 3 in the equation (i), we get
y = –1
? x = 3 and y = –1
3. Since AOB is a Straight Line,
? ?AOB = 180°
? ?AOD + ?COD + ?COB = 180°
? x + 10° + x + x + 20° = 180°
? 3x = 150°
? x = 50°
4. If for one of the solutions of the equation ax + by + c = 0, x is negative and y is positive,
then a portion of the above line definitely lies in the II Quadrant.
As in the II Quadrant the x-axis contains only negative numbers and y-axis contains
only positive numbers.
OR
Let the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y.
Hence, the linear equation is 5x = 2y
5x – 2y = 0.
CBSE IX | Mathematics
Sample Paper 5 – Solution
5. Here the class marks are uniformly spaced,
Therefore,
Class size is the difference between any two consecutive class marks.
? Class Size = 52 – 47 = 5
6. ?PSR = ?RQP = 125° (since PQRS is a parallelogram, then the opposite angles will be
equal)
? ?PQT = 180°…………….. (PQT is a straight line)
? ?PQR + ?RQT = 180°
? 125° + ?RQT = 180°
? ?RQT = 55°
Section B
7. a = 2 + 3
?
11
a
23
?
?
=
1 2 3
2 3 2 3
?
?
??
=
? ?
2
2
23
23
?
?
= 2 – 3
So, a +
1
a
= (2 + 3 ) + (2 – 3 ) = 4
8. (i) I quadrant: (5, 2)
(ii) II quadrant: (–5, 2)
(iii) III quadrant: (–5, –2)
(iv) IV quadrant: (5, –2)
9. Let ‘l’ be the length of the cube.
Now, T.S.A. of the cube = 294 cm
2
…(given)
? 6l
2
= 294
? Side, l = 7 cm
Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm
3
OR
According to the question,
Volume of one matchbox = 4 cm × 2.5 cm × 1.5 cm = 15 cm
3
Volume of 12 such matchboxes = 15 × 12 = 180 cm
3
CBSE IX | Mathematics
Sample Paper 5 – Solution
10. 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.
OR
An equation of the form ax + by + c = 0, where a, b, c are real numbers such that a ? 0
and b ? 0, is called a linear equation in two variables x and y.
11. Perimeter = 36 cm = 10 cm + 2(Length of an equal side)
Length of an equal side = 13 cm
Here,
36
s 18
2
?? , and the sides are 10, 13 and 13.
By Heron's formula,
Area = 18 8 5 5 60 ? ? ? ?
sq. cm
12. (–2x + 5y – 3z)
2
= (–2x)
2
+ (5y)
2
+ (–3z)
2
+ 2(–2x)(5y) + 2(5y)( –3z) + 2(–2x)( –3z)
= 4x
2
+ 25y
2
+ 9z
2
– 20xy – 30yz + 12zx
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
CBSE IX | Mathematics
Sample Paper 5 – Solution
OR
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
a b a a b b a b a a a b
a b b a a b a b b a b b
? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ? ? ?
? ?
? ?
2
2 2 2
2
2 2 2
a a b
a b b
??
?
??
? ?
? ?
2 2 2
2 2 2
a a b
a b b
??
?
??
2 2 2
2 2 2
a a b
a b b
??
?
??
2
2
b
a
?
14. Given: x = 2y + 6 or x - 2y - 6 = 0
We know that if a + b + c = 0, then a
3
+ b
3
+ c
3
= 3xyz
Therefore, we have:
(x)
3
+ (–2y)
3
+ (–6)
3
= 3x(–2y)( –6)
Or, x
3
– 8y
3
– 36xy – 216 = 0
OR
Let p(x) = x
3
+ 3x
2
+ 3x + 1 and g(x) = x + p
Now, g(x) = 0
x + p = 0
x = - p
By the remainder theorem, when p(x) is divided by x + p then the remainder is
p(-p).
p(-p) = (-p)
3
+ 3(-p)
2
+ 3(-p) + 1 = – p
3
+ 3p
2
– 3p + 1
Hence, remainder is – p
3
+ 3p
2
– 3p + 1.
Page 5
CBSE IX | Mathematics
Sample Paper 5 – Solution
CBSE Board
Class IX Mathematics
Sample Paper 5 – Solution
Time: 3 hrs Total Marks: 80
Section A
1.
11
0.011
1000
?
OR
Yes, because 0 can be written as
0
1
which is of the form
p
q
, where p and q are integers
and q ? 0.
2. x + y = 2 ……(i)
and x – y = 4 ……(ii)
Adding (i) and (ii), we get
x = 3
Putting x = 3 in the equation (i), we get
y = –1
? x = 3 and y = –1
3. Since AOB is a Straight Line,
? ?AOB = 180°
? ?AOD + ?COD + ?COB = 180°
? x + 10° + x + x + 20° = 180°
? 3x = 150°
? x = 50°
4. If for one of the solutions of the equation ax + by + c = 0, x is negative and y is positive,
then a portion of the above line definitely lies in the II Quadrant.
As in the II Quadrant the x-axis contains only negative numbers and y-axis contains
only positive numbers.
OR
Let the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y.
Hence, the linear equation is 5x = 2y
5x – 2y = 0.
CBSE IX | Mathematics
Sample Paper 5 – Solution
5. Here the class marks are uniformly spaced,
Therefore,
Class size is the difference between any two consecutive class marks.
? Class Size = 52 – 47 = 5
6. ?PSR = ?RQP = 125° (since PQRS is a parallelogram, then the opposite angles will be
equal)
? ?PQT = 180°…………….. (PQT is a straight line)
? ?PQR + ?RQT = 180°
? 125° + ?RQT = 180°
? ?RQT = 55°
Section B
7. a = 2 + 3
?
11
a
23
?
?
=
1 2 3
2 3 2 3
?
?
??
=
? ?
2
2
23
23
?
?
= 2 – 3
So, a +
1
a
= (2 + 3 ) + (2 – 3 ) = 4
8. (i) I quadrant: (5, 2)
(ii) II quadrant: (–5, 2)
(iii) III quadrant: (–5, –2)
(iv) IV quadrant: (5, –2)
9. Let ‘l’ be the length of the cube.
Now, T.S.A. of the cube = 294 cm
2
…(given)
? 6l
2
= 294
? Side, l = 7 cm
Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm
3
OR
According to the question,
Volume of one matchbox = 4 cm × 2.5 cm × 1.5 cm = 15 cm
3
Volume of 12 such matchboxes = 15 × 12 = 180 cm
3
CBSE IX | Mathematics
Sample Paper 5 – Solution
10. 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.
OR
An equation of the form ax + by + c = 0, where a, b, c are real numbers such that a ? 0
and b ? 0, is called a linear equation in two variables x and y.
11. Perimeter = 36 cm = 10 cm + 2(Length of an equal side)
Length of an equal side = 13 cm
Here,
36
s 18
2
?? , and the sides are 10, 13 and 13.
By Heron's formula,
Area = 18 8 5 5 60 ? ? ? ?
sq. cm
12. (–2x + 5y – 3z)
2
= (–2x)
2
+ (5y)
2
+ (–3z)
2
+ 2(–2x)(5y) + 2(5y)( –3z) + 2(–2x)( –3z)
= 4x
2
+ 25y
2
+ 9z
2
– 20xy – 30yz + 12zx
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
CBSE IX | Mathematics
Sample Paper 5 – Solution
OR
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
a b a a b b a b a a a b
a b b a a b a b b a b b
? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ? ? ?
? ?
? ?
2
2 2 2
2
2 2 2
a a b
a b b
??
?
??
? ?
? ?
2 2 2
2 2 2
a a b
a b b
??
?
??
2 2 2
2 2 2
a a b
a b b
??
?
??
2
2
b
a
?
14. Given: x = 2y + 6 or x - 2y - 6 = 0
We know that if a + b + c = 0, then a
3
+ b
3
+ c
3
= 3xyz
Therefore, we have:
(x)
3
+ (–2y)
3
+ (–6)
3
= 3x(–2y)( –6)
Or, x
3
– 8y
3
– 36xy – 216 = 0
OR
Let p(x) = x
3
+ 3x
2
+ 3x + 1 and g(x) = x + p
Now, g(x) = 0
x + p = 0
x = - p
By the remainder theorem, when p(x) is divided by x + p then the remainder is
p(-p).
p(-p) = (-p)
3
+ 3(-p)
2
+ 3(-p) + 1 = – p
3
+ 3p
2
– 3p + 1
Hence, remainder is – p
3
+ 3p
2
– 3p + 1.
CBSE IX | Mathematics
Sample Paper 5 – Solution
15.
In ?POR and ?QOS
?QPR = ?PQS (given)
OP = OQ (O is the mid-point of PQ)
?POS = ?QOR (given)
?POS + x
o
= ?QOR + x
o
?POR = ?QOS
By ASA congruence rule,
?PQR ? ?QOS
? PR = QS (By CPCT)
16. Let p(z) = az
3
+ 4z
2
+ 3z – 4 and q(z) = z
3
– 4z + a
When p(z) is divided by z – 3, the remainder is given by
p(3) = a × 3
3
+ 4 × 3
2
+ 3 × 3 – 4
= 27a + 36 + 9 – 4
= 27a + 41 ….(i)
When q(z) is divided by z – 3, the remainder is given by
q(3) = 3
3
– 4 × 3 + a
= 27 – 12 + a
= 15 + a ….(ii)
Given that p(3) = q(3).
So, from (i) and (ii), we have
27a + 41 = 15 + a
27a – a = –41 + 15
26a = –26
a = –1
Read More
FAQs on Class 9 Math: Solutions of Sample Practice Question Paper- 5
| 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. In other words, they are the values of the variable that satisfy the equation when substituted into it.
| 2. How can I find the solutions of a quadratic equation? | |
Ans. To find the solutions of a quadratic equation, you can use various methods such as factoring, completing the square, or using the quadratic formula. The method you choose depends on the form of the equation and your preference.
| 3. What is the quadratic formula? | |
Ans. The quadratic formula is a formula used to find the solutions of a quadratic equation of the form ax^2 + bx + c = 0, where a, b, and c are constants. The formula is x = (-b ± √(b^2 - 4ac)) / (2a). It provides the values for x that satisfy the equation.
| 4. Can a quadratic equation have more than two solutions? | |
Ans. No, a quadratic equation can have at most two solutions. This is because a quadratic equation is a second-degree polynomial equation, and a polynomial of degree n can have at most n solutions. In the case of a quadratic equation, since the highest power of the variable is 2, it can have a maximum of two solutions.
| 5. How can I check if my solutions are correct for a quadratic equation? | |
Ans. To check if your solutions are correct for a quadratic equation, you can substitute the values of the solutions back into the equation and see if it holds true. If the equation is satisfied when the solutions are substituted, then the solutions are correct.
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Class 9 Math: Solutions of Sample Practice Question Paper- 5 Notes
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Class 9 Math: Solutions of Sample Practice Question Paper- 5 Class 9 Questions
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The content of Class 9 Math: Solutions of Sample Practice Question Paper- 5 is prepared as per the latest Class 9 syllabus.
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