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## Class 9 : Notes | EduRev

``` Page 1

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)

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)

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)

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)

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

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