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Sample Papers For Class 9

Created by: Indu Gupta

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) 
 (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 
 
 
 
 
 
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