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Mathematics (Maths) Class 9

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 Page 1


  
 
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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