Sample Solution Paper 1 - Math, Class 9 Class 9 Notes | EduRev

Mathematics (Maths) Class 9

Created by: Indu Gupta

Class 9 : Sample Solution Paper 1 - Math, Class 9 Class 9 Notes | EduRev

 Page 1


  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
CBSE Board 
Class IX Mathematics 
Sample Paper 1 – Solution 
Time: 3 hrs  Total Marks: 80 
      
Section A 
 
1. 
2 2 2 2
(5 5)(5 5) (5) ( 5) 25 5 20 a b (a b)(a b) ?? ? ? ? ? ? ? ? ? ? ? ?
??
 
 
2. p(x) = x
3
 + 10x
2
 + px 
(x – 1 ) is the factor of  p(x). 
? x – 1 = 0 
? x = 1 will satisfy p(x) 
? p(1) = 0 
? (1)
3
 + 10(1)
2
 + p(1) = 0 
? 1 + 10 + p = 0  
? p = –11 
 
3. Given, AB = AC  
?  ?B = ?C      ….(1)           ( angles opp. to equal sides are equal) 
 In ? ABC, we have 
? A + ? B + ? C = 180 ?       
? ? A + ? B + ? B = 180 ?   ....[From (1)] 
? 100 ? + 2 ?B = 180 ?           .…( ?A = 100 ?) 
? 2 ?B = 80 ? 
                      ? ?B = 40 ? 
OR 
  The equal angles and the non-equal angle are in the ratio 3:4.  
  Let equal angles be 3x each, therefore non-equal angle is 4x. 
  Angles of a triangle =180  
             ? 3x + 3x + 4x = 180       
             ? 10 x = 180   
             ? x = 18  
  Therefore, 3x = 54 and 4x = 72    
  Angles = 54 , 54 and 72 .  
 
4. Let, the cost of a note book = Rs. x  and the cost of a pen = Rs. y 
According to given statement,  
x = 2y  
i.e. x – 2y = 0  
 
 
Page 2


  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
CBSE Board 
Class IX Mathematics 
Sample Paper 1 – Solution 
Time: 3 hrs  Total Marks: 80 
      
Section A 
 
1. 
2 2 2 2
(5 5)(5 5) (5) ( 5) 25 5 20 a b (a b)(a b) ?? ? ? ? ? ? ? ? ? ? ? ?
??
 
 
2. p(x) = x
3
 + 10x
2
 + px 
(x – 1 ) is the factor of  p(x). 
? x – 1 = 0 
? x = 1 will satisfy p(x) 
? p(1) = 0 
? (1)
3
 + 10(1)
2
 + p(1) = 0 
? 1 + 10 + p = 0  
? p = –11 
 
3. Given, AB = AC  
?  ?B = ?C      ….(1)           ( angles opp. to equal sides are equal) 
 In ? ABC, we have 
? A + ? B + ? C = 180 ?       
? ? A + ? B + ? B = 180 ?   ....[From (1)] 
? 100 ? + 2 ?B = 180 ?           .…( ?A = 100 ?) 
? 2 ?B = 80 ? 
                      ? ?B = 40 ? 
OR 
  The equal angles and the non-equal angle are in the ratio 3:4.  
  Let equal angles be 3x each, therefore non-equal angle is 4x. 
  Angles of a triangle =180  
             ? 3x + 3x + 4x = 180       
             ? 10 x = 180   
             ? x = 18  
  Therefore, 3x = 54 and 4x = 72    
  Angles = 54 , 54 and 72 .  
 
4. Let, the cost of a note book = Rs. x  and the cost of a pen = Rs. y 
According to given statement,  
x = 2y  
i.e. x – 2y = 0  
 
 
  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
5. Arranging the given data in the Ascending order:  
35, 36, 39, 40, 41, 55, 61, 62, 65, 70, 71, 75 
? Range = Maximum value – Minimum value = 75 – 35 = 40  
OR 
The class mark of 45-52 is 
45 52
48.5
2
?
? . 
 
6. In a Parallelogram PQRS, 
?R and ?S are consecutive interior angles on the same side of the transversal SR.  
Therefore, m ?R + m ? S = 180° 
Section B 
 
7.   
  
??
??
??
? ? ? ?
?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
3/ 4 3/ 2
33
42
42
33
33
3 3 3
3 3 3
81 25
16 9
35
23
35
23
23
35
2 3 2 8
125 3 5 5
 
 
OR 
 
2
2
3
0 81
7
27 3
 
 = 
2
3
2 27
31
8
 
  = 
2
3
3 3
91
2
 
 = 
2
3
10
2
 
 = 
9
10
4
 
 = 
9 40
4
 
31
4
 
Page 3


  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
CBSE Board 
Class IX Mathematics 
Sample Paper 1 – Solution 
Time: 3 hrs  Total Marks: 80 
      
Section A 
 
1. 
2 2 2 2
(5 5)(5 5) (5) ( 5) 25 5 20 a b (a b)(a b) ?? ? ? ? ? ? ? ? ? ? ? ?
??
 
 
2. p(x) = x
3
 + 10x
2
 + px 
(x – 1 ) is the factor of  p(x). 
? x – 1 = 0 
? x = 1 will satisfy p(x) 
? p(1) = 0 
? (1)
3
 + 10(1)
2
 + p(1) = 0 
? 1 + 10 + p = 0  
? p = –11 
 
3. Given, AB = AC  
?  ?B = ?C      ….(1)           ( angles opp. to equal sides are equal) 
 In ? ABC, we have 
? A + ? B + ? C = 180 ?       
? ? A + ? B + ? B = 180 ?   ....[From (1)] 
? 100 ? + 2 ?B = 180 ?           .…( ?A = 100 ?) 
? 2 ?B = 80 ? 
                      ? ?B = 40 ? 
OR 
  The equal angles and the non-equal angle are in the ratio 3:4.  
  Let equal angles be 3x each, therefore non-equal angle is 4x. 
  Angles of a triangle =180  
             ? 3x + 3x + 4x = 180       
             ? 10 x = 180   
             ? x = 18  
  Therefore, 3x = 54 and 4x = 72    
  Angles = 54 , 54 and 72 .  
 
4. Let, the cost of a note book = Rs. x  and the cost of a pen = Rs. y 
According to given statement,  
x = 2y  
i.e. x – 2y = 0  
 
 
  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
5. Arranging the given data in the Ascending order:  
35, 36, 39, 40, 41, 55, 61, 62, 65, 70, 71, 75 
? Range = Maximum value – Minimum value = 75 – 35 = 40  
OR 
The class mark of 45-52 is 
45 52
48.5
2
?
? . 
 
6. In a Parallelogram PQRS, 
?R and ?S are consecutive interior angles on the same side of the transversal SR.  
Therefore, m ?R + m ? S = 180° 
Section B 
 
7.   
  
??
??
??
? ? ? ?
?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
3/ 4 3/ 2
33
42
42
33
33
3 3 3
3 3 3
81 25
16 9
35
23
35
23
23
35
2 3 2 8
125 3 5 5
 
 
OR 
 
2
2
3
0 81
7
27 3
 
 = 
2
3
2 27
31
8
 
  = 
2
3
3 3
91
2
 
 = 
2
3
10
2
 
 = 
9
10
4
 
 = 
9 40
4
 
31
4
 
  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
8.   
 
2
2
2
2
2
12
x 2 2x
xx
11
x 2 2 x
xx
11
x 2 x
xx
11
x x 2
xx
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 
 
9.  
a.   Point of the form (a, 0) lie on the x axis. 
      So, the point (–4, 0) will lie on the negative side of the x axis. 
 
b.   (–, +) are the sign of the coordinate of points in the quadrant II. 
       So, the point (–10, 2) lies in the quadrant II. 
 
c.     Point of the form (0, a) lie on the y axis. 
      So, the point (0, 8) will lie on the positive side of y axis. 
 
d.     (+, +) are the sign of the coordinates of points in the quadrant I. 
         So, the point (10, 4) lies in the quadrant I.  
 
 
 
10. ?ACB and ?ACF lie on the same base AC and are between the same parallel lines AC 
and BF. 
? area(?ACB) = area( ?ACF) 
 
11. Length of the box = l = 48 – 8 – 8 = 32 cm 
Breadth of the box = b = 36 – 8 – 8 = 20 cm  
Height = h = 8 cm 
 
Volume of the box formed = l × b × h = 32 × 20 × 8 = 5120 cm
3
 
Page 4


  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
CBSE Board 
Class IX Mathematics 
Sample Paper 1 – Solution 
Time: 3 hrs  Total Marks: 80 
      
Section A 
 
1. 
2 2 2 2
(5 5)(5 5) (5) ( 5) 25 5 20 a b (a b)(a b) ?? ? ? ? ? ? ? ? ? ? ? ?
??
 
 
2. p(x) = x
3
 + 10x
2
 + px 
(x – 1 ) is the factor of  p(x). 
? x – 1 = 0 
? x = 1 will satisfy p(x) 
? p(1) = 0 
? (1)
3
 + 10(1)
2
 + p(1) = 0 
? 1 + 10 + p = 0  
? p = –11 
 
3. Given, AB = AC  
?  ?B = ?C      ….(1)           ( angles opp. to equal sides are equal) 
 In ? ABC, we have 
? A + ? B + ? C = 180 ?       
? ? A + ? B + ? B = 180 ?   ....[From (1)] 
? 100 ? + 2 ?B = 180 ?           .…( ?A = 100 ?) 
? 2 ?B = 80 ? 
                      ? ?B = 40 ? 
OR 
  The equal angles and the non-equal angle are in the ratio 3:4.  
  Let equal angles be 3x each, therefore non-equal angle is 4x. 
  Angles of a triangle =180  
             ? 3x + 3x + 4x = 180       
             ? 10 x = 180   
             ? x = 18  
  Therefore, 3x = 54 and 4x = 72    
  Angles = 54 , 54 and 72 .  
 
4. Let, the cost of a note book = Rs. x  and the cost of a pen = Rs. y 
According to given statement,  
x = 2y  
i.e. x – 2y = 0  
 
 
  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
5. Arranging the given data in the Ascending order:  
35, 36, 39, 40, 41, 55, 61, 62, 65, 70, 71, 75 
? Range = Maximum value – Minimum value = 75 – 35 = 40  
OR 
The class mark of 45-52 is 
45 52
48.5
2
?
? . 
 
6. In a Parallelogram PQRS, 
?R and ?S are consecutive interior angles on the same side of the transversal SR.  
Therefore, m ?R + m ? S = 180° 
Section B 
 
7.   
  
??
??
??
? ? ? ?
?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
3/ 4 3/ 2
33
42
42
33
33
3 3 3
3 3 3
81 25
16 9
35
23
35
23
23
35
2 3 2 8
125 3 5 5
 
 
OR 
 
2
2
3
0 81
7
27 3
 
 = 
2
3
2 27
31
8
 
  = 
2
3
3 3
91
2
 
 = 
2
3
10
2
 
 = 
9
10
4
 
 = 
9 40
4
 
31
4
 
  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
8.   
 
2
2
2
2
2
12
x 2 2x
xx
11
x 2 2 x
xx
11
x 2 x
xx
11
x x 2
xx
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 
 
9.  
a.   Point of the form (a, 0) lie on the x axis. 
      So, the point (–4, 0) will lie on the negative side of the x axis. 
 
b.   (–, +) are the sign of the coordinate of points in the quadrant II. 
       So, the point (–10, 2) lies in the quadrant II. 
 
c.     Point of the form (0, a) lie on the y axis. 
      So, the point (0, 8) will lie on the positive side of y axis. 
 
d.     (+, +) are the sign of the coordinates of points in the quadrant I. 
         So, the point (10, 4) lies in the quadrant I.  
 
 
 
10. ?ACB and ?ACF lie on the same base AC and are between the same parallel lines AC 
and BF. 
? area(?ACB) = area( ?ACF) 
 
11. Length of the box = l = 48 – 8 – 8 = 32 cm 
Breadth of the box = b = 36 – 8 – 8 = 20 cm  
Height = h = 8 cm 
 
Volume of the box formed = l × b × h = 32 × 20 × 8 = 5120 cm
3
 
  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
 
OR 
Given that: 
    Length (l) of cuboid =11 cm 
     Breadth (b) of cuboid =10 cm 
     Height (h) of cuboid = 2 cm 
 
     We know that, 
     Length of the diagonal in a Cuboid = 
2 2 2
l b h ?? 
                                                                       = 
2 2 2
11 10 2 ?? 
                                                                       =  121 100 4 ?? 
                                                                       =  221 4 ? 
                                                                       =  225 
                                                                      = 15 cm 
 
12. Since x = 1, y = 1 is the solution of 9kx + 12ky = 1, it will satisfy the equation. 
? 9k(1) + 12k(1) = 63 
? 9k + 12k = 63 
? 21k = 63 
? k = 3  
 
Section C 
 
13.   
? ? ? ? ?
??
33
25
43 5
35 4
25 343
16 8 7
 
? ? ? ?
? ? ? ?
?
?
??
?
?
??
?
?
?
?
?
33
23
25
3 54
43
5 43
9
3
5
3
54
5
9
3
5
3
9
5
6
3
5
9
57
2 2 7
57
2 2 7
57
27
57
2
 
 
Page 5


  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
CBSE Board 
Class IX Mathematics 
Sample Paper 1 – Solution 
Time: 3 hrs  Total Marks: 80 
      
Section A 
 
1. 
2 2 2 2
(5 5)(5 5) (5) ( 5) 25 5 20 a b (a b)(a b) ?? ? ? ? ? ? ? ? ? ? ? ?
??
 
 
2. p(x) = x
3
 + 10x
2
 + px 
(x – 1 ) is the factor of  p(x). 
? x – 1 = 0 
? x = 1 will satisfy p(x) 
? p(1) = 0 
? (1)
3
 + 10(1)
2
 + p(1) = 0 
? 1 + 10 + p = 0  
? p = –11 
 
3. Given, AB = AC  
?  ?B = ?C      ….(1)           ( angles opp. to equal sides are equal) 
 In ? ABC, we have 
? A + ? B + ? C = 180 ?       
? ? A + ? B + ? B = 180 ?   ....[From (1)] 
? 100 ? + 2 ?B = 180 ?           .…( ?A = 100 ?) 
? 2 ?B = 80 ? 
                      ? ?B = 40 ? 
OR 
  The equal angles and the non-equal angle are in the ratio 3:4.  
  Let equal angles be 3x each, therefore non-equal angle is 4x. 
  Angles of a triangle =180  
             ? 3x + 3x + 4x = 180       
             ? 10 x = 180   
             ? x = 18  
  Therefore, 3x = 54 and 4x = 72    
  Angles = 54 , 54 and 72 .  
 
4. Let, the cost of a note book = Rs. x  and the cost of a pen = Rs. y 
According to given statement,  
x = 2y  
i.e. x – 2y = 0  
 
 
  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
5. Arranging the given data in the Ascending order:  
35, 36, 39, 40, 41, 55, 61, 62, 65, 70, 71, 75 
? Range = Maximum value – Minimum value = 75 – 35 = 40  
OR 
The class mark of 45-52 is 
45 52
48.5
2
?
? . 
 
6. In a Parallelogram PQRS, 
?R and ?S are consecutive interior angles on the same side of the transversal SR.  
Therefore, m ?R + m ? S = 180° 
Section B 
 
7.   
  
??
??
??
? ? ? ?
?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
3/ 4 3/ 2
33
42
42
33
33
3 3 3
3 3 3
81 25
16 9
35
23
35
23
23
35
2 3 2 8
125 3 5 5
 
 
OR 
 
2
2
3
0 81
7
27 3
 
 = 
2
3
2 27
31
8
 
  = 
2
3
3 3
91
2
 
 = 
2
3
10
2
 
 = 
9
10
4
 
 = 
9 40
4
 
31
4
 
  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
8.   
 
2
2
2
2
2
12
x 2 2x
xx
11
x 2 2 x
xx
11
x 2 x
xx
11
x x 2
xx
? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 
 
9.  
a.   Point of the form (a, 0) lie on the x axis. 
      So, the point (–4, 0) will lie on the negative side of the x axis. 
 
b.   (–, +) are the sign of the coordinate of points in the quadrant II. 
       So, the point (–10, 2) lies in the quadrant II. 
 
c.     Point of the form (0, a) lie on the y axis. 
      So, the point (0, 8) will lie on the positive side of y axis. 
 
d.     (+, +) are the sign of the coordinates of points in the quadrant I. 
         So, the point (10, 4) lies in the quadrant I.  
 
 
 
10. ?ACB and ?ACF lie on the same base AC and are between the same parallel lines AC 
and BF. 
? area(?ACB) = area( ?ACF) 
 
11. Length of the box = l = 48 – 8 – 8 = 32 cm 
Breadth of the box = b = 36 – 8 – 8 = 20 cm  
Height = h = 8 cm 
 
Volume of the box formed = l × b × h = 32 × 20 × 8 = 5120 cm
3
 
  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
 
OR 
Given that: 
    Length (l) of cuboid =11 cm 
     Breadth (b) of cuboid =10 cm 
     Height (h) of cuboid = 2 cm 
 
     We know that, 
     Length of the diagonal in a Cuboid = 
2 2 2
l b h ?? 
                                                                       = 
2 2 2
11 10 2 ?? 
                                                                       =  121 100 4 ?? 
                                                                       =  221 4 ? 
                                                                       =  225 
                                                                      = 15 cm 
 
12. Since x = 1, y = 1 is the solution of 9kx + 12ky = 1, it will satisfy the equation. 
? 9k(1) + 12k(1) = 63 
? 9k + 12k = 63 
? 21k = 63 
? k = 3  
 
Section C 
 
13.   
? ? ? ? ?
??
33
25
43 5
35 4
25 343
16 8 7
 
? ? ? ?
? ? ? ?
?
?
??
?
?
??
?
?
?
?
?
33
23
25
3 54
43
5 43
9
3
5
3
54
5
9
3
5
3
9
5
6
3
5
9
57
2 2 7
57
2 2 7
57
27
57
2
 
 
  
 
CBSE IX | Mathematics 
Sample Paper 1 – Solution 
 
     
 
14. Let p(x) = x
3
 + 13x
2
 + 32x + 20 
 p(-1) = –1 + 13 – 32 + 20 = –33 + 33 = 0 
 Therefore (x + 1) is a factor of p(x). 
 On dividing p(x) by (x + 1), we get 
 p(x) ÷ (x + 1) = x
2
 + 12x + 20 
 Thus,  
 x
3
 + 13x
2
 + 32x + 20 = (x + 1)(x
2
 + 12x + 20) 
= (x + 1)(x
2
 + 10x + 2x + 20) 
= (x + 1)[x(x + 10) + 2(x + 10)] 
= (x + 1)(x +2)(x + 10) 
 Hence, x
3
 + 13x
2
 + 32x + 20 = (x + 1)(x +2)(x + 10) 
 
15. LM = MN   (Given) 
     
MLN = MNL (angles opposite equal sides are equal)
MLQ = MNP
LP = QN (Given)
LP + PQ = PQ + QN             (adding PQ on both sides)
 LQ = PN
In LMQ and  NMP
LM = MN
MLQ = MNP
LQ = PN 
 LMQ  NMP         
? ? ?
? ? ?
?
?
??
??
? ? ?          (SAS congruence rule)
 
 
OR 
           In ? ABC and ? PQR 
  B ? = Q ? 
  BC = PQ 
  By Pythagoras theorem, 
     
2 2 2
2 2 2
2
PR PQ QR
10 6 QR
100 36 QR
QR 100 36
QR 64 8cm
??
??
??
??
??
 
     AB = QR 
  Therefore,  
    ? ABC ? ? PQR (SAS and RHS criteria) 
 
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