Maths Past Year Paper SA-2(Set -5) - 2015, Class 10, CBSE Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Created by: Indu Gupta

Class 10 : Maths Past Year Paper SA-2(Set -5) - 2015, Class 10, CBSE Class 10 Notes | EduRev

 Page 1


1 
Strictly Confidential: (For Internal and Restricted Use Only) 
 
Secondary School Examination 
 
March -2015 
 
Marking Scheme--- Mathematics (Outside Delhi) 30/1, 30/2, 30/3  
 
 
 
General Instructions 
 
1. The Marking Scheme provides general guidelines to reduce subjectivity 
and maintain uniformity among large number of examiners involved in the 
marking. The answers given in the marking scheme are the best suggested 
answers. 
2. Marking is to be done as per the instructions provided in the marking 
scheme. (It should not be done according to one’s own interpretation or any 
other consideration.)Marking Scheme should be strictly adhered to and 
religiously followed. 
3. Alternative methods are accepted. Proportional marks are to be awarded. 
4. The Head-Examiners have to go through the first five answer-scripts 
evaluated by each evaluator to ensure that the evaluation has been done as 
per instructions given in the marking scheme. The remaining answer scripts 
meant for evaluation shall be given only after ensuring that there is no 
significant variation in the marking of individual evaluators. 
5. If a question is attempted twice and the candidate has not crossed any 
answer, only first attempt is to be evaluated. Write ‘EXTRA’ with second 
attempt. 
6. A full scale of marks 0 to 90 has to be used. Please do not hesitate to award 
full marks if the answer deserves it. 
7. Separate Marking Scheme for all the three sets has been given. 
8. The Examiners should acquaint themselves with the guidelines given in the 
Guidelines for Spot Evaluation before starting the actual evaluation. 
9. Every Examiner should stay upto sufficiently reasonable time normally 5-6 
hours every day and evaluate 20-25 answer books and should devote 
minimum 15-20 minutes to evaluate each answer book. 
10. Every Examiner should acquaint himself/herself with the marking schemes 
of all the sets. 
Page 2


1 
Strictly Confidential: (For Internal and Restricted Use Only) 
 
Secondary School Examination 
 
March -2015 
 
Marking Scheme--- Mathematics (Outside Delhi) 30/1, 30/2, 30/3  
 
 
 
General Instructions 
 
1. The Marking Scheme provides general guidelines to reduce subjectivity 
and maintain uniformity among large number of examiners involved in the 
marking. The answers given in the marking scheme are the best suggested 
answers. 
2. Marking is to be done as per the instructions provided in the marking 
scheme. (It should not be done according to one’s own interpretation or any 
other consideration.)Marking Scheme should be strictly adhered to and 
religiously followed. 
3. Alternative methods are accepted. Proportional marks are to be awarded. 
4. The Head-Examiners have to go through the first five answer-scripts 
evaluated by each evaluator to ensure that the evaluation has been done as 
per instructions given in the marking scheme. The remaining answer scripts 
meant for evaluation shall be given only after ensuring that there is no 
significant variation in the marking of individual evaluators. 
5. If a question is attempted twice and the candidate has not crossed any 
answer, only first attempt is to be evaluated. Write ‘EXTRA’ with second 
attempt. 
6. A full scale of marks 0 to 90 has to be used. Please do not hesitate to award 
full marks if the answer deserves it. 
7. Separate Marking Scheme for all the three sets has been given. 
8. The Examiners should acquaint themselves with the guidelines given in the 
Guidelines for Spot Evaluation before starting the actual evaluation. 
9. Every Examiner should stay upto sufficiently reasonable time normally 5-6 
hours every day and evaluate 20-25 answer books and should devote 
minimum 15-20 minutes to evaluate each answer book. 
10. Every Examiner should acquaint himself/herself with the marking schemes 
of all the sets. 
2
QUESTION PAPER CODE 30/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1. p = 3 1 m
2. 30
o
1 m
3.
9
1
1 m
4. 120
o
1 m
SECTION - B
5.           ? ? ? ?POR  =  90 – 60 = 30
o
½ m
  
PR 2 OR
2
1
sin30
OR
PRO
o
? ? ? ?
                                     =  PR + QR ½ m
6.              Let AF = AE  =  x
?
  AB  =  6 + x,  AC  =  9 + x,  BC  =  15 ½ m
? ? 54 3 x 9 x 6 15
2
1
? ? ? ? ? ?
1 m
?
  x = 3  
?
  AB  =  9 cm,  AC =  12 cm ½ m
                       and BC  =  15 cm
Q.No. Marks
Page 3


1 
Strictly Confidential: (For Internal and Restricted Use Only) 
 
Secondary School Examination 
 
March -2015 
 
Marking Scheme--- Mathematics (Outside Delhi) 30/1, 30/2, 30/3  
 
 
 
General Instructions 
 
1. The Marking Scheme provides general guidelines to reduce subjectivity 
and maintain uniformity among large number of examiners involved in the 
marking. The answers given in the marking scheme are the best suggested 
answers. 
2. Marking is to be done as per the instructions provided in the marking 
scheme. (It should not be done according to one’s own interpretation or any 
other consideration.)Marking Scheme should be strictly adhered to and 
religiously followed. 
3. Alternative methods are accepted. Proportional marks are to be awarded. 
4. The Head-Examiners have to go through the first five answer-scripts 
evaluated by each evaluator to ensure that the evaluation has been done as 
per instructions given in the marking scheme. The remaining answer scripts 
meant for evaluation shall be given only after ensuring that there is no 
significant variation in the marking of individual evaluators. 
5. If a question is attempted twice and the candidate has not crossed any 
answer, only first attempt is to be evaluated. Write ‘EXTRA’ with second 
attempt. 
6. A full scale of marks 0 to 90 has to be used. Please do not hesitate to award 
full marks if the answer deserves it. 
7. Separate Marking Scheme for all the three sets has been given. 
8. The Examiners should acquaint themselves with the guidelines given in the 
Guidelines for Spot Evaluation before starting the actual evaluation. 
9. Every Examiner should stay upto sufficiently reasonable time normally 5-6 
hours every day and evaluate 20-25 answer books and should devote 
minimum 15-20 minutes to evaluate each answer book. 
10. Every Examiner should acquaint himself/herself with the marking schemes 
of all the sets. 
2
QUESTION PAPER CODE 30/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1. p = 3 1 m
2. 30
o
1 m
3.
9
1
1 m
4. 120
o
1 m
SECTION - B
5.           ? ? ? ?POR  =  90 – 60 = 30
o
½ m
  
PR 2 OR
2
1
sin30
OR
PRO
o
? ? ? ?
                                     =  PR + QR ½ m
6.              Let AF = AE  =  x
?
  AB  =  6 + x,  AC  =  9 + x,  BC  =  15 ½ m
? ? 54 3 x 9 x 6 15
2
1
? ? ? ? ? ?
1 m
?
  x = 3  
?
  AB  =  9 cm,  AC =  12 cm ½ m
                       and BC  =  15 cm
Q.No. Marks
3
7. 4x
2
 + 4bx + b
2
 – a
2
 = 0   
?
  (2 x + b)
2
 – (a)
2
  = 0 ½ m
?
  (2x + b + a)  (2x + b – a)  =  0 ½ m
2
b a
x ,
2
b a
x
?
?
?
? ? ?
½ +½ m
8.
? ? ? ? 167 6d 2a
2
7
4d 2a
2
5
167 S S
7 5
? ? ? ? ? ? ?
24a + 62d  =  334    or   12a + 31d  =  167 .............................(i) ½ m
? ? 47 9d 2a or 235 9d 2a 5 235 S
10
? ? ? ? ? ? ..............(ii) ½ m
Solving (i) and (ii) to get  a = 1,  d = 5.  Hence AP is  1, 6, 11,  ......... ½ + ½ m
9. Here,  AB
2
 + BC
2
  =  AC
2
½ m
?
  (4)
2
 + (p – 4)
2 
 + (7 – p)
2  
 =  (3)
2
 + (– 4)
2
?
  p = 7   or  4 1 m
4 p 7 p since ? ? ? ½ m
10. Using  ar  ( ? ?ABC)  =  0 ½ m
?
  x (7–5) – 5 (5 – y) – 4 (y – 7)  =  0 1 m
2x  –  25 + 5y – 4y + 28  =  0
2x + y + 3  =  0 ½ m
SECTION - C
11. a
14
 = 2 a
8
   ?  a + 13d  =  2  (a + 7d)    ?   a = – d 1 m
a
6
 = – 8       ?  a + 5d = – 8 ½ m
solving to get a = 2,    d = – 2 ½ m
S
20
 =  10 (2a + 19d)  =  10 (4 – 38)  =  – 340 1 m
Page 4


1 
Strictly Confidential: (For Internal and Restricted Use Only) 
 
Secondary School Examination 
 
March -2015 
 
Marking Scheme--- Mathematics (Outside Delhi) 30/1, 30/2, 30/3  
 
 
 
General Instructions 
 
1. The Marking Scheme provides general guidelines to reduce subjectivity 
and maintain uniformity among large number of examiners involved in the 
marking. The answers given in the marking scheme are the best suggested 
answers. 
2. Marking is to be done as per the instructions provided in the marking 
scheme. (It should not be done according to one’s own interpretation or any 
other consideration.)Marking Scheme should be strictly adhered to and 
religiously followed. 
3. Alternative methods are accepted. Proportional marks are to be awarded. 
4. The Head-Examiners have to go through the first five answer-scripts 
evaluated by each evaluator to ensure that the evaluation has been done as 
per instructions given in the marking scheme. The remaining answer scripts 
meant for evaluation shall be given only after ensuring that there is no 
significant variation in the marking of individual evaluators. 
5. If a question is attempted twice and the candidate has not crossed any 
answer, only first attempt is to be evaluated. Write ‘EXTRA’ with second 
attempt. 
6. A full scale of marks 0 to 90 has to be used. Please do not hesitate to award 
full marks if the answer deserves it. 
7. Separate Marking Scheme for all the three sets has been given. 
8. The Examiners should acquaint themselves with the guidelines given in the 
Guidelines for Spot Evaluation before starting the actual evaluation. 
9. Every Examiner should stay upto sufficiently reasonable time normally 5-6 
hours every day and evaluate 20-25 answer books and should devote 
minimum 15-20 minutes to evaluate each answer book. 
10. Every Examiner should acquaint himself/herself with the marking schemes 
of all the sets. 
2
QUESTION PAPER CODE 30/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1. p = 3 1 m
2. 30
o
1 m
3.
9
1
1 m
4. 120
o
1 m
SECTION - B
5.           ? ? ? ?POR  =  90 – 60 = 30
o
½ m
  
PR 2 OR
2
1
sin30
OR
PRO
o
? ? ? ?
                                     =  PR + QR ½ m
6.              Let AF = AE  =  x
?
  AB  =  6 + x,  AC  =  9 + x,  BC  =  15 ½ m
? ? 54 3 x 9 x 6 15
2
1
? ? ? ? ? ?
1 m
?
  x = 3  
?
  AB  =  9 cm,  AC =  12 cm ½ m
                       and BC  =  15 cm
Q.No. Marks
3
7. 4x
2
 + 4bx + b
2
 – a
2
 = 0   
?
  (2 x + b)
2
 – (a)
2
  = 0 ½ m
?
  (2x + b + a)  (2x + b – a)  =  0 ½ m
2
b a
x ,
2
b a
x
?
?
?
? ? ?
½ +½ m
8.
? ? ? ? 167 6d 2a
2
7
4d 2a
2
5
167 S S
7 5
? ? ? ? ? ? ?
24a + 62d  =  334    or   12a + 31d  =  167 .............................(i) ½ m
? ? 47 9d 2a or 235 9d 2a 5 235 S
10
? ? ? ? ? ? ..............(ii) ½ m
Solving (i) and (ii) to get  a = 1,  d = 5.  Hence AP is  1, 6, 11,  ......... ½ + ½ m
9. Here,  AB
2
 + BC
2
  =  AC
2
½ m
?
  (4)
2
 + (p – 4)
2 
 + (7 – p)
2  
 =  (3)
2
 + (– 4)
2
?
  p = 7   or  4 1 m
4 p 7 p since ? ? ? ½ m
10. Using  ar  ( ? ?ABC)  =  0 ½ m
?
  x (7–5) – 5 (5 – y) – 4 (y – 7)  =  0 1 m
2x  –  25 + 5y – 4y + 28  =  0
2x + y + 3  =  0 ½ m
SECTION - C
11. a
14
 = 2 a
8
   ?  a + 13d  =  2  (a + 7d)    ?   a = – d 1 m
a
6
 = – 8       ?  a + 5d = – 8 ½ m
solving to get a = 2,    d = – 2 ½ m
S
20
 =  10 (2a + 19d)  =  10 (4 – 38)  =  – 340 1 m
4
12.
0 3 2 x 2 2 x 3
2
? ? ?
? ? ? ? 0 2 x 3 6 x 0 3 2 x 2 x 2 3 x 3
2
? ? ? ? ? ? ? ? ?
1+1 m
3
2
, 6 x ? ? ? ? x
½ + ½ m
13. Let AL  =  x   
0
60 tan
x
BL
? ?
        Fig. ½ m
m. 1500 x 3
x
3 1500
? ? ? ? 1 m
3
1
30 tan
M L AL
CM
0
? ?
?
?
  1500 + LM  =  1500 (3)  =  4500 1 m
             
?
  LM  =  3000 m.
                
?
  Speed  =  
15
3000
 =  200 m./s.  =  720 Km/hr. . ½ m
14.
4 : 3 PB : AP AB
7
3
AP ? ? ?
1 m
? ? ? ?
7
2
7
8 6
x
4 2, 4 : 3 2 2,
B y) (x, P A
? ?
?
? ?
? ? ?
1 m
7
20
7
8 2 1 –
y ? ?
?
?
½ m
?
?
?
?
?
?
? ?
7
20
,
7
2
P
½ m
15.
? ? ? ?
3
1
blue P ,
4
1
Red P ? ?
? ?
12
5
3
1
4
1
1 orange P ? ? ? ? ?
1½ m
       
? ? 10 balls of no. Total
12
5
? ?
½ m
            
24
5
12 10
balls of no. Total ?
?
? ?
1 m
Page 5


1 
Strictly Confidential: (For Internal and Restricted Use Only) 
 
Secondary School Examination 
 
March -2015 
 
Marking Scheme--- Mathematics (Outside Delhi) 30/1, 30/2, 30/3  
 
 
 
General Instructions 
 
1. The Marking Scheme provides general guidelines to reduce subjectivity 
and maintain uniformity among large number of examiners involved in the 
marking. The answers given in the marking scheme are the best suggested 
answers. 
2. Marking is to be done as per the instructions provided in the marking 
scheme. (It should not be done according to one’s own interpretation or any 
other consideration.)Marking Scheme should be strictly adhered to and 
religiously followed. 
3. Alternative methods are accepted. Proportional marks are to be awarded. 
4. The Head-Examiners have to go through the first five answer-scripts 
evaluated by each evaluator to ensure that the evaluation has been done as 
per instructions given in the marking scheme. The remaining answer scripts 
meant for evaluation shall be given only after ensuring that there is no 
significant variation in the marking of individual evaluators. 
5. If a question is attempted twice and the candidate has not crossed any 
answer, only first attempt is to be evaluated. Write ‘EXTRA’ with second 
attempt. 
6. A full scale of marks 0 to 90 has to be used. Please do not hesitate to award 
full marks if the answer deserves it. 
7. Separate Marking Scheme for all the three sets has been given. 
8. The Examiners should acquaint themselves with the guidelines given in the 
Guidelines for Spot Evaluation before starting the actual evaluation. 
9. Every Examiner should stay upto sufficiently reasonable time normally 5-6 
hours every day and evaluate 20-25 answer books and should devote 
minimum 15-20 minutes to evaluate each answer book. 
10. Every Examiner should acquaint himself/herself with the marking schemes 
of all the sets. 
2
QUESTION PAPER CODE 30/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1. p = 3 1 m
2. 30
o
1 m
3.
9
1
1 m
4. 120
o
1 m
SECTION - B
5.           ? ? ? ?POR  =  90 – 60 = 30
o
½ m
  
PR 2 OR
2
1
sin30
OR
PRO
o
? ? ? ?
                                     =  PR + QR ½ m
6.              Let AF = AE  =  x
?
  AB  =  6 + x,  AC  =  9 + x,  BC  =  15 ½ m
? ? 54 3 x 9 x 6 15
2
1
? ? ? ? ? ?
1 m
?
  x = 3  
?
  AB  =  9 cm,  AC =  12 cm ½ m
                       and BC  =  15 cm
Q.No. Marks
3
7. 4x
2
 + 4bx + b
2
 – a
2
 = 0   
?
  (2 x + b)
2
 – (a)
2
  = 0 ½ m
?
  (2x + b + a)  (2x + b – a)  =  0 ½ m
2
b a
x ,
2
b a
x
?
?
?
? ? ?
½ +½ m
8.
? ? ? ? 167 6d 2a
2
7
4d 2a
2
5
167 S S
7 5
? ? ? ? ? ? ?
24a + 62d  =  334    or   12a + 31d  =  167 .............................(i) ½ m
? ? 47 9d 2a or 235 9d 2a 5 235 S
10
? ? ? ? ? ? ..............(ii) ½ m
Solving (i) and (ii) to get  a = 1,  d = 5.  Hence AP is  1, 6, 11,  ......... ½ + ½ m
9. Here,  AB
2
 + BC
2
  =  AC
2
½ m
?
  (4)
2
 + (p – 4)
2 
 + (7 – p)
2  
 =  (3)
2
 + (– 4)
2
?
  p = 7   or  4 1 m
4 p 7 p since ? ? ? ½ m
10. Using  ar  ( ? ?ABC)  =  0 ½ m
?
  x (7–5) – 5 (5 – y) – 4 (y – 7)  =  0 1 m
2x  –  25 + 5y – 4y + 28  =  0
2x + y + 3  =  0 ½ m
SECTION - C
11. a
14
 = 2 a
8
   ?  a + 13d  =  2  (a + 7d)    ?   a = – d 1 m
a
6
 = – 8       ?  a + 5d = – 8 ½ m
solving to get a = 2,    d = – 2 ½ m
S
20
 =  10 (2a + 19d)  =  10 (4 – 38)  =  – 340 1 m
4
12.
0 3 2 x 2 2 x 3
2
? ? ?
? ? ? ? 0 2 x 3 6 x 0 3 2 x 2 x 2 3 x 3
2
? ? ? ? ? ? ? ? ?
1+1 m
3
2
, 6 x ? ? ? ? x
½ + ½ m
13. Let AL  =  x   
0
60 tan
x
BL
? ?
        Fig. ½ m
m. 1500 x 3
x
3 1500
? ? ? ? 1 m
3
1
30 tan
M L AL
CM
0
? ?
?
?
  1500 + LM  =  1500 (3)  =  4500 1 m
             
?
  LM  =  3000 m.
                
?
  Speed  =  
15
3000
 =  200 m./s.  =  720 Km/hr. . ½ m
14.
4 : 3 PB : AP AB
7
3
AP ? ? ?
1 m
? ? ? ?
7
2
7
8 6
x
4 2, 4 : 3 2 2,
B y) (x, P A
? ?
?
? ?
? ? ?
1 m
7
20
7
8 2 1 –
y ? ?
?
?
½ m
?
?
?
?
?
?
? ?
7
20
,
7
2
P
½ m
15.
? ? ? ?
3
1
blue P ,
4
1
Red P ? ?
? ?
12
5
3
1
4
1
1 orange P ? ? ? ? ?
1½ m
       
? ? 10 balls of no. Total
12
5
? ?
½ m
            
24
5
12 10
balls of no. Total ?
?
? ?
1 m
5
16. r  =  14 cm.  
?
 =  60
0
Area of minor segment  
? sin r
2
1
360
?
r p
2 2
? ?
½ m
2
3
14 14
2
1
360
60
14 14
7
22
? ? ? ? ? ? ? ? ½ m
Approx. cm 17.9 or cm 17.89 or cm 3 49
3
308
2 2 2
?
?
?
?
?
?
? ?
1 m
Area of Major segment
?
?
?
?
?
?
? ? ? 3 49
3
308
r p
2
½ m
2 2
cm 598.10 or cm 3 49
3
1540
?
?
?
?
?
?
? ?
½ m
                Approx. cm 598 or
2
17. Slant height 
? ? ? ? cm. 3.5 2.1 2.8 ) (
2 2
? ? ? ?
½ m
? ?
 tent one for 
3.5 2.1
7
22
4 2.1
7
22
2 canvas of Area ? ? ? ? ? ? ? ?
                    =  6.6  (8 + 3.5)  =  6.6 ? 11.5  m
2
½ m
?
  Area for 100 tents  =  66 ?115 m
2
        
 Cost of 100 tents  =  Rs. 66 ?115 ?100 ½ m
50% Cost  = 33 ? 11500  =  Rs. 379500 ½ m
V alues :  Helping the flood victims 1 m
18. Volume of liquid in the bowl  =  
? ?
3 3
cm 18 p
3
2
? ?
½ m
Volume, after wastage  =  
? ?
3 3
cm
100
90
18
3
p 2
? ?
½ m
Volume of liquid in 72 bottles  =  ? ?
3 2
cm 72 3 p ? ? h
½ m
? ?
? ?
cm. 5.4
72 3 p
10
9
18 p
3
2
h
2
3
?
?
?
? ?
½ + 1 m
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