Maths Past Year Paper SA-2 solution(Set -4) - 2015, Class 10, CBSE Class 10 Notes | EduRev

Past Year Papers For Class 10

Class 10 : Maths Past Year Paper SA-2 solution(Set -4) - 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 (Delhi) 30/1/1, 30/1/2, 30/1/3  
 
 
 
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 (Delhi) 30/1/1, 30/1/2, 30/1/3  
 
 
 
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 30/1/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1.
4
9 –
2.     1 : 3 3 .   
26
21
4.    25
o
1×4 = 4 m
SECTION - B
5.
o
29 AOQ
2
1
ABQ ? ? ? ?
1 m
? ?
o o o o
61 119 – 180 BAT ABQ – 180 ATQ ? ? ? ? ? ? ? 1 m
6. The given quadratic equation can be written as
? ? 0 b – a x 4a – 4x
4 2 2 2
? ? ½ m
? ? ? ? 0 b – a – 2x or
2
2
2
2
?
1 m
? ? ? ?
?
?
?
?
?
?
? ?
? ? ?
2
b a
,
2
b – a
x
0 b – a – 2x b a – 2x
2 2 2 2
2
2 2 2 2
½ m
7.
?
?
?
?
?
?
?
?
?
? ? ?
?
?
?
OT)  to inclined
equally are TQ and (TP 2  1     
TC  TC     
TQ  TP     
TQC and TPC s ? In
1 m
TQC ? TPC ? ? ?
? ? ? ? ? ? ? 4 3 and QC PC ? ? ? ? ? ½ m
Q.No. Marks
Page 3


1 
Strictly Confidential: (For Internal and Restricted Use Only) 
 
Secondary School Examination 
 
March - 2015 
 
Marking Scheme--- Mathematics (Delhi) 30/1/1, 30/1/2, 30/1/3  
 
 
 
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 30/1/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1.
4
9 –
2.     1 : 3 3 .   
26
21
4.    25
o
1×4 = 4 m
SECTION - B
5.
o
29 AOQ
2
1
ABQ ? ? ? ?
1 m
? ?
o o o o
61 119 – 180 BAT ABQ – 180 ATQ ? ? ? ? ? ? ? 1 m
6. The given quadratic equation can be written as
? ? 0 b – a x 4a – 4x
4 2 2 2
? ? ½ m
? ? ? ? 0 b – a – 2x or
2
2
2
2
?
1 m
? ? ? ?
?
?
?
?
?
?
? ?
? ? ?
2
b a
,
2
b – a
x
0 b – a – 2x b a – 2x
2 2 2 2
2
2 2 2 2
½ m
7.
?
?
?
?
?
?
?
?
?
? ? ?
?
?
?
OT)  to inclined
equally are TQ and (TP 2  1     
TC  TC     
TQ  TP     
TQC and TPC s ? In
1 m
TQC ? TPC ? ? ?
? ? ? ? ? ? ? 4 3 and QC PC ? ? ? ? ? ½ m
Q.No. Marks
3
?
?
?
?
? ? ? ? ? ? ? ? ?
PQ of bisector right the is OT
90 4 3 180 4 3 But
o o
½ m
8. The given A.P. is 6, 13, 20, ---, 216
Let n be the number of terms, d = 7, a = 6 ½ m
?
216 = 6 + (n – 1) .7 
   
?    n = 31 ½ m
?
Middle term is 16th ½ m
?
a
16
  = 6 + 15 × 7 = 111 ½ m
9. ABC is right triangle
?
AC
2
 = BC
2  
+
 
 AB
2
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
?
?
?
?
?
?
?
? ? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ? ?
2 2 2 2
2 2 2 2
2 2 2
t – 2 49 t – 2  2 5 A
2 t 6 1 2 t  2 2 B
5 AB 25 2 2  2 – 5 AB
C
C
1 m
? ? ? ?
? ? ? ?
?
?
?
?
?
?
?
? ? ? ?
? ?
? ? ? ? ?
1 t  8 2t 4
8 t – 2 – 2 t
2 t 41 t – 2 9 4
2 2
2 2
1 m
10. Let  P  divide AB in the ratio of k : 1 ½ m
?
?
?
?
?
?
?
? ?
? ? ? ? ?
?
?
?
5
1
 K
3 3K   2 K 8
4
3
1 K
2
1
K 2
1 m
?
Required ratio = 1 : 5 ½ m
Page 4


1 
Strictly Confidential: (For Internal and Restricted Use Only) 
 
Secondary School Examination 
 
March - 2015 
 
Marking Scheme--- Mathematics (Delhi) 30/1/1, 30/1/2, 30/1/3  
 
 
 
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 30/1/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1.
4
9 –
2.     1 : 3 3 .   
26
21
4.    25
o
1×4 = 4 m
SECTION - B
5.
o
29 AOQ
2
1
ABQ ? ? ? ?
1 m
? ?
o o o o
61 119 – 180 BAT ABQ – 180 ATQ ? ? ? ? ? ? ? 1 m
6. The given quadratic equation can be written as
? ? 0 b – a x 4a – 4x
4 2 2 2
? ? ½ m
? ? ? ? 0 b – a – 2x or
2
2
2
2
?
1 m
? ? ? ?
?
?
?
?
?
?
? ?
? ? ?
2
b a
,
2
b – a
x
0 b – a – 2x b a – 2x
2 2 2 2
2
2 2 2 2
½ m
7.
?
?
?
?
?
?
?
?
?
? ? ?
?
?
?
OT)  to inclined
equally are TQ and (TP 2  1     
TC  TC     
TQ  TP     
TQC and TPC s ? In
1 m
TQC ? TPC ? ? ?
? ? ? ? ? ? ? 4 3 and QC PC ? ? ? ? ? ½ m
Q.No. Marks
3
?
?
?
?
? ? ? ? ? ? ? ? ?
PQ of bisector right the is OT
90 4 3 180 4 3 But
o o
½ m
8. The given A.P. is 6, 13, 20, ---, 216
Let n be the number of terms, d = 7, a = 6 ½ m
?
216 = 6 + (n – 1) .7 
   
?    n = 31 ½ m
?
Middle term is 16th ½ m
?
a
16
  = 6 + 15 × 7 = 111 ½ m
9. ABC is right triangle
?
AC
2
 = BC
2  
+
 
 AB
2
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
?
?
?
?
?
?
?
? ? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ? ?
2 2 2 2
2 2 2 2
2 2 2
t – 2 49 t – 2  2 5 A
2 t 6 1 2 t  2 2 B
5 AB 25 2 2  2 – 5 AB
C
C
1 m
? ? ? ?
? ? ? ?
?
?
?
?
?
?
?
? ? ? ?
? ?
? ? ? ? ?
1 t  8 2t 4
8 t – 2 – 2 t
2 t 41 t – 2 9 4
2 2
2 2
1 m
10. Let  P  divide AB in the ratio of k : 1 ½ m
?
?
?
?
?
?
?
? ?
? ? ? ? ?
?
?
?
5
1
 K
3 3K   2 K 8
4
3
1 K
2
1
K 2
1 m
?
Required ratio = 1 : 5 ½ m
4
SECTION - C
11. P is the mid-point of AB
?
    x + 1 = 4   ?   x = 3
        similarly y = 2    ?     B (3, 2) 1 m
        similarly finding C (–1, 2) ½ m
? ? ? ? ? ? ? ? sq.u. 12 24
2
1
2 – 4 – 1 – 4 2 3 2 – 2 1
2
1
ABC ? Area ? ? ? ? ? ? ?
1½ m
12. The given quadratic eqn. can be written as
(k + 1)x
2  
– 2(k – 1)x + 1 = 0 1 m
?
?
?
? ?
? ? ?
3 0,  k                                        
0 3k – k or         0  1) 4(k – 1) – 4(k     roots qual For 
2 2
1 m
?
    Non-zero value of k = 3 : Roots are 
2
1
,
2
1
½+½ m
13.                              Fiqure ½ m
30 y 1 45 tan
y
30
(i)
o
? ? ? ?
1 m
3 10
3
30
3
y
x
3
1
0 3 tan
y
x
(ii)
o
? ? ? ? ? ?
1 m
         
m 3 10 is building of Height ?
½ m
14. Total possible out comes = 36
(i) The possible outcomes are (2, 3), (3, 2), (1, 4), (4, 1) : Number : 4 1 m
9
1
36
4
y Probabilit Required ? ? ?
½ m
Page 5


1 
Strictly Confidential: (For Internal and Restricted Use Only) 
 
Secondary School Examination 
 
March - 2015 
 
Marking Scheme--- Mathematics (Delhi) 30/1/1, 30/1/2, 30/1/3  
 
 
 
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 30/1/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1.
4
9 –
2.     1 : 3 3 .   
26
21
4.    25
o
1×4 = 4 m
SECTION - B
5.
o
29 AOQ
2
1
ABQ ? ? ? ?
1 m
? ?
o o o o
61 119 – 180 BAT ABQ – 180 ATQ ? ? ? ? ? ? ? 1 m
6. The given quadratic equation can be written as
? ? 0 b – a x 4a – 4x
4 2 2 2
? ? ½ m
? ? ? ? 0 b – a – 2x or
2
2
2
2
?
1 m
? ? ? ?
?
?
?
?
?
?
? ?
? ? ?
2
b a
,
2
b – a
x
0 b – a – 2x b a – 2x
2 2 2 2
2
2 2 2 2
½ m
7.
?
?
?
?
?
?
?
?
?
? ? ?
?
?
?
OT)  to inclined
equally are TQ and (TP 2  1     
TC  TC     
TQ  TP     
TQC and TPC s ? In
1 m
TQC ? TPC ? ? ?
? ? ? ? ? ? ? 4 3 and QC PC ? ? ? ? ? ½ m
Q.No. Marks
3
?
?
?
?
? ? ? ? ? ? ? ? ?
PQ of bisector right the is OT
90 4 3 180 4 3 But
o o
½ m
8. The given A.P. is 6, 13, 20, ---, 216
Let n be the number of terms, d = 7, a = 6 ½ m
?
216 = 6 + (n – 1) .7 
   
?    n = 31 ½ m
?
Middle term is 16th ½ m
?
a
16
  = 6 + 15 × 7 = 111 ½ m
9. ABC is right triangle
?
AC
2
 = BC
2  
+
 
 AB
2
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
?
?
?
?
?
?
?
? ? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ? ?
2 2 2 2
2 2 2 2
2 2 2
t – 2 49 t – 2  2 5 A
2 t 6 1 2 t  2 2 B
5 AB 25 2 2  2 – 5 AB
C
C
1 m
? ? ? ?
? ? ? ?
?
?
?
?
?
?
?
? ? ? ?
? ?
? ? ? ? ?
1 t  8 2t 4
8 t – 2 – 2 t
2 t 41 t – 2 9 4
2 2
2 2
1 m
10. Let  P  divide AB in the ratio of k : 1 ½ m
?
?
?
?
?
?
?
? ?
? ? ? ? ?
?
?
?
5
1
 K
3 3K   2 K 8
4
3
1 K
2
1
K 2
1 m
?
Required ratio = 1 : 5 ½ m
4
SECTION - C
11. P is the mid-point of AB
?
    x + 1 = 4   ?   x = 3
        similarly y = 2    ?     B (3, 2) 1 m
        similarly finding C (–1, 2) ½ m
? ? ? ? ? ? ? ? sq.u. 12 24
2
1
2 – 4 – 1 – 4 2 3 2 – 2 1
2
1
ABC ? Area ? ? ? ? ? ? ?
1½ m
12. The given quadratic eqn. can be written as
(k + 1)x
2  
– 2(k – 1)x + 1 = 0 1 m
?
?
?
? ?
? ? ?
3 0,  k                                        
0 3k – k or         0  1) 4(k – 1) – 4(k     roots qual For 
2 2
1 m
?
    Non-zero value of k = 3 : Roots are 
2
1
,
2
1
½+½ m
13.                              Fiqure ½ m
30 y 1 45 tan
y
30
(i)
o
? ? ? ?
1 m
3 10
3
30
3
y
x
3
1
0 3 tan
y
x
(ii)
o
? ? ? ? ? ?
1 m
         
m 3 10 is building of Height ?
½ m
14. Total possible out comes = 36
(i) The possible outcomes are (2, 3), (3, 2), (1, 4), (4, 1) : Number : 4 1 m
9
1
36
4
y Probabilit Required ? ? ?
½ m
5
(ii) The possible outcomes are
(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)
their number is 9
1 m
4
1
36
9
y Probabilit Required ? ? ?
½ m
15. Let a be the first term and d the common difference
S
12
 =  6 [2a + 11d] = 12a + 66d 1 m
S
8
  =  4 [2a + 7d] = 8a + 28d ½ m
S
4
  =  2 [2a + 3d] = 4a + 6d ½ m
3 (S
8
  –  3
4
) = 3 (4a + 22d) = 12a + 66d = S
12
1 m
16. Let OA = OB = r
?
?
?
?
?
?
? ? ? ? ? ? ? ?
7 r                                           
40r 280 r r
7
22
2
r
7
22
40
1 m
2
cm 7 7
7
22
2
1
2
7
2
7
7
22
2
1
area shaded ?
?
?
?
?
?
? ? ? ? ? ? ? ? ?
1 m
      
2 2
cm
4
1
96 cm
4
385
4
5
7 7 ? ?
?
?
?
?
?
? ? or
1 m
17. ? ?ARQ   ~   ? ?ADC ½ m
2 x
12
4
6
x
? ? ? ?
½ m
5 4 4 8 QC
2 2
? ? ?
½ m
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