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 HeadExaminers have to go through the first five answerscripts 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 56 hours every day and evaluate 2025 answer books and should devote minimum 1520 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 HeadExaminers have to go through the first five answerscripts 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 56 hours every day and evaluate 2025 answer books and should devote minimum 1520 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 HeadExaminers have to go through the first five answerscripts 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 56 hours every day and evaluate 2025 answer books and should devote minimum 1520 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 HeadExaminers have to go through the first five answerscripts 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 56 hours every day and evaluate 2025 answer books and should devote minimum 1520 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 midpoint 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 ? Nonzero 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 HeadExaminers have to go through the first five answerscripts 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 56 hours every day and evaluate 2025 answer books and should devote minimum 1520 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 midpoint 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 ? Nonzero 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 mRead More
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