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CBSE 10th Board Class 10 2020 : Mathematics

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Thakkar Tejendra
Swaminarayan Dham International School (SDIS), Gandhinagar

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Strictly Confidential: (For Internal and Restricted use only) Secondary School Examination March 2019 Marking Scheme MATHEMATICS ( SUBJECT CODE -041 ) PAPER CODE: 30/1/1, 30/1/2, 30/1/3 General Instructions: 1. You are aware that evaluation is the most important process in the actual and correct assessment of the candidates. A small mistake in evaluation may lead to serious problems which may affect the future of the candidates, education system and teaching profession. To avoid mistakes, it is requested that before starting evaluation, you must read and understand the spot evaluation guidelines carefully. Evaluation is a 10-12 days mission for all of us. Hence, it is necessary that you put in your best efforts in this process. 2. Evaluation is to be done as per 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. However, while evaluating, answers which are based on latest information or knowledge and/or are innovative, they may be assessed for their correctness otherwise and marks be awarded to them. 3. The Head-Examiner must go through the first five answer books evaluated by each evaluator on the first day, to ensure that evaluation has been carried out as per the instructions given in the Marking Scheme. The remaining answer books meant for evaluation shall be given only after ensuring that there is no significant variation in the marking of individual evaluators. 4. If a question has parts, please award marks on the right-hand side for each part. Marks awarded for different parts of the question should then be totaled up and written in the left-hand margin and encircled. 5. If a question does not have any parts, marks must be awarded in the left hand margin and encircled. 6. If a student has attempted an extra question, answer of the question deserving more marks should be retained and the other answer scored out. 7. No marks to be deducted for the cumulative effect of an error. It should be penalized only once. 8. A full scale of marks 1-80 has to be used. Please do not hesitate to award full marks if the answer deserves it. 9. Every examiner has to necessarily do evaluation work for full working hours i.e. 8 hours every day and evaluate 25 answer books per day. 10. Ensure that you do not make the following common types of errors committed by the Examiner in the past: Leaving answer or part thereof unassessed in an answer book. Giving more marks for an answer than assigned to it. Wrong transfer of marks from the inside pages of the answer book to the title page. Wrong question wise totaling on the title page. Wrong totaling of marks of the two columns on the title page. Wrong grand total. Marks in words and figures not tallying. Wrong transfer of marks from the answer book to online award list. Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly and clearly indicated. It should merely be a line. Same is with the X for incorrect answer.) Half or a part of answer marked correct and the rest as wrong, but no marks awarded. 11. While evaluating the answer books if the answer is found to be totally incorrect, it should be marked as (X) and awarded zero (0) Marks. 12. Any unassessed portion, non-carrying over of marks to the title page, or totaling error detected by the candidate shall damage the prestige of all the personnel engaged in the evaluation work as also of the Board. Hence, in order to uphold the prestige of all concerned, it is again reiterated that the instructions be followed meticulously and judiciously. 13. The Examiners should acquaint themselves with the guidelines given in the Guidelines for spot Evaluation before starting the actual evaluation. 14. Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title page, correctly totaled and written in figures and words. 15. The Board permits candidates to obtain photocopy of the Answer Book on request in an RTI application and also separately as a part of the re-evaluation process on payment of the processing charges. 30/1/1 QUESTION PAPER CODE 30/1/1 EXPECTED ANSWER/VALUE POINTS SECTION A 1. 2. Let the point A be (x, y) 4+y 1+ x = 2 and = 3 2 2 x = 3 and y = 10 Point A is (3, 10) 1 2 1 2 Since roots of the equation x2 + 4x + k = 0 are real 16 4k 0 1 2 k 4 1 2 OR Roots of the equation 3x2 10x + k = 0 are reciprocal of each other 3. Product of roots = 1 1 2 k =1 k=3 3 1 2 tan 2 A = cot (90 2A) 90 2A = A 24 1 2 A = 38 1 2 OR 30/1/1 sin 33 = cos 57 1 2 sin2 33 + sin2 57 = cos2 57 + sin2 57 = 1 1 2 (1) 30/1/1 4. 5. 1 2 Numbers are 12, 15, 18, ..., 99 99 = 12 + (n 1) 3 n = 30 1 2 1 2 AB = 1 + 2 = 3 cm ABC ~ ADE 6. AB2 9 ar (A BC) = = AD 2 1 ar (ADE) ar( ABC) : ar( ADE) = 9 : 1 Any one rational number between 1 2 2 (1.41 approx.) and 3 (1.73 approx.) 1 e.g., 1.5, 1.6, 1.63 etc. SECTION B 7. Using Euclid s Algorithm 7344 = 1260 5 + 1044 1260 = 1044 1 + 216 1044 = 216 4 + 180 216 = 180 1 + 36 180 = 36 5 + 0 1 1 2 1 2 HCF of 1260 and 7344 is 36. OR Using Euclid s Algorithm a = 4q + r, 0 r < 4 a = 4q, a = 4q + 1, a = 4q + 2 and a = 4q + 3. Now a = 4q and a = 4q + 2 are even numbers. 1 1 2 Therefore when a is odd, it is of the form a = 4q + 1 or a = 4q + 3 for some integer q. (2) 1 2 30/1/1 30/1/1 8. an = a21 + 120 = (3 + 20 12) + 120 = 363 1 363 = 3 + (n 1) 12 n = 31 1 or 31st term is 120 more than a21. OR a1 = S1 = 3 4 = 1 1 2 a2 = S2 S1 = [3(2)2 4(2)] ( 1) = 5 1 2 d = a2 a1 = 6 1 2 Hence an = 1 + (n 1) 6 = 6n 7 1 2 Alternate method: Sn = 3n2 4n Sn 1 = 3(n 1)2 4(n 1) = 3n2 10n + 7 1 Hence an = Sn Sn 1 = (3n2 4n) (3n2 10n + 7) 1 2 = 6n 7 1 2 9. K A(1, 3) Let the required point be (a, 0) and required ratio AP : PB = k : 1 P(a, 0) 1 a= 4k + 1 k +1 0= 5k 3 k +1 B(4,5) k= 3 or required ratio is 3 : 5 5 17 Point P is , 8 30/1/1 (3) 0 1 2 1 1 2 30/1/1 10. Total number of outcomes = 8 1 2 Favourable number of outcomes (HHH, TTT) = 2 1 2 Prob. (getting success) = 11. 1 2 2 1 or 8 4 Prob. (losing the game) = 1 1 2 1 3 = . 4 4 Total number of outcomes = 6. (i) Prob. (getting a prime number (2, 3, 5)) = 3 1 or 6 2 (ii) Prob. (getting a number between 2 and 6 (3, 4, 5)) = 12. 1 3 1 or . 6 2 1 System of equations has infinitely many solutions c 3 3 c = = 12 c c 1 2 c2 = 36 c = 6 or c = 6 ...(1) 1 2 Also 3c = 3c c2 c = 6 or c = 0 ...(2) 1 2 From equations (1) and (2) 1 2 c = 6. SECTION C 13. Let us assume 2 be a rational number and its simplest form be integers and b 0. a , a and b are coprime positive b a b So 2 = a2 = 2b2 1 Thus a2 is a multiple of 2 1 2 a is a multiple of 2. Let a = 2 m for some integer m (4) 30/1/1 30/1/1 1 2 b2 = 2m2 Thus b2 is a multiple of 2 1 2 b is a multiple of 2 Hence 2 is a common factor of a and b. This contradicts the fact that a and b are coprimes Hence 14. Sum of zeroes = k + 6 1 Product of zeroes = 2(2k 1) 1 Hence k + 6 = 15. 1 2 2 is an irrational number. 1 2(2k 1) 2 k=7 1 Let sum of the ages of two children be x yrs and father s age be y yrs. y = 3x and y + 5 = 2(x + 10) ...(1) 1 ...(2) 1 Solving equations (1) and (2) x = 15 and y = 45 Father s present age is 45 years. 1 OR Let the fraction be x y x 2 1 = y 3 ...(1) 1 and x 1 = y 1 2 ...(2) 1 Solving (1) and (2) to get x = 7, y = 15. 30/1/1 Required fraction is 7 15 1 (5) 30/1/1 16. 1 2 Let the required point on y-axis be (0, b) (5 0)2 + ( 2 b)2 = ( 3 0)2 + (2 b)2 29 + 4b + b2 = 13 + b2 4b b = 2 1 Required point is (0, 2) 1 2 1 0 OR k= y+ 2x x= Q A(2, 1) P(x, y) 1 2 AP : PB = 1 : 2 4+5 2 8 = 3 and y = = 2 3 3 1 1 + 2 2 B(5, 8) 1 2 Thus point P is (3, 2). Point (3, 2) lies on 2x y + k = 0 6+2+k=0 k = 8. 17. 1 LHS = sin2 + cosec2 + 2sin cosec + cos2 + sec2 + 2cos sec = (sin2 + cos2 ) + cosec2 + sec2 + 1 2sin cos +2 . sin cos = 1 + 1 + cot2 + 1 + tan2 + 2 + 2 1 1 2 1 2 = 7 + cot2 + tan2 = RHS OR 1 cos ec A (1 + tan A + sec A) LHS = 1 + tan A = 1 (tan A + 1 sec A) (1 + tan A + sec A) tan A 1 = 1 [(1 + tan A) 2 sec 2 A] tan A 1 = 1 [1 + tan 2 A + 2 tan A 1 tan 2 A] tan A = 2 = RHS 1 (6) 30/1/1 30/1/1 Alternate method 1 sin A 1 cos A + LHS = 1 + 1 + sin A sin A cos A cos A = (sin A + cos A 1) (cos A + sin A + 1) 1 1 cos A sin A 1 2 = (sin A + cos A) 1 sin A cos A = (1 + 2sin A cos 1) 1 1 sin A cos A 1 2 1 2 = 2 = RHS 18. P TP = TQ 5 4 x T 1 2 Join OT and OQ. M Q O TM PQ and bisects PQ Hence PM = 4 cm Therefore OM = 25 16 = 9 = 3 cm. 1 2 Let TM = x From PMT, PT2 = x2 + 16 From POT, PT2 = (x + 3)2 25 Hence x2 + 16 = x2 + 9 + 6x 25 6x = 32 x = Hence PT2 = PT = 30/1/1 16 3 1 256 400 + 16 = 9 9 20 cm. 3 (7) 1 30/1/1 19. ACB ~ ADC (AA similarity) AC AD = BC CD ...(1) 1 ...(2) 1 Also ACB ~ CDB (AA similarity) AC CD = BC BD Using equations (1) and (2) AD CD = CD BD CD2 = AD BD 1 OR B Correct Figure 1 2 Q AQ2 = CQ2 + AC2 1 BP2 = CP2 + BC2 1 2 C A P AQ2 + BP2 = (CQ2 + CP2) + (AC2 + BC2) = PQ2 + AB2. 20. AC = 1 64 + 36 = 10 cm. Radius of the circle (r) = 5 cm. 1 Area of shaded region = Area of circle Ar(ABCD) = 3.14 25 6 8 1 2 1 = 78.5 48 1 2 = 30.5 cm2. 21. Length of canal covered in 30 min = 5000 m. 1 2 Volume of water flown in 30 min = 6 1.5 5000 m3 1 If 8 cm standing water is needed (8) 30/1/1 30/1/1 then area irrigated = 22. 6 1.5 5000 = 562500 m 2 . .08 1+ 1 2 Modal class is 30-40 1 2 f1 f 0 Mode = l + h 2f1 f 0 f 2 16 10 = 30 + 10 32 10 12 2 = 36. 1 2 SECTION D 23. Let the smaller tap fills the tank in x hrs the larger tap fills the tank in (x 2) hrs. Time taken by both the taps together = 15 hrs. 8 1 1 8 + = x x 2 15 2 4x2 23x + 15 = 0 1 2 (4x 3) (x 5) = 0 Therefore x 3 4 x=5 1 Smaller and larger taps can fill the tank seperately in 5 hrs and 3 hrs resp. 1 2 OR Let the speed of the boat in still water be x km/hr and speed of the stream be y km/hr. Given and 30/1/1 30 44 + = 10 x y x+y 40 55 + = 13 x y x+y (9) ...(i) 1 ...(ii) 1 30/1/1 Solving (i) and (ii) to get x + y = 11 ..(iii) and x y = 5 ...(iv) Solving (iii) and (iv) to get x = 8, y = 3. 1+1 Speed of boat = 8 km/hr & speed of stream = 3 km/hr. 24. S4 = 40 2(2a + 3d) = 40 2a + 3d = 20 1 S14 = 280 7(2a + 13d) = 280 2a + 13d = 40 1 Solving to get d = 2 1 2 and a = 7 1 2 Sn = n [14 + (n 1) 2] 2 = n(n + 6) or (n2 + 6n) 25. LHS = 1 sin A cos A + 1 sin A + cos A 1 Dividing num. & deno. by cos A = tan A 1 + sec A tan A + 1 sec A 1 tan A 1 + sec A = (tan A sec A) + (sec 2 A tan 2 A) 1 = tan A 1 + sec A (tan A sec A) (1 sec A tan A) 1 = 1 1 = = RHS tan A sec A sec A tan A 1 (10) 30/1/1 30/1/1 26. Correct Figure B 1 Let the speed of the boat be y m/min 100 m CD = 2y 60 x C A 30 2y D 100 100 x= x 3 tan 60 = 3= tan 30 = 1 100 = x + 2y = 100 3 3 x + 2y y= 100 3 = 57.73 3 1 1 1 or speed of boat = 57.73 m/min. OR E D h h A 30 80 x 1 Let BC = x so AB = 80 x where AC is the road. 60 B Correct Figure x C tan 60 = 3= and tan 30 = h h=x 3 x 1 1 h = h 3 = 80 x 3 80 x 1 Solving equation to get x = 20, h = 20 3 AB = 60 m, BC = 20 m and h = 20 3 m. 27. 30/1/1 1 Correct construction of ABC. 2 Correct construction of triangle similar to triangle ABC. 2 (11) 30/1/1 Volume of the bucket = 12308.8 cm3 28. Let r1 = 20 cm, r2 = 12 cm 20 cm V= 12308.8 = h 12 cm h 2 2 (r1 + r2 + r1r2 ) 3 h= 3.14 h (400 + 144 + 240) 3 1 12308.8 3 = 15 cm 3.14 784 1 Now l2 = h2 + (r1 r2)2 = 225 + 64 = 289 l = 17 cm. 1 Surface area of metal sheet used = r22 + l (r1 + r2) = 3.14 (144 + 17 32) = 2160.32 cm2. 29. Correct given, to prove, figure and construction 1 1 4=2 2 Correct proof. 30. 2 Class Frequency Cumulative freq. 0-10 f1 f1 10-20 5 5 + f1 20-30 9 14 + f1 30-40 12 26 + f1 40-50 f2 26 + f1 + f2 50-60 3 29 + f1 + f2 60-70 2 31 + f1 + f2 Correct Table 1 40 1 2 Median = 32.5 median class is 30-40. Now 32.5 = 30 + 10 (20 14 f1 ) 12 1 f1 = 3 1 Also 31 + f1 + f2 = 40 1 2 f2 = 6 (12) 30/1/1 30/1/1 OR Less than type distribution is as follows Marks No. of students Less than 5 2 Less than 10 7 Less than 15 13 Less than 20 21 Less than 25 31 Less than 30 56 Less than 35 76 Less than 40 94 Less than 45 98 Less than 50 100 Correct Table 1 1 2 1 1 2 Plotting of points (5, 2), (10, 7) (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98), (50, 100) 30/1/1 Joining to get the curve 1 2 Getting median from graph (approx. 29) 1 2 (13) 30/1/2 QUESTION PAPER CODE 30/1/2 EXPECTED ANSWER/VALUE POINTS SECTION A 1. Let the point A be (x, y) x+3 y+4 =2 = 2 and 2 2 x = 7 and y = 0 1 2 1 2 Point is ( 7, 0) 2. Any one rational number between 2 (1.41 approx.) and 3 (1.73 approx.) 1 e.g., 1.5, 1.6, 1.63 etc. 3. 4. 1 2 Numbers are 12, 15, 18, ..., 99 99 = 12 + (n 1) 3 n = 30 1 2 tan 2 A = cot (90 2A) 90 2A = A 24 1 2 A = 38 1 2 OR 5. sin 33 = cos 57 1 2 sin2 33 + sin2 57 = cos2 57 + sin2 57 = 1 1 2 Since roots of the equation x2 + 4x + k = 0 are real 16 4k 0 1 2 k 4 1 2 (14) 30/1/2 30/1/2 OR Roots of the equation 3x2 10x + k = 0 are reciprocal of each other 6. Product of roots = 1 1 2 k =1 k=3 3 1 2 1 2 AB = 1 + 2 = 3 cm ABC ~ ADE AB2 9 ar (A BC) = = AD2 1 ar (ADE) ar( ABC) : ar( ADE) = 9 : 1 1 2 SECTION B 7. System of equations has infinitely many solutions. 2 3 7 = = k +1 2k 1 4k + 1 1 2 4k 2 = 3k + 3 1 2 k=5 1 2 Also 12k + 3 = 14k 7 k=5 1 2 Hence k = 5. 8. Total number of outcomes = 6. (i) Prob. (getting a prime number (2, 3, 5)) = 3 1 or 6 2 (ii) Prob. (getting a number between 2 and 6 (3, 4, 5)) = 30/1/2 (15) 1 3 1 or . 6 2 1 30/1/2 9. Let the required point be (a, 0) and required ratio AP : PB = k : 1 P(a, 0) 1 K A(1, 3) a= 4k + 1 k +1 0= 5k 3 k +1 B(4,5) k= 3 or required ratio is 3 : 5 5 17 Point P is , 8 10. 1 1 2 Total number of outcomes = 8 1 2 Favourable number of outcomes (HHH, TTT) = 2 1 2 Prob. (getting success) = 11. 0 1 2 1 2 2 1 or 8 4 Prob. (losing the game) = 1 1 2 1 3 = . 4 4 an = a21 + 120 = (3 + 20 12) + 120 = 363 363 = 3 + (n 1) 12 n = 31 1 1 or 31st term is 120 more than a21. OR a1 = S1 = 3 4 = 1 1 2 a2 = S2 S1 = [3(2)2 4(2)] ( 1) = 5 1 2 d = a2 a1 = 6 1 2 Hence an = 1 + (n 1) 6 = 6n 7 1 2 (16) 30/1/2 30/1/2 Alternate method: Sn = 3n2 4n Sn 1 = 3(n 1)2 4(n 1) = 3n2 10n + 7 1 Hence an = Sn Sn 1 12. = (3n2 4n) (3n2 10n + 7) 1 2 = 6n 7 1 2 Using Euclid s Algorithm 7344 = 1260 5 + 1044 1260 = 1044 1 + 216 1044 = 216 4 + 180 216 = 180 1 + 36 180 = 36 5 + 0 1 1 2 1 2 HCF of 1260 and 7344 is 36. OR Using Euclid s Algorithm a = 4q + r, 0 r < 4 a = 4q, a = 4q + 1, a = 4q + 2 and a = 4q + 3. Now a = 4q and a = 4q + 2 are even numbers. 1 1 2 Therefore when a is odd, it is of the form a = 4q + 1 or a = 4q + 3 for some integer q. 30/1/2 (17) 1 2 30/1/2 SECTION C 13. x 50 20 u= x Freq (f) 0-20 10 12 2 24 20-40 30 15 1 15 40-60 50 32 0 0 60-80 70 k 1 k 80-100 90 13 2 26 72 + k x = 53 = 50 + 20 13 + k Correct Table 2 13 + k 72 + k 3k + 216 = 20k 260 k = 28 14. 1 Draw OM AB OAB = OBA = 30 O A fu Class 1 2 21 21 30 M B sin 30 = 1 OM 21 = OM = 2 21 2 C cos 30 = 3 AM 21 = AM = 3 2 21 2 Area of OAB = = 1 1 21 AB OM = 21 3 2 2 2 441 3 cm 2 . 4 1 Area of shaded region = Area (sector ACB) Area ( OAB) = 22 120 441 21 21 3 7 360 4 3 2 2 = 462 441 cm or 271.3 cm (approx.) 4 (18) 1 1 2 30/1/2 30/1/2 15. ACB ~ ADC (AA similarity) AC AD = BC CD ...(1) 1 ...(2) 1 Also ACB ~ CDB (AA similarity) AC CD = BC BD Using equations (1) and (2) AD CD = CD BD CD2 = AD BD 1 OR B Correct Figure 1 2 Q AQ2 = CQ2 + AC2 1 BP2 = CP2 + BC2 1 2 C A P AQ2 + BP2 = (CQ2 + CP2) + (AC2 + BC2) = PQ2 + AB2. 1 16. Let BL = x = BN A N CL = 8 x = CM AC = 12 AM = 4 + x = AN M x B 1 Now AB = AN + NB = 10 x + 4 + x = 10 x 30/1/2 L C x=3 1 BL = 3 cm, CM = 5 cm and AN = 7 cm 1 (19) 30/1/2 Alternate method A z z 10 cm 12 cm CL = CM = y M N x B 1 2 Let BL = BN = x (tangents from external points are equal) AN = AM = z y x L 8 cm y C AB + BC + AC = 2x + 2y + 2z = 30 x + y + z = 15 ...(i) 1 Also x + z = 10, x + y = 8 and y + z = 12 Subtracting from equation (i) 1 1 1 + + 2 2 2 y = 5, z = 7 and x = 3 BL = 3 cm, CM = 5 cm and AN = 7 cm. 17. Length of canal covered in 30 min = 5000 m. 1 2 Volume of water flown in 30 min = 6 1.5 5000 m3 1 If 8 cm standing water is needed then area irrigated = 18. Let us assume 6 1.5 5000 = 562500 m 2 . .08 2 be a rational number and its simplest form be integers and b 0. 1+ 1 2 a , a and b are coprime positive b a b So 2 = a2 = 2b2 1 Thus a2 is a multiple of 2 1 2 a is a multiple of 2. Let a = 2 m for some integer m 1 2 b2 = 2m2 (20) 30/1/2 30/1/2 Thus b2 is a multiple of 2 1 2 b is a multiple of 2 Hence 2 is a common factor of a and b. This contradicts the fact that a and b are coprimes Hence 19. Sum of zeroes = k + 6 1 Product of zeroes = 2(2k 1) 1 Hence k + 6 = 20. 1 2 2 is an irrational number. 1 2(2k 1) 2 k=7 1 1 2 Let the required point on y-axis be (0, b) (5 0)2 + ( 2 b)2 = ( 3 0)2 + (2 b)2 29 + 4b + b2 = 13 + b2 4b b = 2 1 Required point is (0, 2) 1 2 1 k= 0 OR 2x y+ AP : PB = 1 : 2 x= Q A(2, 1) P(x, y) 4+5 2 8 = 3 and y = = 2 3 3 B(5, 8) Thus point P is (3, 2). 1 2 1 1 + 2 2 1 2 Point (3, 2) lies on 2x y + k = 0 6+2+k=0 k = 8. 21. Let sum of the ages of two children be x yrs and father s age be y yrs. y = 3x and y + 5 = 2(x + 10) 30/1/2 1 (21) ...(1) 1 ...(2) 1 30/1/2 Solving equations (1) and (2) x = 15 and y = 45 Father s present age is 45 years. 1 OR Let the fraction be x y x 2 1 = y 3 ...(1) 1 and x 1 = y 1 2 ...(2) 1 Solving (1) and (2) to get x = 7, y = 15. 22. Required fraction is 7 15 1 LHS = sin2 + cosec2 + 2sin cosec + cos2 + sec2 + 2cos sec = (sin2 + cos2 ) + cosec2 + sec2 + 1 2sin cos +2 . sin cos = 1 + 1 + cot2 + 1 + tan2 + 2 + 2 1 1 2 1 2 = 7 + cot2 + tan2 = RHS OR 1 cos ec A (1 + tan A + sec A) LHS = 1 + tan A = 1 (tan A + 1 sec A) (1 + tan A + sec A) tan A 1 = 1 [(1 + tan A) 2 sec 2 A] tan A 1 = 1 [1 + tan 2 A + 2 tan A 1 tan 2 A] tan A = 2 = RHS 1 (22) 30/1/2 30/1/2 Alternate method 1 sin A 1 cos A + LHS = 1 + 1 + sin A sin A cos A cos A = (sin A + cos A 1) (cos A + sin A + 1) 1 1 cos A sin A 1 2 = (sin A + cos A) 1 sin A cos A = (1 + 2sin A cos 1) 1 1 sin A cos A 1 2 1 2 = 2 = RHS SECTION D 23. LHS = sin 2 A / cos 2 A 1/ sin 2 A + 1 1 sin 2 A 1 2 cos A sin 2 A cos 2 A 1 sin 2 A cos 2 A + = sin 2 A cos 2 A sin 2 A cos 2 A 24. 1 = 1 sin A cos2 A 1 = 1 1 2cos2 A 1 2 Here a = 3, an = 83 and Sn = 903 Therefore 83 = 3 + (n 1)d (n 1)d = 80 Also 903 = n = 21 and d = 4 30/1/2 ...(i) n n [2a + (n 1) d] = (6 + 80) = 43n (using (i)) 2 2 (23) 1 1+ 1 2 1 1 2 30/1/2 25. 26. Correct construction of ABC 2 Correct construction of triangle similar to ABC. 2 Class Frequency Cumulative freq. 0-10 f1 f1 10-20 5 5 + f1 20-30 9 14 + f1 30-40 12 26 + f1 40-50 f2 26 + f1 + f2 50-60 3 29 + f1 + f2 60-70 2 31 + f1 + f2 Correct Table 1 40 1 2 Median = 32.5 median class is 30-40. Now 32.5 = 30 + 10 (20 14 f1 ) 12 1 f1 = 3 1 Also 31 + f1 + f2 = 40 1 2 f2 = 6 OR Less than type distribution is as follows Marks No. of students Less than 5 2 Less than 10 7 Less than 15 13 Less than 20 21 Less than 25 31 Less than 30 56 Less than 35 76 Less than 40 94 Less than 45 98 Less than 50 100 (24) Correct Table 1 1 2 30/1/2 30/1/2 Plotting of points (5, 2), (10, 7) (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98), (50, 100) 27. 1 1 2 Joining to get the curve 1 2 Getting median from graph (approx. 29) 1 2 1 4=2 2 Correct given, to prove, figure and construction Correct proof. 2 Volume of the bucket = 12308.8 cm3 28. Let r1 = 20 cm, r2 = 12 cm 20 cm h 2 2 (r1 + r2 + r1r2 ) 3 V= 12308.8 = h 12 cm h= 3.14 h (400 + 144 + 240) 3 12308.8 3 = 15 cm 3.14 784 1 1 Now l2 = h2 + (r1 r2)2 = 225 + 64 = 289 l = 17 cm. 1 Surface area of metal sheet used = r22 + l (r1 + r2) = 3.14 (144 + 17 32) = 2160.32 cm2. 29. Let the smaller tap fills the tank in x hrs the larger tap fills the tank in (x 2) hrs. Time taken by both the taps together = 15 hrs. 8 1 1 8 + = x x 2 15 2 4x2 23x + 15 = 0 1 2 (4x 3) (x 5) = 0 Therefore 30/1/2 1 (25) 30/1/2 x 3 4 x=5 1 Smaller and larger taps can fill the tank seperately in 5 hrs and 3 hrs resp. 1 2 OR Let the speed of the boat in still water be x km/hr and speed of the stream be y km/hr. Given and 30 44 + = 10 x y x+y 40 55 + = 13 x y x+y ...(i) 1 ...(ii) 1 Solving (i) and (ii) to get x + y = 11 ..(iii) and x y = 5 ...(iv) Solving (iii) and (iv) to get x = 8, y = 3. 1+1 Speed of boat = 8 km/hr & speed of stream = 3 km/hr. 30. Correct Figure Let the speed of the boat be y m/min B 100 m A 1 CD = 2y 60 x C 30 2y D 100 100 x= x 3 tan 60 = 3= tan 30 = 1 100 = x + 2y = 100 3 3 x + 2y y= 100 3 = 57.73 3 1 1 1 or speed of boat = 57.73 m/min. (26) 30/1/2 30/1/2 OR E D h h Correct Figure 1 Let BC = x so AB = 80 x A 30 80 x where AC is the road. 60 B x C tan 60 = 3= and tan 30 = h h=x 3 x 1 1 h = h 3 = 80 x 3 80 x 1 Solving equation to get x = 20, h = 20 3 AB = 60 m, BC = 20 m and h = 20 3 m. 30/1/2 (27) 1 30/1/3 QUESTION PAPER CODE 30/1/3 EXPECTED ANSWER/VALUE POINTS SECTION A 1. LCM (x3y2, xy3) = x3y3. 1 2. Numbers are 12, 15, 18, ..., 99 1 2 3. 99 = 12 + (n 1) 3 n = 30 1 2 1 2 AB = 1 + 2 = 3 cm ABC ~ ADE 4. 5. AB2 9 ar (A BC) = = AD 2 1 ar (ADE) ar( ABC) : ar( ADE) = 9 : 1 1 2 Let the point A be (x, y) 4+y 1+ x = 2 and = 3 2 2 x = 3 and y = 10 Point A is (3, 10) 1 2 1 2 Since roots of the equation x2 + 4x + k = 0 are real 16 4k 0 1 2 k 4 1 2 OR Roots of the equation 3x2 10x + k = 0 are reciprocal of each other Product of roots = 1 1 2 k =1 k=3 3 1 2 (28) 30/1/3 30/1/3 6. tan 2 A = cot (90 2A) 90 2A = A 24 1 2 A = 38 1 2 OR sin 33 = cos 57 1 2 sin2 33 + sin2 57 = cos2 57 + sin2 57 = 1 1 2 SECTION B 7. Required numbers are 14, 21, 28, 35, ..., 98. 1 98 = 14 + (n 1) 7 1 2 n = 13 1 2 OR Given Sn = n2 1 2 S1 = a1 = 1 S2 = a1 + a2 = 4 8. a2 = 3 1 2 d = a2 a1 = 2 1 2 a10 = 1 + 18 = 19 1 2 Total number of outcomes = 8 1 2 Favourable number of outcomes (HHH, TTT) = 2 1 2 Prob. (getting success) = 30/1/3 1 2 2 1 or 8 4 (29) 30/1/3 Prob. (losing the game) = 1 9. 1 2 1 3 = . 4 4 P(a, 0) 1 K A(1, 3) B(4,5) a= 4k + 1 k +1 0= 5k 3 k +1 k= 3 or required ratio is 3 : 5 5 1 2 3 1 or 6 2 1 Total number of outcomes = 6. (i) Prob. (getting a prime number (2, 3, 5)) = (ii) Prob. (getting a number between 2 and 6 (3, 4, 5)) = 11. 1 0 17 Point P is , 8 10. 1 2 Let the required point be (a, 0) and required ratio AP : PB = k : 1 3 1 or . 6 2 1 System of equations has infinitely many solutions c 3 3 c = = 12 c c 1 2 c2 = 36 c = 6 or c = 6 ...(1) 1 2 Also 3c = 3c c2 c = 6 or c = 0 ...(2) 1 2 From equations (1) and (2) 1 2 c = 6. 12. Using Euclid s Algorithm 7344 = 1260 5 + 1044 1260 = 1044 1 + 216 1044 = 216 4 + 180 216 = 180 1 + 36 180 = 36 5 + 0 1 1 2 1 2 HCF of 1260 and 7344 is 36. (30) 30/1/3 30/1/3 OR Using Euclid s Algorithm a = 4q + r, 0 r < 4 a = 4q, a = 4q + 1, a = 4q + 2 and a = 4q + 3. 1 1 2 Now a = 4q and a = 4q + 2 are even numbers. Therefore when a is odd, it is of the form 1 2 a = 4q + 1 or a = 4q + 3 for some integer q. SECTION C 13. Let p(x) = 3x3 + 10x2 9x 4. One of the zeroes is 1, therefore dividing p(x) by (x 1) p(x) = (x 1) (3x2 + 13x + 4) 1 = (x 1) (x + 4) (3x + 1) 1 1 2 1 All zeroes are x = 1, x = 4 and x = . 3 14. 1 2 Join OQ, TP = TQ TM PQ and bisects PQ Hence PM = 4 cm. P T M O OM = 25 16 = 9 = 3 cm Let TM = x PT2 = x2 + 16 ( PMT) Q PT2 = (x + 3)2 25 ( POT) Hence x2 + 16 = (x + 3)2 25 = x2 + 9 + 6x 25 6x = 32 x = Hence PT2 = PT = 30/1/3 1 1 16 3 256 400 + 16 = 9 9 20 cm 3 (31) 1 30/1/3 15. Let us assume Let 2+ 3 be a rational number. 5 2+ 3 a = (b 0, a and b are integers) 5 b 3 = 5a 2b b a, b are integers 5a 2b is a rational number b 1 3 is a rational number i.e. which contradicts the fact that Therefore is 16. 1 3 is irrational 2+ 3 is an irrational number. 5 1 LHS = sin2 + cosec2 + 2sin cosec + cos2 + sec2 + 2cos sec = (sin2 + cos2 ) + cosec2 + sec2 + 1 2sin cos +2 . sin cos = 1 + 1 + cot2 + 1 + tan2 + 2 + 2 1 1 2 1 2 = 7 + cot2 + tan2 = RHS OR 1 cos ec A (1 + tan A + sec A) LHS = 1 + tan A = 1 (tan A + 1 sec A) (1 + tan A + sec A) tan A 1 = 1 [(1 + tan A) 2 sec 2 A] tan A 1 = 1 [1 + tan 2 A + 2 tan A 1 tan 2 A] tan A = 2 = RHS 1 (32) 30/1/3 30/1/3 Alternate method 1 sin A 1 cos A + LHS = 1 + 1 + sin A sin A cos A cos A = (sin A + cos A 1) (cos A + sin A + 1) 1 1 cos A sin A 1 2 = (sin A + cos A) 1 sin A cos A = (1 + 2sin A cos 1) 1 1 sin A cos A 1 2 1 2 = 2 = RHS 17. Let sum of the ages of two children be x yrs and father s age be y yrs. y = 3x and y + 5 = 2(x + 10) ...(1) 1 ...(2) 1 Solving equations (1) and (2) x = 15 and y = 45 Father s present age is 45 years. 1 OR Let the fraction be x y x 2 1 = y 3 ...(1) 1 and x 1 = y 1 2 ...(2) 1 Solving (1) and (2) to get x = 7, y = 15. 30/1/3 Required fraction is 7 15 1 (33) 30/1/3 18. 1 2 Let the required point on y-axis be (0, b) (5 0)2 + ( 2 b)2 = ( 3 0)2 + (2 b)2 29 + 4b + b2 = 13 + b2 4b b = 2 1 Required point is (0, 2) 1 2 1 k= 0 OR y+ 2x Q A(2, 1) P(x, y) 1 2 AP : PB = 1 : 2 x= B(5, 8) 4+5 2 8 = 3 and y = = 2 3 3 1 1 + 2 2 1 2 Thus point P is (3, 2). Point (3, 2) lies on 2x y + k = 0 6+2+k=0 k = 8. 19. 1 2 Modal class is 30-40 20. 1 f1 f 0 Mode = l + h 2f1 f0 f 2 16 10 = 30 + 10 32 10 12 2 = 36. 1 2 Length of canal covered in 30 min = 5000 m. 1 2 Volume of water flown in 30 min = 6 1.5 5000 m3 1 If 8 cm standing water is needed then area irrigated = 6 1.5 5000 = 562500 m 2 . .08 (34) 1+ 1 2 30/1/3 30/1/3 21. ACB ~ ADC (AA similarity) AC AD = BC CD ...(1) 1 ...(2) 1 Also ACB ~ CDB (AA similarity) AC CD = BC BD Using equations (1) and (2) AD CD = CD BD CD2 = AD BD 1 OR Correct Figure 1 2 AQ2 = CQ2 + AC2 1 BP2 = CP2 + BC2 1 2 B Q C A P AQ2 + BP2 = (CQ2 + CP2) + (AC2 + BC2) = PQ2 + AB2. 22. AC = 1 64 + 36 = 10 cm. Radius of the circle (r) = 5 cm. 1 Area of shaded region = Area of circle Ar(ABCD) = 3.14 25 6 8 1 2 1 = 78.5 48 1 2 = 30.5 cm2. SECTION D 2 23. 30/1/3 1 1 1 = x2 + + sec = x + 2 4x 2 16x 2 1 (35) 30/1/3 2 2 tan2 = sec 1 = x + 1 tan = x 4x 1 1 x tan = x or 4x 4x 1 2 2 Hence sec + tan = 2x or 24. 1 1 2 2 16x 1 1 2x 1 1 4=2 2 Correct given, to prove, figure, construction Correct proof. 25. 2 Less than type distribution is as follows Daily income Number of workers Less than 220 12 Less than 240 26 Less than 260 34 Less than 280 40 Less than 300 50 Plotting of points (220, 12), (240, 26), (260, 34) (280, 40) and (300, 50) Joining to get curve Correct Table 1 1 2 1 1 2 1 OR x 225 50 fiui 4 2 8 175 5 1 5 200 250 225 12 0 0 250 300 275 2 1 2 300-350 325 2 2 4 Daily expenditure xi No. of households (fi) 100-150 125 150-200 fi = 25 ui = fiui = 7 (36) Correct Table 2 30/1/3 30/1/3 7 Mean = 225 + 50 = 211 25 2 Mean expenditure on food is ` 211. 26. Correct construction of ABC. 2 Correct construction of triangle similar to triangle ABC. 2 Volume of the bucket = 12308.8 cm3 27. Let r1 = 20 cm, r2 = 12 cm 20 cm h 12 cm h 2 2 (r1 + r2 + r1r2 ) 3 V= 12308.8 = h= 3.14 h (400 + 144 + 240) 3 12308.8 3 = 15 cm 3.14 784 1 1 Now l2 = h2 + (r1 r2)2 = 225 + 64 = 289 l = 17 cm. 1 Surface area of metal sheet used = r22 + l (r1 + r2) = 3.14 (144 + 17 32) = 2160.32 cm2. 28. Correct Figure 1 1 Let the speed of the boat be y m/min B CD = 2y 100 m A 60 x C 30 2y 100 100 x= x 3 tan 60 = 3= tan 30 = 1 100 = x + 2y = 100 3 3 x + 2y D y= 100 3 = 57.73 3 or speed of boat = 57.73 m/min. 30/1/3 (37) 1 1 1 30/1/3 OR E D Correct Figure h h Let BC = x so AB = 80 x A 30 80 x 60 x B C 1 where AC is the road. tan 60 = 3= and tan 30 = h h=x 3 x 1 1 h = h 3 = 80 x 3 80 x 1 Solving equation to get x = 20, h = 20 3 AB = 60 m, BC = 20 m and h = 20 3 m. 29. 1 Let the smaller tap fills the tank in x hrs the larger tap fills the tank in (x 2) hrs. Time taken by both the taps together = 15 hrs. 8 1 1 8 + = x x 2 15 2 4x2 23x + 15 = 0 1 2 (4x 3) (x 5) = 0 Therefore x 3 4 x=5 1 Smaller and larger taps can fill the tank seperately in 5 hrs and 3 hrs resp. 1 2 OR Let the speed of the boat in still water be x km/hr and speed of the stream be y km/hr. Given and 30 44 + = 10 x y x+y 40 55 + = 13 x y x+y (38) ...(i) 1 ...(ii) 1 30/1/3 30/1/3 Solving (i) and (ii) to get x + y = 11 ..(iii) and x y = 5 ...(iv) Solving (iii) and (iv) to get x = 8, y = 3. 1+1 Speed of boat = 8 km/hr & speed of stream = 3 km/hr. 30. S4 = 40 2(2a + 3d) = 40 2a + 3d = 20 1 S14 = 280 7(2a + 13d) = 280 2a + 13d = 40 1 Solving to get d = 2 1 2 and a = 7 1 2 Sn = n [14 + (n 1) 2] 2 = n(n + 6) or (n2 + 6n) 30/1/3 1 (39)

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