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ISC Class XII Analysis Of Pupil Performance 2015 : Mathematics

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MATHEMATICS STATISTICS AT A GLANCE Total Number of students who took the examination 44659 Highest Marks Obtained 100 Lowest Marks Obtained 2 Mean Marks Obtained 68.13 Percentage of Candidates according to marks obtained Mark Range Details Number of Candidates Percentage of Candidates Cumulative Number Cumulative Percentage 0-20 1516 3.39 1516 3.39 21-40 2083 4.66 3599 8.06 41-60 12743 28.53 16342 36.59 61-80 14622 32.74 30964 69.33 Range of Marks Obtained 32.74 Percentage of Candidates 35.00 30.67 28.53 30.00 25.00 20.00 15.00 10.00 3.39 4.66 5.00 0.00 0-20 21-40 41-60 Marks Obtained 115 61-80 81-100 81-100 13695 30.67 44659 100.00 B. ANALYSIS OF PERFORMANCE SECTION A [10 3] Question 1 (i) (ii) Find the value of k if M = 1 2 2 and M2 k M I2 = 0 3 Find the equation of an ellipse whose latus rectum is 8 and eccentricity is . (iv) Solve: cos-1(sin cos-1x) = (v) Evaluate: (vi) Evaluate: (vii) The two lines of regressions are 4x + 2y 3 = 0 and 3x + 6y + 5 = 0. Find the correlation co-efficient between x and y. (iii) Using L Hospital s rule, evaluate: lim x 0 x sin x x 2 sin x 2 y2 dy y2 4 , where f(x) = 3, 2 ,0 3 (viii) A card is drawn from a well shuffled pack of playing cards. What is the probability that it is either a spade or an ace or both? and 2 (ix) If 1, (x) Solve the differential equation: sin-1 Comments of Examiners are the cube roots of unity, prove that = + 1 4 (i) A number of candidates wrote M2 as instead of 4 9 5 8 8 13 . Some candidates took k as matrix instead of k as scalar value while some candidates wrote 2 0 I2 as which was not correct. I2 was an identity 0 2 matrix of the order 2 2. (ii) Some candidates took the length of latus rectum as 4a 2b 2 instead of . A few candidates wrote the equation a 116 = Suggestions for teachers Revise all matrix operations in the class. Pay heed to matrix multiplication. Identity matrix, null matrix and their order should be explained thoroughly. The topic of ellipse and hyperbola should be taught separately and then their properties should be compared. Horizontal and vertical ellipse should be explained thoroughly. Relation b2=a2(1-e2) where e<1 should be explained. of hyperbola in place of ellipse i.e. x2 y2 1 instead a2 b2 x2 y2 1 . A few candidates used eccentricity a2 b2 formula, which was incorrect. (iii) Many candidates did this question well but a few used wrong value of cos , while some wrote incorrect 6 -1 conversion of cos x in terms of sin-1 x. A few candidates wrote cos-1x in terms of tan-1x which made the expression complicated. (iv) Some candidates wrote incorrect differentiation of numerator and denominator. They wrote differentiation of 1-cosx as 1-sinx which was not correct. Some wrote the differentiation of sinx as (-cosx). (v) The power of numerator and denominator was equal in the given integral so division was a must or addition and subtraction of constant could also work but many candidates forgot and tried to solve it as it was. (vi) Candidates were able to score marks in this question. (vii) Many candidates answered this question correctly. However, a few candidates wrote incorrect regression coefficient. (viii) Many candidates found this topic difficult. Some did not understand the meaning of either or term in the question. (ix) Many candidates wrote the formula but were not able to apply it correctly. They put the value of w = -1-w 2 and w2 = -1-w, which made the equation very complicated. (x) A number of candidates wrote sin (x + y) as, sin x + sin y which was incorrect. On the other hand, some candidates were not able to substitute x + y = t. A few candidates made calculation mistakes in this question of 117 Teach students derivations of inverse Trigonometric functions. 0 Indeterminate forms i.e. , etc. 0 should be explained properly and revision of differentiation chapter must be done for practice. L Hospital s rule must be taught giving appropriate conditions to deal with different indeterminate forms. Teach students the properties of definite integrals properly and their use in area. Coefficient of regression of lines y on x and x on y should be explained by explaining r = byx bxy and that the value of r should be less than 1; byx and bxy both positive, r will be positive otherwise negative. Theorem either or and theorem AND should be explained properly to students. Number of outcomes and number of favourable outcomes should be explained properly. P(A B) = P(A) + P(B) P(A B); P(A B)= P (A). P(B) and P (A/B)= P( A B) . P( B) Complex numbers should be divided into different parts then explained step by step. Application of cube roots of unity needs to be explained thoroughly. Stress upon the techniques of solving such questions. Differential equations and various forms i.e. separation of variables, homogenous, linear differential equations and their reducible forms need to be revised by doing different types of questions based on them. MARKING SCHEME Question 1. M 2 kM I 2 = 0 1 2 1 2 k 2k 1 0 - - 0 2 3 2 3 2k 3k 0 1 (i) 5 8 k 8 13 2k 2k 1 0 = 0 3k 0 1 5 k 8 2k 1 0 = 8 2k 13 3k 0 1 5 k = 1, 8 2k = 0, 13 3k = 1 k = 4 (ii) 2b a 2 8; e = 1 3 b2 = 4a b2 = a2 4a = a2 9 2 a b2 = 18. Equation of an ellipse: 4 x2 y2 + = 1 or 8x2 + 9y2 = 162 81 18 (iii) cos-1(sin cos-1x) sin(cos-1x) = cos = 6 3 = 6 2 3 2 1-x2 = 3 1 x2 = 4 4 x 1 2 118 (iv) = = x Lt 0 (v) sin x 1 6x 6 2 y2 dy y2 4 y2 4 4 dy y2 4 1 y = 2y - 8. tan 1 c 2 2 y 2y - 4 tan 1 c 2 2 (vi) 3 f ( x) dx 0 = = + (0 0) + 3 3 - 2 3 92 (vii) Let the line of regression of x on y be 4x + 2y - 3 = 0 1 3 y+ 2 4 1 bxy = 2 x= let the line of regression of y on x be 3x + 6y + 5 = 0 y= 1 5 x2 6 byx = 1 2 119 1 1 1 r2 = byx bxy = = 2 2 4 r= , since bxy and byx are negative. (viii) P(E) = (ix) Multiplying numerator and denominator by since (x) sin-1 =x+y Let 120 Question 2 (a) Using properties of determinants, prove that: 1 (b) 2 2 1 2 2 1 Given two matrices A and B 1 2 3 11 A 1 4 1 and B = 1 1 3 2 7 2 2 5 1 1 [5] = (1 + a2 + b2)3 [5] 14 2 , 6 find AB and use this result to solve the following system of equations: x 2y + 3z = 6, x + 4y + z = 12, x 3y + 2z = 1 Comments of Examiners (a) Properties of determinants were not correctly implemented by several candidates. A few expanded the determinants directly without applying any property. They were not able to get zeroes in row or column. Some applied useless properties which did not lead to result. Rows and columns were not correctly identified by several candidates. (b) A few candidates found the product of AB incorrectly. Many did not use the product of AB to solve the equation system. They found A-1 by using matrix inverse method. Several candidates found incorrect cofactors hence their values of x, y, z were incorrect. Some candidates could not obtain adjoint and inverse of a matrix correctly. Suggestions for teachers Plenty of practice must be given in using determinant properties. The idea of obtaining two zeroes in a row or a column is to be taught for easiest simplification. Inverse of a square matrix needs to be taught step by step. Utilisation of the inverse to correctly find the unknown matrix needs to be grasped properly. Product of two matrices needs attention. Sufficient practice is a must. MARKING SCHEME Question 2. (a) Replace C1 C1 bC3, C2 C2 +aC3 and take Replace R3 by R3 bR1 to get Expanding by C1, we get 121 (b) AB= -8 I , Question 3 (a) Solve the equation for x: (b) A, B and C represent switches in on position and A', B' and C' represent them in off position. Construct a switching circuit representing the polynomial ABC + AB C + A B C. Using Boolean Algebra, prove that the given polynomial can be simplified to C(A + B ). Construct an equivalent switching circuit. , x Comments of Examiners (a) Some candidates made mistakes while converting sin-1 to cos-1 or vice versa. Many candidates got incorrect algebraic equation independent from inverse function. As a result they could not solve the equation further. Some candidates applied sin-1 formula but they could not solve further. (b) A few candidates made errors while constructing a switching circuit. They made mistakes while simplifying the given polynomial. They were not able to write distributive law at this step (AB + AB + A B )C while a few wrote B + B = 0 which was incorrect. Some candidates made simplification errors while expanding the Boolean function by applying incorrect properties of Boolean algebra. 122 0 [5] [5] Suggestions for teachers All algebraic and trigonometric laws need to be revised thoroughly before learning inverse trigonometric functions and their operations. Application of formula for inverse trigonometric functions needs attention. Domains and range needs to be explained properly. All properties of Boolean algebra need to be well understood before application. Sufficient practice is a must. MARKING SCHEME Question 3. (a) squaring on both sides x = 13 is the required answer (b) ABC+AB C +A B C = ACB+ACB + A B C =AC ( B+B ) + A B C (B+B =1) = AC + A B C = (A+A B )C =(A+A )(A+B ) C = (A+B ) C 123 Question 4 (a) Verify Lagrange s Mean Value Theorem for the following function: f(x) = 2 sin x + sin2x on [0, ] (b) Find the equation of the hyperbola whose foci are (0, 10) and passing through the point (2, 3). Comments of Examiners (a) Many candidates failed to state all the criteria for application of Lagrange s theorem correctly. The concept of closed or open was not clear to many candidates. A few candidates got confused with Rolle s theorem condition f (a) = f (h). (b) Some candidates did not have proper knowledge of hyperbola and conjugate hyperbola. They wrote incorrect equation of hyperbola, hence got incorrect answer; a few took 2ae = 10 which was incorrect (where 2ae is the distance between the two foci). A few candidates found the value of a & b correctly but substituted incorrectly. [5] [5] Suggestions for teachers Help students enumerate the criteria for mean value theorem correctly. Firstly, the given function has to be continuous in the closed interval, secondly, the derivation of given function needs to exist in open interval and thirdly, f (b) f (a) f (c) = where c b a exists in open interval. Explanation of geometrical interpretation of mean value theorem with the help of figure is a must. Stress upon conics noting details with regard to their sketching and derivation of their equations for standard form as well as for other modified forms. Regular practice of conics is a must. MARKING SCHEME Question 4. (a) f(x) =(2 sinx + sin 2x) is continuous in [ 0, ] f '(x) exists in (0, ) f '(x) =2cos x + 2cos2x All the conditions of Lagrange s Mean Value theorem are satisfied there exist ' c ' in ( 0, ) such that f (c)= ( )- ( ) - 2cosc + 2cos2c = 0 2cos2c + cosc -1=0 cos c = -1, cos c = cos c = (not possible) or cos c = cos 3 0, c= 3 124 f(0) =0, f( ) =0 (b) c = /3 which lies between 0 to , hence, LMV theorem is verified. Foci 0, 10 be = 10 a2 = b2(e2 1) = b2e2 b2 a2 = 10 b2 let the equation be: - =1 1 9 a2 40 + 4 a2 = 10 a2 a4 a4 + 3a2 40 = 0 (a2 + 8) (a2 5) = 0 a2 = -8 or a2 = 5, (a2 can't be negative) a2 = 5, b2 = 5 the required equation is y2 x2 = 5 Question 5 (a) (b) If y = 1 , prove that: [5] Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius 10 cm is a square of side 10 2 cm. Comments of Examiners (a) Some candidates wrote differentiation of cos-1 x incorrectly. Many candidates did not place as y after first differentiation. Second order derivation was incorrectly shown by several candidates. Some made calculation mistakes while simplifying the equation. (b) Many candidates were not able to write the expression in mathematical form. They were not able to express the equation in one variable. A number of candidates made calculation mistakes while differentiating. 125 [5] Suggestions for teachers Differentiation rules for different functions and terms need attention. A through revision is a must. Explain to students the importance of finding the second derivative. They must show the condition of maxima or minima as per the requirement. MARKING SCHEME Question 5. (a) y= dy m e m cos -1 x dx 1 x2 dy differentiating again wrt x. 1 - x2 - my dx = -m 1 - x 2 2 d y dy dy x = -m 2 dx dx dx 1 x2 = - m (-my) = m2y (b) AB = 2x; BC = 2y x2 + y2 = 102 4x2+ 4y2 = 400 2 2 2 2 x 2 y 20 P = 4x + 4y = 4x + 4 100- x D C 10 4 4 =0 0 5 2 10 A = Hence, perimeter is maximum when x = 5 2 y = 5 2 x = y ABCD is square of side 10 2 cm 126 0 2y B 2x Question 6 (a) Evaluate: 1 [5] Find the smaller area enclosed by the circle x2 + y2 and the line x + y = 2. (b) Comments of Examiners (a) Many candidates were not able to integrate the given expression. Some candidates could not decompose the problem into partial fraction. Errors were also made while integrating factors. (b) Many candidates attempted this part correctly by taking arbitrary value of the radius of the circle. Suggestions for teachers Partial fraction rule need to be understood and applied correctly. Methods of proper substitution need attention. A lot of practice of such problems must be given by the teachers. MARKING SCHEME Question 6. (a) I= sec x 1 cos ec x dx = dx = Put t = sin x dt = cos x dx = = - = t = A(1+t)2 + B(1 t2) + c(1 t A(1 2t t 2 ) B Bt 2 C Ct t t 2 ( A B) t (2 A C ) ( A B C ) A B 0, 2 A C 1, A B C 0 Solving equations we get A = , , 127 [5] I= = I= (b) |1 | |1 y The required area: = 4 2 4 = | = 2 4+2 = 2 sq. units 2 2 2 0 x Question 7 (a) Given that the observations are: (9, 4), (10, 3), (11, 1), (12, 0), (13, 1), (14, 3), (15, 5), (16, 8). Find the two lines of regression and estimate the value of y when x = 13 5. [5] (b) In a contest the competitors are awarded marks out of 20 by two judges. The scores of the 10 competitors are given below. Calculate Spearman s rank correlation. [5] Competitors A B C D E F G H I J Judge A 2 11 11 18 6 5 8 16 13 15 Judge B 6 11 16 9 14 20 4 3 13 17 128 Comments of Examiners (a) Many candidates found byx and bxy incorrectly, as a result, the two regression lines were incorrect. Some candidates found the value of y from given value of x by using regression equation of x on y instead of y on x. Several candidates were unable to calculate the correct values of xy, x2, y2, byx and bxy which led to wrong results. (b) Some candidates calculated the ranks incorrectly. Correction factor for d2 was either incorrect or applied incorrectly in the formula for r. Some candidates wrote incorrect formula for spearman s rank correlation. Suggestions for teachers Various methods of finding byx and bxy should be taught giving examples. Students should be careful about the formulae for byx and bxy as well as the regression equation of x on y and that of y on x. Students should be given adequate practice to understand which formula is to be applied when ranks are repeated and when ranks are not repeated. MARKING SCHEME Question 7. (a) y xy x2 y2 9 -4 -36 81 16 10 -3 -30 100 9 11 -1 -11 121 1 12 0 0 144 0 13 1 13 169 1 x 14 3 42 196 9 15 5 75 225 25 16 8 128 256 64 = 100 bxy= = = 9 ( ) = 181 = 1292 ( ) = = 0 596 Line of regression of y on x y = 1 63 (x 12 5) Line of regression of x on y x 12 5 = 0 596 (y ) 129 = 125 = = 12 5 = = 1 125 byx = = = = 1 63 ( ) ( ) x = 0 596 y + 11 83 y, when x = 13 5 y = 1 63 x - 19 25 y = 1 63 13 5 - 19 25 = 2 755 = 2 76 (b) Judge A 2 11 11 18 6 5 8 16 13 15 r=1 6 = 1=1- Judge B 6 11 16 9 14 20 4 3 13 17 10 5.5 5.5 1 8 9 7 2 4 3 8 6 3 7 4 1 9 10 5 2 2 -0.5 2.5 -6 4 8 -2 -8 -1 1 4 0.25 6.25 36 16 64 4 64 1 1 d =196 5 0 194 Question 8 (a) An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replaced into the urn. Otherwise, it is replaced with another ball of the same colour. The process is repeated. Find the probability that the third ball drawn is black. (b) Three persons A, B and C shoot to hit a target. If A hits the target four times in five trials, B hits it three times in four trials and C hits it two times in three trials, find the probability that: (i) Exactly two persons hit the target. (ii) At least two persons hit the target. (iii) None hit the target. 130 [5] [5] Comments of Examiners (a) Many candidates handled this problem well but some could not understand the underlying principle. (b) In some cases, the probability of hitting the target were not found correctly. In part (ii), several candidates could not understand the meaning of at least while some did not apply AND theorem. MARKING SCHEME Question 8. (a) P(E) = P(WWB) + P(WBB) + P(BWB) + P(BBB) = = = Alternate solution: P(E) = P(WWB) + P(WBB) + P(BWB) + P(BBB) = = = (b) P(A) = , P(B) = , P(C) = (i) P ABC P ABC P ABC = = (ii) = P ABC P ABC P ABC P ABC 131 Suggestions for teachers Explain the correct interpretation of such problems. Laws of probability should be taught in detail with plenty of examples and illustrations. Terms such as at least , at most , exact , none should be discussed and problems based on them practiced. = = (iii) P ABC = = Question 9 and | |=1, find the locus of z and illustrate it in the (a) If z = x + iy, Argand Plane. (b) Solve the differential equation: 1 = + 1 [5] 0 when x = 0, y = 1 Comments of Examiners (a) Most of the candidates made mistakes while finding the modulus as well as in the simplification to find the correct values of z. Illustration of z in the Argand plane was incorrectly shown by some candidates. (b) Many candidates substituted y = vx and proceeded further to solve the given equation using the rules of homogeneous equation, which was an incorrect approach. The subsequent integrals were not correctly understood by some. A few candidates did not find the value of C (constant) under the given condition i.e. x=0, y=1. Suggestions for teachers Interpret the locus of a complex number clearly. Explain the concept of Argand plane. The procedure for finding modulus must be revised thoroughly. All forms of integration need rigorous practice. The constant of integration should not be ignored. MARKING SCHEME Question 9. (a) =1 = im (z) = Squaring 4 + 4y + y2 + x2 = 4x2 + 4y2 -4y + 1 3x2 + 3y2 8y 3 = 0 x2+ y2 y 1 = 0 Circle, Centre 0, and r = = [5] 0, 4/3 Type equation here. Re (z) 1 132 (b) Substitute Integrating when SECTION B Question 10 (a) Using vectors, prove that angle in a semicircle is a right angle. [5] (b) Find the volume of a parallelopiped whose edges are represented by the vectors: [5] a 2i 3 j 4k , b i 2 j k , and c 3i j 2k . 133 Comments of Examiners (a) Most of the candidates were unable to proceed with the solution for a vector based geometrical question. Vector symbols were not used by many candidates. Some candidates did not show the arrow in the diagram drawn by them. The dot product of vectors was found incorrectly by some candidates. (b) The concept of scalar triple product was clear to most of the candidates but some wrote dot product first and then cross product, which was incorrect. Some wrote [ a b c ] in determinant form and made mistakes in calculation. Suggestions for teachers Dot product and cross product should be explained well to students. Students must be told to give proper direction to the vectors. Vector algebra in totality needs to be explained well to students, especially the properties of scalar triple product. Combination of dot and cross product in scalar triple product needs thorough understanding as well as rigorous practice. MARKING SCHEME Question 10. (a) Let O be the centre of the circle and AB be the diameter. C is a point on the circumference. Take O as C the origin and let OA a and OC c Therefore, OB a A = = = 0, Where r is radius Therefore, angle ACB is a right angle. (b) The volume of the parallolepiped is: 1 2 1 3 3 1 2 = 2 5 + 3 5 - 4(-5) =2 = 2 5 + 3 5 4 (-5) 1 2 4 1 2 3 1 = 45 cubic units 134 O B Question 11 (a) Find the equation of the plane passing through the intersection of the planes: x + y + z +1 = 0 and 2x 3y + 5z 2 = 0 and the point ( 1, 2, 1). [5] (b) Find the shortest distance between the lines = + 2 + 3 + (2 + 3 + 4 and = 2 + 4 + 5 + (4 + 6 + 8 [5] Comments of Examiners (a) A number of candidates wrote incorrect equation of the plane passing through the intersection of planes. Some made mistakes in calculating the value of . A few candidates applied the condition of perpendicularity in this question which was incorrect. (b) A number of candidates were unable to calculate the correct values of a1 , a2 andb . Some made mistakes in Suggestions for teachers Teach the equation of plane thoroughly. Cartesian and vector of plane should be revised by practicing different types of questions. The concept of parallel and nonparallel lines needs to be explained clearly to students. calculating ( a2 a ). The concepts of skew lines and parallel lines were not clear to many candidates. Some candidates calculated b1 b 0 . They were unable to understand that the given lines are parallel. A few candidates applied wrong formula to calculate the shortest distance between the given lines. MARKING SCHEME Question 11. (a) Equation of plane passing through the intersection of the given planes is: (x + y + z + 1) +k (2x -3y + 5z- 2)=0 If this plane passes through (-1,2,1) then ( -1+2+1+1) +k ( -2 6 + 5 -2) =0 3 =5k K=3/5 5(x+y+z+1)+ 3(2x-3y+5z -2)=0 11x- 4y + 20z -1 =0 (b) Here, a1 + 2 + 3 Or equivalent form and a2 = 2 + 4 + 5 135 shortest distance = 0.415 Question 12 (a) Box I contains two white and three black balls. Box II contains four white and one black balls and box III contains three white and four black balls. A dice having three red, two yellow and one green face, is thrown to select the box. If red face turns up, we pick up box I, if a yellow face turns up we pick up box II, otherwise, we pick up box III. Then, we draw a ball from the selected box. If the ball drawn is white, what is the probability that the dice had turned up with a red face? [5] (b) Five dice are thrown simultaneously. If the occurrence of an odd number in a single dice is considered a success, find the probability of maximum three successes. [5] 136 Comments of Examiners (a) The concept of Bayes theorem was clear to most candidates but some candidates found incorrect probability. While some candidates found conditional probability for the happening of an event incorrectly, even probability of a specific known event was found wrongly by a few candidates. (b) Many candidates were unable to understand the problem correctly. The concept of P (x 3) was not clear to many candidates. Probability distribution theory was incorrectly applied by some candidates. MARKING SCHEME Question 12. (a) P(A) = 3/6 , P(B) = 2/6,P(C) =1/6 Let D be the probability of drawing a white ball. P ( D/A) =2/5, P(D/B) = 4/ 5, P(D /C) = 3/7 P( A/D) = P(A) P(D/A) P(A)P(D/A) +P(B)P(D/B) +P(C) P(D/C) = = = (b) 3/6 2/5 (3/6 2/5+ 2/6 4/5 + 1/6 3/7) ( 6/30 )x(210/113) 42/113 = 0.37 n = 5, p = , q = p(x ) = 1 p(x=4,5) =1=1 = 0.81 137 Suggestions for teachers Teach Bayes theorem with proper explanation and illustration. Pay heed to the laws of total probability. Give adequate practice of Bayes theorem. Revise Binomial theorem in the class thoroughly before teaching the probability distribution theory. Explain each term in the expansion. Train students about the situation of maximum three successes and minimum three successes. SECTION C Question 13 (a) Mr. Nirav borrowed 50,000 from the bank for 5 years. The rate of interest is 9% per annum compounded monthly. Find the payment he makes monthly if he pays back at the beginning of each month. (b) A dietician wishes to mix two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below: Food Vitamin A Vitamin B Vitamin C X 1unit 2 units 3 units Y 2 units 2 units 1 unit [5] [5] One kg of food X costs 24 and one kg of food Y costs 36. Using Linear Programming, find the least cost of the total mixture which will contain the required vitamins. Comments of Examiners (a) Instead of present value of an annuity due, some candidates used the formula for present value of an ordinary Annuity. Number of instalments (n) was not calculated in terms of months, even rate of interest was not calculated per month by a few candidates. Many candidates used wrong formulae. (b) Many candidates took incorrect inequality sign, hence they got incorrect feasible region and their corner points were also incorrect. Some candidates did not show any graphical representation of the inequalities. In some cases, the representation of the problem was not up to the mark, and the work was not systematic resulting in candidates missing the point of minimum cost. 138 Suggestions for teachers Explain the difference between annuities due and ordinary annuities by giving examples. Train students to read the question carefully, understand the meaning of the question and apply the formula accordingly. A thorough and regular practice is a must. Give practice to students in sketching of lines. They should be asked to express the line in intercept form i.e. x/a+ y/b =1, so that sketching is easy. Correct feasible region and its plotting is important. MARKING SCHEME Question 13. (a) Here, P = 50000, i = 0.0075 and n = 60 (1+i)[1-(1+i)-n] Now, P = 50000 = 50000 = . . (1+0.0075)[1-(1+0.0075)-60] (1.0075)[1-(1.0075)-60] A= = 1030.2 Thus, monthly installment should be Rs.1,030.2 (b) Let there be x units of food x and y units of food y. Min z = 24x + 36y Subject to the constraints x + 2y 10 2x + 2y 12 3x+ y 8 x 0, y 0 x 10 0 1 2 y 0 8 5 4 Z(cost) 240 288 204 192 y X 0 (Min. cost) Question 14 (a) A bill for 7,650 was drawn on 8th March, 2013, at 7 months. It was discounted on 18th May, 2013 and the holder of the bill received 7,497. What is the rate of interest charged by the bank? (b) The average cost function, AC for a commodity is given by AC = x + 5 + in terms of output x. Find: (i) The total cost, C and marginal cost, MC as a function of x. (ii) The outputs for which AC increases. 139 , [5] [5] Comments of Examiners (a) A number of candidates calculated discounted days incorrectly. They were not able to calculate the rate of interest. Some tried to find r by using B.G. while others used T.D. A few candidates took the difference of Rs. 7650 and Rs. 7497 as interest and applied the present worth formula which was not correct. Some candidates used formula T.D = A ni instead of B.D = Ani. 1 ni (b) Some candidates wrote incorrect formula of cost function, so their marginal cost was incorrect. Some wrote incorrect differentiation of the expression. Many candidates were not able to answer the second part of the question. They were confused with the maximum minimum condition. They found the derivative and put it equal to zero. Suggestions for teachers Explain Bills of exchange in detail. Differentiate the B.D, T.D and B.G. The procedure for calculating the due date should be taught clearly. The concepts of Marginal Cost, Total Cost and Average Cost should be taught in depth for increasing and decreasing functions by giving sufficient examples. Familiarize students with the different terms used in this question by giving adequate practice. MARKING SCHEME Question 14. (a) (b) Face value of the bill= 7650 = A Discounted value of the bill = 7497 Banker s discount=( 7650 -7497) = 153 Nominal due date is 8th October ( 8th October + 3 days of grace). Legal due date of the bill is 11October Number of unexpired days from 8 May to 11 October is 146 days Banker s discount =Ani 153 = 7650 r (2/5) r =(1/20) = 0.05 r = 5% Cost function C = AC x = (x + 5 + also, 5 For AC to be increasing + 5x + 36 1 > 0 1- Hence, average cost increases if the output x is > 6. 140 0 n =(2/5)year Question 15 (a) Calculate the index number for the year 2014, with 2010 as the base year by the weighted aggregate method from the following data: Commodity A B C D Price in 2010 2014 2 4 5 6 4 5 2 2 [5] Weight 8 10 14 19 (b) The quarterly profits of a small scale industry (in thousands of rupees) is as follows : Year Quarter 1 Quarter 2 Quarter 3 Quarter 4 2012 39 47 20 56 2013 68 59 66 72 2014 88 60 60 67 [5] Calculate four quarterly moving averages. Display these and the original figures graphically on the same graph sheet. Comments of Examiners (a) Some candidates used the weighted average of price relative method instead of weighted aggregate method to calculate the index number. A number of candidates wrote incorrect formula of weighted aggregate method. (b) Some candidates did not calculate centered moving average. Several candidates made mistakes while finding the four yearly moving averages as well as centered moving averages. Plotting of the centered average was inaccurate in a few cases. 141 Suggestions for teachers A thorough and comprehensive practice for calculation of Index number by various method is a must. Students must be advised to read the question carefully so as to work out the question using the correct method. Students must be advised to practice various methods for finding moving averages rigorously. They must be taught to plot a neat graph for both actual and trend. MARKING SCHEME Question 15. (a) Commodity 2010 po 2 5 4 2 A B C D 2014 w 8 10 14 19 p1 4 6 5 2 w 8 10 14 19 The index number for the year 2014 with 2010 as the base year is (b) Year Quarter 2012 1 2 Quarterly profits 39 4 yearly moving total 3 20 4 56 1 68 4 yearly average 4 yearly centered moving average 40.5 47.75 50.75 249 62.25 265 66.25 285 71.25 59 64.25 66 4 72 1 88 2 100 = 125 56.5 68.75 71.375 286 2014 16 50 56 38 160 49.25 203 3 32 60 70 38 200 44.125 191 2 pow 47 162 2013 p1w 71.5 70.75 280 70 275 68.75 60 3 60 4 67 69.375 Correct Graph Note: For questions having more than one correct solution, alternate correct solutions, apart from those given in the marking scheme, have also been accepted. 142 GENERAL COMMENTS: (a) Topics found difficult by candidates in the Question Paper: Determinant properties and their use. Conics (parabola, ellipse, hyperbola) Application of L Hospital s rule. Indefinite Integrals, Definite Integrals. Inverse trigonometric functions. Area of curves. Probability (Both sections) and probability distribution. Differential equations. Complex numbers. Vectors. 3D plane & straight-line. Annuities. Linear programming. Regression lines. (b) Concepts between which candidates got confused: Conics (parabola, ellipse, hyperbola) Open & closed intervals for Mean value theorem. Conversion of inverse trigonometric functions. Regression coefficient byx & bxy and r. Differential equations (Linear & Homogeneous form) Geometrical problem in vectors. Annuity due & ordinary annuity. Banker s discount & banker s gain. Price relative and aggregate method in Index No. Shortest distance between skew lines and parallel lines. Probability distribution (conceptual problem) (c) Suggestions for candidates: Learn to use the easiest method with correct formula for solving a problem. Theorem, rules and laws to be well understood. In each chapter, go through the theory and concepts thoroughly followed by solving the illustrations, examples without looking at their solutions. Revise and practice from previous year s question paper and sample papers. Question paper needs to be read carefully and answered accordingly. Wise choices should be made from the options available. All steps of calculation need to be simplified before proceeding to the next step. Take sufficient rest before the examination. Utilize the reading time properly. 143

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Additional Info : ISC Class XII Analysis Of Pupil Performance 2015 : Mathematics
Tags : ISC Board, Class 11th, Class 12th, NDA/NA Entrance Examination

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