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2013 pupil performance analysis (MY IDEAL - ROHIT SHARMA)

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Ashutosh Rath
De Paul School, Berhampur

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MATHEMATICS STATISTICS AT A GLANCE Total Number of students who took the examination 136056 Highest Marks Obtained 100 Lowest Marks Obtained 7 Mean Marks Obtained 70.12 Percentage of Candidates according to marks obtained Mark Range Details Number of Candidates Percentage of Candidates Cumulative Number Cumulative Percentage 0-20 2180 1.60 2180 1.60 21-40 9813 7.21 11993 8.81 41-60 37598 27.63 49591 36.45 61-80 35662 26.21 85253 62.66 81-100 50803 37.34 136056 100.00 Range of Marks Obtained 37.34 Percentage of Candidates 40.00 35.00 27.63 30.00 26.21 25.00 20.00 15.00 7.21 10.00 1.60 5.00 0.00 0-20 21-40 41-60 Marks Obtained 105 61-80 81-100 MATHEMATICS ANALYSIS OF PERFORMANCE Question 1 2 6 3 2 4 0 (a) Given A = ,B= , C = 2 0 4 0 0 2 Find the matrix X such that A + 2X = 2B + C. [3] (b) At what rate % p.a. will a sum of ` 4000 yield ` 1324 as compound interest in 3 years? [3] (c) The median of the following observations 11, 12, 14, ( x 2), (x + 4), ( x 9) , 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean. [4] Examiners Comments (a) Candidates committed errors in the addition of two matrices and made mistakes in finding A (i.e. negative of a matrix). Concept of change of sign of each element of a matrix to find negative of a matrix was unclear. Division by 2 to get X from 2X was missing, so many wrote incorrect answers. (b) Most candidates answered the question correctly. Some candidates attempted the sum directly without showing steps hence they lost marks. Some candidates took the amount equal to ` 1324 instead of ` 5324. Some failed to express both sides in the correct cube from e.g. 1331/1000 = (11/10)3. (c) Most candidates answered correctly. However a few failed to identify the difference between mean and median and also the median position. Failing to equate x + 4 = 24 the value of x was wrong hence the value of mean was incorrect too. Suggestions for teachers Sufficient dealing of operation of negative integers is necessary e.g. 6 + 4 2 = -4 4 10 2 5 x= 6 2 3 1 Emphasize on reading the question carefully, analyze and then write the given data. Finally they need to write the formula followed by the substitution. 5324 = 4000 ( 1+ r/100 ) . The concept of mean, median and mode for all distribution and methods of finding them must be made clear. Students must be advised on the meaning of the word hence in a particular question. It implies that the second result is a follow up to the first. So an independent working is incomplete that will lead to loss of marks. MARKING SCHEME Question 1 A + 2X 2X = = 2X = 2B + C 2B + C A 3 2 4 0 2 6 2 4 0 0 2 2 0 106 6 4 4 0 2 6 8 0 0 2 2 0 6 4 2 4 0 6 4 10 8 0 2 0 2 0 6 2 = 2X (b) = 2 X= 3 P = ` 4000 CI = ` 1324 A = ` 5324 A = P (1 r ) n 100 5324 = 4000 (1 r )3 100 3 5324 100 r 4000 100 5 1 1331 11 100 r 1000 10 110 100 r 11 110 10 3 (c) 3 100 + r = 110 r = 10% A1 11, 12, 14, (x-2), ( x 4) , (x+9), 32, 38, 47 Hence n 9 n 1 9 1 5th term Median = 2 2 x 4 = Given median = 24 ie x 4 24 x = 20 11,12,14,18,24,29,32,38,47 Substituting x = 20 sum of all the numbers = 225 mean = 225 25 9 Question 2 (a) (b) (c) What number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional? [3] 3 2 If ( x 2) is a factor of the expression 2 x ax bx 14 and when the expression is divided by ( x 3) , it leaves a remainder 52, find the values of a and b. [3] Draw a histogram from the following frequency distribution and find the mode from the graph: Class 0-5 5-10 10-15 15-20 20-25 25-30 [4] Frequency 2 5 18 14 8 5 107 Examiners Comments (a) The initial understanding of proportionality was clear to most candidates and hence wrote 6 x 20 x Or (6 + x)(43 + x) = (20 + x) (15 + x) 15 x 43 x The errors were mostly in multiplication or in simplification. (b) Most candidates committed errors in the first step of application of Remainder and Factor Theorem by writing incorrect simultaneous equations in a and b (c) Most candidates answered the question correctly however a few errors observed were: (i) Incorrect choice of axis. (ii) Histogram drawn with cumulative frequency instead of frequency (iii) Failure to draw the two intersecting lines and vertical line to identify mode. Suggestions for teachers Basic concepts of remainder and factor must be taught with clarity so calculation errors are minimized. These basic but major errors must be highlighted to enable students avoid committing errors in all forms of examination. They must be taught on how to choose the xaxis and y-axis. Concepts of Histogram as well as determination of mode from it should be clearly explained and students should be given additional practice. MARKING SCHEME Question 2 Q.2 (a) (b) Let x be the number 6 x 20 x = 15 x 43 x (6 x)(43 x) 258 49x x2 14x x The required number is (20 x)(15 x) 300 35x x2 42 3 = = = = 3 As x 2 is a factor f(x) = x3 + ax2 + bx + 6 f(x) = 0 2(2)3 + a(2)2 + b(2) 14 = 0 16 + 4a + 2b 14 = 0 4a + 2b = -2 OR 2a + b = -1 2(3) 3+ a(3)2 + b(3) 14 = 52 54 + 9a + 3b = 66 9a + 3b = 12 3a + b = 4 2a + b = -1 (-) (-) (+) 108 a=5 b=4 3a = 4 15 = -11 b = -11 a = 5, b = 11 (c) GRAPH Question 3 (a) Without using tables evaluate 3 cos 80 . Cosec 10 + 2 sin 59 sec 31 . (b) In the given figure, BAD 65 [3] 45 ABD 70 , BDC 45 (i) Prove that AC is a diameter of the circle. (ii) Find ACB [3] (c) AB is a diameter of a circle with centre C = (-2,5). If A = (3,-7). Find (i) the length of radius AC (ii) the coordinates of B. [4] Examiners Comments (a) Candidates committed errors in writing the complementary function correctly. Some skipped steps and wrote a direct complementary result without showing any working which led to loss of marks. (b) Though centre was not given many candidates assumed AC as the diameter and took ABC = 90 .Reasons related to the results was missing which led to loss of marks. (c) Some candidates were incorrect in applying the distance formula ( {(x2 x1)2 + (y2 y1)2}) correctly. Some did not equate (3+x)/2 to -2 or (-7+y)/2 to 5 so as to find x and y. The answer too was not written in the coordinate form. 109 Suggestions for teachers - Stress on working of a particular sum must always be shown, as skipping of relevant working steps leads to loss of marks. The candidates should have a clear concept of trigonometric functions of angles and the corresponding complementary angles. Emphasise on geometry needs to be worked on the basis of given data. Angles must be correctly named and all reasons clearly expressed. Incorrect Substitution was due to concept of integers e.g. +(-2) is -2 and not 2. Insist that the coordinate of a point must always be written in the form (a,b). MARKING SCHEME Question 3 (a) 3 cos 80 . Cosec 10 + 2 sin 59 sec 31 = 3 cos80 . cosec (90 80 ) + 2 sin 59 sec(90 59 ) = 3 cos 80 . sec 80 + 2 sin 59 cosec 59 1 1 3cos80 2sin 59 cos80 sin 59 3+2 5 = (b) = = Given BAD 65 ABD 70 BDC 45 (i) (c) ADB = 180 - (65 +70 ) = 45 (sum of the angles of triangle) ADB + BDC = 45 +45 =90 AC is a diameter (angle in a semicircle is 90 ) (ii) ACB = ADB = 45 (angles in the Same segment). A (3, -7) Centre C = (-2,5) 3 x 2 Or 7 x 5 2 2 3+x = -4 -7 + y = 10 x = -7 y = 17 A (3,-7) C (-2,5) B (x,y) B = ( -7, 17) Length of radius AC = = (3 2)2 ( 7 5) 2 25 144 = 169 = 13 units Question 4 (a) Solve the following equation and calculate the answer correct to two decimal places: x2 5x 10 0. (b) [3] In the given figure, AB and DE are perpendicular to BC. (i) Prove that ABC DEC (ii) If AB = 6cm: DE = 4cm and 110 AC = 15 cm. Calculate CD. (iii) Find the ratio of the area of ABC : area of DEC. (c) [3] Using a graph paper, plot the points A(6, 4) and B(0, 4). (i) Reflect A and B in the origin to get the images A and B . (ii) Write the co-ordinates of A and B . (iii) State the geometrical name for the figure ABA B . (iv) Find its perimeter. [4] Examiners Comments (a) In most cases substitution in formula was correct but Suggestions for teachers candidates made errors in calculations of square root. Explain that the correct calculations Due to rounding off to 2 decimal places at the initial to be done to the least 3 decimal stage the final answer of two decimal places was places so as to obtain the correct incorrect. approximated answer. Instruct Candidates solved the square root by division candidates on how to use 4 figure method however use of a four figure log table to find log tables to find the square root that the square root is permissible. Results obtained did facilitates the calculation process. not consider the 3 decimal places that would lead to Regular practice is necessary the correct answer. specially to write the corresponding (b) Proof of similarity was answered correctly by ratio of sides correctly. Insist on candidates but some made mistakes in writing the writing all ratios in its lowest team. corresponding ratios correctly for which answer to It is very necessary for graphs to be CD was incorrect. In the third part many candidates drawn for questions based on wrote: graphs. Further all coordinate forms Area ABC 36 must be written in word form. . But answers must be written in Area DEC 16 Concepts of negative and positive simplified form, i.e. 9/4 or 9:4. side of both x-axis and y-axis along (c) Various types of errors were committed such as: with plotting points in the graph (i) A few candidates did not use a graph paper paper on the basis of the coordinates of the given points must be made (ii) Some candidates interchanged the ve and +ve very clear. axes of both x and y (iii) Few candidates plotted the points incorrectly (iv) Plotted points (-6,-4) and (0,-4) incorrectly (v) Incomplete figures or named incorrectly. (vi) Perimeter was found in a lengthy process by using distance formula instead of using the graph. MARKING SCHEME Question 4 111 Solve x2 5x 10 0 (a) b b 2 4ac x = 2a 5 25 40 = 2 1 5 65 = 2 = 5 8.062 2 13.062 = or 3.062 2 2 = 6.531 or -1.531 = 6.53 or -1.53 Answer correct to 2 d.p. In ABC and DEC B E 90 and C C ABC ~ DEC (A.A similarity) AB AC DE DC (b) (ii) 6 15 = = DC = 10 cm 4 DC 94 9 : 4 area ABC 6 area DEC 4 A (-6, -4) (iii) (c) 2 3 2 2 B (0,-4) AB A B is a parallelogram B1 Perimeter = 2(6+10) = 32 units Question 5 (a) Solve the following inequation, write the solution set and represent it on the number line: (b) x x 1 1 1 , x R 3 2 3 6 [3] Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is of 8% per annum and Mr. Britto gets ` 8088 from the bank after 3 years, find the value of his monthly instalment. (c) Salman buys 50 shares of face value ` 100 available at ` 132. 112 [3] (i) What is his investment? (ii) If the dividend is 7.5%, what will be his annual income? (iii) If he wants to increase his annual income by ` 150, how many extra shares should he buy? [4] Examiners Comments (a) Errors were in transposition of x terms on one side and constants on the other. Many candidates failed to write the solution in the set form hence they lost marks. There were errors in representation of the solution on the number line. 1 was not marked on the number line. (b) Some candidates took n as 3 instead of 12 x 3 = 36. Many lost marks for using ` 8088 as interest instead of the matured value. In some cases candidates forgot to divide by 12 so as to convert rate per annum to per month. (c) Most candidates answered the question correctly. Some made errors in finding the dividend for taking N.V. as ` 32 instead of ` 100. Some committed errors in shifting the decimal point of 7.5% and ended up writing ` 37.50 instead of ` 375. Suggestions for teachers Emphasize on writing of solutions in a set form. If there is a fractional number in the solution set then it must be indicated. Students must be advised to read the question carefully and identify the given data. The concept that recurring deposits is a monthly deposit hence is monthly and rate is calculated on a per month basis needs to be made very clear. Additional practice sessions of questions on face value, nominal value, market value etc. and hence to apply the formula correctly. Calculation of dividends should be explained thoroughly. MARKING SCHEME Question 5 (a) x x 1 1 3 2 3 x x 4 3 2 3 4 x x 3 2 3 4 3x 2x 3 6 8 5x If correctly transposing all x terms to one side and numbers to the other side. 8 x 5 13 x 5 113 x 1 4 2 6 3 x 9 2 6 0 9 x 2 6 x 3 Sol. Set { x: 1 3 x 3, x R } 5 Let the monthly deposit be ` x OR n = 36 months n(n 1)x 36 37 x P= 2 2 Int = 36 37 x 1 8 2 12 100 = 111x 25 Amt = nx + Int 8088 = 36x + 111x 25 900x 111x 8088 25 1011x 8088 25 x = 8088 25 1011 x = ` 200 (b) (c) 1 13 5 2 3 Let the monthly deposit be `100 P = 36 37 100 2 36 37 100 8 Int = 2 12 100 = 444 Amount = 3600 + 444 = 4044 Amt x 4044 100 8088 8088 100 200 4044 The monthly deposit = `200 `50 x 132 ` 6600 7.5 100 50 ii) The dividend of a share = 100 7.5 100 50 50 shares = 100 = ` 375 iii) ` 7.50 can be earned from 1 share 150 No. of extra shares he should buy = =20 shares 7.5 d extra shares FV Or Extra dividend = 100 7.5 150 = x 100 100 x = 20 shares i) Total investment = = 114 4 Question 6 1 cos A sin A . 1 cos A 1 cos A (a) Show that [3] (b) In the given circle with centre O, ABC = 100 , ACD = 40 and CT is a tangent to the circle at C. Find ADC and DCT. A B O 100 D 40 [3] C (c) T Given below are the entries in a Savings Bank A/c pass book: Date Particulars Withdrawals Deposit Balance Feb 8 B/F - - ` 8500 Feb 18 To self ` 4000 - - April 12 By cash - ` 2230 - June 15 To self ` 5000 - - July 8 By cash - ` 6000 Calculate the interest for six months from February to July at 6 % p.a. 115 [4] Examiners Comments (a) Common errors observed were : (i) Working with both sides together instead of starting with either LHS or the RHS (ii) Squaring of both sides or omitting the square root sign. (iii) Some candidates cross multiplied and then went about proving it. (b) A few candidates made calculation errors in finding DCT. In many cases reasons were not stated to support their answers. (c) It is important to show all essential working. Many candidates were unable to show the completion process of the pass book leading to incomplete answers. Candidates made mistakes in finding the minimum qualifying balance. Some took time as 6 months instead of 1 and did not divide the same by 12 so as to have the rate per month instead of per year. MARKING SCHEME Question 6 (a) L.H.S 1 cos A = 1 cos A 1 cos A 1 cos A 1 cos A 1 cos A = 1 cos 2 A (1 cos A)2 = sin 2 A (1 cos A)2 = sin A 1 cos A 116 Suggestions for teachers Explain the basic rules of an identity and the method of proving it. The identity cannot be changed by squaring or taking the square root. Cross multiplication of terms should be avoided. The rule is to start from one side and prove the other. In problems on geometry it is essential to state reasons to score well. Explain the meaning of monthly balance and why we divide by 12 to find the interest. Mechanical workings without understanding the logic behind the concepts led to such errors, so concepts must be made clear. (b) ADC = 180 ABC = 80 CAD = 180 (80 +40 ) = 60 DCT = CAD = 60 Opp s of cyclic quad are supplementary sum of the s of a is 180 alternate segment theorem A B O 100 D 40 C (c) Date Particulars Withdrawals Feb 8 Feb 18 April 12 June 15 July 8 B/F To self 4000 By cash To self 5000 By cash Principal for the month February T Deposit Balance 2230 6000 = 8500 4500 6730 1730 7730 4500 Principal for the month March = 4500 Principal for the month April = 4500 Principal for the month May = 6730 Principal for the month June = 1730 Principal for the month July = 7730 Total principal for 1 month = ` 29690 Interest = 29690 6 12 100 = ` 148.45 Question 7 (a) In ABC, A(3, 5), B(7,8) and C(1,-10). Find the equation of the median through A. 117 [3] (b) A shopkeeper sells an article at the listed price of ` 1500 and the rate of VAT is 12% at each stage of sale. If the shopkeeper pays a VAT of ` 36 to the Government, what was the price, inclusive of Tax, at which the shopkeeper purchased the article from the wholesaler? (c) [3] In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30 and 60 respectively. Find: (i) the horizontal distance between AB and CD. (ii) the height of the lamp post. [4] Examiners Comments (a) Few candidates made incorrect observations on the slope of the median. It was +6 instead of -6. Some used incorrect points in finding the equation of the median. (b) Some candidates could not identify M.P. as ` 1500. In some cases candidates were unable to calculate the profit of the shopkeeper. (c) Most candidates answered the question correctly, However a few instead of using Tan 60 = 3, tan 30 = 1/ 3 used trigonometric tables which led to long and tedious working. 118 Suggestions for teachers - Insist on writing the coordinates of a point in (a, b) form. To find the equation of a line the point (x1 y1) is a point on the line and not any point from given data. Concepts of VAT needs to be made clear to candidates namely VAT= Output Tax Input Tax. Advise students to avoid using trigonometric tables for special angles. MARKING SCHEME Question 7 (a) Mid point M = 7 1 , 8 10 2 2 = (4, -1) Slope AM = 5 1 6 3 4 Equation of AM y 5 = -6 (x 3) y 5 = -6x +18 6x + y = 23 (b) Out put tax = = = A(3,5) B (7,8) 12% of `1500 12 ` 1500 100 ` 180 Let the price of the article charged by the wholesales be P before tax. Input tax = 12%P M VAT paid by shopkeeper = 12% of his profit 36 = 12 of profit 100 Profit = 36 100 = Rs. 300 12 CP = SP Profit = 1500 300 = `1200 OR 12P 3 P 100 25 VAT = out put Tax input Tax 3 180 36 = P 25 3 P = 180 36 25 = 144 144 25 P = 3 = `1200 Price inclusive of tax = P + input tax 3 P = P 25 = (c) = 1200 + 144 = ` 1344 In the rt-angled triangle ABC AB 3 tan 60 = BC 119 Price inclusive of VAT = 1200 12 1200 100 = 1200 + 144 = ` 1344. C (1,-10) = BC = 60m BC ED = = 60 3 E 60 3 20 3 3 3 = 20 1.732m 60 = 34.64 m the horizontal distance between AB and CD = 34.64 m In the right angled triangle AED AE 1 tan 30 = ED 3 AE = 20 3m 1 AE = 20 3 3 = 20 m the height of the Lamp post is 60 20 = 40 m Question 8 (a) x 3x Find x and y if y 4 y 2 5 = 1 12 . (b) A solid sphere of radius 15 cm is melted and recast into solid right circular cones of [3] radius 2.5 cm and height 8 cm. Calculate the number of cones recast. (c) [3] Without solving the following quadratic equation, find the value of p for which the given equation has real and equal roots: x2 + (p 3) x + p = 0. [4] 120 Examiners Comments (a) A number of candidates failed to show the necessary working of matrix multiplication but went on to 5x 5 write directly prove . This led to loss of 6y 12 marks. (b) Some candidates used incorrect formula for cone or sphere. Some made calculation errors mostly in cases where individual volumes have been calculated instead of forming an equation. (c) Many candidates substituted correctly for discriminant (b2 4ac) but did not equate to zero which is the basic concept of equal roots. Some made mistakes in expanding (p-3)2 or in factorization. Suggestions for teachers Stress on all working related to matrix addition with multiplication being clearly shown. Adequate practice of problems related to equating of volumes of different solids is necessary. Simpler methods of dealing with mensuration problems need to be taught that are time saving and effective too. Regular practice is necessary on nature of roots of quadratic equation and stress must be laid on steps in solving problems. MARKING SCHEME Question 8 (a) x 3x 2 5 y 4 y 1 = 12 2 x 3 x 1 5 2 y 4 y 1 = 12 2 x 3x 5 2 y 4 y = 12 5x 5 6 y = 12 5x 5 6 y 12 x 1 A1 y 2 (b) Volume of the metal obtained by melting sphere = = 4 3 r 3 4 15 15 15cm 3 3 Volume of the right circular cone of radius 2.5cm and height 8 cm 1 r 2 h 3 1 = (2.5) 2 8 3 = 121 4 15 15 15 = 3 1 2.5 2.5 8 3 = 270 Number of cones recast (c) x2 + (p-3) x + p = 0 If roots are equal b2 4 ac = 0 , a = 1, b = (p -3) and c = p (p 3)2 4(p) = 0 Or (p 3)2= 4 p p2 6p + 9 = 4 p p2 10p + 9 = 0 (p 9) (p 1) = 0 p = 9 or 1 Question 9 (a) In the figure alongside, OAB is a quadrant of a circle. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of the shaded [3] portion. (Take = 22 ) 7 (b) 3.5 cm A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box. (c) [3] Find the mean of the following distribution by step deviation method: Class interval Frequency 20-30 30-40 40-50 50-60 60-70 70-80 10 6 8 12 5 9 [4] 122 Examiners Comments (a) Two major errors committed were: (i) Using as 3.14 instead of 22/7 as stated in the question. (ii) In the process of calculation, initial rounding of results led to incorrect answers. (b) Most candidates solved the problem correctly. (c) Some candidates used the direct method instead of step deviation method as asked in the question and hence scored poorly. A number of calculation errors were detected. Suggestions for teachers Students must be advised to read the question carefully and use the data given in the question. Further, candidates must be told not to round off values in the initial stage, this leads to errors in the calculation of the final answer. Stress must be laid on writing the necessary outcomes, followed by explanation of all aspects on questions based on probability. Students must be advised to read the question thoroughly and answer according to the method asked for in the question. MARKING SCHEME Question 9 (a) Area of the quarter r 2 4 1 22 7 7 2 cm = 4 7 2 2 = 9.625cm2 1 OA OD Area of the AOD = 2 1 2 3.5 = 2 = 3.5 cm2 Area of the shaded portion = 9.625 3.5 = 6.125 cm2 circle = (b) Let the box contain x black balls Total number of balls in the box = Probability of drawing a black ball = Probability of drawing a white ball = x x 30 = x = x 30 x x 30 30 x 30 2 30 5 x 30 12 123 3.5 cm (c) Class Interval x =Mid 20 30 30 40 40 50 50 60 60 70 70 80 25 35 45 A 55 65 75 d x A c 2 1 0 1 2 3 f fd 10 6 8 12 5 9 20 6 0 12 10 27 49 26 fd 23 , f 50 , Mean C = 10 = A = 45 fd f C 23 10 50 = 45 + 4.6 = 49.6 Question 10 (a) Using a ruler and compasses only: (i) Construct a triangle ABC with the following data: AB = 3.5 cm, BC = 6 cm and ABC 120 (ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC. (iii) Measure BCP. [4] (b) The marks obtained by 120 candidates in a test are given below: Marks No of candidates 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 5 9 16 22 26 18 11 6 4 3 Draw an ogive for the given distribution on a graph sheet. Use suitable scale for ogive to estimate the following: (i) The median. (ii) The number of candidates who obtained more than 75% marks in the test. (iii) The number of candidates who did not pass the test if minimum marks required to pass is 40. [6] 124 Examiners Comments (a) In problems of construction many candidates did not use a ruler and compass to draw ABC = 120 . Some took an incorrect length for AB and BC. To draw the semicircle some candidates used a ruler to locate midpoint of BC instead of bisecting BC. Some were not clear about the locus of points equidistant from AB and AC. (b) Some candidates made mistakes in computing the c.f. which lead to an incorrect graph. A few plotted the graph with respect to lower boundaries instead of upper boundaries. Some candidates used the given frequency instead of cumulative frequency to draw an Ogive. Suggestions for teachers - Focus on basic locus concepts of points equidistant from 2 arms of the angle or equidistant from 2 points. Special attention must be given to the use of ruler and compass for all types of construction. Students must be told to recheck the c.f. carefully and to see that the last c.f is same as summation f. Ogive is to be drawn with respect to upper boundaries and corresponding cumulative frequency. Ensure students are clear regarding median position n/2 and (n/2 + (n+1)/2)/2. MARKING SCHEME Question 10 (a) (i) Construction of ABC, BC = 6 cm, B = 120 , AB = 3.5 cm. (ii) Bisecting BC and drawing a circle on diameter BC. (iii) Drawing an angle bisector of ABC and marking P on the circumference. (iv) Join PC and measuring BCP = 30 (b) Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70 70 80 80 90 90 100 f 5 9 16 22 26 18 11 6 4 3 Cu.f 5 14 30 52 78 96 107 113 117 120 (i) Median = 44 (ii) Number of candidates who obtained more than 75% = 120 -110 = 10 125 (iii) Number of candidates who did not pass the test = 52 Question 11 (a) In the figure given below, the line segment AB meets X-axis at A and Y-axis at B. The point P(-3,4) on AB divides it in the ratio 2:3. Find the coordinates of A and B. Y B A O X [3] x 4 1 17 . 8 2x 2 (b) Using the properties of proportion, solve for x, given (c) A shopkeeper purchases a certain number of books for ` 960. If the cost per book was ` 8 [3] less, the number of books that could be purchased for ` 960 would be 4 more. Write an equation, taking the original cost of each book to be ` x, and solve it to find the original cost of the books. [4] Examiners Comments (a) Concepts of A(a,0) and B(0,b) points on x-axis and y-axis were incorrect. Some candidates made mistakes in the application of the section formula. A number of candidates fared poorly for not writing the final answer in coordinate form as mentioned in the question. (b) This question was answered correctly by most candidates however a few did not use componendo and dividendo but cross multiplied directly. Some did not write x = 2 and lost marks for accuracy. (c) Most candidates misread the question and took a number of books to be x instead of cost of each book as x as given in the question. Some were unable to factorize x2 8x 1920 = 0. 126 Suggestions for teachers - Explain the basic concepts of points on x and y axis. Together they need to emphasize on writing cords in (a,b) form. - Students must be reminded about use of componendo and dividendo in ratio proportion problems. They must be advised to write results with when square root is taken. Students must be instructed to read the question carefully and before factorizing the equation they must simplify as far as possible. Students must be taught to identify that 960/(x-8) > 960/x. MARKING SCHEME Question 11 (a) A (x,0) 2 P (-3,4) 3 A (0,y) 2y 3 0 The coord of P = 2x0 3x , = (-3,4) 5 5 2y 0 0 3x 3 4 , 5 5 x = -5 , y = 10 A(-5, 0) (A1) and B(0,10) (b) (c) x 4 1 17 . 8 2x 2 Using componendo and dividend x 4 1 2x 2 17 8 x 4 1 2x 2 17 8 (x 2 1)2 25 9 (x 2 1)2 x2 1 5 x2 1 3 3x2 + 3 = 5x2- 5 2x2 = 8 x2 = 4 x = 2 Original cost of each book = x No. of books bought for ` 960 = The cost of each book was ` 8 less, then the cost of a book = 960 x x 8 Difference in the number of books bought = 960 960 4 x 8 x 240 240 1 x 8 x 240x 240(x 8) x(x 8) x2-8x 1920 = 0 x2-48x + 40x - 1920 = 0 (x-48) (x+40) = 0 x = 48 or x = -40 not valid 127 the original cost of the book = ` 48 Topics/Concepts Found Difficult Value Added Tax (VAT) Trigonometry Similarity Probability Choosing of scale for graph as scale was not provided Geometry Locus Construction Rounding off final result Step deviation method of calculation of mean Coordinate geometry , Section formula and points and x and y axis Quadratic equation problem Suggestions For Candidates Reading time must be utilized to make the right choice of questions and to be thorough with the given data Adequate practice must be done on rounding off digits Use graph paper for question based on graphs Use of log table to find square root of numbers Ensure all necessary steps of workings are clearly shown while solving problems. Working for matrix multiplication is essential Adopt methods where calculations are minimal to obtain final answers. 128

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