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Harish
Timpany School, Asilmetta, Visakhapatnam
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MATHEMATICS STATISTICS AT A GLANCE Total Number of students who took the examination 1,43,958 Highest Marks Obtained 100 Lowest Marks Obtained 3 Mean Marks Obtained 69.56 Percentage of Candidates according to marks obtained Mark Range Details 0-20 21-40 41-60 61-80 1877 9594 42477 38331 1.30 6.66 29.51 26.63 1877 11471 53948 92279 1.30 7.97 37.47 64.10 Number of Candidates Percentage of Candidates Cumulative Number Cumulative Percentage Range of Marks Obtained Percentage of Candidates 40.00 35.90 35.00 29.51 26.63 30.00 25.00 20.00 15.00 6.66 10.00 5.00 0.00 1.30 0-20 21-40 41-60 61-80 Marks Obtained 108 81-100 81-100 51679 35.90 143958 100.00 MATHEMATICS ANALYSIS OF PERFORMANCE Question 1 (a) Ranbir borrows ` 20,000 at 12% per annum compound interest. If he repays ` 8400 at the end of the first year and ` 9680 at the end of the second year, find the amount of loan outstanding at the beginning of the third year. (b) (c) Find the values of x , which satisfy the inequation 2x -25 <1 2, x W. Graph the solution set on the number line. 3 6 2 [3] [3] A die has 6 faces marked by the given numbers as shown below: 1 2 3 1 2 3 The die is thrown once. What is the probability of getting (i) a positive integer. (ii) an integer greater than 3. (iii) the smallest integer. [4] Examiners Comments (a) Most candidates committed a number of calculation errors. Some went on to calculate the amount at the end of the third year instead of the beginning of the third year. Others used formulas to find the amount in 2 years and then subtracted the repayment which is completely an incorrect concept. (b) Errors were made in transposing x terms on one side and constants on the other. Many candidates worked out both inequalities simultaneously and hence made major errors both with signs and transpositions, Solutions were not written in the set form, which led to loss of marks. It is necessary to put arrows on both sides of the number line with at least one extra element on each side so as to indicate the continuity of the infinite real number line. 109 Suggestions for teachers Students must be advised to read the question carefully so as to work out the sum in the correct method i.e. find amount at the end of first year ( 22400). Then principal for the second year after repayment ( 22400 8400= 14000).Find amount at the end of second year ( 15680) and finally 15680-9680= 6000 which is the required result. It is advisable to solve inequations of the form -25/6 < 1/2 - 2x/3 2 , by working with the two Inequalities separately to avoid gross errors. Solutions to be represented by set notation. Property of negative numbers must be clearly explained, e.g. -3x > -2 3x< 2 and not 3x>2. (c) Candidates failed to write sample space or listing the probable outcomes. Answers to probability were not given in the simplest form. Some failed to identify the smallest integer from the given set of integers {-3, -2, -1, 1, 2, 3}. Suggestions for teachers It is necessary to teach students the three basic steps of solving a probability sum. (i) Listing the total outcomes and favourable outcomes; (ii) Finding probability by using . ; . (iii) Writing the final answer in the simplest form. MARKING SCHEME Question 1 (a) P1 = `20,000 20000 1 12 = `2400 [M1] 100 A1=Amount at the end of 1st year = 20000+2400 = `22400 Repayment after 1st year = `8400 P2= principal for the 2nd year = 22400 8400 = `14000 [A1] 14000 1 12 =`1680 [M1] I2 = Interest for the second year = 100 A2= Amount at the end of 2nd year = 14000+1680 = `15680 Repayment at the end of 2nd year = `9680 Amount outstanding at the beginning of 3rd year= 15680 9680 = `6000 [A1] I1 = Interest for the first year = (b) 5 1 2x -2 < 6 2 3 2 x 1 17 3 2 6 2 x 3 17 3 6 2 x 10 3 3 x 5 1 2x 2 2 3 1 2x 2 2 3 1 4 2x 2 3 3 2x 2 3 1 2 x 4 correctly transposing all x terms to one side and numbers to the other side [ M1] (any one inequation) -2 1 x <5, x W 4 Solution = { 0, 1, 2, 3,4} (c) S = { 1,2,3, ---------, 25} , n (s) = 25 Let A be the event : getting a prime number 110 A= {2,3,5,7,11,13,17,19,23}, n(A) =9, P(Prime No) = n(A)/ n(S) = 9/25 [B1] Let B be the event: getting a no. divisible by 3 or 4 B= [3,4,6,8,9,12,15,16,18,20,21,24}, n(B) = 12, P(B) = n(B)/n(S) = 12/25 [B1] Let C be the event: getting a factor of 6 C= {1,2,3,6}, n(C) = 4 P(C) = n(C)/ n(S) = 4/25 [B1] Question 2 (a) 2 0 1 2 y Find x, y if + 3 = 2 . 3 1 2 x 1 3 (b) Shahrukh opened a Recurring Deposit Account in a bank and deposited `800 per month [3] for 1 years. If he received `15,084 at the time of maturity, find the rate of interest per annum. (c) [3] Calculate the ratio in which the line joining A(-4, 2) and B(3, 6) is divided by point P(x, 3). Also find (i) x (ii) Length of AP. [4] Examiners Comments (a) Errors were made in finding out the product of the 2 by 2 matrix and the 2 by 1 matrix. 2 0 Candidates wrote the product in the form 3 2 2 0 instead of 3 2 As a result of incorrect multiplication some candidates failed to evaluate x and y. (b) Some candidates considered the given matured value 15084 as the interest earned in the given period. A few took the rate as per annum instead of per month. Rate=r/12. Some substituted n as 1 year instead of 18 months. (c) Though the question was to find the ratio in which P divides join of A and B, many candidates considered P as the midpoint. Some used section formula correctly but equated to 0 instead of 3 as given in the problem. 111 Suggestions for teachers Concepts of multiplication of 2 matrices should be made clearer, e.g., result of the product of a (2X2) matrix and a (2X1) matrix is a (2X1) matrix. It is necessary to stress on reading the question carefully and assimilating the given data before answering it. The given data, e.g., matured value, time, monthly installment etc. It is necessary to identify the given data, write them down and then proceed with working with the noted data. MARKING SCHEME Question 2 Q.2 (a) 2 0 1 2 y 3 1 2 x 3 1 2 3 2 0 6 2 y 3 2 x + 3 6 [M1] [M1] 4 2 y 2x 6 2y = -4, 2x = 6 x= 3, y = -2 [A1] (both correct) (b) p = `800, n=18, M.V = `15084 pn(n 1) 1 R 18x19 1 R x x x x I = =800 x [M1] 2 12 100 2 12 100 = 114R Maturity value = p n + Interest 15084 = 800x 18 + 114R [M1] 15084 14400 684 6% R= 114 114 [A1] (c) The point of intersection is P(x, 3) my 2 ny1 6m 2n 3= [M1] m n m n 3m + 3n = 6m + 2n 3m = n Ratio = m : n = 1 : 3 [A1] 1 3 3 4 9 (i) x= 1 3 4 [A1] 9 ,3 4 A(- P(x,3 B(3,6 2 (ii) AP = 9 49 65 65 2 1 [A1] 4 (2 3) 4 16 16 4 Question 3 (a) Without using trigonometric tables, evaluate sin2 34 + sin 2 56 + 2 tan 18 tan 72 cot2 30 (b) Using the Remainder and Factor Theorem, factorise the following polynomial: 112 [3] x3 + 10 x2 37x + 26. (c) [3] In the figure given below, ABCD is a rectangle. AB = 14cm, BC = 7cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle.( Take = 22/7) A B F G [4] D E C Examiners Comments (a) Some candidates adopted incorrect methods of applying complimentary angles e.g., 34 90 56 34 90 34 or, The correct method is 34 90 34 = 56 34 56 90 56 Or, Many candidates used an incorrect value for cot 30 . (b) According to the question it was necessary to use Remainder and Factor Theorem, however some candidates failed to do so and hence lost marks. After factorizing the given mathematical expression, some did not write the result as the product of three factors. i.e. 10 37 26 1 2 13 Suggestions for teachers The method and concepts of complimentary angles must be made clear with additional revision to avoid the shown errors. It is essential to know the values of the special angles e.g. cot3o , tan60 etc. Emphasize on the use of Remainder and Factor Theorem to find at least one factor. It is essential to see that the final result is given as the product of the factors obtained. Insist that all given values of a question must be rightly used and deviation from given data would result to inaccurate answers. Students must also be instructed to carefully note the given data, e.g. radius, diameter etc. (c) was given to be 22/7 but some candidates took it as 3.14 hence leading to an incorrect result A few candidates took an incorrect radius for the semicircle and the quarter circle. A number of calculation errors were committed by the candidates. 113 MARKING SCHEME Question 3 sin234 + sin2 56 + 2 tan 18 tan 72 - Cot2 30 (a) = sin2 34 + sin2 (90 34 ) + 2 tan 180 tan(90 18 ) [M1] 2 [M1] (any one complementary angle correct) = sin2 34 + cos2 34 + 2 tan 18 cot 18 (b) 3 3 2 = 1 + 2tan 18 x 1/ tan 18 -3 since sin2 + cos2 = 1, sin (90 ) = cos , tan(90 )= cot = 1 + 2 x 1 3 = 0. [A1] f(x) = x3 + 10 x2 - 37x + 26 f(1) = 13 + 10 12 - 37 1 + 26 = 37 37 = 0 [M1] x 1 is a factor. x 1 x 2 11x 26 [M1] x 3 10x 2 37x 26 x3 x2 ( ) (+) 11 x2 37 x 11 x2 11 x (-) (+) - 26 x + 26 - 26 x + 26 (+) (-) 2 x + 11x 26 = x2 + 13x 2x 26 = x(x+13) 2 (x+13) = (x+13) (x 2) [M1] Hence x3 + 10x2 37x + 26 = (x -1) (x 2) (x +13) [A1] (c) Radius of the quarter circle = 7cm, radius of semi circle = 3.5 cm Area of the remaining portion= area of rectangle [area of quarter circle + area of semi circle] 114 =14 x 7 [ 1 22 1 22 7 7 . .7.7 . . . ] 4 7 2 7 2 2 OR [ M1] =98 [38.5+19.25] [M1] subtracting sum = 98 57.75 = 40. 25 cm2 [A1] Question 4 (a) The numbers 6, 8, 10, 12, 13, and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x. (b) [3] In the figure, DBC= 58 . BD is a diameter of the circle. Calculate: (i) BDC (ii) BEC A D (iii) BAC B C [3] E (c) Use graph paper to answer the following questions. (Take 2cm = 1 unit on both axis) (i) Plot the points A( - 4, 2) and B(2, 4) (ii) A is the image of A when reflected in the y-axis. Plot it on the graph paper and write the coordinates of A . (iii) B is the image of B when reflected in the line AA . Write the coordinates of B . (iv) Write the geometric name of the figure ABA B . (v) Name a line of symmetry of the figure formed. 115 [4] Examiners Comments (a) In the calculation of mean, conceptual errors were made, e.g. for 49+x some candidates took it as 49x. Method of finding median was incorrect. Some did not use the given condition mean=median. (b) Some candidates committed errors in this problem for not being clear about the following circle properties: (i) Angles in the same segment are equal e.g. A = D. (ii) Opposite angles of a cyclic quadrilateral are supplementary; e.g. A + E = 180 (iii) Identification of angle in a semicircle equal to 900. e.g. BCD = 90 since BD is the diameter. In number of cases no proper reasons were stated. (c) Most candidates did not use the scale given in the question i.e., 2 cm = 1 unit on both axis. Some marked the coordinate axes incorrectly due to which their plotted points and hence the diagram formed was incorrect. Some made errors in plotting B (2,0) and hence were unable to name the figure correctly. Some did not join the points ABA B to complete the figure. A majority of candidates did not mark/show the line of symmetry. MARKING SCHEME Question 4 rd th Q4 3 term 4 term 10 12 11 (a) median = 2 2 Median = Mean 11 = 6 8 10 12 13 x 6 x= 17 116 Suggestions for teachers Basic concepts of algebraic addition needs to be made more clear, e.g., x+y xy or 4+x 4x. Further the concepts of calculating median of simple distribution needs more drilling. The major problem of geometry is that the steps of working are not supported by valid reasons. Teachers must insist on the following: (i) Naming angles correctly e.g., in the figure cannot be written as B as there are two angles at that point. (ii) Essential working must be shown; (iii) All reasons must be clearly stated. Students must be instructed to read their question carefully and follow all given instructions. Given scales must be used. Stress on correct method of choosing and marking of coordinate axes. It is essential to write the coordinate of all points plotted and name them according to the given question. Train students to draw the line of symmetry name and record the particular line for identification e.g. AA (b) DBC = 58 A BCD = 90 (angle in a semi circle is a rt. D Angle) BDC= 180 (90 + 58 ) = 32 [M1] BEC= 180 - 32 (opposite angles of a cyclic quadrilateral are supplementary) = 148 58 B [A1] C BAC = BDC = 32 [A1] (angles in the same segment are equal) (c) E (i) Plotting points A( 4,2) and B(2,4), on the graph paper [B1] (ii) Reflecting A as A and writing A (4, 2) OR [B1] (iii) Reflecting B as B and writing B ( 2, 0) (iv) Geometric figure : Kite [B1] (v) AA is the line of symmetry [B1] Question 5 (a) A shopkeeper bought a washing machine at a discount of 20% from a wholesaler, the printed price of the washing machine being `18,000. The shopkeeper sells it to a consumer at a discount of 10% on the printed price. If the rate of sales tax is 8%, find: (i) (ii) (b) the VAT paid by the shopkeeper. the total amount that the consumer pays for the washing machine. If x 2 y 2 17 , then find the value of: x 2 y2 8 (i) x:y (ii) (c) [3] x 3 y3 x 3 y3 [3] In ABC, ABC = DAC. AB = 8cm, AC = 4cm, AD = 5cm. (i) Prove that ACD is similar to BCA (ii) Find BC and CD (iii) Find area of ACD : area of ABC B A D 117 C [4] Examiners Comments (a) (b) (c) Suggestions for teachers Most candidates wrote correct answers by finding the discount and subsequently the consumer s price of Concepts of VAT must be made clear to the students. VAT = Output the washing machine. A few calculation errors were Tax Input tax; Or, Tax charged however committed. Some candidates made Tax paid. Much more emphasis on calculation errors in finding the VAT paid by the this concept is essential. shopkeeper. The concept that VAT is the tax charged Students must be thoroughly taught on the additional value of an article was not clear. Number of errors were detected due to insufficient about the properties of ratio and knowledge of properties of ratio proportion. It is necessary for (e.g., compodendo and dividendo). teachers to emphasize that all Some applied comp. & Div. on the one side only, e.g., essential intermediate steps must be . clearly shown. Revision is necessary for proving . triangles similar and writing down , In some cases candidates directly wrote the corresponding proportional without any steps of working . Some were unable to sides and area. Students must be able to evaluate differentiate between congruency A few candidates took the wrong pair of angles in and similarity of triangles. It is proving ACD~ BCA. In some cases corresponding required to explain that proportional proportional sides of the two similar triangles was incorrectly written hence found the values of BC and sides are the sides opposite to the CD. corresponding equal angles. MARKING SCHEME Question 5 (a) Whole saler Printed price =`18000, discount = 20% Sales tax = 8% Sales price = price after 20% discount = 18000 - 20 100 Shopkeeper Consumer Printed price = `18000, discount = 10% Sales tax = 8% Sales price= price after 10% discount = 18000 - x 18000 118 10 100 x18000 = 18000 3600 = `14400 Tax charged by wholesaler = tax paid by Shopkeeper = 8 100 = 18000 1800 = `16200 Tax charged by shopkeeper= tax paid by Consumer = x 14400 = `1152 8 100 x16200 = `1296 (i) Vat paid by shopkeeper = tax charged by shopkeeper tax paid by shopkeeper = 1296 1152 = `144 [A1] (ii) Amount at which consumer bought the washing machine = 16200 + 1296 = `17496 [A1] (b) 2 2 x y 17 2 2 8 x y (i) (ii) x 2 y 2 x 2 y 2 17 8 [M1] x 2 y 2 x 2 y 2 17 8 x 3 y3 125 27 x 3 y3 125 27 x 3 y3 152 76 x 3 y3 98 49 2 25 2 9 2y 2x x 5 y 3 x 3 125 y3 27 x 5 [A1] y 3 [A1] Or x : y = 5 : 3 (c) (i) In ACD and ABC, CAD ABC ( given) ACD ACB ( C common) ACD BCA ( AA test of similarity [B1] A AC AD CD 4 5 CD [M1] or (ii) BC AB AC BC 8 4 BC= 6.4cm [A1] CD = 2.5 cm area ACD : area ABC= 52 : 82 = 25 : 64 [A1] B D C Question 6 (a) Find the value of a for which the following points A(a, 3), B (2, 1) and C(5, a) are collinear. Hence find the equation of the line. (b) Salman invests a sum of money in ` 50 shares, paying 15% dividend quoted at 20% premium. If his annual dividend is ` 600, calculate: (i) the number of shares he bought. 119 [3] (ii) (iii) (c) his total investment. the rate of return on his investment. The surface area of a solid metallic sphere is 2464 cm2. It is melted and recast into solid right circular cones of radius 3.5cm and height 7cm. Calculate: (i) the radius of the sphere. (ii) the number of cones recast. (Take = 22/7) [4] Suggestions for teachers It is necessary to explain to the [3] students that they cannot assume a condition for a question if it is not mentioned. Students must focus on the conditions which are given in the question before working it out. More practice on concepts related to premium and discount is necessary. Students must be advised that no answer must be kept in improper fraction. e.g. Examiners Comments must be written as 12 or 12.5. (a) Concepts of co linearity of points was incorrectly used. Many candidates considered B(2,1) as the midpoint of A(a,3) and C(5,a) hence lost marks. Some lacked the application of finding the equation of a line when two or more points are given. They may use or any other equivalent form of equation. Some used distance formula which is an extremely long method for this particular sum and hence were unable to calculate the correct answer. To solve the question candidates had to find the slopes between the given points and equate them. It is necessary to reinforce students with formulas related to volume and surface area. Students must be instructed to use the value of = 22/7, if it is given in the question and not =3.14. Instead of finding volume of cone and sphere separately students may be advised to use, Volume of Sphere= n volume of cones where n= number of cones. (b) Some candidates made calculation errors hence wrote incorrect answers. Some were unable to find the market value, resulting in incorrect calculations leading to incorrect answers. (c) Some candidates used the wrong formula for surface area of sphere, hence radius of sphere found was incorrect. This led to incorrect number of cones. Some took =3.14 instead of 22/7 as given in the problem. To calculate correctly the given data must be used. MARKING SCHEME Question 6 (a) A(a, 3), B(2, 1) and C(5, a) 1 3 2 Slope of AB = 2 a 2 a OR a 1 a 1 Slope of BC = [M1] 5 2 3 120 2 a 1 2 a 3 2 -6 = 3a a 2 a2 3a 4 = 0 (a 4) (a + 1) = 0 a = 4, a = -1. [A1] When A(4, 3), B(2, 1) ; When A(-1, 3), B(2, 1) 1 3 1 3 eqn y 3 = (x 4) eqn y 3 = (x + 1) 2 4 2 1 or x y = 1. [A1] 2x + 3y = 7 OR (b) Face value = ` 50, div% = 15%, M.V. = 50 + Annual Dividend = 600 (i) (ii) = 20 100 x 50= ` 60 n 50 15 100 n 15 2 No. of shares = n= 80 [B1] Investment = n x M.V. of 1 share = 80 x 60 =` 4800 [B1] Rate of return = A.D. investment 100 600 4800 100 = 12.5% Or = FV MV div% 50 60 15% = 12.5% [B1] (c) surface area of sphere = 4 R2 2464 =4x 22 x R2 7 [M1] R2 = 196 R = 14 cm [A1] No. of cones = Volume of sphere = Volume of cone 4 R 3 4 143 3 3 = = 1 r 2 h 1 (3.5) 2 7 3 3 [M1] = 128 cones [A1] Question 7 (a) Calculate the mean of the distribution given below using the short cut method. Marks 11-20 21-30 No. of 2 6 students 31-40 10 41-50 51-60 12 9 121 61-70 71-80 7 4 [3] (b) In the figure given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8cm, PD = 5cm, PB = 4cm. Find: (i) AB. (ii) the length of tangent PT. T (c) Let [3] A= B A P 3 2 and C = . 1 4 D Find A2 + AC C 4 1 2 1 0 2 , B = 3 2 5B. [4] Examiners Comments (a) Most candidates did not use the shortcut method to calculate mean. It is essential to find the class mark of a group frequency distribution irrespective of the method adopted to calculate its mean. e.g. for direct method we find fx , short cut method d=x-A and step deviation t= and in each case x represents the class mark. (b) Instead of using PT2= PA PB, or PT2= PC PD some candidates wrote PT2= PB X AB or PT2=PD CD. Hence they committed errors resulting in wrong answers. Some correctly proved triangles similar and found the correct answers. Suggestions for teachers Emphasize on all three methods of calculation of mean. For convenience of working guide students to choose one of the class mark as the assumed mean. The various properties of circles in Geometry needs repeated drilling so as to find unknown values using given conditions. Reasons must be clearly stated. More revision of multiplication of two matrices is essential. Stress must also be laid on Matrix addition and subtraction. (c) Many candidates made errors in finding the product A2 or AC. Only a few sfound A2 by squaring each element of the matrix A. Some made calculation errors especially those that involved the negative sign. MARKING SCHEME 122 Question 7 (a) Marks 11 20 21- 30 31- 40 41- 50 51- 60 61- 70 71- 80 Total No. of students 2 6 10 12 9 7 4 f =50 x 15.5 25.5 35.5 45.5 55.5 65.5 75.5 d= x- A - 30 - 20 - 10 0 10 20 30 fd - 60 -120 - 100 0 90 140 120 fd 70 [M1] for x or d column any 4 correct Assumed mean = A= 45.5, Mean = A + (b) PA x PB = PC x PD PA x 4 = 12.8 x 5 fd f = 45.5 + 70 50 [M1] = 45.5 + 1.4 = 46.9 [A1] [M1] 12.8 5 16cm PA = 4 AB= 16 4 = 12cm [A1] Radius of the circle = 6cm PT2 = PA xPB = 16 x 4 PT = 64 = 8cm [A1] T B A D C (c) 2 1 4 1 3 2 A ,B ,C 0 2 3 2 1 4 A2 + AC 5B 2 1 2 1 2 0 2 0 2 0 4 0 2 2 6 1 = 0 0 0 4 0 2 = 1 3 2 4 1 5 2 1 4 3 2 4 4 20 5 0 8 15 10 [M1] [M1] [M1] 5 4 0 7 8 20 0 4 2 8 15 10 = 123 P 3 2 23 = 17 = 8 20 5 4 15 10 3 [A1] 6 Question 8 (a) The compound interest, calculated yearly, on a certain sum of money for the second year is `1320 and for the third year is `1452. Calculate the rate of interest and the original sum of money. (b) [3] Construct a ABC with BC= 6.5 cm, AB= 5.5 cm, AC=5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle. (c) [3] (Use a graph paper for this question.) The daily pocket expenses of 200 students in a school are given below: Number of students Pocket expenses (frequency) (in `) 10 0-5 14 0 10 28 10 15 42 15 20 50 20 25 30 25 30 14 30 35 12 35 40 Draw a histogram representing the above distribution and estimate the mode from the graph. [4] Examiners Comments (a) (b) Many candidates were unable to differentiate between the concept of CI in 2 or 3 years and CI for the 2nd or 3rd year. Here interest for the 2nd and third year were 1320 and 1452 respectively. However answers was incorrect calculated. Some found the rate correctly but were unable to find the original value. Errors were made by candidates for using incorrect measurements of AB, BC, AC to construct triangle ABC. Several candidates did not drop a perpendicular on one of the sides from the point of intersection of the bisectors of the angles so as to get the radius of the in circle. 124 Suggestions for teachers The difference of CI in 2 years and CI during the 2nd year must be made clear to the students. Sufficient practice of inverse problems of CI is essential for better performance. Students must be advised to show all traces of construction in Geometry. And all given measurements must be correctly taken. Emphasize more on scale and choice of axes for all graphs. Adequate revision is necessary to read correct values from the graph. (c) Common errors committed in this particular question are as follows: (i) Incorrect axis. The gaps between the class intervals were not uniformly marked. (ii) Failure to draw the necessary intersecting lines and dropping a perpendicular to identify mode. (iii) Errors were made by candidates in recording the value from the graph. MARKING SCHEME Question 8 (a) Rate of Interest = I3 I 2 1452 1320 100 100 1320 I2 132 = I1 1320 [M1] 100 10% [A1] Let Principal be ` 100 100 1 10 10 = 100 I2 110 1 10 100 11 If the P = 100 , x= ? , I 2 11 I2 1320 Sum of money = x = 1320 100 11 = `12000 (b) Constructing ABC [M1] Constructing the angle bisectors of 2 angles and getting the incircle. [A1] Radius of the incircle = 1.5 cm( .2 ) 125 [A1] [M1] (c) Pocket expenses (in `) 0-5 0 10 10 15 15 20 20 25 25 30 30 35 35 40 Number of students (frequency) 10 14 28 42 50 30 14 12 Correct Bars [M1] Correct x and y-axis [B1] Lines for mode [M1] Mode = 21.4 ( .6) [A1] Question 9 (a) If (x 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x. [3] (b) Solve for x using the quadratic formula. Write your answer correct to two significant figures. (x 1)2 3x + 4 = 0. (c) [3] A page from the savings bank account of Priyanka is given below: Date Particulars Amount Amount withdrawn (`) deposited (`) Balance (`) 4000.00 03/04/2006 B/F 05/04/2006 By cash 2000.00 6000.00 18/04/2006 By cheque 6000.00 12000.00 25/05/2006 To cheque 30/05/2006 By cash 20/07/2006 By self 10/09/2006 By cash 19/09/2006 To cheque 5000.00 7000.00 3000.00 4000.00 6000.00 2000.00 1000.00 10000.00 8000.00 7000.00 If the interest earned by Priyanka for the period ending September, 2006 is ` 175, find the rate of interest. [4] 126 Examiners Comments (a) Some candidates failed to understand the concept of duplicate ratio hence were unable to solve the problem. Some made errors in considering 4:9 as the duplicate ratio of (x-9) : (3x+6). (b) Some candidates wrote (x-1)2 as x2-1 instead of x22x+1 hence lost marks. Candidates while substituting for 5 rounded off at the initial stage hence did not arrive at the correct answer. All approximations as stated in the question must be carried out at the final step. e.g. in this sum after division answer is 3.618 and 1.382 . Hence the answer correct to two significant figures is 3.6 and 1.4. (c) Some candidates made mistakes in identifying the minimum balances for some of the months. This led to incorrect answer. Some took rate as R and not R/12. Since the problem involves interest for 6 months a few took t = 6 instead of 1. MARKING SCHEME Question 9 Q.9 (a) (b) x 9 4 3x 6 9 2 [M1] 81(x 9) = 16(3x + 6) 81x 729 = 48x + 96 33x = 825 X = 25 [A1] [M1] (x - 1)2 3x + 4 = 0 x2 2 x + 1 3x + 4 = 0 x2- 5x + 5 = 0 [M1] ( 5) ( 5) 2 4.1.5 [M1] 2 5 25 20 = 2 5 5 5 2.236 = 2 2 5 2.236 5 2.236 ;x x = 2 2 x= 127 Suggestions for teachers Basic concepts of ratio must be made clear to the students, e.g. duplicate, triplicate, sub duplicate ratio etc. Recapitulation old concepts like etc. is necessary. Further, concepts of approximation of all types need thorough drilling. Teach students to find square root using log tables. This helps them to save time in finding the square root. Concepts of finding interest using must be made clear, i.e. t = 1 as interest is calculated monthly and divided by 12 as rate given is per annum. 7.236 2.764 ; x 2 2 = 3. 618 = 1.382 x = 3.6, = 1.4 [A1] (Both correct) = (c) Month April May June July August September Minimum balance (`) 6,000 7,000 10,000 6,000 6,000 7,000 Total P= 42,000 [M1] for any 4 minimum balance correct P = 42,000, [M1] (all correct) I =175, 1 R S.I.= P x 12 100 42000 1 R 175 = 12 100 [M1] 175 1200 R= = 5% [A1] 42000 Question 10 (a) A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number. (b) [4] The marks obtained by 100 students in a Mathematics test are given below: Marks No of students 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 3 7 12 17 23 14 9 6 5 4 Draw an ogive for the given distribution on a graph sheet. Use a scale of 2cm = 10 units on both axis). Use the ogive to estimate the: (i) median. (ii) lower quartile. (iii) number of students who obtained more than 85% marks in the test. (iv) number of students who did not percentage was 35. pass in the test if the pass [6] 128 Examiners Comments (a) A few candidates failed to identify the two digit and hence were number as 10x+y or 10 unable to form the right quadratic equation. Some calculations were based on a trial and error method without offering any proper explanations. (b) Errors were made in finding CF. Candidates failed to tally the last CF with the total frequency 100. Scale was not taken according to data given in the question. Lines were not shown for identifying the median, quartiles etc. Some candidates selected the x and y axis in the reverse order due to which the orientation of the S curve changed. Suggestions for teachers For a quadratic equation problem it is essential to form a quadratic equation with the given conditions and hence solve to find the unknown number. It must be noted that ogive is a cumulative frequency curve and the plotted points must be joined freehand and not with ruler. Students must be advised to use a scale as advised in the question. Revision and supervision is necessary to avoid such errors. MARKING SCHEME Question 10 (a) Let the unit s digit be x, ten s digit = 6 x 60 6 x +x = Original no. = 10 x x x 6 Reversed no. = 10 x x [B1] (any one correct) Original no + 9 = Reversed no. 60 6 x 9 10x x x [M1] equation 9 = 9x + 1= x - (b) 54 x 6 x [M1] x2 x 6 =0 (x 3) (x + 2) = 0 X = -2 ( not possible) , x = 3 60 60 x 3 = 23 [A1] Original number = x 3 Marks Frequency(f) 0 -10 3 10-20 7 20-30 12 129 c.f. 3 10 22 30-40 17 39 40-50 23 62 50-60 14 76 60-70 9 85 70-80 6 91 80-90 5 96 90-100 4 100 [B1] (for first 6 c.f. correct) [B1] (for smooth curve plotted with upper boundaries) N= 100 N th term, = 50th term. =45( 1) [A1] 2 N th term = 25th term 32( 1) (ii) lower quartile = 4 (i) Median = [A1] (iii) No. of students who obtained more than 85% marks.=6 ( 1) (iv) No. of students who did not pass the test. =29( 1) [A1] [A1] Question 11 (a) In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24cm, OM = 5cm, ON = 12cm. Find the: (i) (ii) length of chord CD. D radius of the circle. N C A (b) [3] O M B Prove the identity (sin + cos ) ( tan + cot ) = sec + cosec . (c) [3] An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45 and 60 respectively. Find the width of the river. Write the answer correct to the nearest whole number. 130 [4] Examiners Comments (a) Most candidates answered the question correctly. Some found CN but did not find the value of CD. (b) Some common errors observed were : (i) Working with both sides together; (ii) Skipping of necessary steps so as to get the answer; (iii) some opened the LHS expression but failed to simplify and come to the RHS. (c) Some candidates did not draw the diagram hence lost marks. Some used tan 600 = 1.732 instead of 3 for which working consumed extra time. A few did not calculate the answer to the nearest whole number. Suggestions for teachers Reasons to Geometry problems are essential and this needs regular drilling and supervision. Ensure that while proving identities students proceed with either LHS or RHS but must not work with both sides simultaneously. Advise students to draw a labeled diagram for all height and distance problem following the given conditions. For values like tan 600 students must consider it as 3 and not 1.732 else working is long and tedious. Students must be advised to read question carefully so as to avoid missing out the sub parts of the question specially rounding off of answers. MARKING SCHEME Question 11 Q.11 AM = BM = (24) ( perpendicular from centre (a) of a circle bisect the chord.) = 12 cm 2 2 C 2 From AOM, AO = 12 + 5 = 169 Radius of the circle = 169 = 13 cm [M1] [A1] A CO = AO = 13 cm From CON, 132 = 122 + CN2 CN2 = 169 - 144 = 25 CN= 5cm, Chord CD = 2 x5 = 10 cm [A1] 131 D N O M B (b) LHS = (sin + cos )( tan + cot ) = ( sin + cos )( sin cos ) cos sin sin cos ) cos sin 1 ) = (sin + cos )( cos sin 1 1 = cos sin = (sin + cos )( 2 = sec + cosec = RHS [M1] 2 [M1] [A1] LHS = RHs (c) CD = AD = 250 m [M1] ( A = 45 ) CD DB 250 [M1] [B1] 3 DB 250 250 3 = DB = 3 3 250 1.732 144.33 = 3 Tan 60 = C E A 45 F 60 60 45 D B AB = 250 + 144. 33 = 394.33 Ans = 394 m [A1] CAO Topics/Concepts Found Difficult Value Added Tax (VAT) Compound Interest inverse problems. Trigonometry Similarity Rounding off final result e.g. significant figures. Theorems on properties of circle. Properties of proportion. Construction of incircle. Short cut method of calculation of mean Coordinate geometry, Section formula and identifying points on x or y axis. Conditions of collinear. Quadratic equation problem 132 Suggestions for Candidates Reading time must be utilized to make the right choice of questions and make oneself familiar with all given data More practice must be done on rounding off of digits Use graph paper for questions based on graphs Use of log table to find square root of numbers Avoid skipping steps. All necessary steps must be clearly shown Working for matrix multiplication is essential Adopt methods where lesser calculation is necessary to get final result Necessary sample space must be written for probability problems. Steps of working is necessary in conversion of trigonometric ratio s of complementary angles. Reasons must be provided for all geometry problems. 133

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