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ICSE Board 2011 CLASS X MATHEMATICS Maximum Marks: 80 Time: 2 1 Hrs 2 ___________________________________________________________ 1. Answer to this paper must be written on the paper provided separately. 2. You will NOT be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper. 3. The time given at the head of this paper is the time allowed for writing the answers. 4. This question paper is divided into two Sections. Attempt all questions from Section A and any four questions from Section B. 5. Intended marks for questions or parts of questions are given in brackets along the questions. 6. All working, including rough work, must be clearly shown and should be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. 7. Mathematical tables are provided. Section A (40 Marks) Ans1 (a). Here, p(x) = x3 + 2x2 kx + 10 For (x-2) to be the factor of p(x) = x3 + 2x2 kx + 10 p(2) = 0 Thus, (2)3 + 2(2)2 k(2) + 10 = 0 8 + 8 - 2k + 10 = 0 k = 13 Thus p(x) becomes x3 + 2x2 13x + 10 Now, (x+5) would be the factor of p(x) iff p(-5) = 0 p(-5) = (-5)3 + 2(-5)2 13(-5) + 10 = -125 + 50 + 65 + 10 =0 So, (x+5) is also a factor of p(x) = x3 + 2x2 13x + 10. (b) Yes, product AB is possible since the number of columns of matrix A is equal to the number of rows of matrix B . (Matrix A is of the order 2 x 2 and B is of the order of 2 x 1) 3 5 The required product AB = 4 2 6 + 20 = 8 8 26 = 0 (c). 2 4 Here Principal, P = 15000 Rate of interest, R = 8% for first year and 10% for second year PXRXT 15000x8x1 Interest for 1st year = = 1200 = 100 100 Amount at the end of first year = 15000 + 1200 = 16200 Kumar repays 6200 Principal for second year = 16200 6200 = 10000 PXRXT 10000x10x1 = 1000 Interest for second year = = 100 100 Amount at the end of second year = 10000 + 1000 = 11000 Ans2 (a) In a deck of cards, for each suit we have three cards with number 3, 6, 9 which are multiples of 3. Thus for four different suits Spade, Heart, Diamond, Club , 3 x 4 = 12 such cards will be removed. Total number of possible outcomes = 52 12 = 40 (i) Each suit has 3 face cards. Four suits (Spade, Heart, Diamond, Club) will have 3 x 4 = 12 face cards. So, required probability will be given by 12 3 P (getting a face card) = = 40 10 (ii) Each suit has 4 (cards with number 2, 4, 8, 10) even numbered cards. Suits Heart and Diamond are of red colour. Thus, two suits will have 2 x 4 = 8 even numbered cards. So, required probability would be given by 8 1 P (getting an even numbered red card) = = 40 5 (b) x 18 =6 x x2 18 =6 x x2 6x 18 = 0 Here a = 1,b = 6 and c = 18 Thus the roots of the equation will be x= b b2 4ac 2a x= ( 6) ( 6)2 4(1)( 18) 2(1) x= 6 108 2 x= 6 6 3 2 x =3 3 3 x = 3 3 1.73 [U sin g, 3 = 1.73] x = 8.19 and 2.19 (c) In AOC, ACO = 30o (Given) OAC = 90o [radius is perpendicular to the tangent at the point of contact] By angle sum property, ACO + OAC + AOC = 180o AOC = 180o (90o + 30o) = 60o Consider AOC and BOC AO = BO (radii) AC = BC (tangents to a circle from an external point are equal in length) OC = OC (Common) AOC is congruent to BOC. (i) BCO = ACO = 30o (ii) AOC = BOC = 60o AOB = AOC + BOC = 120o (iii) We know that, If two angles stand on the same chord, then the angle at the centre is twice the angle at the circumference. AOB and APB stand on the same chord AB. AOB = 2 APB So, APB = 1 AOB =60o 2 Ans 3. (a) P= 2500, n = 2 years = (2 12) months= 24 months Total Principal = 2,500 x 24 = 60,000 Amount = 66,250 Interest = Amount Principal = 66,250 60,000 = Thus, the interest paid by the bank is Rs 6,250. Let r be the rate of interest. n(n + 1) 24 25 N= = = 25 yrs 2 12 2 12 This is equivalent to depositing P N R 100 Interest = 6, 250 = Rs 6,250 2,500 for 25 yrs. 2,500 25 R 100 R = 10 Thus, the rate of interest is 10%. (b) Diameter of the semi circle is 14 cm. ED = AC = 14 cm Therefore, AB = BC = AE = CD = 7 cm Area of the shaded region = Area of semi circle EFD + Area of rectangle AEDC Area of quadrant ABE Area of quadrant BCD r 2 22 1 = 7 7 = 77 cm2 2 72 Area of rectangle AEDC = AC AE = 14 cm 7 cm = 98 cm2 90 2 r 360 Area of quadrant ABE = Area of quadrant BCD = Area of semi circle EFD = 2 = 1 22 7 77 2 = cm 4 7 2 8 77 Area of the shaded region = 77 cm2 + 98 cm2 2 x 8 cm2 = 155.75 cm2 The coordinates of the vertices of ABC are A(1, 3), B(4, b) and C(a, 1). It is known that A(x1, y1), B(x2,y2) and C(x3,y3) are vertices of a triangle, then (c) the coordinates of centroid are G = x1 + x2 + x3 y1 + y2 + y3 , . 3 3 Thus, the coordinates of the centroid of ABC are 1 + 4 + a 3 + b + 1 5 + a 4 + b , = 3 , 3 3 3 It is given that the coordinates of the centroid are G(4, 3). Therefore, we have: 5+a =4 3 5 + a = 12 a=7 4+b =3 3 4+b = 9 b=5 Thus, the coordinates of B and C are (4, 5) and (7, 1) respectively. Using distance formula, we have: BC = (7 4)2 + (1 5)2 = 9 + 16 = 25 = 5 units Ans 4. (a) The given inequation is 2x 5 5x + 4 < 11, where x I 2x 5 5 x + 4 2x 5x 4 + 5 3x 9 x 3 5x + 4 < 11 5x < 11 4 5x < 7 x < 1.4 Since x I, the solution set is {-3, -2, -1, 0, 1}. The solution set can be represented on a number line as follows: (b) 2 2 tan 35 cot 55 sec 40 2 + tan 35 3 cos ec50 cot 55 2 2 tan(90 55 ) cot(90 35 ) sec(90 50 ) = 2 + 3 cos ec50 cot 55 tan 35 Q tan(90 ) = cot 2 2 cot 55 tan 35 cos ec50 = 2 + 3 cot(90 ) = tan cot 55 tan 35 cos ec50 sec(90 ) = cos ec 2 2 = 2 (1) + (1) 3 (1) = 2 +1 3 = 0 (c) The histogram for the given data can be drawn by taking the marks on the x-axis and the number of students on the y-axis. To locate the mode from the histogram, we proceed as follows: i Find the modal class. Rectangle ABCD is the largest rectangle. It represents the modal class, that is, the mode lies in this rectangle. The modal class is 80 90. ii Draw two lines diagonally from the vertices C and D to the upper corners of the two adjacent rectangles. Let these rectangles intersect at point H. iii The x-value of the point H is the mode. Thus, mode of the given data is approximately 83. Section B(40 marks) Ans5 (a) Given cost of washing machine = 15000 (i) Amount of tax collected by manufacturer = 8% of 15000 8 = 15000 = 1200 100 As profit of wholesaler is As trader earns a profit of 1200, VAT to be payed by wholesaler = 8% of 1200 8 = 1200 = 96 100 1800 VAT to be payed by trader = 8% of 1800 8 = 1800 = 144 100 Amount of tax received by government = (1200 + 96 +144) = 1440 (i) Value of machine paid by the consumer = Price charged by manufacturer + Profit of wholesaler + Profit of trader = (15000+1200+1800) = 18000 Tax paid by consumer = 8% of 18000 8 = 18000 = 1440 100 Therefore, amount paid by customer = (b) Volume of solid cone = (18000+1440) = 19440 12 1 22 1 22 r h = 52 8 = 25 8 3 37 37 3 4 22 5 43 4 22 125 Volume of a small sphere = r = = 3 3 7 10 3 7 1000 1 22 25 8 Volume of cone Number of spheres formed = =3 7 = 400 4 22 125 Volume of sphere 3 7 1000 Thus 400 spheres are obtained by melting the solid cone. (c) We know that the diagonals of a parallelogram bisect each other So, coordinates of mid point of BD = x coordinate of B + x coordinate of D y coordinate of B + y coordinate of D , 2 2 5 + 2 8 4 7 = , = ,2 2 2 2 x + 4 y + 7 Now the midpoint of diagonal AC = , 2 2 From (1) and (2), we get ..(1) (2) (ii) 7 x + 4 y + 7 2 , 2 = 2 ,2 Comparing we get, x+4 = 7 and y+7 = 4 Thus x = 3 and y = -3 So, the coordinates of point A are (3, -3) y y1 Equation of a line is given by y y1 = 2 (x-x1) x2 x1 Coordinates of point B and D are (5,8) and (2,-4) respectively. 4 8 Equation of a diagonal BD, y 8 = (x 5) 2 5 y - 8 = 4(x-5) Or 4x y = 12 Ans6. (a) (i) The points A(4,4), B(4,-6) and C(8,0) are plotted on the coordinate plane as follows: (ii) To reflect the points ABC across Y axis we will keep y coordinate as it is and negate the x coordinate. The reflected image of the triangle is shown in blue. (iii) The coordinates of point A , B , C are (-4, 4), (-4,-6) and (-8, 0) respectively. (iv) The figure AA C B BC obtained is a polygon with six sides. Thus such a figure would be called a hexagon. (v) The line of symmetry of AA C B BC would be the y axis. (b) Here rate of interest = 5% Principal Principal Principal Principal Principal Principal Principal Principal Principal Principal for for for for for for for for for for April, 07 = 7350 May, 07 = 11900 June, 07 = 11900 July, 07 = 13400 August, 07 = 13400 September, 07 = 14400 October, 07 = 14400 November, 07 = 15200 December, 07 = 15200 January, 07 = 13200 Principal for Feb, 07 = 13200 Principal for March, 07 = 14150 Total principal for April 2007 to April 2008 = 157700 1 157700 5 Pr incipal Rate Time 12 = Interest paid = = 100 100 Ans7 (a) 3 x + 4 + 3x 5 =9 3x + 4 3x 5 Using componendo and dividendo, 3 x + 4 + 3x 5 + 3 x + 4 3 x 5 3 x + 4 + 3x 5 3 x + 4 + 3 x 5 2 3x + 4 2 3x 5 3x + 4 = = 10 8 5 4 3x 5 Squaring both sides, 3x + 4 25 = 3x 5 16 16(3x + 4) = 25(3x 5) 48 x + 64 = 75x 125 75x 48 x = 64 + 125 27 x = 189 x =7 (b) 2 Given, A = 1 2 At = 5 1 3 5 4 , B = 1 3 2 3 = 9 +1 9 1 657.08 = 657 2 A t .B + B.I = 5 8 1 = 20 3 1 4 2 . + 3 1 3 4 + 3 4 + 0 + 10 + 9 1 + 0 1 4 2 7 = + 1 1 3 17 1 2 7 + 4 = 1 + 3 17 1 3 11 = 2 16 4 1 2 1 . 3 0 0 2 0 + 3 0 1 (c) (i) In ADB and CAB, ADB = CAB ABD = CBA ABC ~ DBA In ADC and BAC, ADC = BAC ACD = ACB DAC ~ ABC ( both 90o) (common angle) (AA similarity criterion) ( both 90o) (common angle) (AA similarity criterion) If two triangles are similar to one triangle, then the two triangles are similar to each other. DAC ~ DBA or CDA ~ ADB (ii) Since the corresponding sides of similar triangles are proportional. CD DA = AD DB AD2 = DB CD AD2 = 18 8 AD = 12 cm (iii) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. So Ar ( ADB) AD2 144 9 = = = Ar ( CDA) CD2 64 4 Thus, the required ratio is 9: 4. Ans8. (a) Class interval Frequency (f) 50 55 55 60 60 65 65 70 70 75 75 80 80 85 85 90 Total 5 20 10 10 9 6 12 8 80 x 52.5 57.5 62.5 67.5 72.5 77.5 82.5 87.5 d=x A= x 67.5 -15 -10 -5 0 5 10 15 20 t= i=5 -3 -2 -1 0 1 2 3 4 d i f t -15 -40 -10 0 9 12 36 32 24 Assumed mean (A) = 67.5 Class size, i = 5 Mean = A + i ft f = 67.5 + 5 24 80 = 67.5 + 1.5 = 69 Thus, the mean of the given data is 69. Modal class is the class corresponding to the greatest frequency. So, the modal class is 55 60. (b) Marks 0 10 10 20 20 30 30 40 40 50 50 60 60 70 70 80 No. of students (f) 5 11 10 20 28 37 40 29 cf 5 16 26 46 74 111 151 180 80 90 90 100 14 6 194 200 The ogive can be drawn as follows: (i) Median marks will be 57.5 as the x coordinate corresponding to n/2 i.e., 100 is 57.5. (ii) The number of students who failed is 46, which is the y coordinate corresponding to 40 marks. (iii) Number of students who secured more than 85 marks (grade one) = Total number of students 184 = 200 184 = 16 Ans 9. (a) (i) Amount invested = 52,000 Face value of share = 100 Discount = 20 Market price = 100 20 = 80 Number of shares = 52,000/ 80 = 650 Dividend % = 8% Total FV = FV of each share Number of shares = 65,000 100 x 650 = D = D% Total FV = D% 8 65, 000 = 5,200 100 Thus, the annual dividend is Thus, (ii) 5,200. Amount at which the shares were sold = Profit earned including his dividend = Profit 120 x 650 = (78,000 52,000) + 78,000 5,200 = 31,200 (b) For constructing the pair of tangents to the given circle following steps will be followed 1. Taking any point O of the given plane as centre draw a circle of 3.5 cm. radius. Locate a point P, 6 cm away from O. Join OP. 2. Bisect OP. Let M be the midpoint of PO. 3. Taking M as centre and MO as radius draw a circle. 4. Let this circle intersect our circle at point Q and R. 5. Join PQ and PR. PQ and PR are the required tangents. We may find that length of tangents PQ and PR are 8 cm each. (c) Consider LHS LHS = ( cos ec A - sin A ) ( sec A - cosA ) sec2 A 1 1 1 = sin A cos A sin A cos A cos 2 A 1 sin2 A 1 cos 2 A 1 = sin A cosA cos2 A cos2 A sin2 A 1 sin A cos A cos 2 A sin A = = tan A = RHS cos A = Ans10 (a). 6 is the mean proportion between two numbers x and y, i.e. 6 = xy So, 36 = xy .(1) It is given that 48 is the third proportional of x and y So, y2 = 48x (2) From (1) and (2), we get 36 y2 = 4 8 y y3 = 1728 Hence, y = 12 (b) Here Principal = 12,000, Rate = 10%, Compound Interest = N R 1 + Compound Interest = P 1 100 N 10 3972 = 12000 1 + 1 100 11 N 3972 = 12000 1 10 N 3972 +1 = 12000 11 10 1331 11 = 1000 10 N 3 N 11 11 = 10 10 N = 3 years (c) Let the height of the building be AB = h and BC = x 3972 In ABC, tan 60o = h x x 3 =h In ADB, ...(1) h x+60 1 h = 3 x+60 tan30o = x+60=h 3 ...(2) From (1) and (2) x+60=x 3 . 3 2x=60 x=30 Thus,h=30 3 = 51.96m Ans 11. (a) Let the radii of the circles with A, B and C as centres be r1, r2 and r3 respectively. According to the given information, AB = 10 cm = r1 + r2 (1) BC = 8 cm = r2 + r3 (2) CA = 6 cm = r1 + r3 (3) Adding equations (1), (2) and (3), 2(r1 + r2 + r3) = 24 (4) r1 + r2 + r3 = 12 Subtracting (1) from (4), r3 = 12 10 = 2 Subtracting (2) from (4), r1 = 12 8 = 4 Subtracting (3) from (4), r2 = 12 6 = 6 Thus, the radii of the three circles are 2 cm, 4 cm and 6 cm. (b) Number of children = x. Share of each child = 480 x If number of children are x + 20, then share of each child = According to the given information: 480 480 = 12 x x + 20 1 1 480 = 12 x x + 20 x + 20 x 480 = 12 x(x + 20) 20 480 = 12 x(x + 20) 480 20 = x(x + 20) 12 x2 + 20x 800 = 0 x2 20x + 40x 800 = 0 x(x 20) + 40(x 20) = 0 (x 20)(x + 40) = 0 x = 20 or x = 40 But, the number of children cannot be negative. Therefore, x = 20 (c) The equation of the line L1 is y = 4. It is given that L2 is the bisector of angle O and O = 90 . Thus, the line L2 makes an angle of 45 with the x-axis. 480 x + 20 Thus, slope of line L2 = tan 45 = 1 The line L2 passes through (0, 0) and its slope is 1. So, its equation is given by y y1 = m(x x1) y 0 = 1(x 0) y=x Now, the point P is the point of intersection of the lines L1 and L2. Solving the equations y = 4 and x = y, we get x=y=4 Thus, the coordinates of the point P are (4, 4).

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