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3rd Preliminary Examination, 2018-19 Mathematics (2 hour and 30 minutes) Grade: X Max Marks: 80 _____________________________________________________________________ Answers to this Paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper. The time given at the head of this Paper is the time allowed for writing the answers. _____________________________________________________________________ SOLUTION Section A Answer 1 a) 1 2 3 4 2 x2 3 X= 5 2x 1 X is a matrix of order 2 x 1. a Let, X = b 1 2 a = 3 3 4 b 5 a 2 b 3 => = 3 a 4 b 5 a 2 b 3 .x 3 => 3 a 4 b 5 x1 3 a 6 b 9 5 2 b 4 b 2 a -4 a = -3 =1 1 X = b) 2 4 a+5 b 4 a 5 b = 8 7 Apply componendo- dividendo, 4 a+5 b+4 a 5 b 4 a 5 b 4 a+5 b 8a 10 b = = 8+7 7 7 15 +1 . 2a 5b a b c) = + 15 +75 2 = a : b = 75 : 2 Let P = (x, y) 2= x + ( 6 ) +11 3 x=1 -5 = y +5+ 8 3 y = -28 Question 2 a) b) 9x2 + px + 1 = 0. b2 4ac = 0 (p)2 4 x 9 x 1 = 0 p2 = 36 p= 6 a=3 ar + ar2 = 90 3r + 3r2 = 90 r2 + r 30 = 0 r2 + 6r 5r 30 = 0 => r (r + 6) 5 (r + 6) = 0 => (r 5) (r + 6) = 0 r = 5, -6 c) 2 tan(90 55) 2 + cot 55 =2 cot 55 2 cot 55 + cot( 90 35) 2 tan 35 tan35 2 -3 tan35 -3 sec(90 50) . cosec 50 cosec 50 . cosec 50 =3 3=0 Answer 3 a) Reflex QPR = 2300. => QLR = 1150. (Angle at the circumference of a circle is half the angle at the centre) Similarly, QMR = 650. 2 QLR + QMR = 2 x 1150 + 600 = 2300 + 650 = 2950 (b) Number of persons to be seated = r 2 3 90 = r area of the floor floor space for one person 2 = 270 Volume of the cone = number of person x air needed for one person 1 3 1 3 c) 94+ x 9 2 r h = 90 x 25 x 270 x h = 2250 h = 25 m = 11 => 94 + x = 99 => x = 99 94 =5 3, 8, 11, 2, 5, 21, 19, 13, 17 Arranging in ascending order, 2, 3, 5, 8, 11, 13, 17, 19, 21 N=9 Median = 9+1 2 th ( ) term = 5th term = 11 Answer 4 a) PLM and PQR 1. PLM = PQR (Corresponding angles) 2. PML = PRQ (Corresponding angles) => PLM ~ PQR (A A similarity) Area of PLM Areaof PQR 2 = LM 2 QR (In similar triangles, ratio between their areas = ratio between squares of their corresponding sides.) Area of PLM 108 = 4 12 36 108 x 81 9 Area of PLM = b) 36 81 = 48 cm2 S.P. = Rs. 110 when the M.P. = Rs. 100 S.P. = Rs. 495 when the M.P. = 100 110 OR Let list price = x x 495 x+ 10 100 x x= 495 = Rs. 450 M.P. = Rs. 450 c) x= 450 Returns of first investment, 1.2 1.4 60 x 100 = 120 147 = 8.5% Returns of second investment, 0.8 6 x 100 = 80 6 = 13.3% The second investment is the better investment. Section B Answer 5 a) 5 1 2x < 2 6 2 3 17 1 2x < 6 2 3 17 1 2x => < 6 2 3 17 3 2 x => < 6 3 20 2 x => > 6 3 -2 11 x 10 x 495 495 x 10 11 = 20 2 x > 6 3 20 x 3 > x 2 x6 => => => 5 >x 1 2 2 x 3 2 x 3 ..(1) 2x 3 2 2- 1 2 3 2 9 2 -2x 9 4 -x 9 4 x - .(2) Combining (1) and (2), we get 9 4 {x : x < 5, x N} Solution set = {0, 1, 2, 3, 4} b) tan 60 = 120 3 x= = 120 x 120 3 3 = 40 3 m tan = 120 y y = 120 m Distance between the two boys = x+ y = 40 3 + 120 = 189.3 m c) a = 81 d = -3 tn < 0 a + (n 1)d < 0 81 3(n 1) < 0 81 < 3 (n 1) 27 < n 1 28 < n n > 28 th 29 term is the first negative term. Answer 6 a) 3, 11, 15, 17, q, q + 2, 27, 31, 40, 42 5 thterm +6 thterm 2 Median = => 23 = q+q +2 2 => 46 = 2q + 2 => 44 = 2q => q = 22 b) Volume of the cylinder = (6)2 x 8 = 288 cm2 Volume of each hemisphere, V1 = 2 3 (23) = 16 3 cm3 V2 = 2 3 (33) = 54 3 18 cm3 V1 + V 2 = 16 3 + 18 = ( 163 +18) = ( 16+54 3 ) Remaining volume of material of the cylinder. = 70 3 V (V1 + V2) 70 3 = 288 - = ( 288 x33 70 ) = 794 3 = 264.7 cm3 = 831.9 cm3 c) 8 p +1 2 p4 17 8 = Let x = P4 x 2 +1 2x => 17 8 = 8x2 + 8 = 34x => 4x2 17x + 4 = 0 => 4x2 16x x + 4 = 0 => 4x (x 4) (x 4) = 0 => (4x 1) (x 4) = 0 = 1 4 x= ,4 Substituting x = P4 P 4 = 22 1 P= ( 22 ) 4 = 2 2 P4 = 2 4 P = 2 Answer 7 a) = , 1 2 1 2 The point A is equidistant from PQ and QR as well as Q and R. b) N.V = Rs. 100 M.V = Rs. 120 Let income per share be x . x 120 x 100 = 6 100 x = = 72 10 120 x 6 100 = 7.2 Rate = 7.2 % Annual income = 7.2 x 250 = Rs. 1800 c) Let one side be x . Hence, length of other side including the right angle is 23 x By pythagoras; theorem. x2 + (23 x)2 = 172 Perimeter = 40 => x2 + 529 46x + x2 = 289 x (23 x)+ hypotenuse = 40 => 2x2 46x + 240 = 0 Hypotenuse = 17 => x2 23x + 120 = 0 => x2 15x 8x + 120 = 0 => x(x 15) 8(x 15) = 0 => (x 8) (x 15) = 0 x = 8, 15 Length of each side is 8 cm, 15 cm and 17 cm. Answer 8 a) QM = QL (pair of tangents drawn from an external) RL = RN Point to a circle are equal in length.) Let the perimeter of PQR be x x = PQ + QR + PR = PQ + QL + RL + PR ( QR = QL + RL) = PQ + QM + RN + PR ( QL = QM and RL = RN) = PM + PN PM = PN (Pair of tangents drawn from an external point to a circle have equal length.) 2PM = x PM = 1 2 1 2 x = (Perimeter of PQR) = PN b) cos 60 sin 45 tan 45 sin 60 1 2 1 1 2 3 2 1 2 1 3 2 2 3 2 2 2 = cos 45 tan 45 tan 45 sin 45 1 1 1 2 = 1 1 1 + 2 2 2 2 2 1 3 3 + 1+ 2 2 2 2 + 1 3 2 2 2 2 c) N = 36 I = 47196 1200 x 36 = 47196 43200 P = 1200 1200 x 3612 x 37 xr 2 1200 1200 x 18 x 37 x r 1200 = 3996 = 3996 = 3996 3996 18 x 37 r= = 6% Answer9 a) Total number of students in a class = 55 No. of girls = 11 No. of boys = 44 A is the event of selecting a boy. P(A) = 44 55 = 4 5 b) LHS cot + cose 1 cot cosec +1 + cose cosec cot = cosec +cot cot +cose 1 cosec + cot 1 cosec + cot cot = 2 = 2 cot + cosec (cose c cot ) cot cosec +1 = cot + cosec = = cos sin + 1 sin cos +1 sin c) Radius of the circumcircle, r = 4 cm. Question 10 a) Given, : AD DB = 4 5 (i) and (ii) : In triangles ADE and ABC. 1. ADE = ABC 2. AED = ACB (Corresponding s) (Corresponding s) ADE ~ ABC AD AB => 4 9 => = = AE AC AE AC (A A similarity) DE = BC DE = BC (iii) In s DEX and BXC, 1. DEX = XCB 2. DEX = XBC DEX ~ BXC Area of DEX Areaof BXC => (Alternate s) (Alternate s) = DE 2 BC 2 = 16 81 (In similar triangles, ratio between their areas = ratio between the squares of their corresponding sides.) b) x+y=4 x y = 2 2x = 6 x=3 3+y=4 y=1 Point of intersection is (3, 1). The equation of the line through (2, 0) and (3, 1) is, x1y1 y 0= 1 0 3 2 Y=x 2 c) (x 2) x2y2 Let f(x) = x3 7x2 + 14x 8 Put x = 1 f(1) = 1 7 + 14 8 f(1) = 0 (x 1) is a factor. x- 1) x2 7x2 + 14x 8 (x2 6x + 8 (-) x2 - x2 - 6x2 + 14x - 8 (-) 6x2 + 6x 8x - 8 8x 8 0 x2 4x 2x + 8 = (x(x 4) 2 (x 4) = (x 2) (x 4) Hence, the factors of f(x) are (x -1), (x 2) and (x 4). Answer 11 a) i) X-axis ii) B (-5, 3) iii) A (-2, 0) b) Class Interval f 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 3 7 8 2 9 c) rc = radius of the cone rc = 6 cm h = 7 cm l= r 2 + h2 = 36+49 = 85 Vc = Volume of the cone = 1 3 1 3 x 62 x 7 = 22 7 12 36 x7 = 264 cm2 x Vhs = Volume of hemisphere = 2 3 = 2 3 x 63 x 22 7 x 216 = 452.6 cm3 V = Volume of solid = Vc + Vhs = 264 + 452.6 = 716.6 cm3 ***************************
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