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ICSE Class X Sample / Model Paper 2024 : Mathematics

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Sample Question Paper 14 (Detailed Solutions) 1. (i) (a) Well-shuffling ensures equally likely outcomes. Total number of outcomes = 52 Now, there are 4 aces, one of each suit. Number of favourable outcomes to the event an ace = 4 4 1 P(an ace) = = 52 13 (ii) (c) The number of atheletes who completed the race in less than 14.6 sec = 2 + 4 + 5 + 71 = 82 (iii) (d) Given, a, 6,18 and b are in continued proportion. a 6 18 = = 6 18 b a 6 = a=2 6 18 18 6 18 18 and = b= b = 54 b 18 6 Hence, a = 2 and b = 54. (iv) (c) Total cost of services = ` (600 + 700 + 800 + 1000) = ` 3100 Amount of GST =10% of ` 3100 3100 10 = = ` 310 100 (v) (a) We have, 3 2 x x 10 10 + 3 2 x x 10 + 10 [adding 10 both sides] 13 2 x x 13 2 x + 2 x x + 2 x [adding 2x both sides] 13 3 x 13 x 3 4.3 x Hence, x { 4 , 3, 2,1} [Q x N ] or x {1, 2, 3, 4 }. (vi) (c) We have, diameter of metallic sphere = 6 cm Radius of metallic sphere, r1 = 3 cm Also, diameter of cross-section of cylindrical wire = 0.2 cm Radius of cross-sections of cylindrical wire, r2 = 01 . cm Let the length of the wire be h cm. Since, metallic sphere is converted into a cylindrical shaped wire of length h cm. Volume of the metal used in wire = Volume of the sphere 4 2 r2 h = r 13 3 2 4 1 h = (3)3 10 3 1 h = 36 100 36 100 h= = 3600 cm = 36 m [Q1m = 100 cm] (vii) (a) Let P(0, k) be any point on Y-axis. Then, PA 2 = PB 2 2 (6 0) + (5 k)2 = ( 4 0)2 + (3 k)2 36 + 25 + k2 10k = 16 + 9 + k2 6k 4 k = 36 k=9 Hence, coordinates of point on Y-axis are (0, 9). (viii) (d) In ACB, BCA = 90 [angle in a semi-circle] AB 2 = AC 2 + BC 2 [by Pythagoras theorem] AB 2 = 4 2 + 32 AB 2 = 16 + 9 AB 2 = 25 AB = 5 5 OA = cm 2 (ix) (c) Let a, a + d , a + 2d and a + 3d be the first four terms of an AP. Given, that first term, a = 2 and common difference, d = 2, then we have an AP as follows 2, 2 2, 2 + 2 ( 2), 2 + 3 ( 2) i.e. 2, 4 , 6, 8 (x) (c) Given, 2 x 2 = 3 x 2x2 3x = 0 x(2 x 3) = 0 x = 0 or 2 x 3 = 0 x = 0 or 2 x = 3 3 x = 0 or x = 2 (writing as ax 2 + bx + c = 0) (factorising left side) (zero product rule) 3 Hence, the roots of the given equation are 0, . 2 (xi) (a) The image of a point (x , y) under the reflection of X-axis is P (x , y), i.e. only the sign of y-coordinate will be changed. (3, 6) AB 4 2 BC 5 2 (xii) (c) Here, = = , = = , DF 6 3 EF 7.5 3 AC 3 2 = = DE 4.5 3 AB BC AC = = DF EF DE So, ABC ~ DFE [by SSS similarity criterion] Hence, figures (i) and (ii) are similar triangles and no other pair of triangles in the given figure are similar. As, (xiii) (d) Given, ax + b is factor of f (x). So, by factor theorem b f = 0, where a 0 a Hence, option (d) is correct. (xiv) (a) We have, a 7 2 c 4 7 b 4 + 1 d = 8 9 a + 2 c 7 4 7 = b + 1 d + 4 8 9 On comparing both sides, we get a + 2= 4 a = 2 c 7 = 7 c =14 b +1= 8 b = 7 d + 4 =9 d =5 Hence, a = 2, b = 7, c =14 and d = 5 (xv) (d) Let r be the common ratio of GP. Then, a7 = 64 ar 7 1 = 64 729r 6 = 64 r 6 = 6 5n 2 + 1205n 90300 = 0 n 2 + 241n 18060 = 0 [dividing by 5] n 2 + 301n 60n 18060 = 0 n(n + 301) 60(n + 301) = 0 (n 60)(n + 301) = 0 n = 60, 301 but n cannot be negative n = 60 Hence, the account was held for 60 months i.e. 5 yr. Common Mistake n cannot be the nagative. As n is the notation used for months . Hence, take positive value only. (ii) Given, the model is drawn to the scale 1 : 500, and the dimensions of the model are : common radius = 3 cm, height of conical part = 4 cm, height of cylindrical part =14 cm 4 cm =10 cm, Therefore, the actual dimensions of the rocket are: common radius, r = 3 500 cm =15 m, height of conical part, h1 = 4 500 cm = 20 m, height of cylindrical part, h2 = 10 500 = 50 m. Now, slant height of rocket, l = r 2 + h12 = 152 + 202 = 625 = 25 m E 64 729 2 2 r 6 = or 3 3 l 6 r= 2 2 or 3 3 Reason If a, ar , ar 2 ......... ar n 1 are in GP, then the an + 1 common ratio is given by r = . an Hence, both are true. 2. (i) Given, money deposited per month, P = ` 400 and interest rate percent per annum, r =10 Let the account be held for n months, then by using the formula n(n + 1) r I =P , we get 2 12 100 n(n + 1) 10 5n(n + 1) I = 400 =` 2 12 100 3 Total money deposited by Beena = (400 n) = ` 400n Total amount at maturity = Total money deposit by Reena + interest 5n(n + 1) = 400n + 3 1200n + 5n(n + 1) 5n 2 + 1205n = =` 3 3 According to given condition, 5n 2 + 1205n = 30100 3 F D C A B (a) The total surface area of the rocket = curved surface area of cone + curved surface area of cylinder + area of base of cylinder = rl + 2 rh2 + r 2 = 15 25 + 2 15 50 + 152 = (375 + 1500 + 225) = 2100 m 2 (b) The total volume of the rocket = volume of conical part + volume of cylindrical part 1 2 = r h1 + r 2h2 3 1 = 152 20 + 152 50 3 20 = 225 + 50 m 3 3 170 = 225 3 =12750 m 3 1 2 1 + tan 2 sec2 (iii) LHS = = = cos 2 2 1 1 + cot cosec sin 2 y1 + y 2 2 0+ b 4= [Q y = 4 , y1 = 0 and y 2 = b] 2 8= b Hence, a = 4 and b = 8 (b) Let the coordinates of the third vertex be (x , y). Then, coordinates of centroid x + x 2 + x 3 y1 + y 2 + y 3 = 1 , 3 3 and 2 = 1 sin 2 sin 2 = = tan = RHS 2 cos 1 cos sin 1 1 tan cos Also, = cos 1 cot 1 sin 2 2 cos sin cos = sin cos sin 2 sin cos sin = cos sin cos 2 sin 2 2 = = ( tan ) = tan cos = RHS 2 1 + tan 1 tan 2 = = tan 1 + cot 2 1 cot 3. (i) Given, Hence proved. x 3 + 3 x 341 = 3 x 2 + 1 91 On applying componendo and dividendo rule, we get (x 3 + 3 x) + (3 x 2 + 1) 341 + 91 = (x 3 + 3 x) (3 x 2 + 1) 341 91 (x + 1)3 432 = (x 1)3 250 (x + 1)3 216 = (x 1)3 125 (x + 1)3 (6)3 = (x 1)3 (5)3 P ( 2, 4) 3 7 + x 5 + 4 + y (2, 1) = , 3 3 x 4 y 1 (2, 1) = , 3 3 Y 6 5 (0,4)B A (6,4) 4 3 2 On taking cube root both sides, we get x +1 6 = x 1 5 5 (x + 1) = 6 (x 1) 5x + 5 = 6x 6 6x 5x = 5 + 6 x =11 (ii) (a) Given, point P( 2, 4) is the mid-point of AB. A (a, 0) On equating the coordinates, we get x 4 y 1 = 2 and = 1 3 3 x 4 = 6 and y 1 = 3 x =10 and y = 2 Hence, the coordinates of third vertex are (10, 2). (iii) Let us take 10 small division =1 unit of both the axes. We can plot the points A(6, 4) and B(0, 4) as follow. (a) Reflect A and B in origin to get the image A and B in this graph. 2 2 y= B (0, b) By mid-point formula, x + x2 x= 1 2 a+ 0 2 = [Q x = 2, x1 = a and x 2 = 0] 2 2 2= a + 0 4 = a a = 4 1 X 6 5 4 3 2 1 1 O 2 3 4 5 6 X 1 2 3 A ( 6, 4) 4 B (0, 4) 5 6 Y (b) The coordinates of A and B are ( 6, 4) and (0, 4). (c) We find that the geometrical name ABA B is parallelogram. (d) Perimeter of a parallelogram ABA B = 2(AB + AB ) = 2(6 + 10) = 32 units (iii) Taking X-axis as class interval and Y-axis as a frequency, draw a histogram from the given data. Y 10 [Q in right angled ABB , AB = AB 2 + BB 2 2 9 2 = 6 + 8 = 100 = 10] = 6000 4500 = `1500 (ii) Given, 13 x 2 + 20 x 92 = 0 On comparing with ax 2 + bx + c = 0, we get a =13, b = 20 and c = 92 Q x= b b 2 4 ac 2a x= 20 (20)2 4 13 ( 92) 2 13 = 20 400 + 4784 26 20 5184 26 20 72 = 26 Taking + ve sign, 20 + 72 x= 26 52 = =2 26 Taking ve sign, 20 72 x= 26 92 = = 3.54 26 x = 2, 3.54 = A 8 4. (i) Investment = ` 9000, market value of one share C Scale : On X-axis 1 cm = 20 units On Y-axis 1 cm = 1 unit E 7 Frequency = `150 The number of shares purchased by Mukul 9000 = = ` 60 150 Annual income = Number of shares Rate of dividend Face value of one share 6 = 60 100 = ` 360 100 Hence, Mukul s annual income = ` 360 50 50% of his shares = 60 = 30 100 Selling price of 30 shares = (200 30) = ` 6000 and cost price of these shares = (150 30) = ` 4500 Mukul s gain in this transaction = SP CP B D 6 5 4 3 2 1 F O 20 40 60 80 100 120 X Class interval In the highest rectangle, draw two straight lines AC and BD which intersect at point E. Through point E, draw a vertical line to meet the X-axis at point F. The abscissa of the point F represents 47 which is required mode. 5. (i) (a) Given, BAC = 40 and AC is a diameter of a circle. So, ABC = 90 [angle in a semi-circle is right angle] We know that the sum of all angles of a triangle is 180 . CAB + ABC + BCA = 180 40 + 90 + BCA = 180 BCA = 180 (90 + 40 ) = 180 130 = 50 (b) Also, BC|| AE EAB + ABC = 180 EAC + 40 + 90 = 180 EAC = 180 130 = 50 E D A C 40 B Here, AEDC is a cyclic quadrilateral. EAO + EDC = 180 50 + EDC = 180 EDC = 130 (c) Since, AC is a diameter of a circle. Therefore, AEC = 90 [angle in a semi-circle is right angle] ECB = 90 [Q sum of all angle of a quadrilateral is 360 ] It implies that line joining BE is the diameter of the circle. Since, AE|| BC and AB is a transversal. EAB + ABC = 180 EAB + 90 = 180 EAB = 90 It is clear that ABCE is a square. So, EB is the angle bisector of AEC. 90 BEC = = 45 2 2 6 3 2 4 0 (ii) Given, A = , B= and C = 2 6 4 0 6 2 3 2 6 4 2B = 2 = 4 0 8 0 Now, we have A + 2 X = 2B + C 2 X = 2B + C A 1 X = (2B + C A) 2 6 4 4 0 2 Now, 2B + C A = + 8 0 6 2 2 6 + 4 4 + 0 2 = 8 + 6 0 + 2 2 2 4 2 6 = 14 2 2 6 2 2 4 + 6 4 = = 14 2 2 6 12 6 6 6 6 10 4 (iii) Let f (x) = 2 x 3 + ax 2 + 7 x b (i) Given, (x + 2) is a factor of f (x), then f ( 2) = 0 2 2 ( 2) + (a) ( 2) + (7) ( 2) b = 0 16 + 4 a 14 b = 0 4 a b 30 = 0 (ii) Also, x + 3 is a factor of f (x) , f ( 3) = 0 2 ( 3)3 + a ( 3)2 + 7 ( 3) b = 0 54 + 9a 21 b = 0 (iii) 9a b 75 = 0 On subtracting Eq. (ii) from Eq. (iii), we get 5a 45 = 0 a = 9 On putting the value of a = 9 in Eq. (ii), we get 36 b 30 = 0 b = 6 Hence, a = 9 and b = 6 m= y 2 y1 x 2 x1 m= 5 3 8 = 6 2 4 m= 2 1 1 = m 2 Let P be the mid-point of line AB. Then, coordinates 2 + 6 3 5 of P , = (4 , 1) 2 2 Slope of the line perpendicular to AB = Y X P A (2, 3) B (6, 5) X Y Thus, the required line passes through P(4 , 1) and 1 having slope . So, its equation is 2 y y1 = m (x x1) 1 y + 1 = (x 4) 2 y + 2 = x 4 2 x 2y 6 = 0 (ii) Given, cosec sin = 5 (cosec sin )2 = ( 5)2 [squaring on both sides] cosec2 + sin 2 2 cosec sin = 5 [Q (a b)2 = a 2 + b 2 2ab] 1 X = (2B + C A) 2 1 4 10 = 2 12 4 2 5 = 6 2 3 6. (i) The slope of line AB, where A(2, 3) and B(6, 5). cosec2 + sin 2 + 2 cosec sin = 5 + 4 cosec sin [adding 4 cosec sin on both sides] (cosec +sin )2 = 5 + 4 1 [Q cosec A sin A =1] (cosec + sin )2 = 9 [taking square root] cosec + sin = 3 But cosec > 0 and sin > 0, so cosec + sin > 0. Hence proved. cosec + sin = 3 (iii) Given, a =121 and d = 117 121 = 4 Let nth term be the first negative term, then an < 0 a + (n 1)d < 0 121 + (n 1)( 4) < 0 [Q an = a + (n 1)d] 121 121 < 4(n 1) < n 1 4 121 125 +1< n <n 4 4 125 1 n> n > 31 4 4 The least positive integral value of n which satisfies 125 is 32. n> 4 Hence, 32nd is the first negative term of the given AP. 7. (i) Box contains 15 balls of which x are blue, so the number of red balls in the box = 15 x 15 x P(red ball) = 15 When 5 red balls are increased, then number of red balls in the box = (15 x) + 5 = 20 x Total number of balls in the box = 15 + 5 = 20 Probability of drawing a red ball in this case 20 x = 20 According to the question, 20 x 15 x = 2 20 15 20 x 2 = (15 x) 4 3 20 x + 15 (x + 1) + 12 (x + 2) 25 60 [multiplying both sides by 60] 20 x + 15 x + 12 x + 15 + 24 1500 47 x 1500 39 47 x 1461 1461 4 x x 31 47 47 The largest value of x = 31 Then, x = 31 , x + 1 = 31 + 1 = 32 and x + 2 = 31 + 2 = 33 Hence, required numbers are 31, 32 and 33. (ii) To calculate the mean and cumulative frequency, the table is given below. Marks ( xi ) Frequency (fi ) fi xi 3 5 15 5 5 7 35 5+7 =12 6 8 48 12+8=20 7 4 28 20+4=24 8 10 80 24+10 = 34 60 3 x = 120 8 x 5 x = 60 x =12 Thus, originally the box contains 12 blue balls and 3 red balls. 3 1 (a) P(red ball) = = 15 5 12 4 (b) P(blue ball) = = 15 5 (c) When 5 red balls are added, then the box will contain 12 blue balls and 3 + 5 i.e. 8 red balls. P(blue ball, if 5 red balls are actually added) 12 2 = = 20 5 (ii) (a) Given, in PQR, QS and RT are medians. So, TS|| QR [Q T , S are the mid-points of PQ and PR, respectively] 1 [by mid-point theorem] (i) TS = QR 2 In TGS and RGQ, GTS = GRQ [alternate angles, as TS|| QR] [alternate angles] GST = GQR and SGT = QGR [vertically opposite angles] TGS ~ RGQ [by AA axiom of similarity] (b) Since, TGS ~ RGQ So, sides of the triangles are proportional. GS TS = GQ QR GS 1 [from Eq. (i)] = GQ 2 2GS = GQ GQ = 2GS Cumulative frequency (cf) 9 6 54 34+6 = 40 10 9 90 40+9 = 49 Total fi = 49 fi xi = 350 We know that Mean (X) = f i x i 350 = = 7.14 fi 49 Here, N = 49, which is odd. N + 1 Median = th observation 2 49 + 1 = th observation 2 50 th observation = 2 = 25th observation = 8 [Q 25 lies in the cumulative frequency 34, whose corresponding marks is 8] Mode = Marks having highest frequency Mode = 8 [Q in the table, highest frequency is 10, whose corresponding marks is 8] (iii) From figure, OR = OS [radii of the circle] S O Q Hence proved. x P 8. (i) Let x , x + 1 and x + 2 be consecutive natural numbers. According to the question, x x +1 x + 2 + + 25 3 4 5 But y R T OSR = ORS [opposite angles of equal sides are equal] OSR = QRP = y [angles in alternate segments are equal] ORS = y [Q ORS = OSR] POR = OSR + ORS [sum of opposite interior angles of a triangle] POR = y + y = 2 y Since, PR is tangent to the circle and OR is radius, therefore, OR PR. ORP = 90 OPR + POR + ORP = 180 [Q sum of angles of a triangle is 180 ] x +2 y +90 = 180 Hence proved. x +2 y = 90 The required ogive is shown on the graph paper as given below. 300 x Reduced list price of the book = ` (x 5) 300 Number of books bought for ` 300 = x 5 Number of books bought for ` 300 = 300 300 According to the question, =5 x 5 x 300 x 300 x + 1500 =5 x (x 5) 1500 =5 x2 5x x 2 5 x 300 = 0 x 2 20 x + 15 x 300 = 0 (1200, 444) F 450 K (1300, 470) G Q (1100, 404) E 400 350 T S (1000, 314) D Number of employees 9. (i) Let ` x be the original list price of the book. Y 500 300 250 I P (900, 194) C 200 150 L 100 50 O (800, 108) B R (700, 40) A 600 J H (600, 0) N M U 700 800 900 1000 1100 1200 1300 X Monthly income x (x 20) + 15 (x 20) = 0 (x 20)(x + 15) = 0 x = 20, 15 x = 20 [Q x = 15 is not possible] Hence, the list price of the book is ` 20. (ii) The cumulative frequency table for continuous distribution is given below Monthly income in ` (class-intervals) Number of employees (frequency) Cumulative frequency 600-700 40 40 700-800 68 108 ( 40 + 68) 800-900 86 194(108 + 86) 900-1000 120 314(194 + 120) 1000-1100 90 404( 314 + 90) 1100-1200 40 444( 404 + 40) 1200-1300 26 470( 444 + 26) Since, the scale on X-axis starts at 600, a kink is shown near the origin on X-axis to show that the graph is drawn to scale beginning at 600. Now, plot the points A(700, 40), B(800,108), C(900,194), D(1000, 314), E(1100, 404), F(1200, 444), G(1300, 470) and H(600, 0). Join these points by a free hand curve. Here, n (number of employees) = 470 (a) Let A be a point on Y-axis representing n 470 frequency = = = 235 2 2 Through I, draw a horizontal line to meet the ogive at P. Through P , draw a vertical line to meet the X-axis at M. The abscissa of the point M represents `930. Hence, the required median is `930. (b) Let the point J on X-axis represent ` 1180. Through J , draw a vertical line to meet the ogive at Q. Through Q, draw a horizontal line to meet the Y-axis at K. The ordinate of the point K represents 436 employees on Y-axis. The number of employees whose income exceeds `1180 = total number of employees number of employees whose income `1180 = 470 436 = 34 The percentage of employees whose income 34 exceeds `1180 = 100 % = 7.23% 470 (iii) (a) For lower quartile (Q1 ) Let L be the point on Y-axis representing n 470 frequency = = = 117.5 4 4 Through L, draw a horizontal line to meet the ogive at R. Through R, draw a vertical line to meet the X-axis at N. The abscissa of the point N represents `815. The lower quartile, Q1 = ` 815 For upper quartile (Q3) Let T be the point on Y-axis representing 3n 3 470 = = 352.5. Through T , draw 4 4 a horizontal line to meet the ogive at S. Through S, draw a vertical line to meet the X-axis at U. The abscissa of the point U. represents `1035. The upper quartile, Q3 = `1035 (b) Inter quartile range = upper quartile lower quartile = `1035 ` 815 = ` 220 B frequency = 10. (i) Since, the numbers are in the ratio 7 : 11, let the required numbers be 7x and 11x. According to the question, (7 x + 7):(11 x + 7) = 2 : 3 7x + 7 2 = 11 x + 7 3 22 x + 14 = 21 x + 21 x=7 7 x = 7 7 = 49 and 11 x = 11 7 = 77 The required numbers are 49 and 77. (ii) Steps of construction (a) Draw a circle of radius 2.5 cm and take a point P outside it. (b) Through P draw a secant PAB which intersects the circle at A and B. (c) Produce AP to a point C such that PA = PC. (d) Draw a semi-circle with CB as diameter. (e) Draw PD CB, intersecting the semi-circle at D. (12 x)m (1 x)m 2 C x 60 A O (g) Join PT and PT . Then, PT and PT are the required tangents. (iii) Let AB be the height of tree. Suppose the tree is broken by the wind at point C and the part CB assumes the position CO and meets the ground at point O. AB =12 m Let AC = x m. Then, CO = CB = (12 x) m Now, in right angled OAC, AC perpendicular sin 60 = Q sin = hypotenuse OC 3 x = 2 12 x 3 Q sin 60 = 2 12 3 3 x = 2 x 2 x + 3 x = 12 3 x(2 + 3) = 12 3 x= D 12 3 2 3 2+ 3 2 3 [multiplying numerator and denominator by (2 3)] T 2.5 cm A C B P T (f) With P as centre and PD as radius draw arcs to intersect the circle of radius 2.5 cm at T and T . x= 12 3 (2 3) 4 3 [Q (a + b)(a b) = a 2 b 2] = 24 3 36 = 24 1.732 36 = 41.569 36 = 5.569 m

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