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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

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