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1 Series HRS Code-30/3 Summative Assessment II Solution Mathematics class 10 CBSE Board 2014 SECTION A All questions are similar in 30/1 set SECTION B All questions are similar in 30/2 set except Q.No Find the value of p so that p x (x-3)+9 = 0 has equal roots. Solution: p x (x-3) + 9 = 0 px 2-3px+9=0 a = p, b=-3p, c=9 b 2-4ac= (-3p) 2-4(p)(9) = 9p 2-36p for having equal roots b 2-4ac = 9p 2-36p = 0 9p 2-36p = 0 9p(p-4) = 0 9p=0 (or) p-4=0 p=0(rejected) or p=4 Therefore, the value of p=4 SECTION C All questions are similar in 30/2 set except Q. No. 20,22,23, Construct a triangle with side 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are 3/5 times the corresponding sides of given triangle. 22. The sum of first seven term of AP is 182. If 4 th and 17 th term are in ratio 1: 5. Find AP Solution: From the given information, a+3d/a+16d=1/5 By solving this equation, 5a+15d=a+16d Page 1
2 4a=d... eq 1 Now as we are given that S 7 = 182, 7/2(2a+6d)=182 By further solving this equation, 7(a+3d) =182 a+3d= By substitution of values from eq1 we get, a + 3 (4a)=26 13a = 26; a=2 By putting in - 1, 4 (2) =d d=8 Therefore, the AP formed will be 2, 10, From the top of 60m high building, the angle of depression of the top and bottom of a tower are 45 0 and 60 0 respectively. Find the height of tower. Let CD be the building with CD = 60m and AB = h m be the height of Tower. CAE = 30 and CBD = 45 As AB = h m CE = (60 h) m So in right angled ACE, Tan 30 0 = CE/AE 1/ 3 = (60-h)/AE Page 2
3 AE = 3(60 h ) In BCD, tan 45 0 = CD/BD 1 = CD/BD CD = BD BD = CD... (2) as BD = AE, so from (1) and (2) CD = 3(60 h ) Given CD = 60m 60 = 3(60 h ) 60/ 3 = (60 h ) 20 3 = (60 h ) 60-20x1.73 = h = h 25.4m = h Hence, the height of tower is 25.4 m. 24. Find a point p on y axis which is equidistant from points A(4,8)and B(-6,6) Solution: A point p on y axis then abscissa is 0 then P (0, y) AP 2 = PB 2 (0 4) 2 + (y 8) 2 = ( 6 0) 2 + (6 y) y 2-16y + 64 = y 2 12y 16 + y 2-16y + 64 = y 2 12y 80-16y = 72 12y = 16y 12y 8 = 4y y = 2 Page 3
4 P (0, 2) SECTION D All questions are similar in 30/2 set except Q.No.31, 32 and (x 4) / (x 5 )/ (x 6) / (x 7 ) = 10/3 32. Five cards the ten, jack,queen king and ace of diamonds are are well suffled with their faces downward. one card are drawn randomly (a)what is the probability that the drawn card is the queen (b)if the queen is drawn and put aside, and a second card id drawn. find the probability that the second card is (i) an ace (ii) a queen Solution: (i) Total number of cards = 5 Total number of queens = 1 P (getting a queen) = 1/5 (ii) When the queen is drawn and put aside, the total number of remaining cards will be 4. (a) Total number of aces = 1 P (getting an ace) = 1/4 (b) As queen is already drawn, therefore, the number of queens will be 0. P (getting a queen) = 0/4=0 34. In fig., A triangle ABC is drawn to circumscribe a circle of radius 4 cm, such that the segment BD and DC are of length 8 cm and 6 cm respectively. Find the side AB and AC Solution: Page 4
5 Firstly, consider that the given circle will touch the sides AB and AC of the triangle at point E and F respectively. Let AF = x Now, in ABC, CF = CD = 6cm (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point C) BE = BD = 8cm (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point B) AE = AF = x (Again, tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point A) Now, AB = AE + EB = x + 8 Also, BC = BD + DC = = 14 and CA = CF + FA = 6 + x Now, we get all the sides of a triangle so its area can be find out by using Heron's formula as: 2s = AB + BC + CA = x x = x Semi-perimeter = s = (28 + 2x)/2 = 14 + x Page 5
6 However, x = 14 is not possible as the length of the sides will be negative. Therefore, x = 7 Hence, AB = x + 8 = = 15 cm AC = 6 + x = = 13 cm Page 6
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