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Class 10 CBSE Pre Board 2015 : Mathematics 1 (Kendriya Vidyalaya (KV) No. 2, Raipur)

10 pages, 31 questions, 20 questions with responses, 29 total responses,

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KENDRIYA VIDYALAYA SANGATHAN; RAIPUR REGION SUMMATIVE ASSESSMENT -II (2014 - 15) CLASS: X MATHEMATICS TIME: 3.00 Hrs. M.M. 90 General instructions: 1. All questions are compulsory. 2. The question paper consists of 31 questions divided into four Sections A, B.C and D. Section A comprises of 4 questions of 1 mark each. Section B comprises of 6 questions of 2 marks each. Section C comprises of 10 questions of 3 marks each and Section D comprises of 11 questions of 4 marks each. 3. There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks each and 2 questions of four marks. You have to attempt only one of the alternatives in all such questions. 4. Use of calculator is not permitted. SECTION A (ONE MARK QUESTIONS) Q1 Find the sum of all even numbers less than 100. Q2 The length of the tangent from a point A to a circle of radius 9 cm is 12 cm. Find the distance of A from the centre of the circle. Q3 The tops of two poles of heights 16m and 10m are connected by a wire. If the wire makes an angle of 300 with the horizontal, then what is the length of the wire? A P B 300 D C Q4 In a circle of diameter 42cm, if an arc subtends an angle of 600 at the centre, find the length of the arc. SECTION B (TWO MARKS QUESTIONS) Q5 Find the nature of roots of the quadratic equation 3x2+2x+5 = 0. Q6 If the numbers n-5, 4n-4 and 5n-1 are in A.P. find the value of n. Q7 Find the value of k for which the distance between points P(1,2) and Q(-2,k) is 5 units. Q8 The length of the minute hand of a clock is 14 cm. Find the area swept by it in 20 minutes. Q9 If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80 , then find POA . Q10 The vertices of a triangle are (-2,0),(2,3) and (1,-3). Is the triangle equilateral, isisceles or scalene? SECTION C (THREE MARKS QUESTIONS) Q 11 Find two consecutive odd positive integers, sum of whose squares is 290. Q 12 If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term. Q 13 In Fig, XY and X Y are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X Y at B. Q 14 Prove that AOB = 90 Two concentric circles are of radii 10cm and 6 cm respectively. Find the length of the chord of the bigger circle that touches the smaller circle. Q15 From the top of a tower 50 m high the angles of depression of the top and bottom of a pole are observed to be 45 and 60 respectively. Find the height of the pole. Q16 Find the value of k for which the points (8,1), (k.-4), (2,-5) are collinear. Q17 A farmer has a circular field of radius 120m. It has a well in the middle of the field of radius 8 m. How much area has he for growing his crops? Q18 Q19 Find the area of the shaded region where ABCD is a square of side 14 cm A 20m deep well with diameter 7m is dug and the earth from digging is evenly spread out to form a platform 22m 14m. Find the height of the platform so formed. Q20 A solid is in the form of a cone standing on a hemisphere with both their radii equal to 4 cm and height of the cone is equal to its radius. Find the volume of the solid in terms of . SECTION D (FOUR MARKS QUESTIONS) Q 21 Find the roots of the equation 2 2 5 + 3 = 0 by the method of completing the square Q 22 A group of enthusiastic persons decided to build a shelter for senior citizens on a rectangular plot of area 300 sq. metres with its length 1m more than twice its breadth. Find the dimensions of the plot. What value is indicated from the question? Q 23 Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000? Q 24 Prove that the lengths of the tangents to a circle drawn from an external point are equal. Q 25 Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ABC = 60 . Then construct a triangle whose sides are Q 26 3 4 of the corresponding sides of the triangle ABC. A man is standing on the deck of a ship, which is 10 m above water level. He observes the angle of elevation of the top of a hill as 60 and the angle of depression of the base of the hill as 30 . Calculate the distance of the hill from the ship and the height of the hill. Q 27 Find the probability that a number selected randomly from the numbers 1,2,3,4,5, 34,35 is a (i) a prime number (ii) a multiple of 7 (iii) a multiple of 2 or 5 (iv) negative integer. Q 28 One card is drwn at random from a well shuffled pack of 52cards. Find the probability of (i) getting a red card (ii) getting a face card (iii) getting an ace (iv) a ten. Q 29 Determine the ratio in which the point P(a, - 2) divides the join of A (- 4, 3) and B (2, - 4). Also find the value of a. Q 30 The cost of fencing a circular field at the rate of Rs 24 per metre is Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m2. Find the cost of ploughing the field (Take = 22/7). Q 31 A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find th cost of the milk which can completely fill the container, at the rate of Rs 20 per litre (Take =3.14). MARKING SCHEME SECTION A (ONE MARK QUESTIONS) Q 1. 2+4+6+ .+98 Here no. of terms = 98/2 = 49 (1/2 mark) 49 So, S = [a+l] = [2 +98] = 2450 2 2 (1/2 mark) B Q2 O A In ABO , OA = 2 + 2 = 122 + 92 Q3 = sin300 6 Q4 1 = 1 (1/2 mark) 2 (1/2 mark) 22 42 600 7 3600 (1/2 mark) (1/2 mark) 2 AD = 12m 2 Length of an arc = = = = 15 cm (1/2 mark) = 22cm (1/2 mark) SECTION B (TWO MARKS QUESTIONS) (1/2 mark) (1 mark) Hence the roots of the given quadratic equation are not real. (1/2 mark) 4n-4-(n-5) = 5n-1-(4n-4) (1 mark) n=1 Q6 3x2+2x+5 = 0, a=3, b=2,c=5 Discriminant = b2- 4ac = 22 4*3*5 = -56 < 0 Q5 (1 mark) Q 7 PQ 2 = 25, (-2-1)2+(k-2)2 = 25 (1 mark) k = 6, -2 Q8 The angle described by a minute hand in 20 minutes, = (1 mark) 20 = 1200 (1 mark) The area = Q9 = 22 7 14 14 1200 3600 = 616 3 2 cm2 = 205.33cm Using angle sum property in triangle PAO (1 mark) POA = 500 (1 mark) Q 10. Let A (-2,0), B(2,3) and C(1,-3) then using distance formula AB = 5 BC = 37 (1 mark) CA = 3 2 (1/2 mark) (1/2+1/2 mark) AB So, Triangle ABC is scalene (1/2 mark) SECTION C (THREE MARKS QUESTIONS) Q 11 Let the smaller of the two consecutive odd positive integers be x. Then, the second integer will be x + 2. ATQ x2 + (x + 2)2 = 290 (1 mark) on solving, we get x = 11 or x = 13 (1 1 marks) 2 But x is given to be an odd positive integer. Therefore, x 13, x = 11. Thus, the two consecutive odd integers are 11 and 13. Q 14 (2 marks) (1 mark) Given , To Prove, Construction (1 mark) Correct Proof Q 13 Writing the correct formula of Sn and calculating common difference d = 10 a20 = 10 + (20 1) 10 = 200, Q 12 (1/2 mark) (2 marks) Drawing correct figure (1/2mark) Using Pythagoras Theorem In rt OCA to get AC = 8cm Writing AB = 2 AC = 16 cm 15 (1mark) (1/2 mark) Correct figure and description (1mark) Using rt ARP to get x = 50 h . Eq (1) Using rt PQB to get x = Using Eq (1) & Eq (2) to get h = = 50 3 1 3 50 3 .Eq (2) ( ) (1/2 mark) (1/2 mark) (1mark) Q16 Since the points are collinear, the area of the triangle formed by the points (8,1), (k.-4), (2, -5) will be equal to 0 (1mark) i.e. [8(-4+5) + k(-5-1) + 2(1+4) =0 so k= 3 Q 17 (2 marks) Area of growing crops = area of field area of the top of the well = (120)2 (8)2 Correct Calculation Q 18 (1mark) = 45056 m2 (2 marks) Area of the shaded portion : Area of square Area of 4 circles 7 Radius of the circle = 2 cm, Area of Square= 14 x 14 = 196 cm2 7 7 (1/2 mark) (1 1 marks) 2 Area of 4 circles = 4 x x 2 x 2 = 154 cm2 Shaded Area = 196 154 = 42cm2 19 (1/2mark) (1/2 mark) Volume of mud dug out = volume of platform Cylinder(well) r= 7/2m, h = 20m, Cuboid: (platform) L= 22m, B = 14 m 2 = Lx Bx H (1/2 mark) Solving and writing the answer : H = 2.5 m. Q 20 (1 mark) (1 Figure 1 marks) 2 (1/2mark) Volume of solid = Volume of conical part + Volume of H.S. part = 1 2 2 r h r 3 3 3 Putting Values & getting V = 64 cm3 (1 mark) (1 1 marks) 2 SECTION D (FOUR MARKS QUESTIONS) Q 21 (2 marks) x = 3/2 , x = 1 Q 22 Dividing by 2 and Solving by the method of completing the squares (1 mark) Let the breadth be x metres. So length = (2x+1) m (1/2mark) x.(2x+1) = 300 , x = 12 or x= is not possible, breadth= 12 m, length = 25m (2 marks) (1/2mark) coorect moral value Savings of Yasmeen are ( in Rs) 32, 36, 40, (1/2mark) Which are clearly in A.P with a = 32, d = 4 Q 23 (1 mark) (1 mark) Sn = n 2a n 1 d , we get 2 2000 = n 2 32 n 1 4 2 Q24 (1mark) n = 25 (1 1 marks) 2 (2 marks) Correct Proof (2 marks) Drawing ABC (1 Constructing the similar triangle Q 25 Given, To Prove, Figure, Construction (2 Q 26 Correct figure & description Using rt PRA to get Using rt ARQ to get x = 10 3 Getting height of hill as h = 40 m Q 27 1 marks) 2 (1mark) 3 x = h -10 Distance of ship from hill as x = 10 3 m or 17.32 1 marks) 2 (1mark) (1/2mark) (1/2mark) (1mark) Total no. of out comes = {1,2,3,4, 35} = 35 , Prime numbers ={2,3,5,7,11,13,17,19,23,29,31} = 11 (i)P (E1)= 11/35 (1 mark) (ii)Multiples of 7 = { 7,14,21,28,35}, P (E2)= 1/7 (1 mark) (iii)Multiples of 2 or 5, P (E3) = 21/35 (1 mark) (iv)P(E4) = 0 (1 mark) Q 28 Total no. of all possible outcomes n = 52 (i)P(red card) = 26 1 , 52 2 (ii)P(face card) = 12 3 , 52 13 (iii)P(ace) = 1 , 13 1 (iv)P(a ten)= 13 (1+1+1+1mark) Let P(a,-2) divides AB in the radio : 1 (1/2mark) Using section formula correctly, we get = 5/2 (2 marks) Putting the value we get Q 29 (1mark) a = 2/7 Required ratio = 5 : 2 Q 30 (1/2mark) Length of the fence (in metres) = 220 (1/2mark) So, circumference of the field = 220 m 1 marks) 2 Therefore, if r metres is the radius of the field, then 2 r = 220, r = 35 m Therefore, area of the field = 22 5 35 m2 (1mark) So, total cost of ploughing the field = Rs 22 5 35 0.50 = Rs 1925 Q 31 (1 (1mark) Volume of frustum = 3 2 2 ( 1 + 2 + 1 2) h (1 mark) Volume = 10449.92 cm3 (2 marks) Convert in to litres = 10449.92/1000 litres = =10.45 litres (1 mark) Calculating the cost =10.45 X 20 = Rs 209 (1 mark)

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