CBSE 10th Board mathematics exam paper 2013
Posted on: 04:58 PM IST Mar 06, 2013
This is the mathematics examination paper which CBSE 10th Board examinees had to solve in 2013. Students are advised to try to solve the questions themselves before consulting the solutions. The solutions have been answered by the http://www.topperlearning.com/ faculty.
Click here for the solutions.####Paper:X Math – CBSE Past Year Paper – 2013 – Set 3Total marks of the paper:90Total time of the paper:3 hrs1. All questions are compulsory.2. The question paper consists of 34 questions divided into four sections A, B, C and D.3. Section A contains 8 questions of 1 mark each, which are multiple choice type question, Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 10 questions of 4 marks each.4. Use of calculators is not permitted.Solutions:1] Let AB be the tower of height 75 m and C be the position of the car.InThus, the distance of the car from the base of the tower is m2] S = {1,2,3,4,5,6}Let event E be defined as ’getting an even number’.n(E) = {1,4,6}3] S = {1, 2, 3,..90}n(S) = 90The prime number less than 23 are 2, 3, 5, 7, 11, 13, 17, and 19.Let event E be defined as ’getting a prime number less than 23’.n(E) = 84] Given: AB, BC, CD and AD are tangents to the circle with centre O at Q, P, S and R respectively.AB = 29 cm, AD = 23, DS = 5 cm and <span style="position:relative;top:2.0pt; msotextraise:2.0pt">B = 90oConstruction: Join PQ.We know that, the lengths of the tangents drawn from an external point to a circle are equal.DS = DR = 5 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> AR = AD – DR = 23 cm – 5 cm = 18 cmAQ = AR = 18 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> QB = AB – AQ = 29 cm – 18 cm = 11 cmQB = BP = 11 cmIn PQB,PQ2 = QB2 + BP2 = (11 cm)2 + (11 cm)2 = 2 (11 cm)2PQ = 11 cm … (1)In OPQ,PQ2 = OQ2 + OP2 = r2 + r2 = 2r2(11 )2 = 2r2<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt">121 = r2<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in">r = 11<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in">Thus, the radius of the circle is 11 cm.5] AP PB (Given)CA AP, CB BP (Since radius is perpendicular to tangent)AC = CB = radius of the circleTherefore, APBC is a square having side equal to 4 cm.Therefore, length of each tangent is 4 cm.6] From the figure, the coordinates of A, B, and C are (1, 3), (1, 0) and (4, 0) respectively. Area of ABC7] Let r be the radius of the circle.From the given information, we have:2<span style="position:relative; top:2.0pt;msotextraise:2.0pt">r – r = 37 cm<span style="fontfamily:"Verdana","sansserif";position:relative;top:3.0pt; msotextraise:3.0pt"> Circumference of the circle = 8] Common difference =<span style="fontfamily:"Verdana","sansserif"; position:relative;top:11.0pt;msotextraise:11.0pt">9] Given: ABCD be a parallelogram circumscribing a circle with centre O.To prove: ABCD is a rhombus.<span style="fontfamily:"Verdana","sansserif";position:relative;top:11.0pt; msotextraise:11.0pt">We know that the tangents drawn to a circle from an exterior point are equal in length.Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.Adding the above equations,AP + BP + CR + DR = AS + BQ + CQ + DS(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)AB + CD = AD + BC2AB = 2BC (Since, ABCD is a parallelogram so AB = DC and AD = BC)AB = BCTherefore, AB = BC = DC = AD.Hence, ABCD is a rhombus.10] Dimension of the rectangular card board = 14 cm <span style="position:relative; top:2.0pt;msotextraise:2.0pt"> 7 cmSince, two circular pieces of equal radii and maximum area touching each other are cut from the rectangular card board, therefore, the diameter of each of each circular piece is = 7 cm.<span style="fontfamily:"Verdana","sansserif";position:relative;top:11.0pt; msotextraise:11.0pt">Radius of each circular piece = cm.<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> Sum of area of two circular pieces = Area of the remaining card board = Area of the card board – Area of two circular pieces= 14 cm 7 cm – 77 cm2= 98 cm2 – 77 cm2 = 21 cm211] Given: AB = 12 cm, BC = 8 cm and AC = 10 cm.Let, AD = AF = x cm, BD = BE = y cm and CE = CF = z cm(Tangents drawn from an external point to the circle are equal in length)<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> 2(x + y + z) = AB + BC + AC = AD + DB + BE + EC + AF + FC = 30 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> x + y + z = 15 cmAB = AD + DB = x + y = 12 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:2.0pt;msotextraise:2.0pt">z = CF = 15 – 12 = 3 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:11.0pt;msotextraise:11.0pt">AC = AF + FC = x + z = 10 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:2.0pt;msotextraise:2.0pt">y = BE = 15 – 10 = 5 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:2.0pt;msotextraise:2.0pt">x = AD = x + y + z – z – y = 15 – 3 – 5 = 7 cm<span style="position:relative; top:11.0pt;msotextraise:11.0pt">12] Three digit numbers divisible by 7 are105, 112, 119,
994This is an AP with first term (a) = 105 and common difference (d) = 7Let an be the last term.an = a + (n – 1)d994 = 105 + (n – 1)(7)7(n – 1) = 889n – 1 = 127n = 128Thus, there are 128 threedigit natural numbers that are divisible by 7.13] <span style="fontfamily:"Verdana","sansserif"; position:relative;top:11.0pt;msotextraise:11.0pt">14] Let E be the event that the drawn card is neither a king nor a queen.Total number of possible outcomes = 52Total number of kings and queens = 4 + 4 = 8Therefore, there are 52 – 8 = 44 cards that are neither king nor queen.Total number of favourable outcomes = 44<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> Required probability = P(E) = 15] <span style="fontfamily:"Verdana","sansserif";position:relative;top:11.0pt; msotextraise:11.0pt">Let the radius and height of cylinder be r cm and h cm respectively.Diameter of the hemispherical bowl = 14 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> Radius of the hemispherical bowl = Radius of the cylinder = r Total height of the vessel = 13 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> Height of the cylinder, h = 13 cm – 7 cm = 6 cmTotal surface area of the vessel = 2 (curved surface area of the cylinder + curved surface area of the hemisphere)(Since, the vessel is hollow)= 1144 cm216] <span style="fontfamily:"Verdana","sansserif";position:relative;top:11.0pt; msotextraise:11.0pt">Height of the cylinder, h = 10 cmRadius of the cylinder = Radius of each hemisphere = r = 3.5 cmVolume of wood in the toy = Volume of the cylinder – 2 <span style="position:relative; top:2.0pt;msotextraise:2.0pt"> Volume of each hemisphere<span style="fontfamily:"Verdana","sansserif"; position:relative;top:62.0pt;msotextraise:62.0pt">17] Radius = 21 cmThe arc subtends an angle of 60o at the centre.(i) l = <span style="position:relative;top:3.0pt;msotextraise: 3.0pt"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:19.0pt;msotextraise:19.0pt"><span style="msotabcount: 1"> <span style="fontfamily:"Verdana","sansserif"; position:relative;top:19.0pt;msotextraise:19.0pt">(ii) <span style="position:relative;top:3.0pt;msotextraise: 3.0pt"><span style="msotabcount: 3"> = 231 cm218] AB and CD are the diameters of a circle with centre O.<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> OA = OB = OC = OD = 7 cm (Radius of the circle)Area of the shaded region= Area of the circle with diameter OB + (Area of the semicircle ACDA – Area of ACD)<span style="fontfamily:"Verdana","sansserif"; position:relative;top:20.0pt;msotextraise:20.0pt">19] Let the yaxis divide the line segment joining the points (4,6) and (10,12) in the ratio k: 1 and the point of the intersection be (0,y).Using section formula, we have:<span style="fontfamily:"Verdana","sansserif"; position:relative;top:80.0pt;msotextraise:80.0pt">Thus, the coordinates of the point of division are <span style="position:relative;top:11.0pt; msotextraise:11.0pt">20] <span style="position:relative;top:11.0pt;msotextraise: 11.0pt">Let AB and CD be the two poles, where CD (the second pole) = 24 m.BD = 15 mLet the height of pole AB be h m.AL = BD = 15 m and AB = LD = hSo, CL = CD – LD = 24 – hIn ACL,<span style="fontfamily:"Verdana","sansserif"; position:relative;top:72.0pt;msotextraise:72.0pt">Thus, height of the first pole is 15.34 m.<span style="position:relative; top:11.0pt;msotextraise:11.0pt">21] (k + 4)x2 + (k + 1)x + 1 = 0a = k + 4, b = k + 1, c = 1For equal roots, dicriminant, D = 0<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> b2 – 4ac = 0 <span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> (k + 1)2 – 4(k + 4) 1 = 0<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> k2 + 2k + 1 – 4k – 16 = 0<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> k2 – 2k – 15 = 0<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> k2 – 5k + 3k – 15 = 0<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> k(k – 5) + 3(k – 5) = 0<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> (k – 5) (k + 3) = 0<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> k = 5 or k = 3Thus, for k = 5 or k = 3, the given quadratic equation has equal roots.22] Sn = 3n2 + 4nFirst term (a1) = S1 = 3(1)2 + 4(1) = 7S2 = a1 + a2 = 3(2)2 + 4(2) = 20a2 = 20 – a1 = 20 – 7 = 13So, common difference (d) = a2 – a1 = 13 – 7 = 6Now, an = a + (n – 1)d<span style="fontfamily:"Verdana","sansserif"; position:relative;top:2.0pt;msotextraise:2.0pt">a25 = 7 + (25 – 1) 6 = 7 + 24 6 = 7 + 144 = 15123] Steps of construction:1. Draw two concentric circle with centre O and radii 4 cm and 6 cm. Take a point P on the outer circle and then join OP.2. Draw the perpendicular bisector of OP. Let the bisector intersects OP at M.3. With M as the centre and OM as the radius, draw a circle. Let it intersect the inner circle at A and B.4. Join PA and PB. Therefore, are the required tangents.24] The given points are A(2,3) B(8,3) and C(6,7).Using distance formula, we have:AB2 = <span style="position:relative;top:9.0pt;msotextraise: 9.0pt"><p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> AB2 = 102 + 0<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> AB2 = 100<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in">BC2 = <p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> BC2 = (2)2 + 42<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> BC2 = 4 + 16<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> BC2 = 20<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> CA = (8)2 + (4)2<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> CA2 = 64 + 16<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> CA2 = 80<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in">It can be observed that:<p class="MsoNormal" style="margintop:10.0pt;marginright:0in;marginbottom: 10.0pt;marginleft:.5in;textindent:.5in">BC2 + CA2 = 20 + 80 = 100 = AB2So, by the converse of Pythagoras Theorem, <span style="fontfamily:"Verdana","sansserif"; position:relative;top:2.0pt;msotextraise:2.0pt">ABC is a right triangle right angled at C.<span style="position:relative; top:11.0pt;msotextraise:11.0pt">25] Diameter of circular end of pipe = 2 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> of circular end of pipe = Area of crosssection = Speed of water = 0.4 m/s = 0.4 <span style="position:relative;top:2.0pt;msotextraise: 2.0pt"> 60 = 24 metre/minVolume of water that flows in 1 minute from pipe = <span style="position:relative; top:3.0pt;msotextraise:3.0pt">Volume of water that flows in 30 minutes from pipe = 30 <span style="position:relative; top:3.0pt;msotextraise:3.0pt">Radius (r2) of base of cylindrical tank = 40 cm = 0.4 mLet the cylindrical tank be filled up to h m in 30 minutes.Volume of water filled in tank in 30 minutes is equal to the volume of water flowed out in 30 minutes from the pipe.<span style="fontfamily:"Verdana","sansserif"; position:relative;top:66.0pt;msotextraise:66.0pt">Therefore, the rise in level of water in the tank in half an hour is 45 cm.26] The group consists of 12 persons.<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> Total number of possible outcomes = 12Let A denote event of selecting persons who are extremely patient<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> Number of outcomes favourable to A is 3.Let B denote event of selecting persons who are extremely kind or honest.Number of persons who are extremely honest is 6.Number of persons who are extremely kind is 12 – (6 + 3) = 3<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> Number of outcomes favourable to B = 6 + 3 = 9.(i)(ii)<span style="fontfamily:"Verdana","sansserif"; position:relative;top:11.0pt;msotextraise:11.0pt">Each of the three values, patience, honesty and kindness is important in one’s life.27] Diameter of upper end of bucket = 30 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> Radius (r1) of upper end of bucket = 15 cmDiameter of lower end of bucket = 10 cm<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> Radius (r2) of lower end of bucket = 5 cmSlant height (l) of frustum =Area of metal sheet used to make the bucket<span style="fontfamily:"Verdana","sansserif"; position:relative;top:7.0pt;msotextraise:7.0pt"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt">Cost of 100 cm2 metal sheet = Rs 10Cost of 545 cm2 metal sheet = Rs.Therefore, cost of metal sheet used to make the bucket is Rs 171.13.28] Given: l and m are two parallel tangents to the circle with centre O touching the circle at A and B respectively. DE is a tangent at the point C, which intersects <i style="msobidifontstyle: normal">l at D and m at E.To prove: Construction: Join OC.Proof: <span style="fontfamily:"Verdana","sansserif";position:relative;top:11.0pt; msotextraise:11.0pt"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:5.0pt;msotextraise:5.0pt">OA = OC (Radii of the same circle)AD = DC (Length of tangents drawn from an external point to a circle are equal)DO = OD (Common side)<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> (SSS congruence criterion)<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> … (1)<span style="fontfamily:"Verdana","sansserif"; position:relative;top:5.0pt;msotextraise:5.0pt"><span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> … (2)Now, AOB is a diameter of the circle. Hence, it is a straight line.<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt">From (1) and (2), we have:2COD + 2<span style="position:relative;top:2.0pt; msotextraise:2.0pt">COE = 180oHence, proved.29] Let the sides of the two squares be x cm and y cm where x > y.Then, their areas are x2 and y2 and their perimeters are 4x and 4y.By the given condition:x2 + y2 = 400
(1)and 4x – 4y = 16 4(x – y) = 16 x – y = 4 x = y + 4 … (2)Substituting the value of x from (2) in (1), we get:(y + 4)2 + y2 = 400 y2 + 16 + 8y + y2 = 400 2y2 + 16 + 8y = 400 y2 + 4y – 192 = 0 y2 + 16y – 12y – 192 = 0 y(y + 16) – 12 (y + 16) = 0 (y + 16) (y – 12) = 0 y = 16 or y = 12Since, y cannot be negative, y = 12.So, x = y + 4 = 12 + 4 = 16Thus, the sides of the two squares are 16 cm and 12 cm.30] <span style="fontfamily:"Verdana","sansserif"; position:relative;top:11.0pt;msotextraise:11.0pt">31] Given: A circle with centre O and a tangent XY to the circle at a point PTo Prove: OP is perpendicular to XY.Construction: Take a point Q on XY other than P and join OQ. Proof: Here the point Q must lie outside the circle as if it lies inside the tangent XY will become secant to the circle.Therefore, OQ is longer than the radius OP of the circle, That is, OQ > OP.This happens for every point on the line XY except the point P.So OP is the shortest of all the distances of the point O to the points on XY.And hence OP is perpendicular to XY.Hence, proved.32] Given AP is 12, 9, 6, …, 21First term, a = 12Common difference, d = 3Let 21 be the nth term of the A.P.21 = a + (n – 1)d <span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> 21 = 12 + (n – 1) 3<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> 33 = (n – 1) <span style="position:relative; top:2.0pt;msotextraise:2.0pt"> 3<span style="fontfamily:"Verdana","sansserif"; position:relative;top:3.0pt;msotextraise:3.0pt"> n = 12Sum of the terms of the AP = S12=If 1 is added to each term of the AP, the sum of all the terms of the new AP will increase by n, i.e., 12.<span style="fontfamily:"Verdana","sansserif"; position:relative;top:2.0pt;msotextraise:2.0pt">Sum of all the terms of the new AP = 54 + 12 = 66<span style="position:relative; top:11.0pt;msotextraise:11.0pt">33] Let AC and BD be the two poles of the same height h m.Given AB = 80 mLet AP = x m, therefore, PB = (80 – x) mIn APC,In BPD,Thus, the height of both the poles is and the distances of the point from the poles are 60 m and 20 m.34] The given vertices are A(x,y), B(1,2) and C(2,1).It is know that the area of a triangle whose vertices are (x1, y1), (x2,y2) and (x3, y3) is given by <span style="fontfamily:"Verdana","sansserif"; position:relative;top:2.0pt;msotextraise:2.0pt">Area of ABCThe area of ABC is given as 6 sq units.<span style="fontfamily:"Verdana","sansserif"; position:relative;top:11.0pt;msotextraise:11.0pt">
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