Mathematics CLASS : X

Transcription

1 SAMPLE PAPER 1 (SA II) MRS.NALNI SARAF KV BANTALAB Mathematics CLASS : X Time: 3hrs Max. Marks: 90 General Instruction:- 1. All questions are Compulsory. 1. The question paper consists of 34 questions divided into 4 sections, A,B,C and D. Section A comprises of 8 questions of 1 mark each. Section-B comprises of 6 questions of 2 marks each and Section- D comprises of 10 questions of 4 marks each. 2. Question numbers 1 to 8 in Section A multiple choice questions where you are to select one correct option out of the given four. 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 each. You have to attempt only one of the alternatives in all such questions. 4. Use of calculator is not permitted. SECTION A Question numbers 1 to 8 carry 1 mark each. For each of the question numbers 1 to 8, four alternative choices have been provided, of which only one is correct. Select the correct choice. Q1. If the numbers n - 2, 4n - 1 and 5n + 2 are in A.P., then the value of n is : (A) 1 (B) 2 (C) 3 (D) 0 Q2.To divide a line segment PQ in the ratio 3 : 4, first a ray PX is drawn so that QPX an acute angle and then at equal distances points are marked on the ray PX such that the minimum number of these points is : (A) 7 (B) 4 (C) 5 (0) 3 P Q3. In the given figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50 with PQ, then POQ is equal to: (A) 100 o (B) 80 (C) 90 (D) 75 Q4.The tops of two poles with heights 25 m and 35 m are connected by a wire, which makes an angle of elevation of 30 o at the top of 25 m pole. Then the length of the wire is: (A)26m (B)35m (C)15m (D)20m O 40 o 50 o 100 o 40 o Q R

2 Q5. One coin is tossed thrice. The probability of getting neither 3 heads nor 3 tails, is: (A) 1/3 (B) ¾ (C) ½ (D) 2/3 Q6.Value of p so that the point (3, p) lies on the line 2x 3y = 5 is : (A) 12 (B) 3 (C) 1/3 (D) ½ Q.7 For a race of 1540 m, number of rounds one have to take on a circular track of radius 35m: (A) 5 (B) 6 7 (C) 7 (D) 10 Q8. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is: (A) 4 cm (B) 3 cm (C) 2 cm (D) 6 cm SECTION-B Question numbers 9 to 14 carry 2 marks each. Q9. Solve the following quadratic equation by factorisation: = 0; (x 0) x 2 x Or Find the roots of the following quadratic equation : (x +3)(x- 1) = 3 x Q10. Find the value of m for which the point with coordinates (3, 5), (m, 6) and ( 1, 15) are collinear. 2 2 Q11. A horse is tethered to one corner of rectangular grass field 40m by 24 m, by a rope 14m long. Over how much area of the field can it graze? Q12. In two concentric circles, prove that a chord of a larger circle which is tangent to smaller circle is bisected at the point of contact. Q13. Find the common difference of an A.P. whose first term is ½ and the 8 th term is 17. Also write its 4 th term. 6

3 Q14. Prove that the lengths of the tangents drawn from an external point to a circle are equal. SECTION-C Q15. Onkar gets pocket money, from his father every week and saves 15 in and on each successive week he increases his saving by 5. (1) Find the amount saved by Omkar in One month (2) Find the amount saved in one year. (3) Which quality of Onkar is referred in the given question? Q16. Find the root of the equation 4-3 = 5, x 0, -3 X (2x +3) 2 Or Find two consecutive positive even integers, the sum of whose squares, the sum of whose squares is 340. Q17. Draw a circle of radius 3 cm. From a point 5 cm away from its center. Construct the pair of tangents of the circle and measure their length. Q18. A card is drawn at random from a well shuffled deck of 52 cards. Find the probability of getting: (i) a king (ii) a king of red suit Or Two dice are thrown at the same time. Determine the probability that the Difference of the number on the two dice is 2. Q19. On seeing a child in first floor of a burning house, a man climbed from the ground along the rope stretched from the top of a vertical tower and tied at the ground. The height of the tower is 24 m and the angle made by

4 1052 the rope to the ground is 30 o. Calculate the distance covered by the man to reach the top of the pole. What do you consider the act done by the man to save the child? Q20. Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape Or An ice-cream cone having radius 5 cm and height 10 cm as shown in figure. Calculate the volume of ice-cream, provided that its 1 part is left unfiled with ice-cream. 6 Q21. A square OABC is inscribed in the quadrant OPBQ. If OA = 10 cm, then find the area of the shaded region. Q22. In the given fig., ABCD is a trapezium with AB DC, AB=18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A,B,C and D have been drawn, then find the area of the shaded region of the figure. Q23. A spherical glass vessel has a cylindrical neck 8cm long, 2cm in diameter, the diameter of the spherical part is 8.5cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm 3. Check whether she is correct, taking the above as the inside measurements, and = D A 7cm 18cm 18cm 32cm B 7cm C Q24. A storage oil tanker consists of a cylindrical portion 7 m in diameter with two hemispherical ends of the same diameter. The oil tanker lying horizontally. If the total length of the tanker is 20m, then find the capacity of the container.

5 SECTION-D Question numbers 25 to 34 carry 4 marks each. Q25. The lower portion of a haystack is an inverted cone frustum and upper part is a cone. Find the total volume of the haystack, If AB = 3 m and CD = 2 m. A bucket of height 8 cm is made up of copper sheet is in the form of frustum of cone with radii of its lower and upper ends as 3 cm and 9 cm respectively. Calculate: (i) The height of the cone of which the bucket is a part. (ii) The volume of the water which can be filled in the bucket. (iii) The area of the copper sheet required to make the bucket. Or Q26. Find four terms in an A.P., whose sum is 20 and the sum of whose squares is 120. Q27. A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards him. If it takes 12 minutes for the angle of depression to change from 30 o to 45, how soon after this, will the car reach the tower? Give your answer to the nearest second. C 30 o D 45 o h ( 3-1) h 3 cm 45 o 30 o h A h B Q28. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

6 Q29. In the figure, OP is equal to the diameter of the circle. Prove that ABP is an equilateral triangle. P Or In the figure, PO QO. The tangents to the circle with centre O at P and Q intersect at a point T. Prove That PQ and OT are right bisectors of each other. Q30. Swati can row her boat at a speed of 5 km/h in still water. If it takes her 1 hour more to row the boat 5.25 km upstream, then to return downstream, find the speed of the stream. Q31. A child's game has 8 triangles of which 5 are blue and rest are red and,10 squares of which 6 are blue and rest are red. One piece is low at random. Find the probability that it is a: (i) triangle (ii) square (iii) square of blue colour (iv) triangle of red colour. Q32. Find the area of the triangle ABC With A(1, -4) and the midpoints of sides through A being (2, -1) and (0, -1) Q33. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices. Q34. Amar and Gugu together have 45 marbles, Both of them lost 5 marbles each and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. While playing in the ground you found a purse containing money and some documents what will you do?

7 SOLUTION SAMPLE PAPER 1 SECTION A Question numbers 1 to 8 carry 1 mark each. For each of the question numbers 1 to 8, four alternative choices have been provided, of which only one is correct. Select the correct choice. Sol.1 (A) 1 [Since Here, n - 2, 4n - 1 and 5n + 2 are in A.P. => 4n (n - 2) = 5n (4n - 1) => 4n n + 2 = 5n + 2-4n + 1 p q => 3n-n = 3-1 => 2n = 2 => n = 1] x Sol.2 (A) 7 Sol.3 (A) 100 [ Here, OPR 90 o and RPQ = 50 OPQ = 90 o 50 o = 40 o OPQ = OQP 40 o Now, POQ = 180 o - 40 o - 40 o = 100 o ] Sol.4 (D)20m [ Here, CE = = 10m and Now, CAE = 30 o AC = cosec 30 o CE AC = CE 2 = 10 2 = 20m Sol.5 (B) ¾ [Here, sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Now, neither 3 heads nor 3 tails = { HHT, HTH, THH, HTT, THT, TTH} Required probability = 6 = 3 ] 8 4 Sol.6 (C) 1 3 [Since (3, p) lies on the line 2x 3y = 5 2(3) 3p = 5

8 3p = 5-6 p = -1 = 1 ] -3 3 Sol.7 (C) 7 [Distance covered in one round = = 220 m 7 Now, number of rounds = 1540 = 7 ] 220 Sol.8 (C) 2 cm [Here, Volume of cylinder = = 16 cm 3 Now, 12 4 r3 = 16 r3 = 16 3 = r = 1 cm Thus, radius of the solid sphere is 1cm and diameter of the sphere is 2cm] SECTION-B Question numbers 9 to 14 carry 2 marks each. Sol = 0 x 2 x = = 0 x 2 x x x x x = = 0 or 2-1 = 0 x x x x Or (x+3)(x-1) = 3 x- 1 3 x 2 +2x-3 = 3x-1 x 2 x 2 = 0 x 2 2x + x 2 = 0 x(x 2) +1(x 2) =0 (x 2 )(x+1) = 0 x 2=0 or x+1=0 x = 2 or x = 1

9 Sol.10 Let the given points be A(3, 5), B(m, 6) and C( 1, 15) are collinear. 2 2 Since given points are collinear. Area of triangle ABC formed by these points is zero m (5-6) = m 5m+ 5 3 = m -10m = m = 0 5m = m = 2 Sol.11 Here, radius (r) = 14 m and = 90 Area of the field in which horse can graze = r 2 = 90 o o 360 o 7 = = 154 m Sol.12 AB is a tangent to the inner circle at point C. OC is the radius drawn at the point of contact. OC AB OCA = OCB = 90 Now, in rt. ed AOC and BOC OA = OB = r OC = OC [common] OCA = OCB [proved above] AOC BOC [by RHS congruency rule] AC = BC [ c.p.c.t] Hence, chord of a larger circle which is tangent to smaller circle is bisected at the point of contact.

10 Sol.13 Here, first term (a) = 1 2 and eighth term a 8 = 17 6 a + 7d = d = d = 17-1 = 14 d = Now, a4 = a + 3d = = 1 +1 = Sol.14 Join PO. TO and TO In PTO and PTO TO = T O = r OP = PO (common) PTO = PTO = 90 o PTO PT O (By RHS congruencey axiom) PT = PT (C.P.C.T) SECTION-C Sol.15 Here, saving after one week is 15 and it increases every week by 5 So, given savings is an A.P. with first term 15 and common difference 5 (i) In one month, we have four weeks a4=a+3d = x 5 = = 30 Now, S4 = 4/3 (15+30) = 30 (ii) In one year, we have 52 weeks S 52 = -52[2a + (n-1)d] = 26[ x 5] 2 [Since Sn = n/2[a + 1 or a n ]

11 = 26 x 285 = 7410 (iii) Savings for future makes individual self dependent. Sol = 5. X (2x +3) 4-3x = 5. (4-3x)(2x + 3) =5x X (2x +3) -6x 2 x + 12 =5x Or x 2 + x -2 = 0 x(x + 2) 1(x+2 =0 x-1 = 0 or x +2 = 0 x = 1 or x = - 2 or 6x 2 + 6x- 12 =0 x 2 +2x-x-2 = 0 (X-1) (x+2) = 0 Let the two consecutive positive even integers be x and x + 2 according to statement of the question, we have X 2 + (X + 2) 2 = 340 X 2 + X 2 + 4x = 0 2x 2 + 4x -336 = 0 X x - 12x 168 =0 (x-12) (X+ 14) =0 X 12 = 0 or x + 14 = 0 X=12orX=-14 (rejecting, given number are positive) Hence, the required integers are 12 and 14.

12 Sol.17 Given: A circle of radius 3 cm and a points P is 5 cm away from its centre. Required: A pair of tangents. Steps of construction: 1. Draw a circle C (0, r) with center O and Radius 3 cm. 2. Take a point P, such that OP = 5 cm, 3. Draw AB, the perpendicular bisector of OP and left it intersects OP in M. 4. With M as centre and PM or MO as radius, draw another circle intersecting the given circle in T and T. 5. Join PT and PT. Thus, PT and PT are the required tangents from point P to the circle C(0, r). Sol.18 (i) Number of kings = 4 Probability (a king) = 4 = (ii) Number of kings of red suit = 2 Probability (a king) = _2_ = Or Total number of elementary events when two dice are thrown = 6 x 6 = 36 Number of favourable outcomes (difference of the numbers on the two dice is 2) ={(1,3), (2,4), (3,1), (3,5), (4,2), (4,6), (5,3), (6,4)} =8 Required probability = 8 = Sol.19 Let AB be the tower of height 24 m and ACB =30 o Distance covered by the man along the rope is CA. Now, in rt. ed CBA CA = cosec 30 o AB CA = AB cosec 30 o = 24 x 2 = 48 m

13 Thus, distance covered by the man is 48 m. Sol.20 Radius of two cones (r) =8 cm Height of two cones (h) =15 cm Slant height (l) = r 2 + h 2 = = 289 = 17cm Now, surface area of the shape so obtained = 2( rl) = = cm 2 Or Radius of the conical and spherical portion (r) = 5 cm Height of the conical portion (h) = 10-5 = 5 cm Now, volume of the ice-cream cone = Vol. of conical portion + Vol. of hemispherical portion = 1 r 2 h + 2 r 3 = 1 r 2 (h+2r) = (5+10) = cm Volume of the ice-cream = = = cm Sol.21 Here, side (OA) of the square = 10 cm Diagonal of the square = 10 2 cm Now, area of the shaded region = Area of quadrant OPBQ Area of square = 90 o o 7 = = = cm 2

14 Sol.22 Here, radii of the arcs with centres A, B, C and D = 7 cm. AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm Now, area of the shaded region = Area of trapezium ABCD - Area of circle with radius 7 cm = 1(18+32) = = 196 cm 2 Sol.23 Radius of cylindrical neck = 2 = 1 cm Height of cylindrical neck = 8 cm Radius of spherical part = 4.25 cm Total volume of water = Volume of spherical part + Volume of cylindrical neck = 4 r 3 + r 2 h = (4.25) (1) = = cm3 = 347 cm3 (approx.) Sol.24 Radius of hemispherical portion = Radius of cylindrical portion = 7 m 2 Total length of the tanker = 20 m Length of cylindrical portion = 20-7 = 13 m Now, Capacity (volume) of the oil tanker = Volume of cylinder + 2 Volume of hemisphere = r 2 h r 3 = = = m 3

15 SECTION-D Question numbers 25 to 34 carry 4 marks each. Sol.25 Here, radius of cone (r) Height of cone (h) = 3 m = 7 m Volume of cone = 1 r2h = = 66 m Height of the frustum = = 3.5 m Now, Volume of the lower portion (frustum of cone) = ( ) 3 7 = 11(9+4+6) = = m 3 Thus, Volume of the haystack = = m 3 Here, r = 3 cm, R = 9 cm and height of the bucket (h) = 8 cm Now, slant height (l) = h 2 + (R-r) 2 = = = 100 = 10 cm (i) Since ABC AB C AB = BC AB B C h = 3 h+8 9 3h = h +8 2h = 8 h = 4 Thus, height of the cone is i.e., 12 cm. (ii) Volume of the bucket = 1 h(r 2 +r 2 +Rr) 3 = ( ) 3 7 = = cm (iii). Surface area of copper sheet= r 2 + (r + R)l = (3+9) = = 2838 = cm

16 Sol.26 Let the four terms in an A.P., be a-3d, a-d, a+d, a+3d According to the statement of the question, we have a-3d+ a-d + a + d + a + 3d = 20 4a = 20 a=5 Also, (a-3d) 2 +(a-d) 2 +(a+d) 2 +(ä+3d) 2 = 120 (5-3d) 2 +(5-d) 2 +(5+d) 2 +(5+3d) 2 = d 2-30d+25+d 2-10d+25+d 2 +10d+25+9d 2 +30d = d 2 = d 2 = 20 d = 1 Hence, the four terms are 5-3, 5-1, 5+1, 5+3 i.ë., 2,4,6,8 Or 5+3, 5+1, 5-1, 5-3 i.e., 8, 6, 4, 2 Sol.27 Let AB be the tower of height h m. ACB= 30 and. ADB = 45 o Now, in rt. ed ADB, we have AB = tan 45 o AB = 1 DB DB AB = DB = h m In rt. ed CBA, we have AB = tan 30 o CB AB = CB tan 30 o h = (CD + DB) 1 3 3h = CD + h CD = ( 3 1)h Speed of the card = ( 3-1)h m/min 12 Time taken by the car to reach the tower after point D = h = 12 = 12( 3+1) = min ( 3-1)h = 16 minutes 24 secs.

17 Sol.28 Given: Let PA and PB be two tangents drawn from an external point P to the circle with centre O. To prove: APB + AOB = 180 o Proof: Because tangent is perpendicular to the Radius of the circle at point of contact. OA AP and OB BP OAP = OBP = 90 o Now, OAP + APB + OBP + AOB = 360 o 90 o + APB + 90 o + AOB = 360 o APB + AOB = 360 o -180 o = 180 o Sol.29 Let r be the radius of the circle. OA = r and OP = 2r [given] [tangent is perpendicular to the radius through the point of contact] In right OPA, we have Sin ( OPA+ = OA = r = 1 OPA = 30 o OP 2r 2 Similarly, OPB = 30 o APB = 30 o + 30 o = 60 o Since PA = PB [ lengths of tangents from an external point are equal] PAB = PBA In APB, we have APB + PAB + PBA = 180 o 60 o + 2 PAB = 180 o PAB = 60 o PAB = 60 o Since all angle are 60 o ABP is equilateral Or Since Tangent to any circle is perpendicular to circle at point of contact. TPO = TQO = 90 o

18 In TPO and TQO, we have OP = OQ OT = OT [radii of a circle] [common side] By RHS congruency, we have TPO = TQO 1 = 2 Again, in PTR and QTR 1 = 2 PT = QT TR = TR [c.p.c.t.] [proved above] [tangents from external point] [common] By SAS congruency, we have PTR = QTR PRT = QRT [c.p.c.t.] But PRT + QRT = 180 o [linear pair] Thus, PRT = QRT.= 90 Also,. PR = QR [c.p.c.t.] PQ and OT are right bisectors of each other. Sol.30 Let Speed of be x km/h. Speed of the boat upstream = (5-x) km/h Speed of the boat downstream = )5+x) km/h Time taken to go upstream = x Time taken to go upstream = x As per question, we have = 1 5-x 5+x x 5+x = 1 (5-x)(5+x)

19 10.5x = 25-x 2 x x -25 = 0 or 2x x -50 = 0 2x x-4x-50 = 0 2(2x+25) -2(2x+25) = 0 (x-2)(2x+25) = 0 x-2 = 0 or 2x +25 = 0 x =2 or x= (speed cannot be negative) x = 2 Hence, speed of the stream is 2 km/h. Sol.31 Here, number of triangles are 8 and number of squares are 10. Total number of outcomes = = 18 (i) Probability (lost card is a triangle) = 8 = (ii) Probability (lost card is a square) = 10 = (iii) Probability (lost card is a square of blue colour) = 6 = (iv)probability (lost card is a triangle of red colour) = 3 = Sol.32 Let the coordinates of B and C be B(x, y) and C(p, q) Now, D(2, -1) is the mid-point of AB 1+x = 2 and -4+Y = x = 3 and y = 2 Thus coordinates of B are B (3, 2) Also, E(0, - 1) is the mid-point of AC 1+P = 0 and -4+q = p=-1 and q = 2 Thus, the coordinates of C are C (-1, 2)

20 Ar(ΔABC) = 1 1 (2-2) + 3(2 + 4) 1(-4 2) 2 = = 1 24 = 12 sq. units 2 2 Sol.33 Let A (-1, 2) and C(3, 2) be the two opposite vertices of a square ABCD and let coordinates of B be B(x, y). Now, AB = BC AB 2 = BC 2 (x + 1) 2 + (y-2) 2 =(3-x) 2 + (2-y) 2 X x+y2+4-4y= 9 +x 2-6x+4+y2-4y 2x + 6x= 9 1 8x = 8 x =1 Also, AB 2 BC 2 = AC 2 (x +1) 2 + (y-2) 2 + (3 x) 2 + (2-y) 2 = (3+1) 2 + (2-2) 2 x x +y y=9+x 2-6x+4+y 2-4y=16 2x 2 +2y 2 4x 8y =16-18 x 2 +y 2-2x-4y=-1 1+y 2-2-4y+1=0 [Put x = 1] y 2-4y=0 y(y-4) = 0 y=0 or y =4 Thus, the point (x,y) is (1, 0 ) or (1,4) Hence, the coordinates of b and d are b (1,0) and d (1,4). Sol.34 Let number of marbles Amar had be x Number of marbles with Gugu = 45- x Both of them lost 5 marbles Number of marbles left with Amar and Gugu be x 5 and 45 x -5 i.e., 40 x respectively According to the statement of the question, we have (x-5) (40-x) =124 -x x =0 x 2-36x-9x +324 =0 (x-9) (x-36)= 0 x-9 = 0 or x-36 = 0 x=9 or x = 36 Thus, either Amar had 9 marbles and Gugu had 36 marbles or Amar had 36 marbles and Gugu had 9 months. Handover the purse containing money and some documents to the concerned teacher.

21 SAMPLE PAPER 2 (SA II) Mathematics CLASS : X MR SANDESH KUMAR BHAT KV 1 JAMMU Time: 3hrs Max. Marks: 90 General Instruction:- 1. All questions are Compulsory. 1. The question paper consists of 34 questions divided into 4 sections, A,B,C and D. Section A comprises of 8 questions of 1 mark each. Section-B comprises of 6 questions of 2 marks each and Section- D comprises of 10 questions of 4 marks each. 2. Question numbers 1 to 8 in Section A multiple choice questions where you are to select one correct option out of the given four. 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 each. You have to attempt only one of the alternatives in all such questions. 4. Use of calculator is not permitted. SECTION A Question number 1 8 are multiple choice type questions carrying 1 mark each. For each question four alternative choices have been given, of which only one is correct. You have to select the correct choice. Q1.If x = -2 is the root of quadratic equation x 2-3x-a=0 then the value of a is (a) -10 (b) 3 (c) 10 (d) -3 Q2. The missing term of the A.P:-3,, 33, 48 is (a) 12 (b) 15 (c) 18 (d) 21 Q3. If angle between two radii of a circle is 110 the angle between the tangents at the end of the radii is (a) 70 (b) 50 (c) 10 (d) 40

22 Q4.From the given figure find the angle of elevation A C 100 C 100 B A (a) 45 (b) 30 (c) 60 (d) 90 Q5. If P (E) = 0.35 then the probability of not E is (a) 0 (b) 0.45 (c) 0.65 (d) 0.53 Q6. The mid- point of a line segment joining the point A (-2, 8) and B (-6, -4) is (a) (-4, -6) (b) (2, 6) (c) (4, 2) (d) (-4, 2) Q7. If the perimeter and the area of a circle are numerically equal then the radius of the circle is (a) 2 units (b) π units (c) 4 units (d) 7 units Q8. Volume of two spheres is in ratio 64:27 the ratio of their radii is (a) 3:4 (b) 4:3 (c) 9:16 (d) 16:9 SECTION B Question numbers 9 14 carry 2 marks each. Q9. Find the value of K for which the equation 2x 2 +Kx+3=0 has two equal roots. Q10. Find the roots of the quadratic equation 6x 2 -x-2=0 by factorization method. Q11. Find the 10 th term of the A.P:- 2, 7, 12 Q12. Two concentric circles of radii 5cm and 3cm. Find the length of the chord of the larger circle which touches the smaller circle. Q13. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4) OR

23 Find the points on the x- axis which is equidistant from (2, -5) and (-2, 9) Q14. In what ratio does the point (-4, 6) divide the line segment joining the points A (-6, 10) and B (3, -8). SECTION C Question numbers carry 3 marks each. Q15. The sum of the 4 th and 8 th term of an A.P is 24 and the sum of the 6 th term and 10 th term is 44. Find the first three terms of the A.P. Q16. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Q17. Draw a pair of tangents to the circle of radius 5cm which are inclined to each other at an angle of 60. Q18. The shadow of a tower standing on a level ground is found to be 40m longer when sun s altitude is 30 than when it is 60. Find the height of the tower. Q19. A child has a block in the shape of a cube with one word written on each face as. learn to serve go come to (i) (ii) The block (cube) is thrown. What is the probability of getting word to? Use all words of the block (cube) to make a meaningful value base sentence Q20. Find the value of k for which the points (7, -2), (5, 1) and (3, k) are collinear. Q21. Show that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) taken in order are the vertices of a square. Q22. A toy is in the form of cone of radius 3.5cm mounted on a hemisphere of same radius. The total height of the toy is 15.5cm. Find the total surface area of the toy. Q23. (i) Find the area of the shaded region given in figure. (ii) Write the four values given in the figure. Which value you consider important in student s life and Why?

24 14cm Hard work Punct uality yyyy Obedi ent Toler ance 14cm 14cm Q24.In a fig. a square OABC is inscribed in a quadrant OPBQ. If OA=20cm. Find the area of the shaded region. SECTION D Question numbers 25 to 34 carry 4 marks each Q25.A plane left 30 minutes later than the schedule time and in order to reach its destination 1500km away in time. It has to increase its speed by 250km/h from its usual speed. Find its usual speed. Q26. Sum of the areas of two squares is 468m 2. If the difference of their perimeter is 24m. Find the sides of the two Squares. Q27. A sum of Rs 700 is to be used for giving seven cash prizes to students of a school for their overall academics performance. If each prize is Rs 20 less than its preceding prize. Find the value of each prizes. Q28. Prove that the length of tangents drawn from an external point to a circle is equal. Q29. Construct a triangle similar to a given triangle ABC with its sides equal to ¾ of the corresponding sides of the triangle ABC (i.e., of scale factor

25 .Q30. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30 which is approaching the foot of the tower with a uniform speed. Six seconds later the angle of depression of a car is found to be 60. Find the time taken by the car to reach the foo of the tower from this point. Q31. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting. (i) (ii) (iii) (iv) A king of red colour. A face card. A spade. A black face card. Q32. A car has two wipers which do not overlap. Each wiper has blade of length 25cm sweeping through an angle of 115. Find the total area cleaned at each sweep of the blades. Q33. Water in a canal 6m wide 15m deep is flowing with a speed of 10km/h. how much area will it irrigate in 30 minutes? If 8cm of standing water is needed. Q34. An open metal bucket is in theshape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet (see Fig ). The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold. (Take = 22/7)

26 SOLUTION SAMPLE PAPER 2 Section A 1. X 2-3x-a=0 Put X= -2 (-2) 2 -(3)(-2)-a=0 4+6-a=0 10-a=0 a=10 2. Let missing term be x -3,x,33,48, a n are in A.P There for a 2 -a 1 = a 3 - a 2 So, x-(-3)=33-x => x+3=33-x => 2x= 33-3 Q 3. => x= In quad.. In Quad. In POQT. P+ Q + T+ O= T + 0 o = 360 o T= 70 o 4.In C B A 5.P ( E) = 0.35,

27 As we know that P( not E) + P ( E) = 1 P( not E) = 1- P ( E) = So, = 0.65 Q6. Let P(x,y) be a mid point Then by mid point formula. X= Y= Q7. Let r is the radius of circle Area of circle = Perimeter of circle Πr 2 = 2Π r r =2 units Q8. Let r be the radius of small sphere. And volume is v Therefore v= 4/3πr 3 Let R be the radius of largest sphere and Volume is V V= 4/3πR 3 Now= = Q.9 As we know that For equal and real roots b 2-4ac=0 k 2-24=0 k 2 =24 => k= ± 24 or ±2 6 Q.10 As 6x 2 +3x-4x-2=0 => 3x (2x+1)-2(2x+1)=0 => (3x-2)(2x+1)=0 => X=3/2 or x=-1/2 Q11. It is given that a=2, n=10 and d=a 2 -a 1 7-2=5 And A n =a+(n-1)d A n =a+9d 2+9(5) => 2+45=47 Section B

28 Q.12 As AB is a tangent of the inner circle From fig. BD 2 +OD 2 = OB 2 BD 2 +9 = 25 BD 2 = 25-9 BD 2 = 16 = 4 cm Q.13 Let C (x,y) be a point. And it is given that A (3, 6) &B (-3, 4) AC 2 =CB 2 [(x-3) 2 +(y-6) 2 ]= [(x+3) 2 +(y-4) 2 ] 12x+4y-20=0 or 3x+y-5=0 OR A (2,-5) B(-2,9) Let P(x, 0) is the point on x axis So, PA=PB By Distance formula ( ) + ( + ) ( + ) + ( ) ( ) + = ( + ) + X x +25 =x x+81 So, x=-7 Q.14 Let line segment A (-6,10) and B (3,-8) is divided by (-4,6) in k:1 => -4=3k-6/k+1 => -4k-4=3k-6 => -4k-3k=-6+4 => -7k=-2 => k=2/7 So, k:1 = -2:7 Or Let ratio = k :1 By using y coordinate of [ ] We get 6=10-8k/k+1 6k+6=10-8k => 6k+8k= k=4 K=2/7 Therefore K: 1=2:7 result

29 SECTION C Q.15 Let a be the first term and d is common difference So accordind to statement a 4 +a 8 =24 and a 6 +a 10 =44 2a+10d=24 (1) 2a+14d=44 (ii) By subtraction (i) from (ii) we get, d =5, and put the value of d in (i) we get a=-13. There fore A.P is : -13,-8,-3.. Q.16 Solution:Take a point Q on XY other than P and join OQ The point Q must lie outside the circle. Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle). Therefore, OQ is longer than the radius OP of the circle. That is, OQ > OP. Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY. Q.17. In Quad. In AOBP. + B + P+ O=360 APB + AOB= 180 AOB=120 Construction of tangents 1.Draw a circle with radius =5cm 2. Draw angle AOB=120 at O. 3. At A and B draw perpendiculars intersecting each other at P. h Q.18 Let height of tower AB =h and AD = x tan 30 =h/(x+40) 1/ =h/(x+40) tan 60 =h/x 3 = h/x h= 3x

30 3 3x=x+40 2x=40 x=20 h=2 3m Q.19 P (E) = favorable outcomes/total outcomes P ( ) = Come to learn go to serve Q.20 A (7,-2) B (5, 1) C (3,3k) => Condition of collinear points => Ar. ABC=0 => 1/2[x 1 (y 2 -y 3 ) +x 2 (y 3 -y 1 ) +x 3 (y 1 -y 2 )]=0 => [7(1-3k) +5(3k+2) +3(-2-1)]=0 => 7-21k+15k+10-9 =0 => 8-6k=0 Therefore K=4/3 Q.21 diagonals of a square are equal. AC = ( + ) + ( + ) BD = ( + ) + ( ) AB 2 + BC 2 = AC 2 D(-4,4) C(-1,-1) ( ( ) + ( ) ) + ( ( + ) + ( + ) ) = ( ) + By the converse of Pythagoras B = 90 0 Therefore quad. ABCD is a square. Q.22 For cone r=3.5 cm, h=12 cm l= =12.5cm 2 Total surface area=πrl+2πr 2 =22/7+3.5*12.5+2*22/7*3.5 =214.5cm 2 Q.23 Side of square = 14 cm, Therefore area = side side= = 196 cm 2 Diameter of circle = 7 cm Radius = 3.5 cm Area of 4 circles = 4 πr 2 = 4 22/ = 154 cm 2 There fore area of shaded portion = area of sq.-area of 4 circles= = 42 cm 2 All the four values are important in students life because these are bases for successful life.

31 Q.24 Join OB, In right angled triangle OAB => OB 2 = OA 2 + AB 2 = (20) 2 + (20) 2 => OB 2 = 800 => OB = 00 = 0 Area of sector= = ( 0 ) = 628 cm 2 Area of shaded region = Area of sector area of square = = 228 cm 2 Let original speed = x km/h Increase speed =(x+250) km/h Acc. to question 1500/x-1500/(x+250)=1/2 X x= X x =0 X x-750x =0 X(x+1000)-750(x+1000)=0 (x+1000)(x-750)=0 So, X= -1000, 750 Rejecting ve speed of 1000 km/h Therefore speed = 750 km/h Section D Q.26 Let side of smaller square =x Perimeter= 4x Perimeter of larger square =24+4x Side=6+x ATQ x 2 + (6+x) 2 = 468 x x 2 +12x -468=0

32 x 2 + 6x 216 =0 x x-12x 216 =0 x(x+18)-12(x+18)= 0 (x+18)(x-12) =0 x=-18 or 12 Reject -18 Sides 12m and 18m Q.27 Let the first prize be of Rs x The next prize will be Rs (x-20),(x-40) x,(x-20),(x-40)... a=x d=-20 Sn=n/2[2a+ (n-1) d] S 7 =7/2[2a-120] =700 a=160 The prizes are Rs160, Rs140, Rs120, Rs100, Rs80, Rs60, Rs40 Q.28 Given: a circle,with centre o. in which PQ, and PR are two tangents. To prove : PQ= PR Construction : Join OP,OQ and OR. Proof : We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P. We are required to prove that PQ = PR. For this, we join OP, OQ and OR. Then OQP and ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP, OQ = OR (Radii of the same circle) OP = OP (Common) Therefore, Δ OQP Δ ORP (RHS) This gives PQ = PR (CPCT)

33 Q.29 1 Steps of Construction 1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. 2. Locate 4 (the greater of 3 and 4 in 3/4) points B1, B2, B3 and B4 on BX so that BB1 = B1B2 = B2B3 = B3B4. 3. Join B4C and draw a line through B3 (the 3rd point, 3 being smaller of 3 and 4 in 3/4) parallel to B4C to intersect BC at C. 4. Draw a line through C / parallel to the line CA to intersect BA at A / Then, triangle A BC ' is the required triangle. Q.30 Let speed of the car=x m/sec Let time taken to reach from D to A =v sec Distance CD=6x Distance AD=V x m tan 30 =h/6x+vx tan 60 =h/vx (v+6)x / 3=vx 3 v=3 Time taken 3 seconds Q.31 P (E) =favorable outcomes/total outcomes P (E) =2/52=1/26 Sample space 4 kings + 4 queen + 4 jacks P (fcard) 12/52=3/13 Favorable outcomes of spade=13 P (spade)=13/52=1/4 Black cards=03+03=6 P(E)=6/52=3/26

34 Q.32 Length of the blade = radius=25 cm Sector angle=115 Area swept by one blade=θπr 2 /360 So, Area swept by two wipers=( )/360 7 =158125/126 cm 2 = cm 2 Q.33 Depth=1.5m Width=6m Volume=l b h Speed=10km/h =10,000m/h Volume of water flowing in 1 Hour = Volume of cuboid of dimensions 10,000 m 6m 1.5 m Now volume of water = l b h =10, m 3 = 90,000 m 3 Volume of water flowing in canal in 30 minutes=45000m 3 Let area of the field being irrigated= x m 2 Area of field height of water= m 3 Area of field 8/100=45000 => Area of field = m 2 Q.34 The total height of the bucket = 40 cm, which includes the height of the base. So, the height of the frustum of the cone = (40 6) cm = 34 cm. Therefore, the slant height of the frustum, l = + ( ) where r1 = 22.5 cm, r2 = 12.5 cm and h = 34 cm. So, l = + ( ) = cm The area of metallic sheet used = curved surface area of frustum of cone + area of circular base + curved surface area of cylinder = [π ( ) + π (12.5) 2 + 2π ] cm 2 = 22/7 ( ) cm 2 = cm 2

35 1 SAMPLE PAPER 3 (SA II) MRS.KIRAN WANGNOO Mathematics CLASS : X Time: 3hrs Max. Marks: 90 General Instruction:- 1. All questions are Compulsory. 1. The question paper consists of 34 questions divided into 4 sections, A,B,C and D. Section A comprises of 8 questions of 1 mark each. Section-B comprises of 6 questions of 2 marks each and Section- D comprises of 10 questions of 4 marks each. 2. Question numbers 1 to 8 in Section A multiple choice questions where you are to select one correct option out of the given four. 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 each. You have to attempt only one of the alternatives in all such questions. 4. Use of calculator is not permitted. SECTION -A Question numbers 1 to 8 carry 1 mark each. For each of the questions 1-8, four alternative choices have been provided of which only one is correct. You have to select the correct choice. Q.1. Which of the following equations has the sum of its roots as 3? (A) x 2 + 3x 5 = 0 (B) -x 2 + 3x + 3 = 0 (C) 2x 2 - x (D) 3x 2-3x 3 = 0 Q.2. The sum of first five multiples of 3 is: (A) 45 (B) 65 (C) 75 (D) 90 Q.3. If radii of the two concentric circles are 15 cm and 17 cm, then the length of each chord of one circle which is tangent to other is (A) 8 cm (C) 30 cm (B) 16 cm (D) 17 cm Q.4. In given fig. PQ and PR are tangents to the circle with centre O such that QPR = 50 0, then OQR is equal to: Q (A) 25 (C) 40 (B) 30 o (D) 50 o P 50 o O R

36 2 Q.5. To draw a pair of tangents to a circle which are inclined to each other at an angle of 100 0, it is required to draw tangents at end points of those two radii of the circle, the angle between tangents should be: (A) 100 (B) 50 (C) 80 0 (D) 200 Q.6. The height of a cone is 60 cm. A small cone is cut off at the top' by a plane parallel to the base and its volume is height from the base at which the section is made is: (A) 15 cm (B) 30 cm the volume of original cone. The (C) 45 cm (D) 20 cm Q.7. A pole 6 m high casts a shadow 2 3 m long on the ground, then the sun's elevation is: (A)60 (B) 45 0 A (C) 30 0 (D) 90 o 6m Q.8. Which of the following cannot be the probability of an event? (A) 1/5 (B) 0.3 (C) 4% (D) 5/4 C 2 3 B SECTION- B Question numbers 9 to 14 carry 2 marks each. Q.9. Two tangents making an angle of 120 with each other, are drawn to a circle of radius 6 cm, then find the length of each tangent. Q.10. If the circumference of a circle is equal to the perimeter of a square then taking = find the ratio of their areas. Q.11. Find the roots of the following quadratic equation: x 2 - x = 0 Q.12 lf the numbers x-2, 4x-1 and 5x+ 2 are in A.P. Find the value of x. Q.13. The tangents PA and PB are drawn from an external point P to a circle with centre O. Prove that AOBP is a cyclic quadrilateral. A O P Q.14. In given fig., a circle of radius 7 cm is inscribed in a square Find the area of the shaded region B

37 3 SECTION C Question numbers 15 to 24 carry 3 marks each. Q.15. How many spherical lead shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm x 11 cm x 12 cm? Q.16. Point P (5, 3) is one of the two points of trisection of the line segment joining the points A(7,- 2) and B (1, - 5) near to A. Find the coordinates of the other point of trisection. Q.17 Show that the point P (- 4, 2) lies on the line segment joining the points A (-4, 6) and B(-4,-6). Q.18. Two dice are thrown at the same time Find the probability of getting Indifferent numbers on both dice. Or A coin is tossed two times. Find the probability of getting atmost One head. Q.19. Find the roots of the equation + = x, 5 Or A natural number, when increased by 12, becomes 160 of its reciprocal. Find the number. Q.20. Find the sum of integers between 100 and 200 that are divisible by 9. Q.21. In given figure two tangents PQ and PR are drawn to a circle with centre O from an external point P. Prove that QPR = 2 OQR. Or Prove that the parallelogram circumscribing a circle is a rhombus. Q.22. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ABC = 60. Then construct a triangle whose sides are ¾ time the corresponding sides of ABC. Q.23. In given fig., OABC is a square inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of shaded region. [Use = 3.14]

38 4 Q.24. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. Or A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire. SECTION D Question numbers 25 to 34 carry 4 marks each. Q.25. A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 600 Find the height of the tower. Q.26. Prove that the points A (4, 3), B (6,4), C (5, -6) and D (3, -7) in that order are the vertices of a parallelogram. Q.27. The points A2, 9), B(a, 5), C(5, 5) are the vertices of a triangle ABC right-angled at B. Find the value of 'a' and hence the area of ABC. Q.28. Cards with numbers 2 to 101 are placed in a box. A card is selected at random from the box. Find the probability that the card which is selected has a number which is a perfect square Q.29. A train travels at a certain average, speed for a distance àf 63 km and then travels a distance of 72 km at an average speed. of 6 km/h more than its original speed. If it takes 3 hours to complete the total journey, what is its.original average speed? Find two consecutive odd positive integers, sum of whose squares is 290. Q.30. A sum of Rs 400 is to be used to give seven cash prizes to students of a school for their Or overall academic performance. If each prize is Rs 40 less than the preceding price, find the value of each of the prize. Q.31. Prove that the lengths of tangents drawn from an external point to a circle are equal. Q.32. A well of diameter 3 m and 14 m deep is dug. The earth, taken out of it, has been evenly spread all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. Comment

39 5 on the importance of water in our daily life. Or 21 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm x 8 cm x 8 cm and then the box is filled with water. Find the volume of water filled in the box. Q.33. The slant height of the frustum of a cone is 4 cm and the circumferences of its circular ends are 18 cm and 6 cm. Find curved surface area the frustum. Q.34. From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45 and 60 respectively. Find height of the tower..how transmission towers are harmful to us?

40 6 SOLUTIONS SAMPLE PAPER 2 (SA II) ANSWERS SECTION -A Question numbers 1 to 8 carry 1 mark each. For each of the questions 1-8, four alternative choices have been provided of which only one is correct. You have to select the correct choice. Ans.1 Sol. (B)3 Ans.2 [ Sum of the roots = -b = -3 = 3 a -1 Sol. (A) 45 [ Required sum = = 45] Ans.3 Sol. (B) 16 cm [ OA 2 = AD 2 + OD = AD AD 2 = AD 2 =64 AD=8cm Ans.4 Sol. (A) 25 AB = 2AD = 16 cm] [ QOR = 180 o 50 o = 130 o OQR = ORQ [ Angle opposite to equal side of a Triangle] 2 OQR = 180 o 130 o = 50 OQR = 25 0 ] Ans.5 Sol. (C) 80 0 [ Angle between the radii = 80] P 50 o Q R O Ans.6 Sol. (C) 45 cm [ ABE ACD. h = r 60 R

41 7 1 r 2 h = 1 R r 2 h = 1 60 R 2 64 h = 60 R 2 64 r h = h 2 h 3 = 60 3 = H = l5cm Thus, height from the base = = 45 cm] Ans.7 Sol. (A) 60 0 [ tan = 6 = 3 = A 6m tan = tan 60 =60 ] C 2 3 B Ans8 Sol. (D) 5/4 [ Probability of an event 1 in any case] Ans.9 Sol. (D) 2 3 cm [ In rt. PQO PQ = Cot 60 0 = 1 OQ 3 PQ = OQ 1 = = 6 3 = 2 3 cm 3 3 Ans.10 Sol 2 r = 4 x Side r = Side 2 SECTION-B

42 8 Area of circle : Area of square = r 2 : r 2 = 4: = 4 : 22 7 = 14:11 2 Ans.11 Sol. Given equation is 2 x 2 - x 3 = x 2 5x -3 = 0 Ans.12 2x 2-6x + x - 3 = 0 2x (x-3) + 1 (x-3)= 0 (2x+1)(x-3)=0 X = 3, X = -½ Sol. x-2, 4x-l and 5x+2 are in A.P. (4x-1)- (x-2) = (5x+2) - (4x-1) 4x-1-x+2 = 5x+2-4x+ 1 2x=2 x=1 Ans.13 Sol. Since angle between the radius and the tangents at the A Point of contact is 90 o PAO + PBO = = 180 or APB + AOB = 180 O P = AOBP is a cyclic quadrilateral B Opposite angles of a quadrilateral are supplementary Ans.14 Sol. Here, Side of the square = Diameter of the circle = 2 x 7 = 14 cm Area of the shaded region = Area of the square - Area of the circle = (14) 2 (22/7) = = 42cm 2

43 9 SECTION-C Ans.15 Sol. Let number of spherical lead shots be n n x (4) r 3 = L x B x H 3 n x 4 x 22 x 3 x 3 x 3 = 9 x 11 x x 11 x 12 x 3 x 7 x 2 x 2 x 2 4 x 22 x 3 x 3 x 3 n = 84 Hence, the required spherical lead shots is 84. Ans.16 Sol. AP = PQ = QB Q is the mid-point of PB. A (7,-2) P (5,-3) Q B (1,-5) Coordinates of Q are Q 5+1, Q (3,-4) Ans.17 Sol. The abscissa (x-coordinate) of the points P, A and B is -4 Points P, A and B lie on the line x = 4 Hence, P A and B are collinear. Ans.18 Sol. Number of outcomes of the sample space when two dice are thrown = 6x 6=36 Number of outcomes of getting same number on both dice = 6 [(1, 1), (2, 2), (3, 3),(4, 4), (5, 5), (6, 6)] Number of favorable outcomes (different numbers on both dice) = 36-6 = 30 Required probability =. -36 = Or All possible outcomes when a coin is tossed twice = HH, HT, TH, UF Favourable outcomes (atmost one head) TT, HT, TH Required probability = 3 4

44 10. Ans.19. Sol. The given equation is = 1 2x-3 x-5 x 5 + 2x - 3 = 1 (2x-3)(x-5) 3x-8 = (2x-3)(x-5) 3x - 8 = 2x 2-13x x 2 16 x + 23 = 0 x = 16 ± ( ) 2 x 2 x = 16 ± 4 x = 16 ± = 16 ± 6 = 4 ± OR Let the number be x According to the statement of the question x + 12 = 160 x x 160 = 0 x 2 +20x-8x-160 = 0 (x-8)(x+20) = 0 = x = 8 or x = -20 (Rejecting - ve value since x is the natural number) Hence, the number is 8. Ans.20. Sol. Integers divisible by 9 between 100 and 200 are 108, 117, 126, 135, 198 a n = 198 a + (n-1) d = (n-1)9 = 198 (n-1)9 = 90 (n-1) = 10 n = 11

45 11 Now S n = n (a + a n ) 2 S 11 = 11 ( ) = 11 x 306 = Ans.21. Sol. Join OR Now. QOR + QPR = 180 o QOR = 180 o - QPR... (i) Also. OQR = ORQ.(ii) [ s opposite to equal sides of a Triangle] OQR + ORQ + QOR = 180 o QOR = OQR - ORQ = 180 o - 2 OQR... (iii) [using (ii)] Now, from (i) and (iii), we have 180 o - QPR 180 o - 2 OQR QPR = 2 OQR Given: A parallelogram ABCD, circumscribes a circle. To prove: ABCD is a rhombus i.e., Proof: AB = BC = CD = DA. Since ABCD is a parallelogram. AB = DC and BC = AD... (i) AP and AS are two tangents from an external point A to the circle. Or AP = AS... (ii) [ Tangents drawn from an external point to the circle are equal] Similarly, we have BP=BQ CR=CQ (iv) and DR = DS (v) Adding (ii), (iii), (iv) and (v), we have (AP + BP) + (CR + DR) = (AS + DS) ± (BQ + CQ)

46 12 AB+CD =AD+BC AB + AB = AD + AD 2AB=2AD [using (i)] AB = AD i.e., adjacent sides of the parallelogram are equal. Thus, all the sides are equal. Hence, ABCD is a rhombus.. Ans.22. Sol. Steps of Construction: 1 Draw a line segment BC = 6 cm 2. At B, construct an angle 60 such that BA = 5 cm.. 3. Join AC, so ABC is the given triangle. 4. Through B, construct an acute angle CBX, (<90 o ). 5. Mark four points B1, B2, B3 and B4, such that BB 1 =B 1 B 2 =B 2 B 3 =B 3 B 4 6. Join B 4 C. 7. Through B 3, draw B 3 C' B 4 C, intersecting BC in C'. 8. Through C', draw C'A' CA, intersecting BA in A'. 9. Hence, A'BC' is the required triangle. Ans.23. Sol. Here, OA = 20 cm and OABC is a square, OA =AB = BC = CO =20 cm OB = OA 2 + AB 2 [by Pythagoras Theorem] = = = 20 2 cm Now, area of the shaded region = Area of quadrant OPBQ - Area of square OABC o = 90 o [ OB=r=20 2 cm] = ¼

47 13 = = 228cm 2 Ans.24. Sol. Edge of the cube = l Radius of the hemisphere = l/ 2 Surface area of the remaining solid = S.A. of cube - S.A. of the top of hemisphere + C.S.A. of hemisphere = 6l 2 - l/2 l/2 + 2 l/2 l/2 = 6l = ¼ l 2 (24+ ) sq. units Volume of the wire = Volume of the copper rod r = ½ ½ 8 r 2 = 8 = r = 1/30 cm Thickness of the wire = Diameter of wire = 2 1 = 1 cm Or Ans.25. SECTION-D Sol. Let us assume the AB be the tower and C is a point 20 m away from the ground Angle of elevation of the top of the tower is 60 o In rt. CBA, AB CB AB 20 = tan 60 o = 3 AB = 20 3 m Hence, the height of the tower is 20 3 m.

48 14 Ans.26. Sol. Here, AB = (6-4) 2 + (4-3) 2 = 4 +1 = 5 units CD = (3-5) 2 + (-7-6) 2 = 4 +1 = 5 units AB = CD Again, BC = (5-6) 2 + (-6-4) 2 = = 101 units AD = (3-5) 2 + (-7-6) 2 = = 101 units BC = AD Now, in quadrilateral of ABCD both pair of opposite sides are equal. Hence, it is a parallelogram. Ans.27.Sol. Given ABC is right-angled at B. By Pythagoras theorem, we have AB 2 + BC 2 = AC 2 (a - 2) 2 + (5-9) 2 + (5-a) 2 + (5-5) 2 - (5-2) 2 + (5-9) 2 a a a 2-10a = a 2-14a-20 = 0 a 2-7a-10 = 0 (a-5)(a-2) = 0 a = 5 or a = 2 Rejecting a = 5, BC reduces to zero. Thus, a = 2 Area of ABC = ½ AB BC = ½ 4 3 = 6 sq. units Ans.28. Sol. Total number of cards in the box = 100 Favorable outcomes (perfect squares) are 4, 9,16,25,36,49,64,81,100

49 15 Required probability = SECTION D Question numbers 29 to 34 carry 4 marks each. Ans.29. Sol. Let the average speed be x km/h; According to the statement of the question = 3 x x+6 63(x+6)+72x = 3 x(x+6) 63x x = 3 x 2 + 6x.. 135x = 3x x 3x 2-117x-378=0. x 2-39x-126 = 0 (x-42)(x+3) = 0 x = 42 or x = -3 (rejecting -ve value because speed cannot be -ve) Hence, original average speed is 42 km/h.. Or Let two consecutive odd positive integers be x, x + 2. According to the statement of the question x 2 + (x + 2) 2 = 290 x 2 +x x = x 2 +4x 286 = 0 x 2 + 2x = 0 x x-11x = 0 (x+13)(x-11) = 0... x = -13 or x = 11.

50 16 (rejecting ve value because x is an odd +ve integer) Numbers are 11 and 13. Ans.30. Sol. Total amount of seven prizes = 1400 Let the value of first prize be x According to given statement, the seven prizes are Now, x-40-x = -40 x, x - 40, x - 80, x - 120,..., x -240 x x + 40= - 40, which is constant. Thus, it is an A.P. with first term (a) as x and common difference (d) as S n = {2a+ (n-1) d} 1400 = {2x+(7-1)(-40)} 400 = 2x 240} 2x = 640 x = 320 Hence, the amount of each prize (in ) is 320, , , , , , i.e., 320, 280, 240, , 120, 80. Ans.31. Sol. Given: A circle C (O, r) with centre O. Through the external point P tangents PT and PT' are drawn. To prove: PT = PT' T Const.: Proof: Join PO, TO and T'O In PTO and PT'O, we have TO =T'O = r hypt. PO = hypt. PO [common] PTO = PT'O = 90 o PTO PT'O [by RHS cong. rule] PT = PT'... [c.p.c.t.]

51 17 Ans.32. Sol. Here, Radius of the well = 3/2 m Depth of the well = 14 m Width of the embankment = 4 m Radius of the embankment = = 5.5 m Let 'h' be the height of the embankment. Volume of the embankment = Volume of the well (cylinder) ( ) h = (1.5) 2 14 ( ) h = ( ) 4 m 28 h = 31.5 h = h = 1.125m Or Radius of sphere = 2 cm Volume of 21 spheres = = 704cm3 Volume of cuboid = 16 x 8 x 8 = 1024 cm 3 Volume of water = = 320 cm 3 Ans.33. Sol. Slant height of the frustum of a cone (l) = 4 cm Circumference of top end = 18 cm 2 R = 18 2 R = 18 R= 18 7 = 63 cm and circumference of bottom end = 6 cm 2 r = 6 2 r = 6 r = 6 7 = 21 cm Curved surface area = l (R + r) = 4 + = 4 = 48 cm 2

52 18 Ans.34. Sol. Let AB be the transmission to fixed on the top of the building of height 20 m. Let AB = h m and.p be a point on ground, such that BPC = 45, APC = 60. In rt. PCB, C = 90 A BC = tan 45 o PC 20 = 1 PC = 20m In rt. PCA, C = 90 AC = tan 60 o PC h+20 = 3 h+20 = h = = 20 ( 3-1) Hence, the height of the tower is 20 ( 3-1)m.

53 1 SAMPLE PAPER 4 (SAII) MR AMIT. KV NANGALBHUR Mathematics CLASS : X Time: 3hrs Max. Marks: 90 General Instruction:- 1. All questions are Compulsory. The question paper consists of 34 questions divided into 4 sections, A,B,C and D. Section A comprises of 8 questions of 1 mark each. Section-B comprises of 6 questions of 2 marks each and Section- D comprises of 10 questions of 4 marks each. Question numbers 1 to 8 in Section A multiple choice questions where you are to select one correct option out of the given four. 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 each. You have to attempt only one of the alternatives in all such questions. Use of calculator is not permitted. Section-A Question number 1 to 10 carry 1 mark each. 1) The discriminant of the quadratic equation.3 3 x x + 3 = 0 is A) 30 B) 36 C) 64 D) 100 2) If a, 2 are three consecutive terms of an A.P., then the value of a. A) 3/5 B) 5/3 C) 4/5 D) 7/5 3) In figure below, ΔABC is circumscribing a circle. he length of BC is A) 10cm B)7cm C) 4cm D) 3cm 4) HCF and LCM of two numbers are 12 and 36 respectively, then product of these two numbers is: A) 36 B)432 C) 342 D) 12 5) Find the perimeter of the given figure, where AED is a semi-circle and ABCD is a rectangle. A) 76cm B)68cm c) 22cm D)54cm

54 2 6) To draw a pair of tangents to a circle which are inclined to each other at an angle of it is required to draw tangents at end points of those two radii of the circle, the angle between which should be A) B) C) 50 0 D) ) Two Tangents making an angle of are drawn to a circle of radius 6cm, then the length of each tangent is equal to A) 2 3 cm B) 2 cm C) 6 3 cm D) 3cm 8) A pole 6m high casts a shadow 2 3 m long on the ground, then the sun s elevation is A) 60 0 B) 90 0 C) 30 0 D) Section-B Question number 9 to 14 carry 2 marks each 9) Solve by factorisation : 7 y 2 6y = 0. 10) Find a point on x-axis which is equidistant from the points (-2,5) and (2, -).3 11) Determine the ratio in which the point P(m, 6) divides the join of A( -4, 3) and B (2,8). 12) Spherical ball of diameter 21 cm is melted and recasted into cubes, each of side 1 cm. find the number of cubes thus formed. 13) A die is thrown once, find the probability of getting (i) a prime number (ii) a number divisible by 2 14) All cards of ace, jack and queen are removed from the deck of playing cards. One card is drawn at random from the remaining cards. Find the probability that the card drawn is :(a) a face card (b) not a face card Section-C Question number 15 to 24 carry 3 marks each 15) Determine the AP whose 3 rd term is 16 and when fifth term is subtracted from 7 th term, we get ) Construct a triangle similar to a given ABC in which AB =4cm, BC =6cm and ABC = 60 0, such that each side of the new triangle is ¾ of the corresponding sides of given ABC.(Steps of Constructions not required)

55 3 17) A F E B D C In the figure above, the incircle of ABC touches the sides BC, CA and AB at D, E and F respectively. If AB = AC, prove that BD = CD. 18) E A 2.8 cm B 1.4cm C F In the fig., find the perimeter of the shaded region where ADC, AEB and BFC are semi-circles on the diameter AC, AB, and BC respectively. 19) Largest sphere is carved out of the cube of side 7cm. find volume of the sphere? 20)The curved surface area of the cone is cm 2. If the radius of the base is 56cm, find its height. 21) The volume of the right circular cylinder of height 7cm is 567 cm 3. Find its curved surface area. 22)An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from his eye is What is height of the chimney? 23)Find the value of p for which the points (-5, 1), (1,p) and (4, -2) are collinear. OR If the coordinates of the mid-points of triangle are (1,2), (0,-1) and (2,-1). Find the coordinates of its vertices. 24) Find the roots of quadratic equation : a 2 b 2 x 2 + b 2 x a 2 x 1 = 0

56 4 Section-D Question number 25 to 34 carry 4 marks each 25) The product of Tanay s age five years ago and his age after 10 years is 16. Find his present age. 26) Find the sum of the first 31 terms of an AP. Whose n th term is given by 3 + 2n/3. 27) OABC is a rhombus whose three vertices A,B and C lie on a circle with centre O. If the radius of the circle is 10cm, find the area of the rhombus. 28) Prove that the lengths of the tangents drawn from an external point to a circle are equal. Use this theorem for the following: If all sides of the parallelogram touches a circle, show that it is a rhombus. 29) From top of a hill the angles of depression of two consecutive kilometre stones due east are found to be 30 0 and Find the height of the hill. 30) A school has 5 houses A, B, C, D and E. In a class having 23 students, 4 are from house A, 8 from house B, 5 from house C, 2 from house D and rest are from E. 1 student is selected at random to be a class monitor. A) Find the probability that the selected student is not from B, C and E. B) find the probability that the student is from house E. C) what moral values the student of this class must share? 31) Find the ratio in which the line 2x + y = 4 divides the join of A(2, -2) and B(3,7). Also find the coordinates of the point of their intersection. 32) A bucket made up of metal sheet is in the form of a frustum of a cone of the height 16cm with radii of its lower and upper ends as 8cm and 20cm respectively. Find the cost of the bucket, if the cost of the metal sheet is Rs. 15 per 100 cm 2. 33) A well with 10m inside diameter is dug 14m deep. Earth taken out of it is spread all around to a width of 5m to form an embarkment. Find the height of the embankment. 34) Sum of the areas of the two squares is 468 m 2. If the difference of their perimeters is 24m, find the sides of the two squares.

57 5 SOLUTION SAMPLE PAPER 4 SECTION A 1) C) 64 Explanation: - Here a =3 3, b = 10, c = 3 D = b 2 4ac = X 3 3 X 3 = = 64 2) D) 7/5 Explanation:- as 4/5, a, 2 are in AP therefore the common difference is same i.e a- 4/5 = 2 a 2a = 2 + 4/5 2a = 14/5 a = 7/5 3) A) 10cm Explanation :- AQ = 4cm ( Tangents from same external point to same circle) AC = AQ + QC or 11 = 4 + QC QC = 7cm PC = QC = 7cm ( Tangents from same external point to same circle) BP = 3cm ( Tangents from same external point to same circle) BC = PC + BP = = 10cm 4) B) 432 Explanation :- Product of two numbers = HCF X LCM = 12 x 36 = 432 5) A) 76cm Explanation :- Required perimeter = AB + BC + CD + r [ here r = ½ X BC] = /7 X 7 = =76 cm 6) D) 80 0 Explanation : As angles between tangents and at centre are supplementary Required angle = ) A) 2 3 cm Explanation :- = 80 0 B A O

58 6 OAB = 120/2 [ the line joining the external point to the centre bisect the angle between the tangents at external point] Therefore OAB = 60 0 Now in AOB Tan 60 0 = OB / AB 3 = 6 / AB ( As tan 60 0 = 3) AB = 6/ 3 = 6 3/( 3 X 3) = 6 3/ 3 = 2 3 cm A 8) A) 60 0 Explanation:- Pole x B C Shadow Let the inclination = x Therefore tan x = AC/BC Tan x = 6/2 3 = 6 3/(2 3X 3) = 6 3/(2X3) = 3 = tan 60 0 Section B 9) 7y 2 6y = 0 Or 7y 2 13y + 7y =0 Or Y( 7y 13) + 7 ( 7y 13) =0 Or ( 7y 13) ( y + 7) = 0 Or ( 7y 13) = 0 or ( y + 7) = 0 A(-2,5) Y = 13/ 7 or y = ) Let the point on X axis be P(a,0) P(a,0) X-axis According to question AP = PB [(a+2) 2 + (-5) 2 ] = [( a-2) ] [ a a +25] = [ a a + 9] B (2, -.3) a 2 + 4a -a 2 + 4a = a = -16

59 7 a = -2 (Ans) 11) A(-4,3) K P(m,6) 1 B(2,8) Let required ratio = k : 1 Using section formula ( on y coordinate) {[mx 1 + nx 2 ]/[m+n], [my 1 + ny 2 ]/[m+n]} 6 = (8k +3)/(k+1) Or 6(k+1) = 8k + 3 Or 6k + 6 = 8k +3 Or 6k 8k = 3 6-2k = -3 K = 3/2 Therefore reqired ratio = 3:2 12) Let the number of cubes formed = N N X vol of cube = vol of sphere N X 1 3 = 4/3 X N = 4/3 X 22/7 X 21/2 X21/2 X21/2 N = 38808/8 N = ) Total outcomes = 6[ 1,2,3,4,5,6] i) Total prime numbers =3[2,3,5] P(E) = 3/6 = ½ ii) Total numbers divisible by 2 =3[ 2,4,6] P(E) = 3/6 = ½ 14) Number of cards removed = 12 [4A +4J +4Q] Remaining cards = =40 a) Number of face cards = 4 [kings] P( E) = 4/40 = 1/10 b) Probability[not a face card] = 1 P( E) = 1 1/10 = 9 /10 SECTION C 15) Given a 3 = 16 i.e a + 2d = (1) Also a 7 a 5 = 12 a + 6d ( a + 4d) =12 a +6d a 4d = 12 2d = 12 d = 6 Put d = 6 in (1)

60 8 a + 2X6 = 16 a + 12 = 16 a = 4 therefore required AP : 4, 10, 16, 22.. Q16 16) Given AB = AC Therefore AF + FB = AE + EC { from fig.) AF AE + FB = EC AF AF + FB = EC { AF =AE, lengths of two tangents from same external point } FB = EC { lengths of two tangents from same external point } BD = DC { lengths of two tangents from same external point } 17) Required perimeter = perimeter(semi-circle ADC) + perimeter(semi-circle AEB) + perimeter(semi-circle BFC) = X 4.2/2 + X 2.8/2 + X 1.4/2 = = 13.2 cm