MATHS MCQ SOLVED AND UNSOLVED

44 CBSE Sample Paper Mathematics Standard Class X (Term I) 19. One equation of a pair of dependent linear equations is - 5x + 7y = 2. The second equation can be (a) 10x + 14y + 4 = 0 (b) - 10x - 14y + 4 = 0 (c) - 10x + 14y + 4 = 0 (d) 10x - 14y = - 4 20. A letter of English alphabets is chosen at random. What is the probability that it is a letter of the word ‘MATHEMATICS’? (a) 4/13 (b) 9/26 (c) 5/13 (d) 11/26 Section B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. If sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are (a) 2 (b) 3 (c) 4 (d) 5 22. Given below is the graph representing two linear equations by lines AB and CD respectively. What is the area of the triangle formed by these two lines and the line x = 0? Y A4 3 2C 1 X¢ B X –4 –3 –2 –1 0 1 2 3 4 –1 D –2 –3 –4 Y¢ (a) 3 sq units (b) 4 sq units (c) 6 sq units (d) 8 sq units Latest CBSE SAMPLE PAPER 23. If tan a + cot a = 2, then tan 20 a + cot 20 a is equal to (a) 0 (b) 2 (c) 20 (d) 2 20 24. If 217x + 131y = 913, 131x + 217y = 827, then x + y is (a) 5 (b) 6 (c) 7 (d) 8 25. The LCM of two prime numbers p and q ( p > q) is 221. Find the value of 3p - q. (a) 4 (b) 28 (c) 38 (d) 48 26. A card is drawn from a well shuffled deck of cards. What is the probability that the card drawn is neither a king nor a queen? (a) 11/13 (b) 12/13 (c) 11/26 (d) 11/52 27. Two fair dice are rolled simultaneously. The probability that 5 will come up at least once is (a) 5/36 (b) 11/36 (c) 12/36 (d) 23/36

CBSE Sample Paper Mathematics Standard Class X (Term I) 45 28. If 1 + sin 2 a = 3 sin a cosa, then values of cot a are (a) - 1, 1 (b) 0, 1 (c) 1, 2 (d) - 1, - 1 29. The vertices of a parallelogram in order are A(1, 2), B (4, y), C(x, 6) and D(3, 5). Then, (x, y) is (a) (6, 3) (b) (3, 6) (c) (5, 6) (d) (1, 4) 30. In the given figure, ÐACB = ÐCDA, AC = 8 cm and AD = 3 cm, then BD is C 8 cm A 3 cm D B (a) 22/3 cm (b) 26/3 cm (c) 55/3 cm (d) 64/3 cm 31. The equation of the perpendicular bisectors of line segment joining points A(4, 5) and B(- 2, 3) is (a) 2x - y + 7 = 0 (b) 3x + 2 y - 7 = 0 (c) 3x - y - 7 = 0 (d) 3x + y - 7 = 0 32. In the given figure, D is the mid-point of BC, then the value of cot y is cot x A x y CD B (a) 2 (b) 1/2 (c) 1/3 (d) 1/4 33. The smallest number by which 1/13 should be multiplied, so that its decimal expansion terminates after two decimal places is (a) 13/100 (b) 13/10 (c) 10/13 (d) 100/13 Latest CBSE SAMPLE PAPER 34. Sides AB and BE of a right triangle, right angled at B are of lengths 16 cm and 8 cm respectively. The length of the side of largest square FDGB that can be inscribed in the DABE is A 16 cm D F B G E 8 cm (b) 16/3 cm (a) 32/3 cm (c) 8/3 cm (d) 4/3 cm

46 CBSE Sample Paper Mathematics Standard Class X (Term I) 35. Point P divides the line segment joining R(- 1, 3) and S(9, 8) in ratio k :1. If P lies on the line x - y + 2 = 0, then value of k is (a) 2/3 (b) 1/2 (c) 1/3 (d) 1/4 36. In the figure given below, ABCD is a square of side 14 cm with E, F, G and H as the mid-points of sides AB, BC, CD and DA respectively. The area of the shaded portion is A EB HF DG C (a) 44 cm2 (b) 49 cm2 (c) 98 cm2 (d) 49p / 2 cm2 37. Given below is the picture of the Olympic rings made by taking five congruent circles of radius 1cm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1cm. Total area of all the dotted regions assuming the thickness of the rings to be negligible is B A Latest CBSE SAMPLE PAPER (a) 4(p / 12 - 3 / 4) cm2 (b) (p / 6 - 3 / 4) cm2 (c) 4(p / 6 - 3 / 4) cm2 (d) 8 (p / 6 - 3 / 4) cm2 38. If 2 and 1 are the zeroes of px 2 + 5x + r, then 2 (a) p = r = 2 (b) p = r = - 2 (c) p = 2 and r = - 2 (d) p = - 2 and r =2 39. The circumference of a circle is 100 cm. The side of a square inscribed in the circle is (a) 50 2 cm (b) 100 / p cm (c) 50 2 / p cm (d) 100 2 / p cm 40. The number of solutions of 3x + y = 243 and 243x - y = 3 is (a) 0 (b) 1 (c) 2 (d) infinite

CBSE Sample Paper Mathematics Standard Class X (Term I) 47 Section C Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted. 41-45 are based on Case Study-1. CASE STUDY 1 The figure given alongside shows the path of a diver, when she takes a jump from the diving board. Clearly, it is a parabola. Annie was standing on a diving board, 48 ft above the water level. She took a dive into the pool. Her height (in ft) above the water level at any time ‘t’ in seconds is given by the polynomial h(t) such that h(t) = - 16t 2 + 8t + k 41. What is the value of k? (b) - 48 (d) 48/(- 16) (a) 0 (c) 48 42. At what time will she touch the water in the pool? (a) 30 s (b) 2 s (c) 1.5 s (d) 0.5 s 43. Rita’s height (in ft) above the water level is given by another polynomial p(t) with zeroes - 1 and 2. Then, p(t) is given by (a) t 2 + t - 2 (b) t 2 + 2 t - 1 Latest CBSE SAMPLE PAPER (c) 24t 2 - 24t + 48 (d) - 24t 2 + 24t + 48 44. A polynomial q(t) with sum of zeroes as 1 and the product as - 6 is modelling Anu’s height (in ft) above the water at any time t ( in s). Then, q(t) is given by (a) t 2 + t + 6 (b) t 2 + t - 6 (c) - 8t 2 + 8t + 48 (d) 8t 2 - 8t + 48 45. The zeroes of the polynomial r(t) = - 12t2 + (k - 3)t + 48 are negative of each other. Then, k is (a) 3 (b) 0 (c) - 1.5 (d) - 3

48 CBSE Sample Paper Mathematics Standard Class X (Term I) 46-50 are based on Case Study-2. CASE STUDY 2 A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf. It is rectangular in shape-100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 m (4 yards) apart, and the lower edge of the crossbar must be 2.14 m (7 ft) above the ground. Each team plays with 11 players on the field during the game including the goalie. Positions you might play include: Forward As shown by players A, B, C and D. Midfielders As shown by players E, F and G. Fullbacks As shown by players H, I and J. Goalie As shown by player K. Using the picture of a hockey field below, answer the questions that follow 8 7 6A 5F B H4 3 Latest CBSE SAMPLE PAPER 2 60 yards K I1 E O 0 –6 –5 –4 –3 –2 –1 1 2 3 45 6 7 8 9 10 11 12 13 1415 J –1 C –2 –3 G D –4 –5 100 yards –6 46. The coordinate of the centroid of DEHJ are (a) (- 2 / 3, 1) (b) (1, -2 / 3) (c) (2 / 3, 1) (d) (-2 / 3, -1) 47. If a player P needs to be at equal distances from A and G, such that A, P and G are in straight line, then position of P will be given by (a) (-3 / 2,2) (b) (2 , -3 / 2 ) (c) (2 ,3 / 2 ) (d) (-2, -3) 48. The point on X-axis equidistant from I and E is (a) (1/ 2 ,0) (b) (0, -1 / 2 ) (c) (-1/ 2,0) (d) (0,1 / 2 ) 49. What are the coordinates of the position of a player Q such that his distance from K is twice his distance from E and K,Q and E are collinear? (a) (1, 0) (b) (0, 1) (c) (-2 ,1) (d) (-1,0) 50. The point on Y-axis equidistant from B and C is (a) (-1,0) (b) (0, -1) (c) (1,0) (d) (0, 1) ANSWERS 1. (b) 2. (a) 3. (b) 4. (d) 5. (a) 6. (d) 7. (b) 8. (c) 9. (a) 10. (d) 11. (b) 12. (c) 13. (b) 14. (a) 15. (d) 16. (b) 17. (b) 18. (a) 19. (d) 20. (a) 21. (c) 22. (c) 23. (b) 24. (a) 25. (c) 26. (a) 27. (b) 28. (c) 29. (a) 30. (c) 31. (d) 32. (b) 33. (a) 34. (b) 35. (a) 36. (c) 37. (d) 38. (b) 39. (c) 40. (b) 41. (c) 42. (b) 43. (d) 44. (c) 45. (a) 46. (a) 47. (c) 48. (a) 49. (b) 50. (d)

CBSE Sample Paper Mathematics Standard Class X (Term I) 49 SOLUTIONS 1. (b) We have, least composite number is 4 and Let ABCD be a rhombus. the least prime number is 2. We have, AC = 32 cm and BD = 24 cm. Since, diagonals of a rhombus bisects each \\LCM (4, 2) = 4 other at right angle. \\ AB 2 =OA2 + OB 2 and HCF (4, 2) = 2 [by Pythagoras theorem] So, LCM (4, 2) : HCF (4, 2) = 4 :2 = 2 : 1 AB2 = 162 + 12 2 2. (a) We know that, if a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 coincide, then [Q OA = 1 AC and OB = 1 BD] a1 = b1 = c1 2 2 a2 b2 c2 = 256 + 144 = 400 So, if 5x + 7y - 3 = 0 and 15x + 21y - k = 0 coincide, then \\ AB = 20 cm Now, area of a rhombus = 1 AC ´ BD 5 = 7 = -3 15 21 -k 2 Þ 1=1=3 = 1 ´ 32 ´ 24 33 k \\ 1=3 2 3k Þ k =9 = 384 cm2 3. (b) From the figure, we have Also, area of a rhombus = Side ´ Altitude Distance of the girl from the starting point \\ Side ´ Altitude = 384 = OB Þ Altitude = 384 = 19.2 cm N 20 B 5. (a) When two fair coins are tossed, then 150 m possible outcomes are HH, HT,TH,TT. The favourable outcomes of getting at the most W O 200 m A E one head are HT,TH,TT. \\Required probability = Favourable number of outcomes Total number of outcomes =3 S 4 Now, In DOAB, 6. (d) Given, AB2 = 4 By Pythagoras theorem, PQ2 9 OB2 = OA2 + AB2 Þ AB = 2 Latest CBSE SAMPLE PAPER = (200)2 + (150)2 PQ 3 = 40000 + 22500 We know that, if two triangles are similar, then the ratio of corresponding altitudes is equal to = 62500 the ratio of corresponding sides. \\ OB = 62500 = 250 \\ AM = AB = 2 PN PQ 3 So, required distance is 250 m. 4. (d) 7. (b) We have, D C 2 sin2 b - cos2 b = 2 Þ 2 sin2 b - (1 - sin2 b) = 2 32Ocm 24 cm AB [Q cos2 q + sin2 q = 1] Þ 2 sin2 b - 1 + sin2 b = 2 Þ 3 sin2 b = 3 Þ sin2 b = 1 Þ sin2 b = sin2 90° Þ b = 90°

50 CBSE Sample Paper Mathematics Standard Class X (Term I) 8. (c) We have, 44.123 = 44123 14. (a) We have, 1000 Ratio of angles of DABC are 1 : 1 : 2 Now, prime factors of denominator = (10)3 \\ ÐA = x, ÐB = x = (2 ´ 5)3 and ÐC = 2 x \\ Prime factors are 2, 5. We know that, sum of all angles of a triangle is 9. (a) Line x = a is a line parallel to Y-axis and 180°. y = b is a line parallel to X-axis. So, both lines \\ ÐA + ÐB + ÐC = 180° are intersecting line. Þ x + x + 2 x = 180° 10. (d) Distance of the point A(- 5, 6) from the Þ 4x = 180° origin O(0, 0) is given as OA = (- 5 -0)2 + (6 - 0)2 Þ x = 45° [from distance formula] So, ÐA = 45°, ÐB = 45° and ÐC = 90° = (- 5)2 + (6)2 Now, sec A - tan A = 25 + 36 cosec B cot B = 61 units = sec 45° - tan 45° cosec 45° cot 45° 11. (b) We have, a2 = 23 = 2 -1 25 21 Þ a = 23 will = 1-1 5 =0 Since, 23 is a irrational number, so 23 15. (d) We have, 5 Radius of circle, r = 0.7m also be a irrational number. So, a is an irrational number. Distance travelled in one revolution 12. (c) We know that, = Circumference of the circle = 2 pr = 2 ´ 22 ´ 0.7 LCM (a, b) ´ HCF (a, b) = a ´ b 7 \\LCM (x, 18) ´ HCF (x, 18) = x ´ 18 = 4.4 m Þ 36 ´ 2 = x ´ 18 Now, Number of revolutions = Total distance travelled Þ x=4 Distance travelled in one revolution = 176 = 40 13. (b) We have, 4.4 tan A = 3 16. (b) We know that, if two triangles are similar then the ratio of their perimeter is equal to Latest CBSE SAMPLE PAPER Þ tan A = tan60° [Q tan60° = 3] the ratio of their corresponding sides. Þ A = 60° Now, DABC ~ DDEF \\ Perimeter of DABC = BC Again, in DABC Perimeter of DDEF EF ÐA + ÐB + ÐC = 180° Þ AB + BC + CA = BC Þ 60° + 90° + ÐC = 180° Perimeter of DDEF EF Þ ÐC = 30° Þ Perimeter of DDEF = ( AB + BC + CA) ´ EF Now, cos Acos C - sin Asin C BC = (3 + 2 + 2.5) ´ 4 = cos60° cos30° - sin60°sin30° 2 =1´ 3 - 3´ 1 = 7.5 ´ 2 = 15 cm 22 22 = 3- 3 44 =0

CBSE Sample Paper Mathematics Standard Class X (Term I) 51 17. (b) In DABC, 20. (a) Number of possible outcomes DE|| BC = Number of letters of English alphabets \\ Ð1 = Ð3 and Ð2 = Ð4 [corresponding angles] = 26 A Favourable outcomes are letters of the word ‘MATHEMATICS’ D1 2E i.e. M, A, T, H, E, I, C, S \\Number of favourable outcomes = 8 3 4 So, required probability = 8 = 4 B C 26 13 Now, in DADE and DABC Ð1 = Ð3 21. (c) We have, HCF of two numbers is 81. Ð2 = Ð4 So, numbers are 81x and 81y, where x and y are coprime. By AA similarity criterion, DADE ~ DABC Now, according to the question \\ AD = DE [Q AB = AD + BD] 81x + 81y = 1215 AB BC Þ 3 = DE Þ x + y = 15 3 + 4 14 [divide both sides by 81] Þ DE = 3 ´ 14 Þ DE = 6 cm So, two coprime numbers whose sum is 15 are 7 (1, 14), (2, 13), (4, 11) and (7, 8). 18. (a) We have, So, there are 4 possible number of pairs. 4 tanb = 3 22. (c) Required area = Area of DACD Þ tanb = 3 = 1 ´ Base ´ Altitude 2 4 = 1´6´2 Now, 4 sinb - 3 cosb 2 4 sinb + 3 cosb [Base = AD = 4 - (- 2 ) = 6 and 4 sinb - 3 cosb height = Distance of point C = cosb cosb from Y-axis = 2] 4 sinb + 3 cosb = 6 sq units cosb cosb 23. (b) We have, [Q dividing numerator and tan a + cot a = 2 denominator by cosb] Þ tan a + 1 =2 tan a = 4 tanb - 3 é tan q = sin q ù 4 tanb + 3 êëQ cos q ûú Þ tan2 a + 1 = 2 tan a Þ tan2 a - 2 tan a + 1 = 0 4´ 3 -3 Þ (tan a - 1)2 = 0 Latest CBSE SAMPLE PAPER =4 Þ tan a = 1 4´ 3 +3 Again, cot a = 1 = 1 = 1 4 tan a 1 = 3 -3 3+3 Now, tan20 a + cot20 a = (1)20 + (1)20 =1 + 1=2 =0 24. (a) We have, 217x + 131y = 913 19. (d) Second equation can be obtained by 131x + 217y = 827 … (i) multiplying or dividing the first equation by … (ii) a non-zero number. On adding Eqs. (i) and (ii), we get 348x + 348y = 1740 Now, - 5x + 7y = 2 Þ 348(x + y) = 1740 Þ - 5x + 7y - 2 = 0 Þ - 2(- 5x + 7y - 2 ) = 0 [multiply by -2] Þ 10x - 14y + 4 = 0 Þ x + y = 1740 = 5 348 Þ 10x - 14y = - 4

52 CBSE Sample Paper Mathematics Standard Class X (Term I) 25. (c) We know that, 30. (c) LCM of two prime numbers C = Product of the numbers. \\ p ´ q = 221 8 cm Þ p ´ q = 13 ´ 17 Þ p = 17, q = 13 [Q p > q] \\ 3 p - q = 3 ´ 17 - 13 A 3 cm D B = 51 - 13 = 38 In DACD and DABC 26. (a) We know that, there are 4 kings and ÐA = ÐA (common) 4 queens in a well shuffled deck of 52 cards. ÐADC = ÐACB (given) \\ Required probability = 52 - 8 = 44 = 11 So, by AA similarity criterion, 52 52 13 DACD ~ DABC 27. (b) When two fair dice are rolled simultaneously, then total number of possible \\ AC = AD outcomes will be 36. AB AC Now, favourable outcome are those in which Þ 8 = 3 Þ AB = 64 5 will come up atleast once i.e. (5, 1), (5, 2), AB 8 3 (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5). Þ AD + BD = 64 3 \\Total number of favourable outcomes = 11 Þ BD = 64 - AD \\Required probability = 11 3 36 = 64 - 3 = 55 cm 33 28. (c) We have, 1 + sin2 a = 3 sin a cos a 31. (d) Let l be the perpendicular bisector of line segment joining the points A(4, 5) Þ sin2 a + cos2 a + sin2 a = 3 sin a cos a and B(- 2 , 3). Let P(x, y) be any point of l. [Q sin2 q + cos2 q = 1] P (x, y) Þ 2 sin2 a - 3 sin a cos a + cos2 a = 0 Þ 2 sin2 a - 2 sin a cos a -sin a cos a + cos2 a = 0 Þ 2 sin a(sin a - cos a) A (4, 5) B (–2, 3) - cos a(sin a - cos a) = 0 l Þ (sin a - cos a)(2 sin a - cos a) = 0 So, P will be equidistant from A and B. Þ sin a = cos a or 2 sin a = cos a \\ PA = PB Þ (x - 4)2 + (y - 5)2 = (x + 2 )2 + (y -3)2 Latest CBSE SAMPLE PAPER Þ 1 = cot a or 2 = cot a Þ (x - 4)2 + (y - 5)2 = (x + 2 )2 + (y - 3)2 Þ x2 - 8x + 16 + y2 -10y + 25 é cos q = cot qùûú ëêQ sin q = x2 + 4x + 4 + y2 - 6y + 9 Þ - 12 x - 4y + 28 = 0 \\ cot a = 1 or 2 Þ 3x + y - 7 = 0 which is required equation. 29. (a) We know that, diagonals of a 32. (b) parallelogram bisects each other. So, A Mid-point of AC = Mid-point of BD x Þ æ1+ x,2 + 6 ö = æ 4 +3, y + 5ö y ç ÷ ç ÷ è2 2 øè2 2 ø Þ 1 + x = 7 and 4 = y + 5 22 2 [comparing x and y coordinate] Þ x = 6 and y = 3 \\ (x, y) is equal to (6, 3). CD B

CBSE Sample Paper Mathematics Standard Class X (Term I) 53 In DABC, cot y = Base = AC Þ 16 - x = x x 8-x Perpendicular BC Þ (16 - x) (8 - x) = x2 and in DADC, cot x = Base = AC Þ 128 - 8x - 16x + x2 = x2 Perpendicular CD Þ 24x = 128 Þ x = 128 æ AC ö ç ÷ 24 \\ cot y = è BC ø Þ x = 16 cm cot x æ AC ö 3 ç÷ 35. (a) Since, P(x1 , y1 ) divides the line segment è CD ø joining R(- 1, 3) and S(9, 8) in the ratio k :1. = CD = CD BC 2 CD k:1 [Q D is the mid-point of BC, so CD = BD] R (–1, 3) P (x1, y1) S (9, 8) =1 2 Then, P(x1 , y1 ) = æ 9k - 1, 8k + 3 ö çç k +1 k +1 ÷÷ 33. (a) (a) 1 ´ 13 = 1 = 0.01, which is è ø 13 100 100 terminate after two decimal places. [from section formula] (b) 1 ´ 13 = 1 = 0.1, which is terminate after Also, P lies on the line x - y + 2 = 0, 13 10 10 one decimal place. So, P satisfies it \\ 9k - 1 - 8k + 3 + 2 = 0 (c) 1 ´ 10 = 10 = 0.05917..., which is not 13 13 169 k +1 k +1 terminate. Þ 9k - 1 - 8k - 3 + 2k + 2 = 0 (d) 1 ´ 100 = 100 = 0.5917..., which is not 13 13 169 Þ 3k - 2 = 0 terminate. Þ k =2 34. (b) Let the side of the square be x cm. 3 36. (c) Let the side of square be a. Then, AF = AB - FB = (16 -x) cm A EB and GE = BE - BG = (8 - x) cm A HF 16 cm D Latest CBSE SAMPLE PAPER F DG C Area of shaded region G = Area of semi-circle B 8 cm E + (Area of half square Now, in DAFD and DDGE ÐF =ÐG = 90° [BGDFis a square] - Area of two quadrants) ÐADF = ÐDEG = 1 pæç a ö2 + 1 a2 -2 ´ 1 pæç a ö2 [corresponding angles, as FD||BE] ÷ ÷ 2 è2 ø 2 4 è2 ø So, by AA similarity criterion, DAFD ~ DDGE [Q Radius of semi-circle] = 1 pa2 + 1 a2 - 1 pa2 \\ AF = FD DG GE 8 28 = 1 a2= 1 (1[4Q)2a = 14cm] 22 = 98 cm2

54 CBSE Sample Paper Mathematics Standard Class X (Term I) 37. (d) We have, OA = OB = AB = 1 cm D r C r O a OB A A aB \\ DOAB is an equilateral triangle of side 1 cm. From figure, we have, AB2 + BC2 = AC2 \\Required area = 8 ´ Area of one segment Þ a2 + a2 = (2 r)2 = 8 é60° ´ p(1)2 - 3 ´ (1)2 ù Þ 2 a2 = 4r2 êë360 4 ú Þ a2 = 2 r2 û [Q q = 60°, r = 1 cm, side of triangle = 1 cm] Þ a= 2r ép 3 ù cm 2 Þ a = 2 ´ 100 = 50 2 cm = 8 ëê6 - 4 2p p ú û 38. (b) We have, 2 and 1 are the zeroes of the 40. (b) We have, 2 3x + y = 243 polynomial px2 + 5x + r. Þ 3x + y = 35 \\ Sum of zeroes = - Coefficient of x Þ x + y =5 …(i) Coefficient of x2 Again, 243x - y = 3 Þ 35( x - y) = 3 = -(5) p Þ 5(x - y) = 1 Þ x-y=1 Þ 2 + 1 = -5 …(ii) 2p 5 Þ 5 = -5 On adding Eqs. (i) and (ii), we get 2p 2x = 5 + 1 5 Þ p=-2 Þ 2 x = 26 and product of zeroes = Constant term 5 Coefficient of x2 Þ x = 13 =r 5 p On putting, x = 13 in Eq. (i), we get Latest CBSE SAMPLE PAPER Þ 2´ 1= r 5 2p y = 5 - x = 5 - 13 = 12 Þ 1= r 55 p \\ x = 13 Þ r=p 5 = -2 and y = 12 5 39. (c) Let the radius of the circle be r cm and side of the square be a cm. So, number of solution is one. Then, 2 pr = 100 41. (c) Initially, when t =0, h(t) = 48 ft Þ r =100 cm \\ h(0) = 48 2p Þ - 16(0)2 + 8(0) + k = 48 Þ k = 48

CBSE Sample Paper Mathematics Standard Class X (Term I) 55 42. (b) When Annie touches the pool, her height E (2, 1), F (1, 5), G (1, - 3), H (- 2, 4), h(t) = 0 ft. Let the time be t s. I (- 1, 1), J (- 2 , - 2 ), K (- 4, 1) \\ h(t) = 0 46. (a) Centroid of DEHJ Þ - 16t2 + 8t + k = 0 æ 2 -2 -2 ,1+ 4 -2 ö ç ÷ Þ - 16t2 + 8t + 48 = 0 [Q k = 48] = Þ 2 t2 -t - 6 = 0 [divide by - 8] è3 3ø Þ 2t2 - 4t + 3t - 6 = 0 = æ - 2 , 1÷ö ç è3 ø Þ 2 t(t - 2 ) + 3(t -2 ) = 0 47. (c) If P needs to be at equal distance from Þ (t - 2 ) (2 t + 3) = 0 A(3, 6) and G(1, - 3) such that A, P and G are Þ t =2 , - 3 collinear, then P will be the mid-point of AG. 2 Þ t =2 s \\ Coordinate of P = æ 3 + 1, 6 -3 ö = çæ2 , 3 ö ç ÷ ÷ è 2 2 ø è 2ø [Q t can’t be negative] 48. (a) Let the point of X-axis which is equidistant from I (-1,1) and E(2 ,1) be 43. (d) We have, - 1 and 2 are the zeroes of p(t). P (x,0). Then, \\ p(t) = l (t2 - (sum of zeroes) t PI = PE + product of zeroes) Þ (x + 1)2 + (0 - 1)2 = (x -2 )2 + (0 -1)2 = l (t2 -(- 1 + 2 )t + (- 1) ´ 2 ) = l (t2 - t - 2) Þ (x + 1)2 + 1 = (x - 2 )2 + 1 Þ (x + 1)2 = (x - 2 )2 Now, when t =0, height = 48 Þ x2 + 2x + 1 = x2 - 4x + 4 \\ p(0) = 48 Þ 6x = 3 Þ l(02 - 0 - 2 ) = 48 Þ l =- 24 Þ x=1 \\ p(t) = -24 (t2 - t - 2 ) 2 = -24t2 + 24t + 48 So, required point is æ 1 , 0 ÷ö. ç è2 ø 44. (c) We have, q(t) = b(t2 - (Sum of zeroes) t 49. (b) Let the coordinates of the position of the player Q be (x, y). Since distance of Q from + Product of zeroes) K(- 4, 1) is twice the distance from E(2 , 1) and = b(t2 - t - 6) K,Q and E are collinear. So Q divides the line segment KE in 2 : 1. Now, when t = 0, height = 48 2:1 \\ q(0) = 48 Þ b(02 - 0 - 6) = 48 K (–4, 1) Q (x, y) E (2, 1) Þ b=-8 \\Coordinates of Q Latest CBSE SAMPLE PAPER \\ q(t) = - 8t2 + 8t + 48 45. (a) We have, r(t) = - 12 t2 + (k - 3)t + 48 = æ 2 ´ 2 + 1´(-4) , 2 ´1 + 1 ´ 1 ö çç 2 +1 2 + 1 ÷÷ è ø Since, zeroes of the polynomial r(t) are negative of each of other. = (0, 1) \\ Sum of zeroes = 0 50. (d) Let the point on Y-axis which is equidistant from B (4, 3) and C (4, - 1) be Þ - (Coefficient of t) = 0 Coefficient of t2 T (0,y). Then Þ -(k - 3) = 0 -12 TB = TC Þ k -3 =0 Þ (4 -0)2 + (3 -y)2 = (4 -0)2 + (- 1 - y)2 Þ k =3 Þ 16 + 9 - 6y + y2 = 16 + 1 + 2 y + y2 Solutions (46-50) Þ - 8y = - 8 From the given figure, we have coordinates Þ y =1 of A (3, 6), B (4, 3), C (4, - 1), D (3, - 4), \\ Required point is (0, 1).







CBSE Sample Paper Mathematics Standard Class X (Term I) 59 SAMPLE PAPER 1 MATHEMATICS (Standard) A Highly Simulated Practice Questions Paper for CBSE Class X (Term I) Examination Instructions 1. The question paper contains three parts A, B and C. 2. Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 3. Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 4. Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions. 5. There is no negative marking. Roll No. Maximum Marks : 40 Time allowed : 90 minutes Section A Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 1. The product of a non-zero rational and an irrational number is (a) always rational (b) rational or irrational (c) always irrational (d) zero 2. 2 times the distance between (0, 5) and (-5, 0) is ……… (a) 10 (b) 8 (c) 17 (d) 14 3. The circumference of a circle of diameter 21 cm is (a) 66 cm (b) 69 cm (c) 63 cm (d) 68 cm 4. If a, b are the zeroes of the polynomial f (x) = x 2 - p(x + 1) - c such that (a + 1)(b + 1) = 0, then c is equal to (a) 1 (b) 0 (c) -1 (d) 2 5. Which of the equation has solution as x = 2, y = 1? SAMPLE PAPER 1 (a) 2 x + 7y = 11 (b) 4x -2 y = 5 (c) x -3y = 5 (d) 3x - 4y = 8 6. 1192 - 1112 is (a) prime number (b) composite number (c) an odd prime number (d) an odd composite number 7. The probability that a non-leap year has 53 Sunday’s, is (a) 2 (b) 5 (c) 6 (d) 1 7 7 7 7

60 CBSE Sample Paper Mathematics Standard Class X (Term I) 8. If 3 is one zero of the polynomial f (x) = 9x 2 - 3(a - 1)x + 5, then the value of a is (a) 81 (b) 95 (c) 40 (d) None of these 5 9 9 9. In which quadrant does the point (-3, 5) lie? (a) I (b) II (c) III (d) IV 10. The sum of two numbers is 137 and their difference is 43. The situation can be algebrically represented as (a) x - y = 137, x + y = 180 (b) 2(x + y) = 137, 2(x - y) = 43 (c) x + y = 137,x - y = 43 (d) x + y = 43, x - y = 137 11. If the sum of the circumference and the radius of a circle is 51 cm, then the radius of the circle (in cm) is (a) 154 (b) 44 (c) 14 (d) 7 12. If DABC ~ DDEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of DDEF is 25 cm, then the perimeter of DABC is (a) 36 cm (b) 30 cm (c) 34 cm (d) 35 cm 13. If a = 23 ´ 3, b = 2 ´ 3 ´ 5, c = 3n ´ 5 and LCM (a, b, c) = 23 ´ 32 ´ 5, then n = (a) 1 (b) 2 (c) 3 (d) 4 14. A paper is in the form of a rectangle ABCD in which AB = 18 cm and BC = 14 cm. A semi circular portion with BC as diameter is cut off. Find the area of the remaining paper. (a) 175 cm2 (b) 165 cm2 (c) 145 cm2 (d) None of these 15. If the zeroes of the quadratic polynomial x 2 + (m + 1)x + n are 4 and 5, then (a) m = -20, n = -80 (b) m = -20, n = 80 (c) m = -10, n = 20 (d) m = 20, n = -10 16. If Qçæ a , - 4 ö is the mid-point of the segment joining the points P(6, - 5) and R(2, - 3), ÷ è3 ø then the value of ‘a’ is (a) 12 (b) - 6 (c) - 12 (d) - 4 17. If am ¹ bl then the pair of equations ax + by = c and lx + my = n (a) has a unique solution (b) has no solution (c) has infinitely many solutions (d) may or may not have a solution 18. One card is drawn at random from a well-shuffled deck of 52 cards. What is the SAMPLE PAPER 1 probability of getting a face card? (a) 1 (b) 3 (c) 3 (d) 4 26 26 13 13 19. The coordinates of the point which is reflection of point (-3, 5) in X-axis are (a) (3, 5) (b) (3, -5) (c) (-3, - 5) (d) (-3, 5) 20. 2.13113111311113…… is (b) a rational number (d) None of these (a) an integer (c) an irrational number

CBSE Sample Paper Mathematics Standard Class X (Term I) 61 Section B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. The diameter of a wheel is 1.26 m. How far will it travel in 400 revolutions? (a) 2670 m (b) 2880 m (c) 1980 m (d) 1584 m 22. If p - q + r = 0 then a zero of the polynomial px 2 + qx + r is (a) 0 (b) 1 (c) -1 (d) None of these 23. For the pair of linear equations 47x + 31y = 18 and 31x + 47y = 60 the value of x + y is (a) 1 (b) 0 (c) -4 (d) 7 24. If LCM = 350, product of two numbers is 25 ´ 70, then their HCF = 5. (a) 12 (b) 15 (c) 5 (d) 10 25. The distance of the point (2, 11) from the X-axis is (a) 11 units (b) 2 units (c) 13 units (d) 12 units 26. A bag contains 3 red and 7 black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball? (a) 3 (b) 7 10 10 (c) 1 (d) None of these 10 27. The area of a quadrant of a circle whose circumference is 44 cm is (a) 24 cm 2 (b) 28 cm 2 (c) 35.5 cm 2 (d) 38.5 cm 2 28. If one zero of the polynomial p(x) = (k 2 + 9)x 2 + 9x + 6k is the reciprocal of the other zero, then k is (a) - 2 (b) 3 (c) 2 (d) - 3 29. Two vertices of a triangle are (- 3, 5) and (7, - 4). If its centroid is (2, - 1), then the third vertex is (a) (2, 4) (b) (- 2 , 4) (c) (2 , - 4) (d) (- 2, - 4) 30. A number x is chosen at random from the numbers - 3, - 2, - 1, 0, 1, 2, 3 the probability that|x|< 3 is (b) 2 (c) 3 (d) 1 SAMPLE PAPER 1 (a) 5 7 7 7 7 31. There are deer and peacock in the zoo. By counting heads, they are 52, the number of their legs is 176. Number of peacock are (a) 12 (b) 16 (c) 20 (d) 36 32. Which of the following rational numbers have terminating decimal? (i) 16 (ii) 5 (iii) 2 (iv) 7 50 18 21 250 (a) i and ii (b) ii and iii (c) i and iii (d) i and iv

62 CBSE Sample Paper Mathematics Standard Class X (Term I) 33. The minute hand of a clock is 6 cm long. Then, the area of the face of the clock described by the minute hand in 35 minutes is (a) 265 cm2 (b) 266 cm2 (c) 264 cm2 (d) None of these 34. If the sum of the zeroes of the quadratic polynomial kx 2 + 2x + 3k is equal to their product, then k is equal to (c) - 1 (b) - 2 3 (a) 1 (d) 2 33 3 35. Find the value of a so that the point (3, a) lies on the line represented by 2x - 3y = 5. (a) 1 (b) 1 (c) 1 (d) 1 3 5 4 6 36. The graphical representation of x - 2y + 4 = 0 and 3x + 4y + 2 = 0 will be (a) intersecting (b) parallel (c) coincident (d) None of these 37. In DABC, ÐB = 90° and BD ^ AC. If AC = 9 cm and AD = 3 cm, then BD is equal to (a) 2 2 cm (b) 3 2 cm (c) 2 3 cm (d) 3 3 cm 38. If the radius of a circle is increased by 25% then its circumference will increases by (a) 25% (b) 50% (c) 75% (d) 100% 39. The value of 4 sin 2 60° + 3 tan 2 30° - 8 sin 45° cos 45° is (a) 0 (b) 1 (c) 2 (d) 5 40. The smallest rational number by which 1 should be multiplied so that its decimal 3 expansion terminates after one place of decimal, is (a) 3 (b) 1 (c) 3 (d) 3 10 10 100 Section C Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted. 41-45 are based on Case Study-1. Case Study 1 Two hotels are at the ground level on either side of a mountain. On moving a certain distance towards the top of the mountain two huts are situated as shown in the figure. The ratio between the distance from hotel 2 to hut Band that of hut B to mountain top is 3 : 7. Mountain top SAMPLE PAPER 1 10 miles Hut-B Ground level Hut-A Hotel-1 Hotel-2

CBSE Sample Paper Mathematics Standard Class X (Term I) 63 41. What is the ratio of the perimeters of the triangle formed by both hotels and mountain top to the triangle formed by both huts and mountain top? (a) 5 : 2 (b) 10 : 7 (c) 7 : 3 (d) 3 : 10 42. The distance betweeen the hotel 1 and hut A is (a) 2.5 miles (b) 29 miles (c) 4.29 miles (d) 1.5 miles 43. If the horizontal distance between the hut A and hut B is 8 miles, then the distance between the two hotels is (a) 2.4 miles (b) 11.43 miles (c) 9 miles (d) 7 miles 44. If the distance from mountain top to hut A is 5 miles more than that of distance from hotel 2 to mountain top, then what is the distance between hut B and mountain top (a) 3.5 miles (b) 6 miles (c) 5.5 miles (d) 4 miles 45. Which property of geometry will be used to find the distance between hut B and mountain top? (a) Congruent of triangles (b) Thales theorem (c) pythagoras theorem (d) None of these 46-50 are based on Case Study-2. Case Study 2 Children were playing a game by making some right angled triangles on the plane sheet of paper. They took a right angled triangle with two of its sides AC = 25 cm, BC = 20 cm and ÐABC = 90°. With the help of right angled triangle, solve the following questions. A 25 cm 90° C 20 cm B 46. Using the above data, the value of sin A is (a) 12 (b) 3 (c) 5 (d) 4 13 5 13 5 47. Using the above data, the value of sin C is (d) 13 5 (a) 12 (b) 13 (c) 3 13 12 5 (d) 3 5 48. Using the above data, the value of tan C is (d) 12 (a) 12 (b) 3 (c) 4 13 SAMPLE PAPER 1 5 4 5 (d) 9 49. Using the above data, the value of cos A is 13 (a) 3 (b) 3 (c) 4 5 4 5 50. Using the above data, the value of tan A + sin A is tan A - sin A (a) 9 (b) 4 (c) 12 4 9

OMR SHEET SP 1 Roll No. Sub Code. Student Name Instructions Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet. Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read by the software.  Correct Incorrect Incorrect Incorrect Do not write anything on the OMR Sheet. Multiple markings are invalid. 1 18 35 2 19 36 3 20 37 4 21 38 5 22 39 6 23 40 7 24 41 8 25 42 9 26 43 10 27 44 11 28 45 12 29 46 13 30 47 14 31 48 15 32 49 16 33 50 17 34 Check Your Performance Score Percentage = Total Correct Questions 100 Total Questions: × Total Correct Questions: Total Questions Less than 60% > Average (Revise the concepts again) If Your Score is Greater than 60% but less than 75% > Good (Do more practice) Above 75% > Excellent (Keep it on)

CBSE Sample Paper Mathematics Standard Class X (Term I) 65 Answers 1. (c) 2. (a) 3. (a) 4. (a) 5. (a) 6. (b) 7. (d) 8. (b) 9. (b) 10. (c) 11. (d) 12. (d) 13. (b) 14. (a) 15. (c) 16. (a) 17. (a) 18. (c) 19. (c) 20. (c) 21. (d) 22. (c) 23. (a) 24. (c) 25. (a) 26. (b) 27. (d) 28. (b) 29. (c) 30. (a) 31. (b) 32. (d) 33. (d) 34. (b) 35. (a) 36. (a) 37. (b) 38. (a) 39. (a) 40. (a) 41. (b) 42. (c) 43. (b) 44. (a) 45. (b) 46. (d) 47. (c) 48. (b) 49. (a) 50. (b) SOLUTIONS 1. Product of a non-zero rational and an irrational Here, the expression has more than two factor. number is always irrational. Hence, it is a composite number. e.g. 3 ´ 2 = 3 2 (irrational) 7. A non-leap year has 365 days. 44 In 365 days, there are 52 weeks and 1 day. In 52 weeks, the number of Sundays will be 52. 2. Given points are (0, 5) and (- 5, 0). 1 remaining day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Let the distance between the points be d. Saturday. By distance formula, We can have any one of these days out of 7 days. d = (-5 - 0)2 + (0 - 5)2 Hence, out of these 7 outcomes, the favourable outcome is 1 d = 25 + 25 = 50 = 5 2 units Therefore, probability of getting 53 Sundays in a non-leap year = 1 \\ 2 d = 5 2 ´ 2 = 10 units 7 3. Radius of the circle, r = 21 cm 8. Given, 3 is one zero of the polynomial 2 f (x) = 9x2 -3 (a - 1)x + 5 Now, circumference of circle = 2 pr \\ f (3) = 0 = 2 ´ 22 ´ 21 = 66 cm Þ 0 = 9(3)2 - 3(a - 1)3 + 5 72 Þ 0 = 81 - 9a + 9 + 5 4. Given, a, b are the zeroes of the polynomial Þ 9a = 95 \\ a = 95 f (x) = x2 - p(x + 1) - c 9 = x2 - px - p - c = x2 - px - ( p + c) 9. Given point is (-3, 5) \\ a + b = - Coefficient of x = - -p = p Coefficient of x2 1 Y and ab = Constant term = -( p + c) = -(p + c) Coefficient of x2 1 Now, it is given that (a + 1)(b + 1) = 0 Þ ab + a + b + 1 = 0 Þ -( p + c) + p + 1 = 0 II I Þ -c + 1 = 0 (–, +) (+, +) \\ c=1 X¢ X 5. Put x = 2 , y = 1 in given equations, we get III IV (–, –) (+, –) (a) LHS = 2(2 ) + 7(1) = 11 = RHS Y¢ SAMPLE PAPER 1 (b) LHS = 4(2 ) -2(1) = 6 ¹ RHS (c) LHS = 2 -3(1) = -1 ¹ RHS The point has negative x-coordinate and (d) LHS = 3(2 ) - 4(1) = 2 ¹ RHS positive y-coordinate. Only 2 x + 7y = 11 satisfies the given values. Hence, it lies in II quadrant. Hence, 2 x + 7y = 11 is the required equation. 10. Let the two numbers be x and y, where x > y 6. We know that, Given, sum of numbers = 137 a2 - b2 = (a + b)(a - b) \\ x + y = 137 And difference of numbers = 43 \\ 1192 - 1112 = (119 + 111)(119 - 111) \\ x - y = 43 = 230 ´ 8 = 2 4 ´ 5 ´ 23

66 CBSE Sample Paper Mathematics Standard Class X (Term I) 11. Let r be the radius of the circle. Then, Þ 4 + 5 =-m-1 circumference = 2 pr cm Þ m = - 10 According to the question, and product of zeroes = Coefficient term = n Coefficient of x2 1 2 pr + r = 51 Þ r æ 44 + 1ö÷ = 51 Þ 4´ 5=n ç è7 ø Þ n = 20 44 + 7 Þ r æ ö = 51 æ x1 + x2 y1 + y2 ö ç ÷ 16. Mid-point = ç , ÷ è7ø è2 2ø Þ r = 7 ´ 51 = 7 cm 51 = æ 6 + 2 , - 5 - 3 ö ç ÷ è2 2 ø 12. Given, DABC ~ DDEF, AB = 9.1, DE = 6.5 cm æ a , 4 ö (4, 4) Þ ç - ÷ = - and perimeter of DDEF = 25 cm è3 ø \\ Perimeter of DABC = AB \\ a =4 Perimeter of DDEF DE 3 Perimeter of DABC = 9.1 Þ a = 12 25 6.5 17. Given equations can be written as Þ Perimeter of DABC = 91´ 25 = 35 cm 65 ax + by - c = 0 13. Given, a = 2 3 ´ 3, and lx + my - n = 0 b =2´3´ 5 As, am ¹ bl and c = 3n ´ 5 a¹ b \\ LCM (a, b, c) = 2 3 ´ 3n ´ 5 lm On comparing with the given LCM, we get Condition for the pair of equations has unique solution, n =2 a¹ b 14. Area of rectangle ABCD = AB´ BC lm = 18 ´ 14 = 252 cm2 Hence, ax + by = c and lx + my = n has a unique solution. DC 18. In 52 cards, the king, queen and Jack are three face cards in each suit. 14 cm So, total face cards = 12 \\Probability of getting face card = 12 = 3 52 13 AB 19. When the point is reflected in X-axis, its 18 cm x-coordinate remains same and y-coordinate changes by negative sign. Radius of semi-circle (r) = 1 BC = 1 ´ 14 = 7cm 22 Hence, the point will be (-3, - 5). SAMPLE PAPER 1 Area of semi-circle = 1 pr2 20. The given decimal is non-terminating and 2 non-repeating decimal. = 1 ´ 22 ´ 72 = 77 cm 2 27 Hence, it must be an irrational number. 21. Radius of the wheel, r = 1.26 = 0.63 m Area of remaining portion = Area of rectangle 2 - Area of semi-circle = 252 - 77 = 175 cm 2 [Q diameter = 2 (radius)] 15. Given 4 and 5 are the zeroes of the quadratic Distance travelled in one revolution is equal to polynomial x2 + (m + 1)x + n. the perimeter of the wheel. \\Distance = 2 pr = 2 ´ 22 ´ 0.63 = 3.96 m \\Sum of zeroes = - Coefficient of x Coefficient of x2 7 = - (m + 1) \\Distance travelled in 400 revolutions 1 = 400 ´ 3.96 = 1584 m

CBSE Sample Paper Mathematics Standard Class X (Term I) 67 22. Let p(x) = px2 + qx + r be the polynomial 28. Given, p(x) = (k 2 + 9)x2 + 9x + 6k \\ p(-1) = p(-1)2 + q(-1) + r Let one zero be a then other zero is 1 . p(-1) = p - q + r a According to the question, Product of zeroes = Constant term p - q + r = 0, so clearly -1 is the zero of the Coefficient of x2 polynomial. Þ a × 1 = 6k 23. Given, 47x + 31y = 18 …(i) a k2 + 9 and 31x + 47y = 60 …(ii) Þ k2 + 9 = 6k On adding Eqs. (i) and (ii), we get Þ k2 -6k + 9 = 0 Þ (k - 3)2 = 0 Þ k = 3 78x + 78y = 78 On dividing both sides by 78, we get 29. Let the coordinates of the third vertex be (x, y), x + y =1 then centroid of triangle is 24. We know that æ x - 3 + 7 , y + 5 - 4 ö = (2 , - 1) LCM × HCF = Product of two numbers ç ÷ Given LCM = 350, è3 3ø Product of two numbers = 25 ´ 70 \\ 350 ´ HCF = 25 ´ 70 Þ æ x + 4 , y + 1 ö = (2 , - 1) Þ HCF = 25 ´ 70 = 5 ç ÷ 350 è3 3ø Þ x + 4 = 2, y + 1 = - 1 33 Þ x + 4 = 6, y + 1 = - 3 25. Plot the point (2, 11) in coordinate axes. Þ x =2,y = - 4 Y \\Coordinates of third vertex is (2 , - 4). 2 units (2, 11) 30. Total numbers = 7 11 units Number x such that |x |< 3 are - 2 , - 1, 0, 1, 2. OX \\Total numbers of x such that |x |< 3 = 5 \\Probability = 5 7 From the above figure, the required distance is 31. There are deer and peacocks in a zoo. By 11 units. counting heads they are 52. 26. Total number of balls in bag = 3 + 7 = 10 The number of their legs is 176. Total number of black balls in bag = 7 Let there be x deer and y peacock. Probability of getting a black ball Then, 4x + 2 y = 176 … (i) = Total number of black balls in bag Total number of balls in bag and x + y = 52 … (ii) =7 Multiply by 2 in Eq. (ii), we get 10 2 x + 2 y = 104 …(iii) 27. Given, circumference of circle = 25 cm Subtract Eq. (iii) from Eq. (i), we get We know that circumference of circle = 2 pr Þ 44 = 2 ´ 22 ´ r Þ 44 = 44 ´ r 4x + 2 y = 176 77 2 x + 2 y = 104 \\ r = 7 cm -- - Area of a quadrant of a circle 2 x = 72 SAMPLE PAPER 1 = 1 ´ p´ r´ r 4 Þ x = 36 = 1 ´ 22 ´ 7 ´ 7 \\ y = 52 - 36 = 16 [from Eq. (ii)] 47 32. (i) 16 = 24 (ii) 5 = 5 = 11 ´ 7 = 38.5 cm2 50 ´ 52 18 ´ 32 2 2 2 (iii) 2 = 2 (iv) 7 = 7 21 3 ´ 7 250 53 ´ 2 Only i and iv have denominator in the form of 2 m ´ 5n, hence i and iv have terminating decimal.

68 CBSE Sample Paper Mathematics Standard Class X (Term I) 33. Let r be the radius of the clock. 37. Given, AC = 9 cm, AD = 3 cm Angle described by the minute hand in CD = AC - AD = 9 - 3 = 6 cm 60 minutes = 360° B Angle described by the minute hand in 35 minutes = æ 360 ´ 35 ö° = 210° ç ÷ è 60 ø Required area = Area of sector with central angle of 210° 3 cm æ q pr2 ö 2 AD C ç ÷ 9 cm = ´ cm è 360° ø In DABC and DADB = 210° ´ p´ 6 ´ 6 ÐBAC = ÐBAD 360° ÐABC = ÐADB [each 90° angle] \\ DABC ~ DADB = 210° ´ 22 ´ 6 ´ 6 360° 7 [by AA similarity criterion] … (i) In DABC and DBDC, = 66 cm2 ÐABC = ÐBDC [each 90° angle] 34. Given polynomial = kx2 + 2 x + 3k ÐACB = ÐBCD If ax2 + bx + c is a polynomial then its sum of \\ DABC ~ DBDC roots is given by - b and product of zeroes is [by AA similarity criterion] … (ii) a From Eqs. (i) and (ii) DADB~ DBDC given by c. \\ BD = AD a CD BD \\ Sum of zeroes of given polynomial = - 2 BD2 = AD× CD = 3 ´ 6 = 18 k BD = 3 2 cm Product of zeroes = 3k = 3 38. Let original radius be R cm. Then original k circumference = (2 pR) cm It is given that, sum of zeroes New radius = 125 % of R cm = product of zeroes = æ 125 ´ R ö cm = 5R cm ç ÷ Þ -2 =3 è 100 ø 4 k Þ k = -2 3 35. Given point (3, a) lies on the line 2 x - 3y = 5 New circumference = æ 2 p ´ 5 R ö cm ç ÷ è 4ø Q (3, a) satisfies the given equation. = 5pR cm \\ 2´3 -3´ a = 5 2 Þ 6 -3a = 5 Þ 1 = 3a Increase in circumference Þ a=1 = æ 5pR - 2 pR ö = pR cm 3 ç ÷ è2 ø 2 Percentage increase = æ pR ´ 1 ´ 100 ö÷% ç 36. On comparing the given equation with è 2 2 pR ø standard equation, we get = 25% a1 = 1, b1 = -2, c1 = 4 SAMPLE PAPER 1 39. Since, sin60° = 3 , tan30° = 1 and a2 = 3, b2 = 4, c2 = 2 23 For unique solution, and cos 45° = sin 45° = 1 a1 ¹ b1 2 a2 b2 \\ 4 sin2 60° + 3 tan2 30° - 8 sin 45°cos 45° Here 1 ¹ -2 æ 3 ö 2 æ 1 ö2 1´ 1 34 = 4 ´ çç 2 ÷÷ ç ÷ 2 2 ø + 3 ´ -8´ è è 3ø So, graphical representation of given lines are = 4´3 + 3 - 8 =3 +1-4 =0 intersecting. 4 32

CBSE Sample Paper Mathematics Standard Class X (Term I) 69 40. Now, 1 ´ 3 = 1 = 0.1 45. Thales theorem will be used to find the 3 10 10 distance between hut B and mountain top. \\ 1 should be multiplied by 3/10, so that its Solutions (46-50) 3 46. In DABC, ÐB = 90° decimal expansion terminates after one place of decimal. A Solutions (41-45) 41. Let DABC be the triangle formed by both hotels 25 cm and mountain top. DCDE is the triangle formed by both huts and mountain top. 90° Clearly DE|| AB and So C 20 cm B DABC ~ DDEC [By AAA similarity criterion] sin A = BC = Perpendicular side AC Hypotenuse C 7 = 20 = 4 25 5 D E 47. Use Pythagoras theorem A ( AB)2 + ( BC)2 = ( AC)2 3 AB = ( AC)2 -( BC)2 B = (25)2 -(20)2 Perimeter of DABC = BC = CE + BE Perimeter of DDEF EC EC = 625 - 400 = 225 = 15 = 7 + 3 = 10 \\ sin C = AB = 12 = 3 77 AC 25 4 48. tan C = AB 42. Since, DE||AB, therefore, é AD = CD´ EBù BC CD = CE êë CE úû Þ tan C = 15 = 3 AD EB 20 4 49. cos A = AB = 15 = 3 Þ 10 = 7 AC 25 5 AD 3 Þ AD = 10 ´ 3 7 = 4.29 miles 43. Since, DABC ~ DDEC 50. tan A + sin A = BC + BC AB AC BC = AB tan A - sin A BC - BC EC DE AB AC [Q Corresponding sides of similar triangles are proportional] 20 + 20 = 15 25 Þ 10 = AB 78 20 - 20 15 25 Þ AB = 80 = 11.43 miles 7 = 25 + 15 SAMPLE PAPER 1 25 -15 44. Given, DC = 5 + BC = 40 = 4 Clearly, BC = 10 - 5 = 5 miles 10 Now, CE = 7 ´ BC = 7 ´ 5 = 3.5 miles 10 10

70 CBSE Sample Paper Mathematics Standard Class X (Term I) SAMPLE PAPER 2 MATHEMATICS (Standard) A Highly Simulated Practice Questions Paper for CBSE Class X (Term I) Examination Instructions 1. The question paper contains three parts A, B and C. 2. Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 3. Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 4. Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions. 5. There is no negative marking. Roll No. Maximum Marks : 40 Time allowed : 90 minutes Section A Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 1. The smallest number by which 27 should be multiplied so as to get a rational number is (a) 27 (b) 3 3 (c) 3 (d) 3 2. A quadratic polynomial can have at most ______ zeroes. (a) 0 (b) 1 (c) 2 (d) infinite 3. The point of intersection of the coordinate axes is (a) X-axis (b) Y-axis (c) origin (d) (1, 2) 4. In a DABC, it is given that AB = 3 cm, AC = 2 cm and AD is the bisector of ÐA. Then, BD : DC = SAMPLE PAPER 2 A Ö3 cm 2 cm (a) 3 : 4 B DC (d) 3 : 2 (b) 9 : 16 (c) 4 : 3

CBSE Sample Paper Mathematics Standard Class X (Term I) 71 5. æ x2 y2 ö If x a secq cosf, y= b secq sin f and z ctan q, then ç a2 b2 ÷ is equal to = = ç + ÷ è ø (a) æ + z2 ö (b) æ - z2 ö æ z2 ö (d) z2 çç1 c2 ÷÷ çç1 c2 ÷÷ (c) çç c2 -1÷÷ c2 è ø è ø è ø 6. A card is selected from a deck of 52 cards. The probability of its being a black face card is (a) 3 (b) 3 (c) 2 (d) 1 26 13 13 2 7. If A = 2n + 13 , B = n + 7, where n is a natural number then HCF of A and B is (a) 2 (b) 1 (c) 3 (d) 4 8. The sum and product of zeroes of a quadratic polynomial are respectively 1 and - 2. 3 Then the corresponding quadratic polynomial is (a) 4x2 + x - 4 (b) x 2 - 4x - 4 (c) 4x2 - 4x - 1 (d) 3x2 - x - 6 9. In the given figure P(5, - 3) and Q(3, y) are the points of trisection of the line segment joining A(4, 7) and B(1, - 5). Then y equals A(4, 7) P(5, –3) Q(3, y) B(1, – 5) (a) 2 (b) 4 (c) -4 (d) - 5 2 10. In the given figure, PQ||BC, find AQ. C B 4 cm 6 cm P Q 6 cm A (a) 3.5 cm (b) 4.5 cm (c) 9 cm (d) 9.5 cm 11. If x cosq = 1 and tan q = y, then x 2 - y2 is (a) 2 (b) - 1 (c) 3 (d) 1 SAMPLE PAPER 2 12. A girl calculates the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, then the total number of tickets she bought is (a) 40 (b) 240 (c) 480 (d) 750 13. From the following rational number, which decimal expansion is terminating, is (a) 2 (b) 11 (c) 17 (d) 6 15 160 60 35

72 CBSE Sample Paper Mathematics Standard Class X (Term I) 14. If zeroes a and b of a polynomial x 2 - 7x + k are such that a - b = 1, then the value of k is (a) 21 (b) 12 (c) 9 (d) 8 15. If the point P(x, y) is a equidistant from L(5, 1) and M(-1, 5), then the relation between x and y is (a) 3x = 2y (b) x = y (c) 2x = 3y (d) 3x = 6y 16. In DABC it is given that, AB = BD if ÐB = 60° and ÐC = 60°, then ÐBAD is equal to AC DC A BD C (a) 30° (b) 40° (c) 45° (d) 50° 17. If sin A = cos A, 0°< A< 90°, then A is equal to (a) 30° (b) 45° (c) 60° (d) 90° 18. The probability that it will rain tomorrow is 0.3. What is the probability that it will not rain tomorrow? (a) 0.3 (b) 0.2 (c) 0.7 (d) 0.07 19. 13 is a 1250 (a) terminating decimal fraction (b) non-terminating decimal fraction (c) upto 2 decimal fraction (d) None of these 20. The value of sin 60°+ cot 45° - cosec 30° is sec 60° - cos 30° + tan 45° (a) 4 3 -9 (b) 4 3 + 9 (c) 9 3 - 4 (d) 9 3 + 4 33 33 33 33 Section B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. The product of the HCF and LCM of the smallest prime number and the smallest composite number is SAMPLE PAPER 2 (a) 2 (b) 4 (c) 6 (d) 8 22. If a and b are zeroes of the quadratic polynomial 2x 2 + kx + 4 and if a 2 + b2 = 8, then k is equal to (a) ± 4 3 (b) ± 3 3 (c) ± 2 3 (d) ± 3 23. If 2x + 3y = 7 and (a + b)x + (2a - b)y = 21 has infinite solutions, then (a) a = 1, b = 5 (b) a = 5, b = 1 (c) a = - 1, b = 5 (d) None of these

CBSE Sample Paper Mathematics Standard Class X (Term I) 73 24. The coordinates of one of the points of trisection of the line segment joining the points P(7, - 2) and Q(1, - 5) are (a) æ - 13 , - 1 ö (b) (3, 5) (c) æ - 13 , -3 ö (d) (-5, -3) ç ÷ ç ÷ è 3 3ø è3 ø 25. A piece of wire 20 cm is bent into the form of an arc of a circle subtending an angle of 60° at its centre, then the radius of the circle will be (in cm) (a) 30 (b) 40 (c) 50 (d) 60 p p p p 26. In an isosceles triangle PQR, if PR = QR and PQ2 = 2PR 2, then ÐR is (a) acute angle (b) obtuse angle (c) right angle (d) None of these 27 If xcosec2 30° sec2 45° = tan 2 60° - tan 2 30°, then x is equal to 8 cos2 45° sin 2 60° (a) 1 (b) -1 (c) 2 (d) 0 28. Two dice are thrown simultaneously. Then the number of possible outcomes for getting the sum from 3 to 10 is (a) 32 (b) 30 (c) 34 (d) 38 29. The sum of powers of prime factors of 196 is (a) 1 (b) 2 (c) 4 (d) 6 30. If a and b are the zeroes of the polynomial p(x) = 4x 2 + 3x + 7, then the value of a + b is ba (a) 47 - 47 (c) - 28 (d) 28 28 (b) 47 47 28 31. The distance between the points P(2, - 3) and Q(10, y) is 10 then the value of y is (a) 3, - 9 (b) 2, 7 (c) 1, 3 (d) 3, 9 32. In the given figure, if ABCD is a rhombus, then the value of x is DC x x–1 O 45 AB (a) 3 (b) 4 (c) 5 (d) 6 33. If sin q - cosq = 0, then sin 4 q + cos4 q is equal to SAMPLE PAPER 2 (a) 1 (b) 3 (c) 1 (d) 1 4 2 4 34. A letter is chosen at random from the English alphabets Find the probability that the letter chosen succeeds V. (a) 2 (b) 5 (c) 1 (d) 1 13 26 26 2

74 CBSE Sample Paper Mathematics Standard Class X (Term I) 35. Which of the following rational numbers have terminating decimal? (i) 16 (ii) 5 (iii) 2 (iv) 7 25 18 21 250 (a) (i) and (ii) (b) (ii) and (iii) (c) (i) and (iii) (d) (i) and (iv) 36. A quadratic polynomial whose one zero is 5 and product of the zeroes is 0, is (a) x2 - 5 (b) x 2 - 5x (c) 5x2 + 1 (d) x2 + 5x 37. If Pçæ a , 4 ö is the mid-point of the line segment joining the points Q(- 6, 5) and R(- 2, 3), è 3 ÷ ø then the value of a is (a) - 4 (b) - 12 (c) 12 (d) - 6 38. The sides of a triangle are 30, 70 and 80 units. If an altitude is droped upon the side of length 80 units, the larger segment cut off on this side is (a) 62 units (b) 63 units (c) 64 units (d) 65 units 39. If x sin q = 1 and cot q = y, then which of the following is correct? (a) x2 + y2 = 1 (b) x 2 - y 2 = 1 (c) y 2 - x 2 = 1 (d) None of these 40. There are five cards in which the numbers are written as nine, ten, jack, queen and king of hearts. These cards are well shuffled with their face downwards, one card is then picked up at random. The probability that the drawn card is a king, is (a) 1 (b) 2 (c) 3 (d) 4 5 5 5 5 Section C Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted. 41-45 are based on Case Study-1. Case Study 1 A earing is a small piece of jewellery which has a hook/pin at the back side so that it can be fastened on ears. Designs of some earing are shown below. Observe them carefully. SAMPLE PAPER 2 ABC Design A Earing A is made with platinum wire in the form of a circle with diameter 28 mm. The wire used for making 4 diameters which divide the circle into 8 equal parts. Desion B Earing B is made two colours platinum and silver. Outer parts is made with platinum The circumference of silver part is 88 mm and the platinum part is 7 mm wide everywhere. Observe the above designs and answer the following questions.

CBSE Sample Paper Mathematics Standard Class X (Term I) 75 Refer to Design A 41. The total length of platinum wire required is (a) 180 mm (b) 200 mm (c) 250 mm (d) 280 mm 42. The area of each sector of earing is (a) 44 mm2 (b) 52 mm2 (c) 77 mm2 (d) 68 mm2 Refer to Design B 43. The circumference of outer part platinum is (a) 48.49 mm (b) 82.20 mm (c) 72.50 mm (d) 132 mm 44. The difference of areas of platinum and silver parts is (a) 245 p mm2 (b) 44 p mm2 (c) 147 p mm2 (d) 64 p mm2 45. A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 168 p mm? (a) 2 (b) 3 (c) 4 (d) 5 46-50 are based on Case Study-2. Case Study 2 Palak went to a mall with her mother and enjoy rides on the giant wheel and play hoopla (a game in which you throw a ring on the items kept in stall and if the ring covers any object completely you get it). The number of times she played hoopla is half the number of times she rides the giant wheel. If each ride costs ` 3 and a game of hoopla costs ` 4 and she spent ` 20 in the fair. Based on the given information, give the answer of the following questions 46. The representation of given statement algebraically is (a) x - 2 y = 0 and 3x + 4y = 20 (b) x + 2 y = 0 and 3x - 4y = 20 (c) x - 2 y = 0 and 4x + 3y = 20 (d) None of these 47. Graphically, if the pair of equations intersect at one point, then the pair of equations is (a) consistent (b) Inconsistent (c) Consistent or inconsistent (d) None of these 48. The intersection point of two lines is SAMPLE PAPER 2 (a) (-4, -2) (b) (4, 3) (c) (2, 4) (d) (4, 2) 49. Intersection points of the line x - 2y = 0 on X and Y-axes are (a) (2, 0), (0, 1) (b) (1, 0), (0, 2) (c) (0, 0) (d) None of these 50. Intersection points of the line 3x + 4y = 20 on X and Y-axes are (a) æ 20 , 0 ö÷, (0, 5) (b) (2, 0), (0, 1) c) (5, 0), æç0, 20 ö (d) None of these ç ÷ è3 ø è 3ø

OMR SHEET SP 2 Roll No. Sub Code. Student Name Instructions Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet. Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read by the software.  Correct Incorrect Incorrect Incorrect Do not write anything on the OMR Sheet. Multiple markings are invalid. 1 18 35 2 19 36 3 20 37 4 21 38 5 22 39 6 23 40 7 24 41 8 25 42 9 26 43 10 27 44 11 28 45 12 29 46 13 30 47 14 31 48 15 32 49 16 33 50 17 34 Check Your Performance Score Percentage = Total Correct Questions 100 Total Questions: × Total Correct Questions: Total Questions Less than 60% > Average (Revise the concepts again) If Your Score is Greater than 60% but less than 75% > Good (Do more practice) Above 75% > Excellent (Keep it on)

CBSE Sample Paper Mathematics Standard Class X (Term I) 77 Answers 1. (c) 2. (c) 3. (c) 4. (d) 5. (a) 6. (a) 7. (b) 8. (d) 9. (c) 10. (c) 11. (d) 12. (c) 13. (b) 14. (b) 15. (a) 16. (a) 17. (b) 18. (c) 19. (a) 20. (a) 21. (d) 22. (a) 23. (b) 24. (c) 25. (d) 26. (c) 27. (a) 28. (a) 29. (c) 30. (b) 31. (a) 32. (c) 33. (c) 34. (a) 35. (d) 36. (b) 37. (b) 38. (d) 39. (b) 40. (a) 41. (b) 42. (c) 43. (d) 44. (a) 45. (c) 46. (a) 47. (a) 48. (d) 49. (c) 50. (a) SOLUTIONS 1. As 27 = 3 ´ 3 ´ 3 = 3 3 6. We have, Total number of outcomes are 52. So, if we multiply it by 3 it will become There are 6 black face cards in a deck of 3 3´ 3 =3´3 =9 52 cards. i.e a rational number. So, favourable number of outcomes are 6. 2. The degree of a quadratic polynomial is 2. \\Required probability = 6 = 3 There are 2 zeroes of quadratic polynomial. 52 26 The maximum number of zeroes of a 7. We have, A = 2 n + 13, B = n + 7 polynomial is equal to its degree, so quadratic polynomial can have at most 2 zeroes. Here, we put n = 1, 2 , 3,K . Thus, we get A = 2 ´ 1 + 13 = 15, B = 1 + 7 = 8 3. Since, X-axis and Y-axis intersect at (0, 0). So, A = 2 ´ 2 + 13 = 17, B = 2 + 7 = 9 point of intersection of the coordinate axes is origin. A = 2 ´ 3 + 13 = 19, B = 3 + 7 = 10 Here, we find that A and B are coprime. 4. We know that the bisector of an angle of a Hence, HCF ( A, B) = 1 triangle divides the opposite side in the ratio of the sides containing the angle. 8. Sum of zeroes = 1 3 \\ BD : DC = AB: AC Product of zeroes = - 2 = 3 :2 \\The required polynomial is 5. We, have k [x2 - (sum of zeroes) x sec q cos f = x a …(i) + (product of zeroes] sec q sin f = y …(ii) b …(iii) = k æ x2 - 1 x - 2 ö ç ÷ and tan q = z è3 ø c Put k = 3 Required polynomial is 3x2 - x - 6 9. From the figure PQ = QB …(i) Squaring and adding Eqs. (i) and (ii), we get The distance between P and Q, Q and B are PQ = (5 - 3)2 + (-3 - y)2 x2 + y2 = sec2 q cos2 f + sec2 q sin2 f a2 b2 and QB = (3 -1)2 + (y + 5)2 = sec2 q(cos2 f + sin2 f) = sec2 q [using distance formula] [Q cos2 A + sin2 A = 1] On putting the values in Eq. (i), we get = 1 + tan2 q (5 - 3)2 + (- 3 - y)2 = (3 - 1)2 + (y + 5)2 SAMPLE PAPER 2 Þ (2 )2 + (3 + y)2 = (2 )2 + (y + 5)2 [Q sec2 A - tan2 A = 1] Þ 4 + 9 + 6y + y2 = 4 + y2 + 25 + 10y æ + z2 ö [from Eq. (iii)] = çç1 c2 ÷÷ ø Þ 6y + 13 = 29 + 10y è æ x2 y2 öæ z2 ö Þ 4y = - 16 \\ ç + b2 ÷ = çç 1 + c2 ÷÷ Þ y=-4 ç a2 ÷ ø è øè

78 CBSE Sample Paper Mathematics Standard Class X (Term I) 10. In DABC, we have 15. The distance between P and Lis PL = (x - 5)2 + (y - 1)2 PQ|| BC Þ AQ = AP [by Thales Theorem] = x2 + 25 - 10x + y2 + 1 - 2 y QC PB Þ AQ = 6 The distance between P and M is 64 PM = (x + 1)2 + (y - 5)2 Þ AQ = 6 ´ 6 = 9 cm = x2 + 1 + 2 x + y2 + 25 - 10y 4 11. We have, As PM = PL Þ PM2 = PL2 [given] x cos q = 1 \\ x2 + 26 - 10x - 2 y + y2 Þ cos q = 1 = x2 + y2 + 2 x - 10y + 26 x Þ -12 x = -8y Þ x = 1 = sec q and y = tan q cos q Þ 3x = 2y [divide by 4] 16. Given, in DABC, AB = BD Now, x2 - y2 = sec2 q - tan2 q = 1 [Q sec2 q = 1 + tan2 q] AC DC ÐB = 60° and ÐC = 60° 12. Given, total number of tickets sold = 6000 We know that sum of angles of a triangle is 180°. Let she bought x tickets. In DABC, ÐA + ÐB + ÐC = 180° Then, probability of her winning the first prize Þ ÐA + 60° + 60° = 180° =x 6000 Þ ÐA = 180° - 120° = 60° [Given] Now, AB = BD Þ 0.08 = x 6000 AC DC Þ x = 0.08 ´ 6000 = 480 Therefore, AD bisects BC. Hence, she bought 480 tickets. [By angle bisector theorem] Then, ÐBAC = 1 ÐA = 30°. 13. (a) 2 = 2 , which is not in the form of 15 3 ´ 5 2 Hence, the value of ÐBAD is 30°. 2 m ´ 5n, so it is non terminating. 17. Given, sin A = cos A, 0° < A < 90° Þ sin A = 1 (b) 11 = 11 cos A 160 25 ´ Þ tan A = 1 5 Þ tan A = tan 45° \\ A = 45° The denominator of is in the form of 2 m ´ 5n. Hence, its decimal expansion is terminating. 14. Given polynomial is x2 - 7x + k and a - b = 1 Sum of zeroes, a + b = - b 18. Let A be the event of rain tomorrow. a Then, P( A) = 0.3 Þ a + b = - (- 7) 1 We know that, Þ a +b=7 P( A) + P( A) = 1 Product of zeroes, Then, probability that it will not rain tomorrow a×b = c SAMPLE PAPER 2 a = 1 -0.3 = 0.7 19. Since, the factors of the denominator 1250 is of Þ a×b = k the form 21 ´ 54. Now, (a + b)2 = (a - b)2 + 4 × a ×b Þ \\ 13 is a terminating decimal. (7)2 = 1 + 4k 1250 Þ 49 - 1 = 4k Now, 13 = 13 = 13 ´ 2 3 \\ k = 48 = 12 1250 21 ´ 54 24 ´ 54 4 = 104 = 104 = 0.0104 104 10000

CBSE Sample Paper Mathematics Standard Class X (Term I) 79 20. We have, sin60°+ cot 45° - cosec 30° [Q Condition for infinite solution ; a1 = b1 = c1 ù sec60° - cos30° + tan 45° a2 b2 ú c2 û 3 + 1 -2 Þ 2 = 3 =1 =2 a + b 2a - b 3 2 - 3 +1 \\ 2 = 3 and 3 = 1 2 a + b 2a - b 2a - b 3 3 +2 -4 =2 = 3 -2 Þ 4a - 2 b = 3a + 3b and 2 a - b = 9 Þ a = 5b and 2 a - b = 9 4- 3 +2 6- 3 \\ a = 5, b = 1 2 24. Let P(7, -2 ) and Q(1, - 5) be the given points and A and B are the points of trisection. = 3 -2 ´6 + 3 6- 3 6+ 3 [rationalise the denominator] P(–7, –2) 1:2 2:1 = ( 3 -2 )(6 + 3 ) A B Q(1, –5) 62 -( 3 )2 The coordinates of point A are [Q(a + b)(a - b) = a2 - b2 ] æ 1´ 1 + 2 ´ ( -7) , 1´ (-5) + 2(-2 ) ö ç 1 + 2 1+2 ÷ = 6 3 + 3 - 12 - 2 3 ç ÷ 36 - 3 è ø = 4 3 -9 or æ 1 - 14 , -5 - 4 ö or Aæç - 13 , -9 ö 33 ç 1+2 1+2 ÷ è 3 3 ÷ ç ÷ ø 21. Smallest prime number = 2 è ø Smallest composite number = 4 or æ - 13 , - 3 ö Since, Product of HCF and LCM ç ÷ è3 ø = Product of numbers \\HCF ´ LCM = 2 ´ 4 = 8 25. We have, 22. We have, l = 20 cm, q = 60° a and b are the zeroes of 2 x2 + kx + 4, then a + b = - Coefficient of x = - k Now, we know that Coefficient of x2 2 q and ab = Constant term =4 =2 l = ´ 2 pr Coefficient of x2 2 360° Þ 20 = 60° ´ 2 ´ p ´ r 360° Þ r = 20 ´ 6 = 60 cm 2p p Now, it is given that 26. Given, PR = QR and PQ2 = 2 PR2 a2 + b2 = 8 Þ ( PQ)2 = PR2 + PR2 Þ ( PQ)2 = (QR)2 + ( PR)2 Þ (a + b)2 - 2 ab = 8 Þ æ - k ö2 -2 ´ 2 =8 ç ÷ è2 ø P Þ k 2 = 12 SAMPLE PAPER 2 4 Þ k 2 = 48 Þ k =±4 3 23. Given, 2 x + 3y - 7 = 0 and QR (a + b)x + (2 a - b)y - 21 = 0 has infinite By converse of pythagoras theorem, DPQR is solutions, then right angle triangle at angle R. 2 = 3 = -7 a + b 2 a - b - 21

80 CBSE Sample Paper Mathematics Standard Class X (Term I) 27. We have, æ 9 - 14 ö ç ÷ xcosec230°sec2 45° = è 16 4 ø 8 cos2 45°sin2 60° = tan2 60° - tan2 30° æ7ö ç÷ è4ø x(2 )2( 2 )2 3 )2 -æç 1 ö2 Þ =( è ÷ æ 9 - 56 ö ö2 ç÷ 8çæ 1 ø÷öæèçç 3 ÷÷ 3ø = è 16 ø è 2 2 ø æ7ö Þ 8x = 3 - 1 Þ 8x = 8 ç÷ 3 3 33 è4ø Þ x =3´8 =1 = - 47 83 28 28. Number of possible outcomes to get the sum as 31. Given, PQ = 10 3 = 2 {i.e. (2 , 1), (1, 2 )} Þ PQ2 = (10)2 [squaring both side] Number of possible outcomes to get the sum as 4 = 3 {i.e. (2 , 2 ), (1, 3), (3, 1)} = 100 Number of possible outcomes to get the sum as From the distance formula 5 = 4 {i.e. (3, 2 ), (2 , 3), (4, 1), (1, 4)} PQ = (10 - 2 )2 + (y + 3)2 Number of possible outcomes to get the sum as 6 = 5 {i.e. (5, 1), (1, 5), (3, 3), (4, 2 ), (2 , 4)} On squaring both sides, we get Number of possible outcomes to get the sum as (10 -2 )2 + (y + 3)2 = 100 [Q PQ2 = 100] 7 = 6 {i.e. (4, 3), (3, 4), (6, 1), (1, 6), (5, 2 ), (2 , 5)} Þ (y + 3)2 = 100 - 64 = 36 Þ y2 + 9 + 6y = 36 Number of possible outcomes to get the sum as Þ y2 + 6y - 27 = 0 8 = 5 {i.e. (4, 4), (6, 2 ), (2 , 6), (5, 3), (3, 5)} Þ y2 + 9y - 3y - 27 = 0 Number of possible outcomes to get the sum as Þ y(y + 9) - 3(y + 9) = 0 9 = 4 {i.e. (6, 3), (3, 6), (5, 4), (4, 5)} Þ (y - 3) (y + 9) = 0 Number of possible outcomes to get the sum as Þ y = 3, y = - 9 10 = 3 {i.e. (5, 6), (6, 4), (4, 6)} 32. Since ABCD is a rhombus, then \\ The total posible outcomes OA = OB = 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 = 32 OC OD 29. We have, [Q Diagonals of rhombus bisect each other] Þ 4 =5 2 196 2 98 x -1 x 7 49 77 Þ 4x = 5x - 5 Þ x=5 1 \\ 196 = 2 2 ´ 72 33. We have, sin q - cos q = 0 So, sum of powers of prime factors = 2 + 2 = 4 Þ sin q = cos q 30. We have, a and b are the zeroes of the Þ sin q = 1 polynomial p(x) = 4x2 + 3x + 7, then cos q é tan q = sin q ù a + b = - 3 and ab = 7 êëQ cos q úû 44 Þ tan q = 1 a b a2 + b2 SAMPLE PAPER 2 Now, += Þ tan q = tan 45° [Q tan 45° = 1] b a ab = (a + b)2 - 2 ab Þ q = 45° ab Now, sin4 q + cos4 q = sin4 45° + cos4 45° æ -3 2 - 2çæ 7 ö = æ 1 ö4 + æ 1 ö4 ç ÷ ç ÷ ç ÷ ö è 2ø è 2ø ÷ =è 4 ø è4ø 7 =1 + 1=1 4 42 4

CBSE Sample Paper Mathematics Standard Class X (Term I) 81 34. Total number of outcomes = 26 A Letter succeeding V are W , X, Y , Z. Then, total number of favourable outcomes = 4 30 units 70 units \\Required probability h units = Number of favourable outcomes B (80 – x) D x units C Total number of possible outcomes Þ Required probability = 4 = 2 units 26 13 80 units 35. (i) 16 = 16 Let us assume AB = 30 units, AC = 70 units and 25 (5)2 ´ 2 0 BC = 80 units. (ii) 5 = 5 Let h is the height of altitude which cuts BC 18 2 ´ 32 into two parts. (iii) 2 = 2 Let us asume the first part be x so that the 21 3 ´ 7 second part will be (80 - x). (iv) 7 = 7 Now, in DABD 250 2 ´ 53 h2 = 302 - (80 - x)2 [use pythagoras theorem] Hence (i) and (iv) having terminating decimal Also, in DACD, because denominators are the form of 2 m ´ 5n . h2 = 702 - x2 36. Let a and b are the zeroes of quadratic polynomial. Þ 302 -(80 - x)2 = 702 - x2 Þ 900 - (6400 + x2 - 160x) = 4900 - x2 According to the question, Þ 160x = 10400 a =5 Þ x = 65 units a×b =0 39. We have, x sin q = 1 Þ b=0 Þ x= 1 Now, required quadratic polynomial sin q = x2 - (a + b)x + a ×b Þ x = cosec q = x2 - (5 + 0) x + 0 = x2 - 5x and y = cot q 37. We have, Pæç a , 4 ö is the mid-point of Q(- 6, 5) Now, we know that ÷ cosec2 q - cot2 q = 1 è3 ø Þ x2 - y2 = 1 and R(- 2 , 3), then 40. Total number of cards = 5 æ a , 4 ö = æ - 6 - 2 , 5 + 3 ö Total number of king = 1 ç ÷ ç ÷ è3 ø è 2 2ø \\ The required probability = Number of favourable outcomes [using section formula] Total number of outcomes =1 Þ æ a , 4 ö = (- 4, 4) 5 ç ÷ è3 ø On comparing the x-coordinate, we get Solutions (41-45) SAMPLE PAPER 2 \\ a =-4 41. Let r be the radius and d be the diameter. 3 Þ a = - 12 Here, total length of platinum wire used 38. Given, a triangle with sides 30, 70 and = 2 pr + 4d é r = 28 = 14ùûú 80 units. = 2 ´ 22 ´ 14 + 4 ´ 28 ëêQ 2 Altitude is droped upon the side of length 7 80 units. = 88 + 112 = 200 mm

82 CBSE Sample Paper Mathematics Standard Class X (Term I) 42. Here, the radius of the sector = 28 = 14 mm Then, y = x and 3x + 4y = 20 2 2 Since, circle is divided into 8 sectors. Rewrite the above equations to represent Therefore, angle of each is 360° = 45°. algebraically 8 x -2y =0 …(i) Area of each sector = 45° ´ 22 ´ 14 ´ 14 3x + 4y = 20 …(ii) 360° 7 = 1 ´ 22 ´ 2 ´ 14 47. If the pair of equations intersect at only one point. Then the pair of equations has a unique 8 solution and hence consistent. = 77 mm2 48. From Eq. (i), x = 2 y 43. Let r be the radius of the silver part. Given, circumference of silver part = 88 mm Put x = 2 y in Eq. (ii), we get \\ 2 pr = 88 r = 88 ´ 7 = 14 mm 3(2 y) + 4y = 20 2 ´ 22 Radius of outer part, Þ 10y = 20 R = 7 + 14 = 21 mm Þ y =2 Circumference of outer part Þ x =2´2 =4 Hence, the intersection point is (4, 2). = 2 pR = 2 ´ 22 ´ 21 49. Put y = 0 in the given equation. 7 x -0 =0Þ x =0 = 132 mm \\ The point is (0, 0). Put x = 0 in the given equation 44. Let R be the radius of the platinum part and r be the radius of the silver part. x - 2 y = 0, Then, R = 21 and r = 14 0 -2y =0 \\Required difference of area of platinum and Þ y =0 silver parts = pR2 - pr2 \\ The point is (0, 0). = p (21)2 - p(14)2 Hence, we conclude that given line intersect X = p (441 - 196) and Y axes at only (0, 0). = 245 p mm2 50. Put y = 0 in the given equation, we get 45. Circumference of the brooch B = 2 p´ 21 = 42 p mm 3x + 0 = 20 Number of revolutions = 168p ¸ 42 p = 4 x = 20 3 Solutions (46-50) \\The point is æ 20 , 0 ÷ö. 46. Let x be the number of rides on the giant wheel ç and y be the numebr of hoopla played by è3 ø Palak. Put x = 0 in the given equation, we get 0 + 4y = 20 Þ y=5 \\ The point is (0, 5). SAMPLE PAPER 2

CBSE Sample Paper Mathematics Standard Class X (Term I) 83 SAMPLE PAPER 3 MATHEMATICS (Standard) A Highly Simulated Practice Questions Paper for CBSE Class X (Term I) Examination Instructions 1. The question paper contains three parts A, B and C. 2. Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 3. Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 4. Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions. 5. There is no negative marking. Roll No. Maximum Marks : 40 Time allowed : 90 minutes Section A Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 1. Given figure shows the graph of the polynomial f (x) = ax 2 + bx + c Y f(x) = ax2 + bx +c P X X¢ O –b, –D 2a 4a Y¢ (a) a < 0, b < 0 and c > 0 (b) a > 0, b < 0 and c > 0 (c) a < 0, b < 0 and c < 0 (d) a > 0, b > 0 and c > 0 2. The sum of the numerator and denominator of a fraction x is 8. If the denominator is y increased by 1, the fraction becomes 1. The situation can be represented algebraically as SAMPLE PAPER 3 2 (a) x + y = 8 and x + 1 = 1 (b) x + y = 8 and x + 1 = 1 y2 y2 (c) x = 8 and x = 1 (d) x + y = 8 and x = 1 y y+1 2 y+1 2 3. If tan a = 3 and tan b = 1 , 0< a, b< 90°, then the value of cot(a + b) is 3 (a) 3 (b) 0 (c) 1 (d) 1 3

84 CBSE Sample Paper Mathematics Standard Class X (Term I) 4. The distance of the point (- 3, 8) from the X-axis is (a) 3 units (b) - 3 units (c) 8 units (d) 5 units 5. The decimal expansion of the rational number 53 will terminate after how many 2453 places of decimal? (a) 1 (b) 2 (c) 3 (d) 4 6. The perimeter of a square circumscribing a circle of radius a cm is A GB F OH D EC (a) 8a cm (b) 4a cm (c) 2a cm (d) 16a cm 7. A box contains cards numbered 6 to 50. A card is drawn at random from the box, the probability that the drawn can has a number, which is a perfect square, is (a) 1 (b) 2 9 3 (c) 1 (d) 7 6 18 8. A boy walks 8 m due East and 6 m due South. How far is he from the starting point? (a) 31 m (b) 26 m (c) 62 m (d) 10 m 9. The graph of x 2 + 1 = 0 (a) Intersecting X-axis at two distinct points (b) Touches X-axis at a point (c) Neither touches nor intersect X-axis (d) Either touches or intersect X-axis 10. If the pair of linear equations 3x + y = 3 and 6x + ky = 8 does not have a solution, then the value of k is (a) 2 (b) -3 (c) 0 (d) 1 SAMPLE PAPER 3 11. In DPQR, if PS is the internal bisector of ÐP meeting QR at S and PQ = 13 cm, QS = (3 + x) cm, SR = (x - 3) cm and PR = 7 cm, then find the value of x. (a) 9 cm (b) 10 cm (c) 13 cm (d) 12 cm 12. Two concentric circles form a ring. The inner and outer circumference of the ring are 50 2 m and 75 3 m respectively. Find the width of the ring. 77 (a) 1 m (b) 2 m (c) 3 m (d) 4 m

CBSE Sample Paper Mathematics Standard Class X (Term I) 85 13. In the given figure, line BD||CE. If AB = 1.5 cm, BC = 6 cm and AD = 2 cm. Find DE. E D 6 cm C 2 cm A 1.5 cm B (a) 6 cm (b) 8 cm (c) 4 cm (d) None of these 14. A child has a die whose six faces show the number as given below. 12 2 3 4 6 The die is thrown once the probability of getting 5 is (a) 1 (b) 2 (c) 0 (d) 1 6 5 15. Which of the following is not a polynomial? (a) 2 x -3 - 5 + 3x -1 (b) x 3 + 2 x - 9 (c) (x -2 )2 + 3x (d) (2 x + 10) (x2 -25) ´ x+5 16. The age of a daughter is one third the age of her mother. If the present age of mother is x yr, then the age (in yr) of the daughter after 15 yr will be (a) x + 15 (b) x + 15 (c) x + 5 (d) x - 5 3 3 3 17. Check the relation between the following triangles PA 58° 83° C 39° R 58° QB (a) similar by SAS (b) similar by AAA (c) similar by SSS (d) similar by ASS 18. If sin x + cosec x = 2, then sin 19 x + cosec 20x is equal to SAMPLE PAPER 3 (a) 2 19 (b) 2 20 (c) 2 (d) 2 39 19. Evaluate 8 3 cosec2 30° sin 60° cos 60° cos2 45° sin 45° tan 30° cosec3 45°. (a) 8 (b) 4 3 (c) 8 3 (d) 16 3 20. The diameter of the wheel of a bus is 1.4 m. The wheel makes 10 revolutions is 5 s. The speed of the vehicle (in km/h) is (a) 30 (b) 31 (c) 31.68 (d) 35

86 CBSE Sample Paper Mathematics Standard Class X (Term I) Section B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. If a and b are the zeroes of a polynomial f (x) = px 2 + qx + r. Then, a + b + 2 is equal to ba q2 q2 q2 (d) None of these (a) (b) r2 p (c) p2 r rp 22. If HCF (306, 657) = 9, what will be the LCM (306, 657)? (a) 12338 (b) 22338 (c) 23388 (d) 22388 23. The area of the triangle formed by the lines 2x + 3y = 12, x - y - 1 = 0 and X-axis (as shown in figure), is 2x + 3y – 12 = 0 Y B(0, 4) D (3, 2) X¢ O C(1, 0) (6, 0)A X E(0, –1) x – y – 1 = 0 Y¢ (a) 7 sq units (b) 5 sq units (c) 6.5 sq units (d) 6 sq units 24. In the following figure, LM||BC and LN||CD, then which of the following relation is true? B M C AL N D (a) AM = AN (b) ML = AL (c) Both (a) and (b) (d) None of these AB AD BC AC 25. From the given figure, the value of 25 (sin 2 q + 2 cos2 q - tan q) is SAMPLE PAPER 3 P 10 cm 8 cm Rq 6 cm Q (a) 2 (b) - 2 (c) 3 (d) - 3 3 3 2 2

CBSE Sample Paper Mathematics Standard Class X (Term I) 87 26. If the radius of a circle is diminished by 10%, then its area is diminished by ……… (a) 29% (b) 19% (c) 15% (d) 9% 27. If DBAC is triangle with ÐA = 90°. From A, a perpendicular AD is drawn on BC. Which one of the following is correct? (a) Only DABC ~ DDAC (b) Only DDAC ~ DDBA (c) Only DABC ~ DDBA (d) All of these 28. A card is drawn from a pack of cards numbered 2 to 53. The probability that the number of the card is a prime number less than 20 is (a) 2 (b) 4 (c) 5 (d) 8 13 13 13 13 29. For what value of k, does - 4 is a zero of the polynomial x 2 - x - (2k + 2) ? (a) 7 (b) 8 (c) 9 (d) 10 30. The probability of passing a certain test is x . If the probability of not passing is 7, then 24 8 x is equal to (a) 2 (b) 3 (c) 4 (d) 6 31. To place a pole vertical on the ground a guy attach a wire of length 26 m to it at a point 10 m away from its foot. Then, the length of pole will be (a) 10 m (b) 28 m (c) 20 m (d) 24 m 32. A two-digit number, where ten’s digit is greater than ones digit is obtained by either multiplying sum of the digits by 8 and adding 1 or by multiplying the difference of digits by 13 and adding 2. The number is (a) 14 (b) 51 (c) 41 (d) 49 33. Two players Sania and Deepika play a tennis match. If the probability of Sania winning the match is 0.68, then the probability of Deepika winning the match is (a) 0.32 (b) 0.38 (c) 0.42 (d) 0.48 34. In an equilateral triangle DABC, G is the centroid. Each side of the triangle is 6 cm. The length of AG is (a) 2 2 cm (b) 3 2 cm (c) 2 3 cm (d) 3 3 cm 35. The value of æ 1 + 1 öæ 1 - 1 ö is ç ÷ç ÷ è cosq cot q ø è cosq cot q ø (a) 0 (b) -1 (c) 1 (d) 2 36. The area of a circular field is 13.86 hectares, then the cost of fencing at the rate of SAMPLE PAPER 3 ` 4.40 per metre is (a) ` 5800 (b) ` 5808 (c) ` 5600 (d) ` 5500 37. The value of tan 30° + cot 60° is tan 30°(sin 30° + cos 60°) (a) 2 (b) 6 (c) 4 (d) 5 38. If the point C(k, 4) divides the join of points A(2, 6) and B(5, 1) in the ratio 1 : 3, then the value of k is (b) 29 (c) 11 (d) 9 (a) 11 4 4 4

88 CBSE Sample Paper Mathematics Standard Class X (Term I) 39. In the figure ABDCA represents a quadrant of a circle of radius 7 cm with centre A. Find the area of the shaded portion. C D E SAMPLE PAPER 3 A 7 cm B 2 cm (a) 14 cm2 (b) 24.5 cm2 (c) 38.5 cm2 (d) 31.5 cm2 40. If 1 is a zeroes of the polynomial x 2 + kx - 5, then the sum of the zeroes is 24 (a) 2 (b) - 2 (c) 1 (d) - 1 2 2 Section C Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted. 41-45 are based on Case Study-1. Case Study 1 A vegetable seller has 420 potatoes and 130 tomatoes. He wants to stack them in such a way that each stack has same number and they take up the least area of the tray. Based on the above information of a vegetable shop, answer the following questions 41. The total number of vegetables are (a) 420 (b) 130 (c) 550 (d) 290 42. The product of exponents of the prime factors of total number of vegetables is (a) 2 (b) 3 (c) 5 (d) 6 43. What is the number of vegetables that can be placed in each stack for this purpose? (a) 45 (b) 40 (c) 10 (d) 35 44. The sum of exponents of the prime factors of the number of vegetables that can be placed in each stack for this purpose is (a) 5 (b) 2 (c) 4 (d) 6 45. What is the total number of rows in which they can be placed? (a) 15 (b) 25 (c) 35 (d) 55

CBSE Sample Paper Mathematics Standard Class X (Term I) 89 46-50 are based on Case Study-2. Case Study 2 Class X students of a school in Gomtinagar have been alloted a rectangular plot of a land for gardening activity. Sapling of roses are planted on the boundary at a distance of 1m from each other. There is a triangular grassy fountain in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot. BC 6 P R 5 4 3 2 Q 1 A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 D Considering A as origin answer the following questions 46. What are the coordinates of P? (b) (6, 4) (d) (5, 4) (a) (4, 6) (c) (4, 5) 47. What are the coordinates of R? (b) (5, 6) (d) (7, 4) (a) (6, 5) (c) (6, 0) 48. The distance between points Q and R is (b) 2 3 units (d) 10 3 units (a) 13 units (c) 13 units 49. DPQR is a/an (b) equilateral triangle (d) None of these (a) right angled triangle (c) scalene triangle 50. The centroid of DPQR is (a) æ 13 , 13 ö (b) æ 13 , 13 ö ç ÷ ç ÷ è2 2 ø è3 3ø (c) (13, 13) (d) None of these SAMPLE PAPER 3

OMR SHEET SP 3 Roll No. Sub Code. Student Name Instructions Use black or blue ball point pens and avoid Gel & Fountain pens for filling the OMR sheet. Darken the bubbles completely. Don’t put a tick mark or a cross mark, half-filled or over-filled bubbles will not be read by the software.  Correct Incorrect Incorrect Incorrect Do not write anything on the OMR Sheet. Multiple markings are invalid. 1 18 35 2 19 36 3 20 37 4 21 38 5 22 39 6 23 40 7 24 41 8 25 42 9 26 43 10 27 44 11 28 45 12 29 46 13 30 47 14 31 48 15 32 49 16 33 50 17 34 Check Your Performance Score Percentage = Total Correct Questions 100 Total Questions: × Total Correct Questions: Total Questions Less than 60% > Average (Revise the concepts again) If Your Score is Greater than 60% but less than 75% > Good (Do more practice) Above 75% > Excellent (Keep it on)

CBSE Sample Paper Mathematics Standard Class X (Term I) 91 Answers 1. (b) 2. (d) 3. (b) 4. (c) 5. (d) 6. (a) 7. (a) 8. (d) 9. (c) 10. (a) 11. (b) 12. (d) 13. (b) 14. (c) 15. (a) 16. (a) 17. (b) 18. (c) 19. (c) 20. (c) 21. (a) 22. (b) 23. (b) 24. (c) 25. (a) 26. (b) 27. (d) 28. (a) 29. (c) 30. (b) 31. (d) 32. (c) 33. (a) 34. (c) 35. (c) 36. (b) 37. (a) 38. (c) 39. (d) 40. (b) 41. (c) 42. (a) 43. (c) 44. (b) 45. (d) 46. (a) 47. (a) 48. (c) 49. (c) 50. (b) SOLUTIONS 1. From the figure, the graph of polynomial p(x) 6. Let ABCD be the square that inscribes a circle with centre O, touching it at the points E, F, G is a parabola open upwards. Therefore, a > 0, and H. y = ax2 + bx + c, cuts Y-axis at P. Given, the radius of the circle = a cm On putting x = 0 in y = ax2 + bx + c, we get Thus, OE = OF = OG = OH = a cm Hence, OG + OE = AD = BC y=c and OF + OH = AB = DC So, each side of the square = 2 a cm Hence, the coordinates of P are (0, y)or (0, c). It is Hence, perimeter of the square = 4 (2 a) = 8a cm clear that P lies on positive Y-axis, therefore c >0. 7. Given numbers are 6, 7, 8, 9, ……, 50. Total number of possible outcomes Also, x-coordinate of vertex should be positive = 50 - 6 + 1 = 45 i.e. - b > 0 Favourable outcomes = Perfect square numbers between 6 to 50 i.e. 9, 16, 25, 36, 49 2a = 32, 42, 52, 62, 72 Since, a > 0 therefore b should be negative \\Total number of favourable outcomes = 5 i.e. b < 0. So, the required probability 2. We have numerator and denominator of the = Number of Favourable Outcomes fraction be x and y respectively. Total Number of Possible Outcomes Sum of numerator and denominator = 8 = 5 =1 i.e. x + y = 8 45 9 Denominator is increased by 1 i.e New, denominator = y + 1 \\ New fraction is x = 1 y +1 2 3. We have, tan a = 3 Þ a = 60° [Q tan60° = 3] …(i) 8. Let A be the starting point. Again, tanb = 1 A 8m B N \\ 3 b = 30° é tan30° = 1 ù …(ii) W E ëêQ 3 ûú 6m Adding Eqs. (i) and (ii), we get S a + b = 60° + 30° = 90° C \\ cot(a + b) = cot90° In DABC, ÐB = 90° = 0 [Q cot90° = 0] AC2 = AB2 + BC2 4. The given point is (- 3, 8). SAMPLE PAPER 3 The distance of any point (x, y) from X-axis is [by pythagoras theorem] equal to absolute value of y-coordinate. Þ AC2 = 82 + 62 \\ Distance (- 3, 8) from X-axis = 8 units. = 64 + 36 = 100 = 102 5. 53 = 53 ´ 5 Þ AC = 10 m 2453 24 ´ 54 9. We have, x2 + 1 = 0 265 265 = (2 ´ 5)4 = 104 = 0.0265 x2 = -1 It means, x cannot be real. Thus, the rational number terminates after 4 decimal places.

92 CBSE Sample Paper Mathematics Standard Class X (Term I) Hence, the graph is neither touches nor \\ 2 = 1.5 intersect X-axis. 2 + DE 1.5 + 6 10. Given, 3x + y = 3 and 6x + ky = 8 Þ 2 = 1.5 On comparison the above equation with 2 + DE 7.5 standard equation, we get a1 = 3, b1 = 1 and c1 = -3 Þ 2 =1 and a2 = 6, b2 = k and c2 = -8 2 + DE 5 Since, given equations have no solution. \\ a1 = b1 ¹ c1 Þ 2 + DE = 10 a2 b2 c2 Þ DE = 8 cm Þ 3 = 1Þ k =2 6k 14. On the given dice, 5 is not printed. So, its impossible event. Probability of impossible 11. Since, PS is the internal bisector of ÐP and it event is 0. meets QR at S. 15. A polynomial is an algebraic expression P containing two or more terms. Also, it requires 7 cm the exponents of the variables in the each term should be positive integer. (a) 2 x- 3 - 5 + 3x-1, it is not polynomial, SAMPLE PAPER 3 because it has negative power in second term. 13 cm (b) x3 + 2 x - 9, it is a polynomial. (c) (x - 2 )2 + 3x = x2 + 4 - 4x + 3x Q (3 + x) cm S (x – 3) cm R = x2 - x + 4, it is a polynomial. \\ PQ = QS (d) 2 x + 10 ´ (x2 - 25) = 2 (x + 5) ´ (x2 - 25) PR RS x+5 (x + 5) Þ 13 = 3 + x = 2 x2 - 50, it is a polynomial. 7 x -3 16. It is given, present age of mother = x yr Þ 13(x - 3) = 7 (3 + x) Since age of daughter is one third the age of Þ 13x - 39 = 21 + 7x mother. Þ 6x = 60 \\ Present age of daughter = x yr Þ x = 10 cm 3 After 15 yr age of the daughter = æ x + 15 ö yr ç ÷ 12. Let the inner and outer radii be r and R. è3 ø Then, 2 pr = 50 2 = 352 17. In DABC, 77 ÐA + ÐB + ÐC = 180° Þ r = æ 352 ´ 7 ´ 1ö =8m [sum of all angles of a triangle is 180°] ç ÷ 83° + 58° + ÐC = 180° è 7 22 2 ø Þ ÐC = 180° - (83° + 58°) and 2 pR = 75 3 = 528 Þ ÐC = 39° 77 Now, In DPQR, Þ R = æ 528 ´ 7 ´ 1ö = 12 m ÐP + ÐQ + ÐR = 180° ç ÷ Þ 58° + ÐQ + 39° = 180° è 7 22 2 ø Þ ÐQ = 180° - (58° + 39°) Þ ÐQ = 83° \\Width of the ring = ( R - r) = (12 - 8) m = 4 m Now, In DABC and DQPR, 13. In DADB and DAEC, ÐA = ÐQ = 83° ÐB = ÐP = 58° ÐADB = ÐAEC ÐC = ÐR = 39° \\ DABC ~ DQPR [corresponding angles as BD||CE] [by AAA similarity criterion] ÐABD = ÐACE [corresponding angles as BD||CE] ÐA = ÐA [common] \\ DADB~ DAEC [by AAA similarity criterion] Þ AD = AB [by CPCT] 18. We have, sin x + cosec x = 2 AE AC

CBSE Sample Paper Mathematics Standard Class X (Term I) 93 Þ sin x + 1 = 2 = 1 ´ 5 ´ 2 = 5 sq units sin x 2 Þ sin2 x + 1 = 2 sin x 24. In DABC, Þ (sin x -1)2 = 0 ML|| BC Þ sin x = 1 \\ By basic proportionality theorem, we have Þ cosec x = 1 AM = AL …(i) \\ sin19 x + cosec20x = (1)19 + (1)20 AB AC =1 + 1 =2 Again, in DADC, 19. We have, 8 3 cosec230°sin60°cos60° NL|| DC cos2 45°sin 45°tan30°cosec3 45° \\ By basic proportionality theorem, we have =8 3 (2 )2 æ 3 ö æ 1 ö æ 1 2 1 öæ 1 ö ( 2 )3 AN = AL …(ii) çç 2 ÷÷ ç 2 ÷ ç ÷ç ÷ AD AC ø è ø è öæ ÷ç 2 øè 2ø è 3ø è =8 3´ 4´ 3 ´ 1 ´ 1 ´ 1 ´ 1 ´ 2 2 =8 3 From Eqs. (i) and (ii), we get 2 22 2 3 AM = AN 20. The distance covered in one revolution is equal AB AD to the circumference of the wheel. Again, in DAMLand DABC Circumference of the wheel = p ´ d = p´ 1.4 m ÐAML = ÐABC [corresponding angles] = 22 ´ 1.4 = 4.4 m 7 ÐALM = ÐACB [corresponding angles] \\ Distance covered in 10 revolutions \\ By AA similar criterion, DAML~ DABC = 10 ´ 4.4m = 44 m \\ By CPCT, we have \\ Speed = Distance traveled = 44 m/s AM = ML …(iii) Total time 5 AB BC = 44 ´ 60 ´ 60 = 31.68 km/h From Eqs. (i) and (iii), we have 5 1000 ML = AL 21. Given, quadratic polynomial is px2 + qx + r. BC AC Since, a and b are the zeroes of quadratic 25. In DPQR, ÐQ = 90°, PQ = 8 cm, QR = 6 cm polynomial. \\ a + b = -q and a ×b = r and PR = 10 cm \\ sin q = Perpendicular = PQ = 8 = 4 pp Hypotenuse PR 10 5 a2 + b2 + 2ab (a + b)2 ab cos q = Base = QR = 6 = 3 Now, + + 2 = = Hypotenuse PR 10 5 ba ab ab æ q ö2 and tan q = Perpendicular = PQ = 8 = 4 çç - p ÷÷ q2 q2 Base QR 6 3 =è ø = ´ p = æ r ö p2 r pr \\ 25(sin2 q + 2 cos2 q - tan q) ç ÷ ç ÷ éæ 4 ö2 3 ö2 4ù è p ø = 25 êç ÷ + 2 ´ æ ÷ - ú ç ëè 5 ø è5ø 3û 22. We know that, LCM ´ HCF = Product of two numbers = 25 é16 +2 ´9 - 4ù êë25 25 3 ûú \\ LCM (306, 657) ´ HCF (306, 657) = 306 ´ 657 SAMPLE PAPER 3 Þ LCM ´ 9 = 306 ´ 657 = 25 é 48 + 54 - 100 ù Þ LCM = 22338 êë 25 ´ 3 ûú 23. From the figure, the required triangle is DACD. = 25 é102 - 100 ù = 2 Here, base of triangle = 6 -1 = 5 ëê 25 ´ 3 ûú 3 Height of triangle = 2 We know that, 26. Let radius of the circle be x. Area of DACD = 1 ´ base × height 2 Given that the radius is diminished by 10%. So, new radius = x - x = 9x 10 10