MATHS MCQ SOLVED AND UNSOLVED

144 CBSE Sample Paper Mathematics Standard Class X (Term I) Answers 1. (b) 2. (b) 3. (d) 4. (b) 5. (b) 6. (b) 7. (c) 8. (c) 9. (a) 10. (a) 11. (a) 12. (d) 13. (c) 14. (c) 15. (a) 16. (c) 17. (c) 18. (b) 19. (a) 20. (a) 21. (c) 22. (c) 23. (a) 24. (c) 25. (a) 26. (d) 27. (d) 28. (d) 29. (c) 30. (c) 31. (d) 32. (a) 33. (c) 34. (b) 35. (c) 36. (a) 37. (c) 38. (a) 39. (c) 40. (a) 41. (c) 42. (a) 43. (b) 44. (a) 45. (d) 46. (d) 47. (b) 48. (c) 49. (c) 50. (c) SOLUTIONS 1. The system of given equations will have 4. We have, cos A + sin A cot A infinitely many solutions, when = æ cos A ´ sin A÷ö + sin A êéë\\cot A = cos Aù a1 = b1 = c1 ç sin A ûú a2 b2 c2 è cos A ø On comparision the given equation with = sin A + sin A = 2 sin A standard equation, we get 5. Since, DABC ~ DPRQ a1 = k, b1 = 1, c1 = - k 2 Therefore by CPCT, Consider, a2 = 1, b2 = k, c2 = -1 AB = PR k = 1 = -k 2 BC = RQ 1 k -1 and AC = PQ k = 1 Þ k2 =1 1k 6. The point (3, 5) is shown in the figure Þ k =±1 Y and 1 = k 2 3 unit (3, 5) k1 Þ k3 =1 Þ k =1 5 unit X Hence, the common solution is k = 1. 2. Given, The probability of happening of an event is From the figure, the distance is 5 units. P( E) = p 7. Let d1 and d2 be external and internal diameters. and probability of non-happening of an event Given, d1 = 12 m and d2 = 8 m Area of the path can be given by the difference is P( E) = q of the area of internal and external circles. We know that, P( E) + P( E) = 1 Þ p+ q=1 3. Let r be the radius of the circle. d1 Area of circle = pr2 d2 Area of sector OAPB = x ´ pr2 SAMPLE PAPER 7 360° \\Area of the circular path = p éæ d1 ö2 - æ d2 ö2 ù êç 2 ÷ ç 2 ÷ú Given, area of sector OAPB ëè ø è øû = 5 ´ Area of circle 12 = péêçæ 12 ö2 - æ 8 ö2 ù = p[(6)2 - (4)2 ] ëè 2 ÷ ç 2 ÷ú On putting the values, we get ø è øû x ´ pr2 = 5 ´ pr2 = p{36 - 16} = 20 p m2 360° 12 Þ x =5 8. ( 7 - 3) - 7 = 7 - 3 - 7 = - 3 360° 12 Since, - 3 is a rational number. Þ x = 5 ´ 360° = 150° So, ( 7 - 3) - 7 is a rational number. 12

CBSE Sample Paper Mathematics Standard Class X (Term I) 145 9. Total number of possible outcomes = 49 Thus, distance travelled in 500 revolutions = 500 ´ 3.96m = 1980 m [Q Numbers 0 and 50 are not included in the outcomes] 14. Given, cosec2q(1 + cos q)(1 - cos q) = l Favourable outcomes are {4, 18, 12 ..., 48} i.e. 12. Þ cosec2q (1 - cos2 q) = l [Q a2 - b2 = (a - b)(a + b)] Required probability = Number of favourable outcomes Þ cosec2q ´ sin2 q = l Total number of outcomes [Q1 - cos2 q = sin2 q] = 12 49 cosec2q 1 é sin q 1 ù sin2 cosec Þ ´ = l êQ = ú ë û 10. We have, q q x × tan 45° × cot60° = sin30°cosec 60° Þ l=1 Þ x´ 1. 1 = 1 ´ 2 Þ x = 1 ´ 3 15. Total outcomes = 52 32 3 3 Favourable outcomes of getting a king = 4 \\ x =1 Probability = 4 = 1 11. Let P(x, 0) be the point of intersection of X-axis 52 13 with the line segment joining A(3, 6) and B(12 , - 3) which divides the line segment AB in 16. The distance between A and B is the ratio l : 1. AB = (k - 2 )2 + (3 - 3)2 Using section formula, [By distance formule] [given] (x, 0) = æ 12 l + 3 , - 3l + 6 ö \\ (k - 2 )2 = 5 ç l+1 l+1 ÷ ç ÷ Squaring both sides, we have (k - 2 )2 = 25 è ø Now, equating the y component on both sides, k -2 =± 5 we get Þ k =2 ± 5 Þ k = 7 or - 3 - 3l + 6 = 0 Þ l = 2 l+1 1 17. Given, 4 tan q = 3 So, X-axis divides AB in the ratio 2 : 1. Þ tan q = 3 = BC 4 AB 12. The given equation can be written as x -y =-1 …(i) Let BC = 3k and AB = 4k …(ii) and 3x + 2 y = 12 In right DABC, On multiplying Eq. (i) by 2, we get …(iii) AC = AB2 + BC2 2x -2y = -2 = (4k )2 + (3k )2 Adding Eqs. (ii) and (iii), we get = 16k 2 + 9k 2 5x = 10 \\ x =2 = 25k 2 = 5k Putting x = 2 in Eq. (i), we get C 2 -y = -1 Þ - y = - 3 or y = 3 3k \\ x = 2 and y = 3 q SAMPLE PAPER 7 13. Diameter of the wheel = 1.26 m A 4k B Radius of the wheel, \\ sin q = BC = 3 and cos q = AB = 4 r = 1.26 = 0.63 m AC 5 AC 5 2 Now, 4 sin q - cos q = 4´ 3 - 4 Distance travelled in one revolution 4 sin q + cos q 4´ 5 + 5 3 4 = Perimeter of the wheel 55 = 2 pr = 2 ´ 22 ´ 0.63 = 3.96 m = 12 - 4 = 8 = 1 7 12 + 4 16 2

146 CBSE Sample Paper Mathematics Standard Class X (Term I) 18. By definition of Pythagores theorem, Þ ( PR + PQ)(1) = 9 [Q PR - PQ = 1] In a right-angled triangle, the square of the Þ PR + PQ = 9 hypotenuse is equal to the sum of square of other two sides containing right angle. On solving PR + PQ = 9 and PR - PQ = 1, we get 19. The given pair of linear equations will be PR = 5 cm and PQ = 4 cm parallel, when a1 = b1 ¹ c1 \\ sin R = PQ = 4 a2 b2 c2 PR 5 Here, on comparison the given equations with So, sin2 R + cosec R standard equation, we get = æ 4 ö2 + 1 = 16 + 5 = 64 + 125 = 189 a1 = 3, b1 = - 1, c1 = - 5 ç ÷ a2 = 6, b2 = - 2, c2 = - p è 5 ø 4 / 5 25 4 25 ´ 4 100 \\ 3 = -1 ¹ -5 Þ 1 ¹ 5 6 -2 -p 2 p 23. Total number of outcomes = 40 Þ p ¹ 10 Multiples of 5 from 1 to 40 are (5, 10, 15, 20, 25, 30, 35, 40) 20. If two angles of a triangle are equal to the So, number of favourable outcomes = 8 corresponding two angles of another triangle \\ Required probability then in such case two triangles are similar by AA similarity criteria. = Number of favourable outcomes Total noumber of outcomes 21. Given, DAPB is a right triangle, right angled = 8 =1 at P. 40 5 \\ AB2 = AP2 + BP2 …(i) 24. In DABC and DPQR, The distance between vertices of a triangle are ÐC = ÐR = 35° AB = (11 - 5)2 + (- 5 - 3)2 ÐB = ÐQ = 90° \\ DABC ~ DPQR [by AA similarity criterian] AP = (12 - 5)2 + (y - 3)2 CR BP = (12 - 11)2 + (y + 5)2 35° 35° \\From Eq. (i), 90° 90° (11 - 5)2 + (- 5 - 3)2 = (12 - 5)2 + (y - 3)2 A BP Q + ( 12 - 11)2 + (y + 5)2 25. Given, x + y -4 =0 Þ 62 + (- 8)2 = 72 + y2 - 6y + 9 + 1 + y2 + 10y + 25 and 2 x + ky - 3 = 0 Þ 2 y2 + 4y + 84 = 100 Here, on comparison the above equation with standard equation, we get Þ y2 + 2y - 8 = 0 a1 = 1, b1 = 1, c1 = - 4 Þ (y + 4) (y - 2 ) = 0 Þ y = - 4 or 2 a2 = 2 , b2 = k , c2 = - 3 22. Since, DPQR is right angled triangle. For no solution, P a1 = b1 ¹ c1 a2 b2 c2 4 cm 5 cm Þ 1 = 1 ¹ -4 SAMPLE PAPER 7 Q 3 cm R 2 k -3 From Pythagoras theorem, Þ 1 = 1 and 1 ¹ 4 PR2 = PQ2 + QR2 2k k3 Þ QR2 = PR2 - PQ2 Þ k = 2 and k ¹ 3 Þ k = 2 Þ 32 = PR2 - PQ2 4 Þ PR2 - PQ2 = 9 [Q QR = 3] 26. In DOAB, ÐAOB = 60° [given] Þ ( PR + PQ) ( PR - PQ) = 9 and OA = OB {Q a2 - b2 = (a + b)(a - b)} \\ DAOB is an equilateral triangle.

CBSE Sample Paper Mathematics Standard Class X (Term I) 147 So, Area of triangle = 3 ´ 52 = 25 3 cm2 31. In this figure, if D is mid-point of BC, then 44 DC = DB [Q Area of equilateral D = 3 (side)2] 4 In DACD, cot x = AC …(i) and In DACB, CD …(ii) Area of shaded portion = Area of circle - Area of triangle cot y = AC CB = pr2 - 25 3 Since, D is the mid-point of BC 4 Þ BC = 2 CD = æ 22 ´ 5 ´ 5 - 25 3 ö cm 2 çç 7 4 ÷÷ Dividing Eq. (ii) by Eq. (i), we get è ø AC = 78.57 - 10.82 = 67.75 cm2 cot y = CB = CD = CD = 1 cot x AC CB 2 CD 2 27. Let the points A (0, 0), B(2, 0) and C(0, 2) be the vertices of the required triangle. CD \\ cot y = 1 Then, the distance between points is AB = (2 - 0)2 + (0 - 0)2 = 4 + 0 = 2 cot x 2 BC = (0 -2 )2 + (2 -0)2 = 4 + 4 = 8 = 2 2 32. In DABC and DPQR, CA = (0 - 0)2 + (0 - 2 )2 = 0 + 4 = 2 3 = 5 =4 =2 Q \\ Required perimeter = 2 + 2 2 + 2 4.5 7.5 6 3 \\ AC = AB = BC PR PQ RQ = (4 + 2 2 ) units So, by SSS similarity criteria, DABC ~ DPQR. 28. The given rational number is 43 33. According to the situation, 24 ´ 53 As the denominator is of the form 2 n ´ 5m P k:1 Q where m and n are integers. (6,–8) (–3,10) R So, it has terminating decimal expansion (–1,6) Now, 43 = 43 ´ 5 = 215 By section formula, 24 ´ 53 24 ´ 54 (2 ´ 5)4 æ -3k + 6 , 10k - 8 ö = (-1,6) 215 215 çç k +1 k +1 ÷÷ = 104 = 10000 = 0.0215 è ø Þ - 3k + 6 = - 1 and 10k - 8 = 6 29. Total possible outcomes are k +1 k +1 {HHH, HHT, HTH,THH,TTH,THT, HTT,TTT} On further solving, we get i.e. 8 - 3k + 6 = - k - 1 Let E be the event of getting exactly 1 head. and 10k - 8 = 6k + 6 Þ k = 7 and k = 7 \\ Outcomes favourable to E are {TTH,THT, HTT} i.e. 3 22 \\ P( E) = Favourable outcomes = 3 \\ Required ratio = 7 : 2 Total possible outcomes 8 30. The condition for parallel lines is a1 = b1 ¹ c1 34. There are 13 letters in the word SAMPLE PAPER 7 a2 b2 c2 ‘ASSASSINATION’ out of which one letter can be chosen in 13 ways. On comparing the given equations with Hence, total number of outcomes = 13 standard equation, we get There are 6 vowels in the given word. a1 = 3, b1 = 4, c1 = 5 Hence, required probability = 6 Here, a2 = 6, b2 = 8, c2 = 9 13 Þ But given that, 6 = 6 a1 = 3 = 1 , b1 = 4 = 1 and c1 = 5 a2 6 2 b2 8 2 c2 9 2 x + 1 13 a1 = b1 ¹ c1 a2 b2 c2 Þ 2 x + 1 =13 Þ 2 x = 12 Hence, the equation represents parallel lines. Þ x =6

148 CBSE Sample Paper Mathematics Standard Class X (Term I) 35. Given, side of square = 6 cm 40. Let the radius of inner track be r. \\ 2 pr = 352 Þ r = 352 6 cm 2p \\Diameter of circle (d) = side of square Let the radius of outer track be R. \\ 2 pR = 396 Þ R = 396 = 6 cm \\Radius of circle (r) = d = 6 = 3 cm 2p 22 Width of the track \\Area of circle = pr2 = p(3)2 = 9p cm 2 = R - r = 396 - 352 = 44 2p 2p 2p 36. In DABC and DPQR, = 44 ´ 7 = 7 m 2 ´ 22 ÐA = ÐP = 53º Solutions (41-45) and AC = BC é 6 = 4 =2ù 41. Prime factors of 616 = 2 3 ´ 7 ´ 11 PR RQ êëQ 4 6 3 ûú 2 616 Here two sides are proportional and one angle 2 308 in equal but this angle is not between the sides. 2 154 7 77 So, triangles are not similar. 11 11 1 37. Let the present age of father be x years and 42. Factors of 32 = 2 5 present age of son be y years. Factors of 616 = 2 3 ´ 7 ´ 11 According to the problem, \\HCF (32, 616) = 2 3 = 8 43. Factors of 56 = 2 3 ´ 7 x = 6y …(i) Factors of 1000 = 2 3 ´ 53 \\LCM(56,1000) = 2 3 ´ 53 ´ 7 = 7000 After 4 yr, x + 4 = 4(y + 4) 44. Prime factor of 1000 = 2 3 ´ 53 x + 4 = 4y + 16 45. HCF of 56 and 1000 is the required number of column in which army can march Þ x - 4y = 12 …(ii) Factors of 56 = 2 3 ´ 7 Factors of 1000 = 2 3 ´ 53 Put x = 6y in Eq. (ii), we get \\ HCF (56, 1000) = 2 3 = 8 6y - 4y = 12 Solutions (46-50) Þ 2 y = 12 46. The shape of the path traced shown in the given figure is the form of parabola. Þ y =6 47. The graph of parabola opens downwards, and x = 6y = 36 if a < 0. \\Present age of son = 6 yr Present age of father = 36 yr 38. Central angle of a semicircle = 360° = 180° 2 Now, the supplementary angle will be = 180° - 180° = 0° SAMPLE PAPER 7 48. In the given graph, we see that curve intersect the X-axis at four points. Hence, number of zeroes in the given polynomial are 4. 39. Here, AQ : BQ = 2 : 1. 49. The given curve intersect the X-axis at points x = -3, - 1, 2 and 4. A P Q B Hence, four zeroes in the given graph are (7, –2) (5, –3) (3, y) (1, –5) - 3, - 1, 2, 4. Then, by section formule, 50. Option (a) is biquadratic polynomial. Option (b) is cubic polynomial. y = 2 ´ (- 5) + 1 ´ (- 2 ) = - 10 - 2 = -12 = - 4 Option (c) is quadratic polynomial. 2 +1 33 Option (d) is linear polynomial.

CBSE Sample Paper Mathematics Standard Class X (Term I) 149 SAMPLE PAPER 8 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. 1242 is a decimal 343 (a) non-terminating repeating (b) non-terminating non repeating (c) terminating (d) None of the above 2. A quadratic polynomial having 5 and - 3 as zeroes is (a) x 2 - 2 x - 15 (b) x 2 - 2 x + 15 (c) x 2 + 2 x + 15 (d) x 2 + 2 x - 15 3. The value of x in the given figure is P 10 cm 35° 35° 8 cm SAMPLE PAPER 8 Q 5 cm S x cm R (a) 5 cm (b) 4 cm (c) 2 cm (d) 3 cm 4. If the circumference of two circles are in the ratio of 3 : 4, then the ratio of their areas is (a) 9 : 16 (b) 16 : 9 (c) 9 : 17 (d) 7 : 17

150 CBSE Sample Paper Mathematics Standard Class X (Term I) 5. The coordinates of the point which divides the line segment joining the points (4, - 3) and (8, 5) in the ratio 1 : 3 internally are (a) (4, 3) (b) (7, 3) (c) (3, 5) (d) (5, - 1) 6. The value of æèç 11 - 11 ø÷ö is cot 2 cos 2 q q (a) 11 (b) 0 (c) 1 (d) - 11 11 7. HCF of (23 ´ 32 ´ 5), (22 ´ 33 ´ 52 ) and (24 ´ 3 ´ 53 ´ 7) is (a) 30 (b) 48 (c) 60 (d) 105 8. The difference between the circumference and the radius of a circle is 37 cm. The area of the circle is (a) 149 cm2 (b) 154 cm2 (c) 121 cm2 (d) 169 cm2 9. If a and b are the zeroes of f (x) = 2x 2 + 8x - 8, then (a) a + b = ab (b) a + b > ab (c) a + b < ab (d) a + b + ab = 0 10. It is given that DABC ~ DDFE, ÐA = 50°, ÐC = 30°, AB = 10 cm, AC = 15 cm and DF = 8 cm. Then, which of the following is true? (b) DE = 12 cm and ÐF = 100° (a) DE = 12 cm and ÐF = 50° (d)EF = 12 cm and ÐD = 30° (c) EF = 12 cm and ÐD = 100° 11. If tan q = 3, then cos2 q - sin 2 q is equal to (c) - 7 25 4 (a) 7 (b) 1 (d) 4 25 25 12. C is the mid-point of PQ, if P is (4, x),C is (y, - 1) and Q is (- 2, 4), then x and y respectively are (b) - 6 and 2 (c) 6 and - 1 (d) 6 and - 2 (a) - 6 and 1 13. The radius of a wheel is 0.25 m. The number of approximate revolutions it will make to travel a distance of 11 km, is ____ . (a) 5000 (b) 7000 (c) 6000 (d) 1000 14. Which of the following is the decimal expansion of an irrational number? (a) 4.761 (b) 0.32 (c) 5.010010002… (d) 6.030303… 15. In two triangles DABC and DDEF, ÐA = ÐE and ÐB = ÐF. Then, AB is equal to AC (a) DE (b) ED (c) EF (d) EF SAMPLE PAPER 8 DF EF ED DF 16. The sum and product of zeroes of a quadratic polynomial are 0 and 3 respectively. The quadratic polynomial is (a) x2 - 3 (b) x 2 + 3 (c) x2 - 3 (d) None of these 17. If 16 cot x = 12, then sin x - cos x equal sin x + cos x (a) 1 (b) 3 (c) 2 (d) 0 7 7 7

CBSE Sample Paper Mathematics Standard Class X (Term I) 151 18. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is (a) 22 : 7 (b) 14 : 11 (c) 7 : 22 (d) 11 : 14 19. The centre of a circle is (2a, a - 7). The value of a, if the circle passes through the point (11, - 9) and has diameter 10 2 units is (a) 3 or 6 (b) 5 or 3 (c) 7 or 4 (d) None of these 20. ABCD is a trapezium such that BC||AD and AB = 4 cm. If the diagonals AC and BD intersect at O such that AO = OB = 1, then CD is equal to OC OD 2 (a) 7 cm (b) 8 cm (c) 9 cm (d) 6 cm Section B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. In the given figure, AD = 3 cm, BD = 4 cm and CB = 12 cm, then sin q equals A 90° q 90° D CB (a) 3 (b) 5 (c) 4 (d) 12 4 13 3 5 22. A race track is in the form of a circular ring whose outer and inner circumferences are 396 m and 352 m respectively. The width of the track is (a) 63 m (b) 56 m (c) 7 m (d) 3.5 m 23. If a and b are the zeroes of the polynomial x 2 - 5x + c and a - b = 3, then c is equal to (a) 0 (b) 1 (c) 4 (d) 5 24. The least number which when divided by 18, 24, 30 and 42 will leave in each case the same remainder 1, would be (a) 2520 (b) 2519 (c) 2521 (d) None of these 25. The coordinates of the point P which divides the joining the points A(2, 5) and B(3, - 5) in the ratio 2 : 3, are (a) (1, 0) (b) çèæ152 , 1÷øö (c) (3, 0) (d) (0, 1) 26. If sin A + cosec A = 2, then sin 2 A + cosec2 A is equal to SAMPLE PAPER 8 (a) 2 (b) 3 (c) 1 (d) 4 27. A man goes 15 m due West and then 8 m due North. How far is he from the starting point? (a) 12 m (b) 17 m (c) 18 m (d) 24 m 28. If in DABC, DE||BC, then (a) AD = AE (b) AB = AE (c) AB = AC (d) None of these AB AC AD AC AD AD

152 CBSE Sample Paper Mathematics Standard Class X (Term I) 29. If a pair of linear equations is consistent, then the lines will be (a) always coincident (b) parallel (c) always intersecting (d) intersecting or coincident 30. If a and b are the zeroes of the polynomial f (x) = x 2 - p(x + 2) - q, then (a + 1) (b + 1) is equal to (b) 1 - q - p (c) q (d) 1 + q (a) q - 1 31. If P(- 2, 5) and Q(3, 2) are two points. The coordinates of the point R on PQ such that PR = 3QR are (a) Rèæç 4 , 3÷øö (b) Ræçè 7 , 141÷öø 3 4 (c) Rçèæ31 , 7øö÷ (d) None of these 32. The LCM and HCF of two non-zero positive numbers are equal, then the numbers must be (a) prime (b) coprime (c) composite (d) equal 33. The area of the shaded region in the given figure, if AC = 24 cm, BC = 10 cm and O is the centre of the circle, is [take p = 3.14] A O (a) 128.56 cm 2 BC (c) 248.16 cm 2 (b) 145.33 cm 2 (d) None of these 34. If in an equilateral triangle, the length of the median is 3 cm, then the length of the side of equilateral triangle is (a) 1 cm (b) 2 cm (c) 3 cm (d) 4 cm 35. If DABC is right angled at C, then the value of sin(A + B) is (a) 0 (b) 1 (c) 1 (d) 3 2 2 SAMPLE PAPER 8 36. If one of the zeroes of a quadratic polynomial of the form x 2 + ax + b is the negative of the other, then it (a) has no linear term and the constant term is negative (b) has no linear term and the constant term is positive (c) can have a linear term but the constant term is negative (d) can have a linear term but the constant term is positive 37. The area of the largest circle that can be drawn inside the given rectangle of length ‘a’ cm and breadth ‘b’ cm (a > b) is (a) 1 p b2 cm 2 (b) 1 p b2 cm 2 (c) 1 p b2 cm 2 (d) p b2 cm 2 2 3 4

CBSE Sample Paper Mathematics Standard Class X (Term I) 153 38. If X = 28 + (1 ´ 2 ´ 3 ´ 4 ´¼ ´ 16 ´ 28) and Y = 17 + (1 ´ 2 ´ 3 ´¼ ´ 17), then which of the following is true? (a) X is a prime number (b) Y is a prime number (c) X - Y is a prime number (d) X - Y is a composite number 39. The value(s) of y for which the distance between the points P(y, 3) and Q(9, 10) is (10 - y) units, is (a) 10 (b) 20 (c) - 15 (d) None of these 40. If cos 30°× tan 30° = sin a and a < 90°, then the value of tan 2a is (a) 1 (b) 3 (c) 1 (d) 0 3 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 Vikas is working with TCS, and he is sincer and dedicated to his work. He pay all his taxes on time and invest the some amount of his salary in funds for his future. He invested some amount at the rate of 12% simple interest and some other amount at the rate of SAMPLE PAPER 8 10% simple interest. He received yearly interest of ` 130. But, if he interchange the amounts invested, he would have received ` 4 more as interest. Now, answer the following questions 41. Consider the amount invested at 12% be p and at 10% be q. Then, formulate the required linear equation for first condition. (a) 10 p + 12 q = 13400 (b) 8 p + 15q = 12000 (c) 12 p + 10q = 13000 (d) 5 p + 6q = 6000 42. Now, formulate the linear equation for the second condition? (a) 12 p + 10q = 13000 (b) 10 p + 12 q = 13400 (c) 5 p + 6q = 12000 (d) 8 p + 15q = 6000

154 CBSE Sample Paper Mathematics Standard Class X (Term I) 43. What is value of the variable p? (a) ` 700 (b) ` 500 (c) ` 300 (d) ` 150 44. What is the value of the variable q? (a) ` 700 (b) ` 500 (c) ` 300 (d) ` 150 45. If the rate of interest changes to 15% for first amount and 12% for second amounts, how much amount of money is earned? (a) ` 170 (b) ` 150 (c) ` 148 (d) ` 159 46-50 are based on Case Study-2. Case Study 2 Sanjeev sees a game in a fair. He is keenly interested to play it. He asks the rules of game from the owner. Owner says that you will move this spinner first, if it stops on a prime number, then you are allowed to pick a marble from a bag which contain 14 white and 10 black marbles. Prizes are given when a white marble is picked randomly. 2 4 8 7 13 5 10 11 Based on the above information, answer the following questions: 46. What is the probability that Sanjeev will not be allowed to pick a marble from the bag? (a) 3 (b) 5 (c) 1 (d) 1 8 8 2 47. What is the probability that Sanjeev will be allowed to pick a marble from a bag? (a) 3 (b) 1 (c) 5 (d) 3 8 2 8 7 48. The probability that Sanjeev will pick a black marble from the bag is (a) 7 (b) 5 (c) 1 (d) 2 12 12 2 3 49. The probability that Sanjeev will get prize is (a) 3 (b) 3 (c) 5 (d) 7 5 4 12 12 SAMPLE PAPER 8 50. The probability that Sanjeev will pick a red marble from the bag is (a) 1 (b) 0 (c) 5 (d) 7 12 12

OMR SHEET SP 8 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)

156 CBSE Sample Paper Mathematics Standard Class X (Term I) Answers 1. (b) 2. (a) 3. (b) 4. (a) 5. (d) 6. (d) 7. (c) 8. (b) 9. (a) 10. (b) 11. (a) 12. (a) 13. (b) 14. (c) 15. (c) 16. (b) 17. (a) 18. (b) 19. (b) 20. (b) 21. (b) 22. (c) 23. (c) 24. (c) 25. (b) 26. (a) 27. (b) 28. (a) 29. (d) 30. (b) 31. (b) 32. (d) 33. (b) 34. (b) 35. (b) 36. (a) 37. (c) 38. (d) 39. (c) 40. (b) 41. (c) 42. (b) 43. (b) 44. (a) 45. (d) 46. (a) 47. (c) 48. (b) 49. (d) 50. (b) SOLUTIONS 1. 1242 = 1242 = æèç 240 , -4 øö÷ = (5, - 1) 343 73 4 Here, Denominator has one factor namely 7, 6. We have, which is other than 2 and 5. 11 - 11 = 11éëê 1 q - 1ù So, given rational number is a non-terminating cot2 q cos2 q cot 2 cos2 q úû non-repeating decimal. = 11[tan2 q - sec2 q] 2. Let a = 5 and b = - 3 \\Required polynomial is x2 - (a + b)x + ab é sec2 q - tan2 q = 1 ù êQ tan2 q - sec2 q = - ú = x2 - (5 - 3)x + 5(- 3) = x2 - 2 x - 15 êë Þ 1ûú 3. In DPSQ and DPSR, = 11 ´ (- 1) = - 11 ÐQPS = ÐSPR = 35° [Q PS ^ QR] 7. Given, (2 3 ´ 32 ´ 5), (2 2 ´ 33 ´ 52 ) and ÐPSQ = ÐPSR (2 4 ´ 3 ´ 53 ´ 7) From AA similarity \\HCF of above expressions = 2 2 ´ 3 ´ 5 DPSQ ~ DPSR = 4 ´ 3 ´ 5 = 60 \\ PQ = SQ PR SR 8. Let r be the radius of the circle. Þ 10 = 5 Given, circumference of circle 8x - radius of the circle = 37. Þ x = 4 cm Þ 2 pr - r = 37 Þ r(2 p - 1) = 37 4. Let r1 and r2 are radii of two circle. Then Þ r = 37 p- according to the given condition, 2 1 2 pr1 = 3 = 37 = 37 ´ 7 = 7 cm 2 pr2 4 2 èçæ272 öø÷ 37 - 1 Þ r1 = 3 r2 4 \\ Area of circle = pr2 = 22 ´ (7)2 = 154 cm2 7 pr12 èçæ r1 ö÷ø 2 æèç 3 ø÷ö 2 9 pr22 r2 4 16 Now, = = = 9. Given, 5. Given, a and b are the zeroes of f (x) = 2 x2 + 8x - 8 SAMPLE PAPER 8 (x1 , y1 ) = (4, - 3) and (x2 , y2 ) = (8, 5) \\a + b = - Coefficienct of x = - (8) = - 4 Coefficient of x2 2 Let (x, y) be the coordinates of the point which Constant term -8 divides the line joining the points (x1, y1 ) and and a ×b = Coefficient of x2 = 2 = - 4 (x2 , y2 ) in ratio m: n = 1 : 3 internally. çæè + + öø÷ So, (x, y) = mx2 + nx1 , my2 + ny1 So, a + b = a ×b m n m n 10. Given, DABC ~ DDFE = çæè 1(8) + 3(4) , 1(5) + 3(- 3)÷øö 1 + 3 1+3

CBSE Sample Paper Mathematics Standard Class X (Term I) 157 A D = 2 ´ 22 ´ 0.25 = 1.57 7 50° 15 cm Q Number of revolutions 10 cm = Distance covered by wheel 8 cm Circumference of wheel = 11 ´ 1000 ~- 7000 (approximate) B 30° C F 30° E 1.57 Then, ÐA = ÐD = 50° 14. An irrational number has non repeating non terminating decimal expansion. ÐC = ÐE = 30° \\ ÐB = ÐF = 180° - (30° + 50° ) = 100° Here, (a) is terminating decimal expansion and (b) and (d) are non-terminating but repeating Also, AB = AC decimal expansion. DF DE Þ 10 = 15 \\(c) is non terminating non-repeating decimal 8 DE expansion. Þ DE = 15 ´ 8 = 12 cm 10 So, (c) is the decimal expansion of irrational number. Hence, DE = 12 cm and ÐF = 100° 15. In DABC and DEFD, 11. We have, tanq = AB = 3 AC 4 ÐA = ÐE and ÐB = ÐF [given] \\ DABC ~ DEFD Let AB = 3k and BC = 4k, where k is positive constant. [by AA similarity criterion] \\ AB = AC Þ AB = EF \\ AC2 = BC2 + AB2 EF ED AC ED Þ AC2 = 9k 2 + 16k 2 = 25k 2 16. Given, Sum of zeroes = 0 Þ AC = 5k C and product of zeroes = 3 5k We know that, 3k f (x) = x2 - (a + b)x + a ×b = x2 - 0 × x + Aq B So, quadratic polynomial is (x2 + 3 ). 3 4k \\ cosq = AB = 4k = 4 17. Let DABC be the right triangle such that AC 5k 5 ÐB = 90° and ÐC = x and sinq = BC = 3k = 3 A AC 5k 5 cos2 q sin2 q æèç 4 ö÷ø 2 èæç 3 ÷øö 2 16 9 7 5 5 25 25 25 \\ - = - = - = 4k 5k 12. Given, C is the mid-point of PQ i.e. P(4, x) and B xC 3k Q(- 2 , 4). 4-2 x+ 2 2 Given, cot x = 12 = 3 Therefore (y, - 1) = æèç , 4 ø÷ö 16 4 Þ (y, 1) = çèæ1, x + 4 ÷øö \\ cot x = BC = 3 2 AB 4 On equating the coordinate, we get Let BC = 3k and AB = 4k, where k is positive SAMPLE PAPER 8 y = 1 and -1 = x + 4 constant. 2 \\ AC = AB2 + BC2 \\ x = - 6 and y = 1 [use Pythagoras theorem] = 16k 2 + 9k 2 1:1 P C Q = 25k 2 = 5k (4, x) (y, –1) (–2, 4) So, sin x = AB = 4k = 4 13. The distance covered in one revolution is equal AC 5k 5 …(i) to the circumference of the wheel. …(ii) Circumference of wheel = 2 pr cos x = BC = 3k = 3 AC 5k 5

158 CBSE Sample Paper Mathematics Standard Class X (Term I) sin x - cos x 4 - 3 4-3 1 Þ a2 - 8a + 15 = 0 [divide by 5] sin x + cos x 5 + 5 4+3 7 Now, = 4 3 = = Þ a2 - 5a - 3a + 15 = 0 55 [by factorization method] [Using Eqs. (i) and (ii)] Þ a(a - 5) - 3(a - 5) = 0 Þ (a - 5)(a - 3) = 0 Hence, the value of sin x - cos x is 1. Þ a = 3, 5 sin x + cos x 7 Hence, the required values of a are 5 and 3. 18. Let r be the radius of a circle and a be the side 20. In DAOB and DCOD, we have ÐAOB = ÐCOD of a square. Given, Perimeter of a circle = Perimeter of a [vertically opposite angles] square 2 pr = 4a Þ a = pr …(i) AO = OB = 1 [given] \\ 2 OC OD 2 Area of circle = pr2 AD Area of square = a2 O Now, Area of a circle = pr2 = pr2 BC Area of a sqaure (a)2 çèæ pr ÷øö 2 2 \\DAOB~ DCOD [by SAS similarity criterion] [From Eq. (i)] \\ AB = AO = pr2 CD OC p2r2 / 4 Þ 4 = 1 Þ CD = 8 cm = 4 = 4 = 28 = 14 CD 2 p 22 / 7 22 11 21. We have, AD = 3 cm, BD = 4 cm and CB = 12 cm \\The required ratio will be 14 : 11. In DABD, 19. By using the condition, AB2 = BD2 + AD2 Distance between the centre (2 a, a - 7) and the [from Pythagoras theorem] point P(11, - 9) = Radius of circle. AB2 = 42 + 32 AB = 25 = 5 m C (2a, a–7) Now, In DABC Radius AC2 = AB2 + BC2 P (11, –9) = (5)2 + (12 )2 \\ Radius of circle = (11 - 2 a)2 + (- 9 - a + 7)2 In DABC, = 25 + 144 …(i) = 169 = 13 cm [Q Distance between two points (x1 , y1 ) [from Pythagoras theorem] and (x2 , y2 ) = (x2 - x1 )2 + (y2 - y1 )2 ] ÐB = 90º sinq = AB = 5 Given that, length of diameter = 10 2 \\Length of radius = length of diameter BC 13 SAMPLE PAPER 8 2 22. Let R and r be the outer and inner radii of the = 10 2 = 5 2 track respectively. 2 Given, outer and inner circumference be 396 m Put this value in Eq. (i), we get and 352 m. 5 2 = (11 - 2 a)2 + (- 2 - a)2 R Squaring on both sides, we get r 50 = (11 - 2 a)2 + (2 + a)2 Þ 50 = 121 + 4a2 - 44a + 4 + a2 + 4a \\ 2 pR = 396 …(i) Þ 5a2 - 40a + 75 = 0 and 2 pr = 352 …(ii) On subtracting Eq. (ii) from Eq. (i), we get

CBSE Sample Paper Mathematics Standard Class X (Term I) 159 2 pR - 2 pr = 396 - 352 C N 8m WE Þ 2 p( R - r) = 44 S \\Width of the track = outer radius - inner radius = R - r= 44 = 44 =7 m 2p èæç2 ´ 22 ÷öø B 15 m A 7 The distance between starting point A and end 23. Given, a - b = 3 … (i) point C is \\ Given polynomial is x2 - 5x + c and AC = ( AB)2 + ( BC)2 a , b are the zeroes of the polynomial. [Apply pythagoras theorem] AC = (15)2 + (8)2 = 225 + 64 \\ a + b = - Coefficient of x = - (- 5) …(ii) Coefficient of x2 = 289 = 17 m Adding Eqs. (i) and (ii), we get 28. Again, in DABC, DE|| BC 2a = 8 AD = AE Þ a=4 \\ DB EC [from thales theorem] Þ b = 5-4=1 Þ DB + 1 = EC + 1 AD AE And a ×b = constant term coefficient of x2 Þ AD + DB = AE + EC AD AE Þ 4´1= c Þ c= 4 1 24. The factors of the given numbers are A 18 = 2 ´ 32 24 = 2 3 ´ 3 DE 30 = 2 ´ 3 ´ 5 42 = 2 ´ 3 ´ 7 BC LCM (18, 24, 30, 42) = 2 3 ´ 32 ´ 5 ´ 7 Þ AB = AC = 2520 AD AE As the remainder is 1. Þ AD = AE Hence, the required number is 2520 + 1 = 2521 AB AC 25. Let the coordinates of P be (x, y). Using section formula, 29. If a pair of linear equations is consistent, then it has unique solution and infinitely solutions. (x, y) = èæç mx2 + nx1 , my2 + my1 ö÷ø m + n m+ n For unique solution a1 ¹ b1 = èçæ 2(3) + 3(2 ) , 2(- 25)++33(5)ö÷ø a2 b2 2 + 3 \\The lines will be intersecting. = èæç 12 , 5 ÷öø 5 5 For infinitely solutions, a1 = b1 = c1 = èçæ 12 , 1÷øö a2 b2 c2 SAMPLE PAPER 8 5 \\The lines will be coincident. 26. Given, sin A + cosec A = 2 Squaring both sides, we get 30. Given, a and b are the zeroes of the sin2 A + cosec2 A + 2 sin A× cosec A = 4 polynomial Þ sin2 A + cosec2 A + 2 × sin A× 1 = 4 sin A f (x) = x2 - p(x + 2 ) - q Þ sin2 A + cosec2 A = 4 - 2 = 2 = x2 - px - (2 p + q) (- p) 1 \\ a + b = - Coefficient of x = - = p Coefficient of x2 27. Let A be starting point of man. A man goes 15 m West and 8 m North. and a ×b = - (2 p + q)

160 CBSE Sample Paper Mathematics Standard Class X (Term I) Now, (a + 1) (b + 1) = ab + a + b + 1 = 265.33 - 120 = - (2 p + q) + p + 1 = 145.33 cm2 = - 2 p - q + p + 1 = 1q = - p- q+1 34. Let a be the side of equilateral triangle. =1- p- q Median is also the altitude of an equilateral 31. Given, coordinate of P(- 2 , 5) and coordinate triangle. A of Q(3, 2 ). aa Since PR = 3QR 31 Altitude P R Q (–2, 5) (3, 2) B a/2 D a/2 C Therefore point R divides PQ in the ratio 3 : 1. In DADC, Then, coordinates of R are (Altitude)2 + çèæ a ÷øö 2 = a2 2 R = é èæç 1 ´ (- 2) + 3 ´ 3 ÷öø, çèæ 1 ´ 5 + 3 ´ 2 ö÷ø úùû ê 3+ 1 1 + 3 ë ( 3 )2 + a2 = a2 4 [by section formula] Þ 12 + a2 = a2 R = é æèç - 2+ 9 øö÷, èæç 5 + 6 øö÷ ù 4 ëê 4 4 ûú Þ 12 + a2 = 4a2 = çèæ 7 141ø÷ö 4, Þ 3a2 = 12 32. Given, LCM and HCF are equal. Þ a2 = 4 Let two non-zero positive number are p and q. Þ a=±2 Then, HCF ( p, q) = LCM ( p, q) [given] [Q a = - 2 is not possible] Let HCF ( p, q) = k \\ Side of triangle = 2 cm Þ p = ka 35. Given, A and q = kb Where a and b are natural numbers. 90° B Q HCF ´ LCM = product of numbers C k ´ k = ka ´ kb In DABC is right angled at C, ÐC = 90° Þ a´b=1 \\a = b = 1 as they are natural number. We know that Hence, p = q or the number must be equal. ÐA + ÐB + ÐC = 180° 33. Given, AC = 24 cm, BC = 10 cm SAMPLE PAPER 8 We know that angle in a semi-circle = 90° [sum of interior angles of triangle] ( AB)2 = ( AC)2 + ( BC)2 ÐA + ÐB = 180° - 90° = 242 + 102 ÐA + ÐB = 90° sin(ÐA + ÐB) = sin90° = 1 = 576 + 100 = 676 cm 36. If one of the zeroes of the equation is negative So, AB = 26 cm, of the other, then the zeroes will be m and- m. Hence, sum of zeroes = m - m = - a Þ a = 0 AO = 13 cm = radius of the circle Thus, there will be no linear term involved in Area of shaded region = Area of semicircle the equation. Also, the constant term = b = product of zeroes - Area of DABC = pr2 - 1 ´ BC ´ AC = - m2, will be negative. 22 = p ´ (13)2 - 1 ´ 10 ´ 24 22 = 3.14 ´ 169 - 120 2

CBSE Sample Paper Mathematics Standard Class X (Term I) 161 37. The largest circle that can be drawn inside a Þ 12 p + 10q = 13000 rectangle is possible when rectangle becomes a 42. Vikas received ` 4 extra, if he interchange the investment amount. square. \\According to the situation, \\Diameter of the circle = Breadth of rectangle 10 p + 12 q = 130 + 4 100 100 =b \\Radius of the circle = b / 2 Þ 10 p + 12 q = 13400 Hence, area of the circle = pr2 = p ´ b2 cm 2 43. Given, from above situation 4 38. Given, 12 p + 10q = 13000 …(i) X = 28 + (1 ´ 2 ´ 3 ´ 4 ´ ¼ ´ 16 ´ 28) Y = 17 + (1 ´ 2 ´ 3 ´ ¼ ´ 17) 10 p + 12 q = 13400 …(ii) X can be rewritten as Adding Eqs. (i) and (ii), we get X = 28 + [28(1 ´ 2 ´ 3 ´ ¼ ´ 16)] = 28 [1 + (1 ´ 2 ´ 3 ´ ¼ ´ 16)] 22 p + 22 q = 26400 Similarly, Y = 17 + [17(1 ´ 2 ´ ¼ ´ 16)] Þ p + q = 1200 …(iii) = 17[1 + (1 ´ 2 ´ 3 ´ ¼ ´ 16)] Now, subtracting both Eqs. (i) and (ii), we get Hence, both X and Y are composite number and X - Y is also a composite number. 2 p - 2 q = - 400 39. Given : P(y, 3) and Q(9, 10), d = 10 units Þ p - q = - 200 …(iv) Now, on adding Eqs. (iii) and (iv), we get 2 p = 1000 Þ p = ` 500 We know that 44. Put the value of ‘p’ in Eq. (iii) D= (x2 - x1 )2 + (y2 - y1 )2 p = 1200 - 500 = ` 700 10 - y = (9 - y)2 + (10 - 3)2 45. Now, the rate of interest is 15% and 12%. 10 - y = 81 + y2 - 18y + 49 \\ 15 p + 12 q = earned money 100 100 10 - y = 130 + y2 - 18y Þ earned money = 15 ´ 500 + 12 ´ 700 100 100 Squaring both side (10 - y)2 = 130 + y2 - 18y = 75 + 84 100 + y2 - 20y = 130 + y2 - 18y = ` 159 - 20y + 18y = 130 - 100 - 2 y = 30 Solutions (46-50) y = - 30 = - 15 46. Total numbers on the spinner = 8 2 Non-prime numbers are 4, 8, 10 i.e. 3 SAMPLE PAPER 8 40. Given, cos30°×tan30° = sina So, P(non-prime number) = 3 Þ 3 ´ 1 = sina 8 23 47. Required probability = 1 - 3 = 8 - 3 = 5 Þ sina = 1 Û a = 30° 88 8 2 48. Total number of marbles = 10 + 14 = 24 \\ tan2a = tan(2 ´ 30° ) Number of black marbles = 10 = tan60° = 3 \\ P(getting a black marble) = 10 = 5 24 12 Solutions (41-45) 49. Number of white marbles = 14 41. Vikas received ` 130 as profit. \\According to the situation, \\ P(getting a prize) = 14 = 7 24 12 12 p + 10 q = 130 100 100 50. There is no red marble in the bag. \\ P (getting a red marble) = 0

162 CBSE Sample Paper Mathematics Standard Class X (Term I) SAMPLE PAPER 9 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 value of HCF (8, 9, 25) ´ LCM (8, 9, 25) is (a) 50 (b) 1800 (c) 1810 (d) 1500 2. If (0, 0), (3, 3) and (3, p) are vertices of an equilateral triangle, then value of p is (a) - 3 (b) 3 (c) 3 (d) - 3 3. If DABC ~ DDEF, such that DE = 2AB and BC = 8 cm, then the length of EF is (a) 8 cm (b) 16 cm (c) 18 cm (d) 26 cm 4. If in a lottery, there are 5 prizes and 20 blanks, then the probability of getting a prize is (a) 2 (b) 4 (c) 1 (d) 1 5 5 5 5. The sum and the product of the zeroes of polynomial 6x 2 - 5 respectively are SAMPLE PAPER 9 (a) 0 and - 6 (b) 0 and 6 (c) 0 and 5 (d) 0 and - 5 5 5 6 6 6. For the equation y = a + b, where a and b are real numbers, if y = 1 when x = - 1 and y = 5 x when x = - 5, then a + b equals (a) - 1 (b) 0 (c) 11 (d) 10 7. If C(- 2, 1) is the mid-point of the line segment joining A(- 6, p) and B(2, p + 6), then the value of p is (a) 2 (b) - 2 (c) 0 (d) 4

CBSE Sample Paper Mathematics Standard Class X (Term I) 163 8. Name the criteria of similarity by which following triangles are similar. R C 4.5 3 54° 54° 7.5 Q A5 BP (a) SSS (b) SAS (c) AAA (d) ASA 9. If LCM of (a, b) = 53 and HCF of (a, b) = 12, then product of a and b is (a) 636 (b) 666 (c) 696 (d) 646 10. If P(E) is 0.75, what is P(not E)? (a) 0.35 (b) 0.25 (c) 0 (d) 1 11. If ax + by + c = 0, where a, b and c are real numbers, then for which condition the equation is said be a linear equation in two variables x and y. (a) a ¹ b (b) a2 = b2 (c) a2 + b2 = 0 (d) a2 + b2 ¹ 0 12. The number of polynomials having zeroes as - 2 and 5 is (a) 1 (b) 2 (c) 3 (d) more than 3 13. The perpendicular distance of the point P(4, 2) from the Y-axis is (a) 4 units (b) 6 units (c) 2 units (d) 8 units 14. The values of x and y in the given figure are 4 y3 x 7 (a) 10 and 14 (b) 21 and 84 (c) 21 and 25 (d) 10 and 40 15. A card is drawn from a deck of 52 cards. The event E is that card which is not an ace of heart. The number of outcomes favourable to E is (a) 4 (b) 13 (c) 48 (d) 51 16. Graphically ax + by + c = 0 represents a line. Every solution of the equation is a point (a) on the line representing (b) not on the line representing it (c) on the X-axis (d) on the Y-axis 17. For any two similar triangles which of the following statements are valid (a) their sides are proportional (b) their sides are equal SAMPLE PAPER 9 (c) their sides are parallel (d) None of these 18. If a and 1 are the zeroes of the polynomial ax 2 + bx + c, then value of c is a (a) 0 (b) a (c) - a (d) 1 19. The number 13 have decimal expansion 3125 (a) terminating (b) non-terminating repeating (c) non-terminating non-repeating (d) non-terminating

164 CBSE Sample Paper Mathematics Standard Class X (Term I) 20. In the figure, M is the mid-point of line LN. Then, x + y is equal to P(5, 2) (7, 3)L M(x, y) N(1, 4) (a) 7.5 (b) 3.5 (c) 4.5 (d) 5.5 Section B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. If DABC and DDEF are similar triangles such that ÐA = 57° and ÐE = 73°, then ÐC is equal to (a) 50° (b) 60° (c) 70° (d) 80° 22. A quadratic polynomial, whose zeroes are - 2 and 4, is (a) x2 - 2x + 8 (b) x 2 + 2 x + 8 (c) x2 - 2 x - 8 (d) 2 x 2 + 2 x - 24 23. The sum of the digits of a two-digit number is 9. If 27 is added to it, the digits of the number get reversed. The number is (a) 27 (b) 72 (c) 45 (d) 36 24. A school has five houses A, B, C, D and E. The class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single student is selected at random for the class monitor. The probability that the selected student is not from A, B and C is (c) 8 (a) 4 (b) 6 23 (d) 17 23 23 23 25. The points (- 5, 0), (5, 0) and (0, 4) are the vertices of (a) right triangle (b) equilateral triangle (c) isosceles triangle (d) scalene triangle 26. In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required, if in each room the same number of participants are to be seated and all of them being in the same subject. (a) 210 (b) 21 (c) 12 (d) 3780 27. DABC is a right triangle, right-angled at A and AD ^ BC. Then, BD is equal to DC (a) æ AB ö2 (b) AB (c) æ AB ö2 (d) AB ç ÷ AC ç ÷ AD è AC ø è AD ø SAMPLE PAPER 9 28. ……… is a solution of pair of equations 3x - 2y = 4 and 2x + y = 5. (a) x = 2 and y = 1 (b) x = 2 and y = 2 (c) x = 3 and y = 1 (d) x = 3 and y = 2 29. If 2 and 3 are zeroes of polynomial 3x 2 - 2kx + 2m, then the value of k and m are, respectively (a) 9 and 15 (b) 15 and 9 (c) 9 and 15 (d) 15 and 9 2 2 2 30. If four vertices of a parallelogram taken in order are (- 3, - 1), (a, b) , (3, 3) and (4, 3), then a : b is equal to (a) 1 : 4 (b) 4 : 1 (c) 1 : 2 (d) 2 : 1

CBSE Sample Paper Mathematics Standard Class X (Term I) 165 31. In a simultaneously throw of a pair of dice. The probability of getting a double is (a) 5 (b) 1 (c) 1 (d) 1 12 3 6 2 32. Two alarm clocks ring their alarms at regular intervals of 50 s and 48 s. If they first beep together at 12 noon, at what time will they beep again for the first time? (a) 12 : 20 pm (b) 12 : 12 pm (c) 12 : 11 pm (d) None of these 33. (cos4 A - sin 4 A) is equal to (a) 1 - 2 cos 2 A (b) 2 sin2 A - 1 (c) sin2 A - cos 2 A (d) 2 cos 2 A - 1 34. It is given that, DABC ~ DEDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm, then the sum of the remaining sides of the triangles is (a) 23.05 cm (b) 16.8 cm (c) 6.25 cm (d) 24 cm 35. The point of intersection of the line 3x - 2y = 6 and the X-axis is (a) (2, 0) (b) (0, - 3) (c) (- 2, 0) (d) (0, 3) 36. A fair dice is rolled. Probability of getting a number x such that 1 £ x £ 6 is (a) 0 (b) > 1 (c) between 0 and 1 (d) 1 37. If the points A(- 2, 5) and B(4, 3) are equidistant from the Y-axis, then the coordinates of that point are (a) (0, - 1) (b) (0, 1) (c) (1, 0) (d) None of these 38. Which of the following has a non-terminating repeating decimal expansion? (a) 3 (b) 13 (c) 7 (d) 29 8 125 80 343 39. If one of the zeroes of the quadratic polynomial (k - 1)x 2 + kx + 1 is - 3, then the value of k is (b) - 4 (c) 2 (d) - 2 (a) 4 3 3 3 3 40. In an equilateral triangle DABC , if AD ^ BC, then (a) 2 AB2 = 3 AD2 (b) 4 AB2 = 3 AD2 (c) 3 AB2 = 4 AD2 (d) 3 AB2 = 2 AD2 Section C Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted. SAMPLE PAPER 9 41-45 are based on Case Study-1. Case Study 1 While eating sandwich, Chetna jokingly remarked that she can find out the value of any trigonometric ratio, if just one ratio is known to her, as the sandwich is a right triangle. A 12 cm 6 cm C 6Ö3 cm B

166 CBSE Sample Paper Mathematics Standard Class X (Term I) On the basis of above information answer the following question. 41. The value of ÐC is equal to (b) 60° (d) None of these (a) 30° (c) 45° 42. The value of ÐA is equal to (b) 60° (d) None of these (a) 30° (c) 45° 43. The value of tan C is (b) 1 3 (a) 3 (d) None of these (c) 1 44. The value of tan C ´ tan A is (a) 2 (b) 1 2 (c) 1 (d) None of these 45. If 3 tan A = 4, then the value of 3 sin A + 2 cos A is 3 sin A - 2 cos A (a) 4 (b) 11 (c) 7 (d) 3 15 15 46-50 are based on Case Study-2. Case Study 2 Mr. Jay purchase a design which he has to decorate. In the design there are three semi-circles A, B and C having diameter 3 cm each, another larger semi-circle having diameter 9 cm and a circle D of diameter 4.5 cm D E AC B On the basis of above information, answer the following questions. 46. Total area of three semi-circles A, B and C is (a) 10.6 cm2 (b) 3.53 cm2 (c) 7.06 cm2 (d) 21 cm2 (d) 31.82 cm2 SAMPLE PAPER 9 47. Area of circle D is (b) 63.6 cm2 (c) 15.91 cm2 (d) 81 cm2 (d) 22.98 cm2 (a) 8 cm2 (d) 24.75 cm2 48. Area of largest semi-circle is (a) 15.91 cm2 (b) 31.82 cm2 (c) 63.64 cm2 49. Find the area of unshaded region. (a) 12.38 cm2 (b) 10.6 cm2 (c) 15.91 cm2 50. Find the area of shaded region. (a) 31.82 cm2 (b) 22.98 cm2 (c) 12.375 cm2

OMR SHEET SP 9 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)

168 CBSE Sample Paper Mathematics Standard Class X (Term I) Answers 1. (b) 2. (a) 3. (b) 4. (c) 5. (d) 6. (c) 7. (b) 8. (b) 9. (a) 10. (b) 11. (d) 12. (d) 13. (a) 14. (b) 15. (d) 16. (a) 17. (a) 18. (b) 19. (a) 20. (a) 21. (a) 22. (c) 23. (d) 24. (b) 25. (c) 26. (b) 27. (a) 28. (a) 29. (b) 30. (b) 31. (c) 32. (a) 33. (d) 34. (a) 35. (a) 36. (d) 37. (b) 38. (d) 39. (a) 40. (c) 41. (a) 42. (b) 43. (b) 44. (c) 45. (d) 46. (a) 47. (c) 48. (c) 49. (d) 50. (c) SOLUTIONS 1. Prime factorization of 8 , 9 and 25 are 6. Given, y = a + b …(i) 8 =2 ´2 ´2 x 9 =3´3 Put x = - 1 and y = 1 in above equation. and 25 = 5 ´ 5 1=a + b Þ a -b=1 \\HCF (8, 9, 25) = 1 -1 and LCM (8, 9, 25) = 2 3 ´ 32 ´ 52 = 1800 \\ HCF (8, 9, 25) ´ LCM (8, 9, 25) = 1 ´ 1800 Put x = - 5 and y = 5 in the given equation. 5 = a + b Þ 25 = 5a - b = 1800 -5 …(ii) 2. Let the points be A(0, 0), B(3, 3 ) and C(3, p), Subtract Eq. (i) from Eq. (ii) which are the points of equilateral triangle. 4a = 24 \\ AB = AC Þ (3 - 0)2 + ( 3 - 0)2 = (3 - 0)2 + ( p - 0)2 Þ a =6 \\ b=6 -1=5 \\ a + b = 6 + 5 = 11 [by distance formula] 7. Given, C(-2 , 1) is the mid-point of line segment 9 + 3 = 9 + p2 Þ joining A(-6, p) and B(2 , p + 6). Þ p2 = 3 Þ \\ æ -6 + 2 , p+ p+6 ö = (-2 , 1) As, p ¹ p=± 3 ç ÷ \\ è2 2ø 3, otherwise two points will be same. p = - 3. [by section formula] Þ 2p+6 =1 Þ 2p+6 =2 3. Since, DABC ~ DDEF \\ AB = BC 2 DE EF Þ AB = 8 Þ 2 p = - 4 Þ p = -2 Þ 2 AB EF 8. In DABC and DPQR, é 3 =5 = 2ù ëêQ 4.5 7.5 3 ûú EF = 2 ´ 8 = 16 cm AC = AB PR PQ 4. Total number of outcomes = 5 + 20 = 25 and ÐA = ÐP = 54° \\ DABC ~ DPQR (by SAS similarity criterion) Number of favourable outcomes = 5 9. We know that, Product of two numbers a and b = Product of their LCM and HCF SAMPLE PAPER 9 \\Required probability = 53 ´ 12 = 636 = Number of favourable outcomes 10. Given, P( E) = 0.75 Total number of outcomes Since, P( E) + P (not E) = 1 = 5 =1 Þ P (not E) = 1 - P( E) = 1 - 0.75 25 5 = 0.25 5. Given polynomial is 6x2 - 5 Sum of zeroes = - Coefficient of x Coefficient of x2 11. Equation ax + by + c = 0 is said to be a linear =-0 =0 equation in two variable, if a and b are 6 simultaneously not equal to zero i.e. a2 + b2 ¹ 0 Product of zeroes = Constant term = -5 As, a2 + b2 = 0 only, if a = 0 and b = 0 Coefficient of x2 6

CBSE Sample Paper Mathematics Standard Class X (Term I) 169 12. Given, zeroes of a polynomial are - 2 and 5. 20. From the figure, mid-point of LN is given by \\ Required polynomial (x, y) = æ 7 + 1, 3 + 4 ö = (4, 3.5) = k [x2 - (sum of zeroes)x ç ÷ è2 2 ø + product of zeroes] = k [x2 - (- 2 + 5)x + (- 2 ´ 5)] Now, value of x + y = 4 + 3.5 = 7.5 = k [x2 - 3x - 10], where k Î R. 21. Given, DABC ~ DDEF, Hence, infinite number of polynomials exist. ÐA = 57° and ÐE = 73° AD 13. Firstly, plot the point P(4, 2 ) 57° Y 73° 2 4 units P(4, 2) B CE F 1 2 units According to the question, X¢ O 1 2 3 4 5 6 7 X Y¢ DABC ~ DDEF Hence, the distance from P to the X-axis is Then, ÐA = ÐD = 57° 4 units. ÐB = ÐE = 73° 14. Given, ÐC = ÐF y We know that, sum of all the angles of a triangle is equal to 180°. 4x \\ ÐA + ÐB + ÐC = 180° 37 Þ 57° + 73° + ÐC = 180° Here, x = 3 ´ 7 = 21 and y = 4 ´ x = 4 ´ 21 = 84 Þ ÐC = 180° - 130° = 50° 15. In a deck of 52 cards, there are 13 cards of heart 22. Let zeroes of given polynomial be and 1 of them is ace of heart. So, number of favourable outcomes to E a = - 2 and b = 4. = 52 -1 = 51 Then, a + b = - 2 + 4 = 2 and ab = - 2 ´ 4 = - 8 Now, quadratic polynomial is x2 - (a + b) x + ab = x2 - 2 x - 8 16. The solution set of the line ax + by + c = 0 is 23. Let the digit at unit’s place be y and the digit at collinear with the other points on the given line. ten’s place be x. Hence, the solution is on the given line which represents it. So, the number will be 10x + y. 17. Two triangles are said to be similar, if According to first condition, (i) their corresponding angles are equal. x + y =9 …(i) (ii) their corresponding sides are proportional. According to second condition, Coefficient of x 10y + x = 10x + y + 27 Coefficient of x2 18. Sum of zeroes = - Þ 9y = 9x + 27 Þ y = x + 3 [divide by 9] …(ii) Þ a+ 1 =-b Put the value of y in Eq. (i), we get aa Constant term x + x +3 =9 Coefficient of x2 Product of zeroes = Þ 2x =6Þ x =3 Þ a´ 1 = c \\ y =3 +3 =6 SAMPLE PAPER 9 aa So, the number is 10 ´ 3 + 6 i.e. 36. Þ 1= cÞc=a a 24. Total number of students = 23 Number of students in house A = 4 19. The prime factorisation of the denominator Number of students in house B = 8 must be in the form of 2 n ´ 5m, then only the Number of students in house C = 5 rational number have terminating decimal Total number of students in house A, B and expansion. C = 4 + 8 + 5 = 17 \\Remaining students = 23 - 17 = 6 Here, 3125 = 55 \\ 13 has terminating decimal expansion. 3125

170 CBSE Sample Paper Mathematics Standard Class X (Term I) Thus, number of outcomes favourable to the Similarly, DCAB~ DADB given event = 6 \\Required probability (by AA similarity criterion) = Number of favourable outcomes = 6 Þ AB = BC Total number of outcomes 23 BD AB 25. Let the points be A(- 5, 0), B(5, 0) and C(0, 4) Þ AB2 = BC × BD …(ii) The distance between the points A and B, B and Dividing Eq. (ii) by Eq. (i), we get C, C and A are AB2 = BC × BD = BD AC2 BC × DC DC AB = (5 + 5)2 + (0 - 0)2 28. Given, 3x - 2 y = 4 …(i) = 102 + 0 = 100 = 10 units and 2 x + y = 5 …(ii) BC = (0 - 5)2 + (4 - 0)2 Multiply Eq (ii) by 2, we get = (- 5)2 + 42 = 25 + 16 4x + 2 y = 10 …(iii) = 41 units and CA = (- 5 - 0)2 + (0 - 4)2 Adding Eqs (i) and (iii), we get = (-5)2 + (-4)2 7x = 14 Þ x = 2 = 25 + 16 = 41 units From Eq. (ii), 4 + y = 5 Þ y = 1 Here, BC = CA \\ x = 2 and y = 1 \\DABC is an isosceles triangle. 29. For a quadratic polynomial ax2 + bx + c, the sum 26. Prime factorization of 60, 84 and 108 are of the zeroes = - b and product of the zeroes = c 60 = 2 2 ´ 3 ´ 5 aa 84 = 2 2 ´ 3 ´ 7 108 = 2 2 ´ 33 Here, given polynomial is 3x2 - 2 kx + 2 m and \\HCF (60, 84, 108) = 2 2 ´ 3 = 12 zeroes are 2 and 3 Therefore, in each room 12 participants can be seated. Thus, 2 + 3 = - (- 2 k ) \\Number of rooms required 3 = Total number of participants Þ 5 = 2 k Þ k = 15 and 2 ´ 3 = 2 m 12 32 3 = 60 + 84 + 108 = 252 = 21 Þ 6 ´ 3 = 2 mÞ m = 9 12 12 Thus, k = 15 and m = 9 27. In a right triangle DABC, right-angled at A and 2 AD ^ BC. 30. Let points be A(-3, -1), B(a, b), C(3, 3)and D(4, 3). In DCAB and DCDA, So, coordinates of the mid-point of AC = C coordinates of the mid-point of BD [Q in parallelogram, diagonals bisect each other] Þ æ -3 + 3 , -1 + 3 ö = æ a + 4 , b + 3 ö ç ÷ ç ÷ è2 2 øè2 2 ø Þ (0, 1) = æ a + 4 , b + 3 ö ç ÷ è2 2ø D Þ a + 4 = 0 and b + 3 = 1 22 SAMPLE PAPER 9 Þ a = - 4 and b = - 1 Now, a = -4 = 4Þ a:b =4:1 b -1 1 AB 31. When two dice are tossed. Total possible ÐCAB = ÐCDA (each 90º) outcomes = 36 i.e. n(S) = 36 ÐACB = ÐDCA (common) and total favourable outcomes (doublet) \\ DCAB~ DCDA = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} (by AA similarity criterion) i.e. n( E) = 6 Þ BC = AC \\Required probability = 6 = 1 AC DC 36 6 Þ AC2 = BC × DC …(i)

CBSE Sample Paper Mathematics Standard Class X (Term I) 171 32. The prime factorisation of 50 and 48 are So, the probability of getting a number that is 50 = 2 ´ 52 more than or equal to 1 and less than or equal 48 = 2 4 ´ 3 to 6 is \\LCM (50, 48) = 2 4 ´ 3 ´ 52 = 1200 P( E) = Outcomes of Event E Total number of outcomes Here, LCM of 50 s and 48 s = 1200 s =6 =1 6 i.e. 20 min The first beep is at 12 noon (given) 37. Let point on Y-axis are P(0, y). \\Again, the beep is at 20 min later PA2 = PB2 or 12 : 20 pm Then, 33. Consider, cos4 A - sin4 A Þ (0 + 2 )2 + (y - 5)2 = (0 - 4)2 + (y - 3)2 = (cos2 A)2 - (sin2 A)2 Þ 4 + y2 + 25 - 10y = 16 + y2 + 9 - 6y = (cos2 A - sin2 A)(cos2 A + sin2 A) Þ 4y = 29 - 25 [Q a2 - b2 = (a - b) (a + b)] Þ y = 4 =1 = (cos2 A - sin2 A) [Q cos2 q + sin2 q = 1] 4 = cos2 A - (1 - cos2 A) [Q sin2 q = 1 - cos2 q] Hence, required coordinates onY-axis are (0, 1). = cos2 A - 1 + cos2 A = 2 cos2 A - 1 38. A fraction can be expressed in the form 34. Given, DABC ~ DEDF terminating decimal, if denominator can be expressed in the form of 2 n ´ 5m, where n and m E are integers. A Here, 8 =23 12 cm 125 = 53 5 cm 7 cm 80 = 2 4 ´ 5 343 = 73 B CD 15 cm F So, 29 has non-terminating repeating decimal 343 Since, DABC ~ DEDF expansion. We know that, two triangles are said to be 39. Given, (- 3) is the zeroes of the polynomial similar, if their corresponding sides are in (k - 1)x2 + kx + 1 proportion. So, (- 3) must satisfy the equation \\ 5 = 7 = BC (k - 1)x2 + kx + 1 = 0 12 EF 15 Þ (k - 1) (- 3)2 + k(- 3) + 1 = 0 I II III Þ 9(k - 1) - 3k + 1 = 0 Þ 9k - 9 - 3k + 1 = 0 On taking I and II ratios, we get Þ 6k = 8 Þ k=4 5 = 7 Þ EF = 7 ´ 12 = 84 = 16.8 cm 12 EF 55 3 On taking I and III ratios, we get 40. Given, DABC is an equilateral A triangle and AD ^ BC 5 = BC 12 15 \\ AB = BC = AC Þ BC = 5 ´ 15 = 25 = 6.25 cm 12 4 Now, sum of the remaining sides of triangle ÐABC = ÐBAC = ÐACB = 60° SAMPLE PAPER 9 = EF + BC = 16.8 + 6.25 = 23.05 cm and ÐADB = ÐADC = 90° B D C In DABD, 35. Given, 3x - 2 y = 6 …(i) ÐBAD + ÐABD + ÐADB = 180° Equation of X-axis is y = 0 Put y = 0 in Eq. (i), [angle sum property of triangle] 3x - 2(0) = 6 Þ x = 2 ÐBAD = 180° - (90° + 60°) = 30° Hence, required point of intersection is (2, 0). Similarly for DACD, 36. On rolling a dice at once the numbers that are ÐCAD + ÐACD + ÐADC = 180° more than or equal to 1 and less than or equal to 6 are 1, 2, 3, 4, 5, 6. [angle sum property of triangle] ÐCAD = 180° - 90° - 60° = 30°

172 CBSE Sample Paper Mathematics Standard Class X (Term I) Now, In DABD and DACD, = 3 tan A + 2 = 4 + 2 [Q3 tan A = 4] AB = AC 3 tan A - 2 4 - 2 ÐBAD = ÐCAD = 30° [given] = 6 =3 2 and AD = AD (common side) Thus, DABD and DACD are congruent by SAS Solutions (46-50) congruency criterion. 46. In the given figure, \\ BD = DC [by CPCT] The, radius of each semi-circle = 3 cm Þ 1 BC = 1 AB 2 22 Total area of three semi-circles = 3 pr2 Now, DABD is a right angled triangle. 2 Therefore, AB2 = BD2 + AD2 = 3 ´ 1 ´ 22 ´ 3 ´ 3 é r = 3 ù 2 7 22 ëêQ 2 úû [by Pythagoras theorem] Þ AB2 = æ 1 AB÷ö2 + AD2 = 594 = 10.6 cm2 ç 56 è2 ø Þ AB2 = AB2 + AD2 47. In the given figure, 4 The radius of circle, D = 4.5 = 2.25 cm 2 Þ AB2 - AB2 = AD2 \\Area of circle, D = pr2 = 22 ´ 2.25 ´ 2.25 4 7 = 111.375 = 15.91 cm2 Þ 4 AB2 - AB2 = 4 AD2 7 Þ 3 AB2 = 4 AD2 Solutions (41-45) A In DABC, ÐB = 90° 12 cm 6 cm 48. In the given figure, AB = 6 cm The radius of largest semi-circle = 9 = 4.5 cm BC = 6 3 cm 2 AC = 12 cm CB \\Area of largest semi-circle 6Ö3 cm = pr2 = 22 ´ 9 ´ 9 = 1782 = 63.64 cm2 sin C = AB = 6 7 2 2 28 41. AC 12 Þ 49. Area of unshaded region sin C = 1 2 = 2 ´ area of semi-circle A + area of circle D = 2 ´ 1 ´ 22 ´ 3 ´ 3 + 22 ´ 2.25 ´ 2.25 Þ sin C = sin30° 2 7 22 7 \\ ÐC = 30° = 99 + 111.375 = 7.07 + 15.91 = 22.98 cm2 42. sin A = BC AC 14 7 Þ sin A = 6 3 Þ 12 50. The area of shaded region sin A = 3 2 = area of region E + area of region B - area of region A - area of region C Þ sin A = sin60° - area of region D \\ Ð A = 60° = 1 p(4.5)2 + 1 p(1.5)2 43. tan C = AB = 6 = 1 22 SAMPLE PAPER 9 BC 6 3 3 - 1 p(1.5)2 - 1 p(1.5)2 - p æ 4.5 ö2 ç ÷ 22 è2 ø 44. tan C ´ tan A = AB ´ BC = 1 BC AB = 1 p(4.5)2 - 1 p(1.5)2 - 1 ´ p ´ (4.5)2 2 2 22 45. We have, 3 tan A = 4 and 3 sin A + 2 cos A 3 sin A - 2 cos A = 1 ´ ì ( 4.5)2 - (1.5)2 - (4.5)2 ü pí ý 2î 2þ Divide by cos A in numerator and denominator = 1 ´ p [20.25 - 2.25 - 10.125] 3 sin A + 2 cos A 2 = cos A cos A = 1 ´ 22 ´ 7.875 = 12.375 cm2 3 sin A - 2 cos A 27 cos A cos A

CBSE Sample Paper Mathematics Standard Class X (Term I) 173 SAMPLE PAPER 10 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. If a and b be the zeroes of the polynomial p(x) = x 2 - 5x + 2, find the value of 1 + 1 - 3ab ab (a) - 3 (b) - 5 (c) - 7 (d) - 9 2 2 2 2 2. If a man goes 24 m towards East and then 10 m towards North. How far he is starting from? (a) 26 m (b) 17 m (c) 18 m (d) None of these 3. Given that HCF (96, 404) is 4, then the LCM (96, 404) is (a) 9187 (b) 9230 (c) 9696 (d) 10387 4. In what ratio does the point P(3, 4) divided the line segment joining the points A(1, 2) and B(6, 7)? (b) 2 : 3 (c) 3 : 4 (d) 1 : 1 SAMPLE PAPER 10 (a) 1 : 2 5. The maximum value of 1 is cosec q (a) 0 (b) - 1 (c) 1 (d) 3 2 6. A wire is in the shape of a circle of radius 21 cm. It is bent in the form of a square.

174 CBSE Sample Paper Mathematics Standard Class X (Term I) The side of the square is æ use p = 22 ö ç ÷ è 7ø (a) 22 cm (b) 33 cm (c) 44 cm (d) 66 cm 7. The probability of getting a bad egg from a lot of 400 eggs is 0.035. The number of bad eggs in the lot is (a) 7 (b) 14 (c) 21 (d) 28 8. The zeroes of the polynomial f (x) = x 2 - 2 2x - 16 are (a) 2 and - 2 (b) 4 2 and - 2 2 (c) - 4 2 and 2 2 (d) 4 2 and 2 2 9. In a DABC it is given that AB = 6 cm, AC = 8 cm and AD is the bisector of A. Then, BD : DC is A SAMPLE PAPER 10 8 cm 6 cm BD C (a) 3 : 4 (b) 9 : 16 (c) 4 : 3 (d) 3 : 2 10. If x = 3 sec2 q - 1 and y = tan 2 q - 2, then x - 3y is equal to (a) 3 (b) 4 (c) 8 (d) 5 11. If the point (x, y) is equidistant from the points (2, 1) and (1, - 2), then (a) x + 3y = 0 (b) 3x + y = 0 (c) x + 2y = 0 (d) 3x + 2y = 0 12. If the radius of a circle is 7 cm, then the area of the circle is p (a) 72 cm 2 (b) 49 cm 2 (c) 36 cm 2 (d) 56 cm 2 13. A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. The probability that the drawn ball is red, is (a) 5 (b) 3 (c) 7 (d) 1 15 15 15 15 14. A quadratic polynomial whose one zero is 6 and sum of the zeroes is 0, is (a) x2 - 6x + 2 (b) x 2 - 36 (c) x 2 - 6 (d) x2 - 3 15. In the figure, values of x in terms of a, b and c are L a P 35° b x c K M 35° N (a) ab (b) ac a+b b+c

CBSE Sample Paper Mathematics Standard Class X (Term I) 175 (c) bc (d) ac a+b a+b 16. In the figure, the area of the shaded portion is (use p = 3.14) A O 8 cm 6 cm BC (a) 15.25 cm2 (b) 12.75 cm2 (c) 18.05 cm2 (d) 20.60 cm2 17. sec A is equal to (b) 1 (c) 1 1 + cot 2 A cosec A 1 + cot 2 A (d) (a) 1 cot A cot A 18. The probability of guessing the correct answer to certain question is p . If the 12 probability of not guessing the correct answer to same question is 3, then value of p is 4 (a) 3 (b) 4 (c) 2 (d) 1 19. The value of k, for which the system of equations x + (k + 1)y = 5 and (k + 1)x + 9y = 8k - 1 has infinitely many solutions, is (a) 2 (b) 3 (c) 4 (d) 5 20. 1 + cot 2 A is equal to 1 + tan 2 A (a) tan2 A (b) sec2 A (c) cosec2 A - 1 (d) 1 - sin2 A 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 zeroes of the polynomial f (x) = ax 2 + bx + c, then 1 + 1 is equal to a2 b2 b2 - 2ac b2 - 2ac b2 + 2ac b2 + 2ac (a) a2 (b) c2 (c) a2 (d) c2 22. If a circular grass lawn of 35 m in radius has a path 7 m wide running around it on the outside, then the area of the path is (a) 1450 m 2 (b) 1576 m 2 (c) 1694 m 2 (d) 3368 m 2 SAMPLE PAPER 10 23. A game consists of tossing a coin 3 times and noting the outcomes each time. If getting the same result in all the tosses is a successes, then probability of losing the game is (a) 3 (b) 1 (c) 3 (d) 1 4 4 8 24. Which of the following has a terminating decimal expansion? (a) 23 (b) 17 (c) 8 (d) 3 200 9 75 35

176 CBSE Sample Paper Mathematics Standard Class X (Term I) 25. The perimeter of the triangle formed by the points (0, 0), (2, 0) and (0, 2) is (a) (1 - 2 2 ) units (b) (2 2 + 1) units (c) (4 + 2 ) units (d) (4 + 2 2 ) units 26. In a right triangle DABC right-angled at B, if P and Q are points on the sides AB and AC respectively, then (a) AQ2 + CP2 = 2( AC2 + PQ2 ) (b) 2( AQ2 + CP2 ) = AC2 + PQ2 (c) AQ2 + CP2 = AC2 + PQ2 (d) AQ1 + CP = 1 ( AC + PQ) 2 27. In the following figure, OABC is a square of side 7 cm. OAC is a quadrant of a circle with O as centre. The area of the shaded region is OC 7 cm (a) 10.5 cm2 A B (d) 11.5 cm2 (b) 38.5 cm2 (c) 49 cm2 28. If both zeroes of the quadratic polynomial x 2 - 2kx + 2 are equal in magnitude but opposite in sign, then value of k is (c) 1 (d) -1 (a) 0 (b) 1 2 29. DABC is an equilateral triangle with each side of length 4 p. If AD ^ BC, then value of AD is (a) 3 (b) 2 3 p (c) 2 p (d) 4p 30. The value of sin q 3-q2-sicno3sqq, where q is acute angle, is 2 cos (a) cosec q (b) cot q (c) tan q (d) sin q cos q 31. In a circle of diameter 42 cm, if an arc subtends an angle of 60° at the cente where p = 22, then what will be the length of arc (in cm) 7 (a) 22 (b) 21 (c) 24 (d) 27 32. Seven face cards are removed from a deck of cards and the cards are well shuffled. Then, the probability of drawing a face card is SAMPLE PAPER 10 (a) 1 (b) 1 (c) 2 (d) 3 8 9 7 4 33. If x = p secq and y = q tan q, then (b) x 2 q2 - y 2 p2 = pq (d) x 2 q2 - y 2 p2 = p2 q2 (a) x 2 - y 2 = p2 q2 (c) x 2 q2 - y 2 p2 = 1 pq 34. The point which divides the line segment joining the points (7, - 6) and (3, 4) in ratio 1 : 2 internally lies in the

CBSE Sample Paper Mathematics Standard Class X (Term I) 177 (a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant 35. If the distance between A(k, 3) and B(2, 3) is 5 units, then the value of k is (a) 5 (b) 6 (c) 7 (d) 8 (d) 1 and 3 36. The zeroes of the polynomial f (x) = x 2 + x - 3 are 2 4 (a) - 1 and 3 (b) 1 and - 3 (c) - 3 and 1 22 22 2 37. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, - 5) and R(- 3, 6). Then, the coordinates of P are (a) (16, 8) (b) (14, 7) (c) (18, 9) (d) (10, 5) 38. If DABC ~ DDEF such that DE = 6 cm, EF = 4 cm, DF = 5 cm and BC = 8 cm, then perimeter of DABC is (a) 18 cm (b) 20 cm (c) 12 cm (d) 30 cm 39. The centroid of the triangle whose vertices are (3, - 7), (- 8, 6) and (5, 10) is (a) (0, 3) (b) (0, 9) (c) (1, 3) (d) (3, 5) 40. If tan q + sin q = m and tan q - sin q = n, then m2 - n2 is equal to (a) mn (b) m n (c) 4 mn (d) None of these 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 Prabhat is planning to buy a house and the layout is given below. The design and the measurement has been made such that areas of two bedrooms and garden together is 95 sq m. 5m p 2 q 2m Bedroom 1 Garden 5m Bath Bedroom 2 Room Living Room SAMPLE PAPER 10 15 m Based on the above information, answer the following questions:

178 CBSE Sample Paper Mathematics Standard Class X (Term I) 41. Which is the correct equation in two variables from this situation. (a) p + q = 13 (b) p + q = 15 (c) p + q = 17 (d) p + q = 19 42. Find the length of the outer boundary of the layout. (a) 86 m (b) 45 m (c) 34 m (d) 54 m (d) 75 m 2 43. Find the area of each bedroom and garden in the layout. (a) Bedroom 20 m 2 and garden 25 m 2 (b) Bedroom 50 m 2 and garden 55 m 2 (c) Bedroom 30 m 2 and garden 35 m 2 (d) Bedroom 40 m 2 and garden 45 m 2 44. Find the area of living room in the layout. (a) 85 m 2 (b) 65 m 2 (c) 45 m 2 45. Find the area of the plot. (b) 3 41 m2 2 (a) 3 41 m2 (d) None of these (c) 41 m2 46-50 are based on Case Study-2. Case Study 2 Indian Water Department were carrying out periodic inspection of water tanks at every water pump to check carbonate number. They found some carbonate number at both the water tanks at Shyam’s water pump and asked him to empty them immediately and to cover them with a lid. The water tanks have capacities of 420 L and 130 L. Shyam had no other option to empty the two filled water tanks with the help of a bucket. 420 Liters 130 Liters 46. The maximum capacity of the bucket he should use so that no water remains in the tanks is SAMPLE PAPER 10 (a) 10 L (b) 13 L (c) 130 L (d) 420 L 47. When the HCF (420, 130) is expressed as a linear combination of 420 and 130 i.e. HCF (420, 130) = 420x + 130y, then values of x and y satisfying the above relation are (a) 3 and 1 (b) - 4 and 13 (c) 4 and - 13 (d) 2 and 3 48. The HCF of the smallest composite number and the smallest even number is (a) 1 (b) 2 (c) 4 (d) 8

OMR SHEET SP 10 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)

180 CBSE Sample Paper Mathematics Standard Class X (Term I) 49. The greatest number which divides 285 and 1249 leaving remainders 9 and 7, respectively is (a) 138 (b) 276 (c) 1242 (d) 2484 50. The HCF of the smallest and largest two-digit numbers is (a) 0 (b) 1 (c) 2 (d) 990 Answers 1. (c) 2. (a) 3. (c) 4. (b) 5. (c) 6. (b) 7. (b) 8. (b) 9. (a) 10. (c) 11. (a) 12. (b) 13. (c) 14. (b) 15. (b) 16. (a) 17. (d) 18. (a) 19. (a) 20. (c) 21. (b) 22. (c) 23. (a) 24. (a) 25. (d) 26. (c) 27. (a) 28. (a) 29. (b) 30. (c) 31. (a) 32. (b) 33. (d) 34. (d) 35. (c) 36. (b) 37. (a) 38. (d) 39. (a) 40. (c) 41. (a) 42. (d) 43. (c) 44. (d) 45. (b) 46. (a) 47. (b) 48. (b) 49. (a) 50. (b) SOLUTIONS 1. Given, a and b are the zeroes of the polynomial Þ 4 ´ LCM = 96 ´ 404 Þ LCM = 24 ´ 404 = 9696 p(x) = x2 - 5x + 2 4. Let P(3, 4) divides the line segment joining \\Sum of zeroes = - Coefficient of x Coefficient of x2 A(1, 2 ) and B(6, 7) in the ratio k : 1. \\x-coordinate of P is 6k + 1 = 3 a + b = - (-5) = 5 1 k +1 and product of zeroes = Constant term [by section formula] Coefficient of x2 Þ 6k + 1 = 3k + 3Þ 3k = 2 Þ k =2 ab = 2 = 2 1 3 Now, 1 + 1 - 3ab = a +b 5. We know that, - 3ab sin q = 1 ab ab cosec q = 5 - 3(2 ) = - 7 22 2. In DABC, sin q is maximum when q = 90° BN i.e. sin90° = 1 Therefore, maximum value of 1 is 1. 10 m WE cosec q A 24 m CS 6. As, length of wire is same so, both figures have same perimeter. SAMPLE PAPER 10 ÐA = 90° CB2 = CA2 + AB2 \\Circumference of circle = Perimeter of the square. [by Pythagoras theorem] Þ CB2 = 242 + 102 Let r be the radius of the circle and a is the side of the square. = 576 + 100 = 676 \\ SB = 26 m So, 2 pr = 4a 4a = 2 ´ 22 ´ 21 3. Given, HCF (96, 404) is 4 Þ [Q r = 21 cm] We know that, HCF ´ LCM = Product of two numbers 7 Þ 4a = 132 Þ a = 33 cm 7. Here, total number of eggs = 400 Probability of getting a bad eggs = 0.035

CBSE Sample Paper Mathematics Standard Class X (Term I) 181 Þ Number of bad eggs = 0.035 \\ P(red ball) = 7 Total number of eggs 15 Þ Number of bad eggs = 0.035 14. Given, Sum of the zeroes = 0 400 Let a and b be the zeroes of the quadratic Number of bad eggs = 400 ´ 0.035 polynomial. Then, = 400 ´ 35 = 14 1000 a +b=0 8. The zeroes of f (x) = x2 - 2 2 x - 16 are Þ 6 +b=0 f(x) = 0 Þ b=-6 i.e. x2 - 2 2 x - 16 = 0 Þ x2 - 4 2 x + 2 2 x - 16 = 0 \\Required polynomial = x2 - (a + b)x + a ×b Þ x(x - 4 2 ) + 2 2 (x - 4 2 ) = 0 = x2 - 0 × x + 6(- 6) = x2 - 36 Þ (x + 2 2 )(x - 4 2 ) =0 15. In DKNP and DKML, Þ x = 4 2 or x = - 2 2 ÐKNP = ÐKML = 35° [given] 9. Given, AD is the angle bisector of ÐA. ÐK = ÐK [common] By angle bisector theorem, bisector of an angle divides the opposite side in the same ratio as of \\ DKNP ~ DKML two remaining sides. \\ BD = AB [by AA similarity criterion] DC AC Þ PN = KN Þ BD = 6 LM KM DC 8 [Q corresponding sides of similar Þ BD = 3 triangles are proportional] DC 4 Þ x= c Þx= c 10. We have, x = 3 sec2 q - 1 and y = tan2 q - 2 a KN + NM a c + b Consider, x - 3y = (3 sec2 q - 1) - 3(tan2 q - 2 ) = 3 sec2 q - 1 - 3 tan2 q + 6 Þ x = ac = 3(sec2 q - tan2 q) + 5 b+ c = 3 ´ 1 + 5 [Q sec2 A - tan2 A = 1] =3 + 5 =8 16. DABC is right-angled triangle 11. Let the points be P(x, y), A(2 , 1) and B(1, - 2 ). \\In DABC, AC2 = AB2 + BC2 [using Pythagoras theorem] Þ AC2 = (8)2 + (6)2 = 100 Þ AC = 10 cm Now, Area of shaded portion Since, P is equidistant from A and B. = Area of half circle - Area of DABC \\ AP = BP Þ AP2 = BP2 Þ (x - 2 )2 + (y - 1)2 = (x - 1)2 + (y + 2 )2 = 1 pr2 - 1 ´ BC ´ AB 22 [by distance formula] = p (5)êéë2Q-r21= ´O6A´=81 AC = 5 cmúùû Þ x2 - 4x + 4 + y2 - 2y + 1 2 2 = x2 - 2x + 1 + y2 + 4y + 4 = 39.25 - 24 = 15.25 cm2 Þ 2x + 6y = 0 17. We know that, sec A = 1 é sec q = 1ù cos A êëQ cos q úû Þ x + 3y = 0 [divide by 3] Divide numerator and denominator by sin A SAMPLE PAPER 10 12. Radius of the circle, r = 7 cm 1 p Þ sin A = cosec A = 1 + cot2 A \\ Area of the circle = pr2 = pæç 7 ö2 = 49 cm 2 cos A cot A cot A ÷ è pø sin A [Q cosec2 q - cot2 q = 1] 13. Given, Number of black balls = 5 18. Given, the probability of guessing the correct answer to certain question is p and the Number of red balls = 7 12 Number of white balls = 3 probability of not guessing the correct answer Total balls = 5 + 7 + 3 = 15

182 CBSE Sample Paper Mathematics Standard Class X (Term I) to same question is 3. 22. Let r = radius of inner circle and R = radius of 4 outer circle. We know that, R 35 m P( E) + P( E) = 1 7m Þ p + 3 =1 r 12 4 Þ p =1- 3 = 1 Area of the path = Area of outer circle - area of inner circle. 12 4 4 \\Area of path = pR2 - pr2 = p( R2 - r2 ) Þ p = 12 = 3 = 22 [(42 )2 - (35)2 ] 4 7 [Q R = 42 and r = 35] 19. For infinite many solutions, a1 = b1 = c1 = 22 (42 - 35) (42 + 35) a2 b2 c2 7 [Q a2 - b2 = (a - b) (a + b)] Þ 1 =k +1= -5 k + 1 9 - (8k - 1) = 22 ´ 7 ´ 77 7 So, 1 = 5 k + 1 8k - 1 = 1694 m2 Þ 8k - 1 = 5k + 5 23. Let H and T represent Head and Tails Þ 3k = 6Þ k = 6 = 2 respectively. 3 The possible outcomes when a coin is tossed 3 times 20. We have, 1 + cot2 A {( HHH, HHT, HTH, THH, HTT, 1 + tan2 A THT, TTH, TTT)} Total number of possible outcomes = 8 = 1 + cot2 A é tan q = 1 ù Possible outcomes of winning = 2 ëêQ cot q ûú So, possible outcomes of losing = 8 -2 = 6 1 + 1 \\Required probability = 6 = 3 cot 2 A 84 24. 23 = 23 cot2 A (1 + cot2 A) = 200 2 3 ´ 52 17 = 17 cot2 A + 1 9 32 8= 8 = cot2 A 75 3 ´ 52 = cosec2 A - 1 3= 3 35 5 ´ 7 [Q1 + cot2 q = cosec2q] As we know that, if the denominator is in the 21. We have, a and b are zeroes of the polynomial form of 2 m ´ 5n, where m and n are integers, then the number has terminating decimal f (x) = ax2 + bx + c expansion. Hence, in the given option, only option (a) has \\ Sum of zeroes (a + b) = - b …(i) denominator in the form of 2 m ´ 5n. a …(ii) 25. Let A(0, 0), B(2 , 0) and C(0, 2 ) be the vertices and product of zeroes (ab) = c a of DABC. \\ AB = (2 - 0)2 + (0 - 0)2 Now, 1 + 1 b2 + a2 = (a + b)2 - 2 ab a2 b2 = (ab)2 a 2b2 [Q(a + b)2 = a2 + b2 + 2 ab] SAMPLE PAPER 10 æ - b ö2 -2 c ç ÷ =è aø a æ c ö2 ç÷ è a ø [using Eqs. (i) and (ii)] b2 - 2c b2 - 2 ac b2 a2 a a2 = = = - 2 ac c2 c2 c2 a2 a2

CBSE Sample Paper Mathematics Standard Class X (Term I) 183 = 4 + 0 = 2 units \\a + (- a ) = - (- 2 k ) Þ 0 = 2 k [by distance formula] 1 BC = (0 - 2 )2 + (2 - 0)2 é Sum of zeroes = - Coefficient of x ù Coefficient of x2 = 4+4= 8 êQ ú = 2 2 units ë û and CA = (0 - 0)2 + (0 - 2 )2 Þ k =0 = 0 + 4 = 2 units \\Required perimeter = AB + BC + CA 29. Given an equilateral triangle DABC in which, = (2 + 2 2 + 2 ) units A = (4 + 2 2 ) units 4p 4p 26. Given, right triangle DABC right-angled at B. Join PQ, PC and AQ BDC A AB = BC = CA = 4 p and AD ^ BC. P Since, in equilateral triangle, perpendicular bisects the side. C QB \\ BD = DC or BD = 1 BC = 1 (4 p) = 2 p In DABQ, by pythagoras theorem, …(i) 22 AQ2 = QB2 + AB2 In DADB, AB2 = AD2 + BD2 Similarly, In DPCB, [by Pythagoras theorem] PC2 = BC2 + PB2 (4 p)2 = AD2 + (2 p)2 Þ 16 p2 = AD2 + 4 p2 Þ 16 p2 - 4 p2 = AD2 Þ 12 p2 = AD2 … (ii) \\ AD = 2 3 p and In DACB, 30. We have, sin q - 2 sin3 q AC2 = AB2 + BC2 2 cos3 q - cos q … (iii) sin q(1 - 2 sin2 q) and In DPQB, … (iv) = PQ2 = PB2 + QB2 cos q(2 cos2 q - 1) Adding Eqs. (i) and (ii), we get = tan q × [1 - 2(1 - cos2 q)] (2 cos2 q - 1) AQ2 + PC2 = QB2 + AB2 + BC2 + PB2 [Q sin2 A = cos2 A - 1] AQ2 + PC2 = AC2 + PQ2 (2 cos2 q - 1) [from Eqs. (iii) and (iv)] = tan q × (2 cos2 q - 1) = tan q 27. Given, OABC is a square with side 7 cm and 31. Given, diameter = 42 cm OAC is a quadrant of a circle with centre O. Þ r = 42 = 21 cm 2 Now, Area of shaded portion = Area of square and q = 60° - Area of quadrant O SAMPLE PAPER 10 = (Side)2 - 1 pr2 60° 4 q = (7)2 - 1 ´ 22 ´ 7 ´ 7 \\Length of arc = ´ 2 pr 47 360° = 49 - 77 = (49 - 38.5) 2 = 60° ´ 2 ´ 22 ´ 21 = 22 cm = 10.5 cm2 28. As, the polynomial is x2 - 2 kx + 2 and its zeroes are equal but opposite in sign. Let the zeroes of the polynomials are a and - a. 360° 7 32. Total number of possible outcomes

184 CBSE Sample Paper Mathematics Standard Class X (Term I) = 52 - 7 = 45 36. The zeroes of the polynomial f (x) = x2 + x - 3 Remaining number of face cards 4 = 12 - 7 = 5 are given by f (x) = 0 So, favourable number of outcomes = 5 Þ x2 + x - 3 = 0 \\ Required probability 4 Þ 4x2 + 4x - 3 = 0 = Number of favourable outcomes Þ 4x2 + 6x - 2x - 3 = 0 Total number of possible outcomes [splitting middle term] = 5 =1 45 9 Þ 2 x(2 x + 3) - 1(2 x + 3) = 0 33. Given, x= psec q Þ æ x ö2 = sec2 q … (i) Þ (2 x - 1) (2 x + 3) = 0 çç p ÷ … (ii) è ÷ Þ 2 x - 1 = 0 and 2 x + 3 = 0 ø Þ x = 1 and x = - 3 y = qtan q Þ æ y ö2 = tan2 q 22 ç q ÷ ç ÷ 37. Let the coordinates of P be (x, y). è ø \\ sec2 q - tan2 q = x2 - y2 Then, x = 2 y (given) … (i) p2 q2 Since, P is equidistant from Q(2 , - 5) [subtract Eq. (ii) from Eq.(i)] and R(- 3, 6) Since, sec2 q - tan2 q = 1 \\ PQ = PR (2 - x)2 + (- 5 - y)2 = (- 3 - x)2 + (6 - y)2 x2 y2 \\ p2 - q2 =1 [by distance formula] Þ x2q2 - y2 p2 = p2q2 Squaring both sides, we get 34. Given points are (7, - 6) and (3, 4). (2 - 2 y)2 + (- 5 - y)2 = (- 3 - 2 y)2 + (6 - y)2 Let, x1 = 7, x2 = 3, y1 = - 6, y2 = 4, m = 1 and n = 2 [using (i)] By section formula, the coordinate of the point 4 + 4y2 - 8y + 25 + y2 + 10y which divides the line segment joining the = 9 + 4y2 + 12 y + 36 + y2 - 12 y Þ 5y2 + 2 y + 29 = 5y2 + 45 points (7, - 6) and (3, 4) in the ratio 1 : 2 Þ 2 y = 16 Þ y = 8 Hence, the coordinates of P are (16, 8). internally are æ 1 ´ 3 + 2 ´ 7 , 1 ´ 4 + 2 ´ (- 6) ö çç 1 + 2 1 + 2 ÷÷ è ø 38. Given, DABC ~ DDEF æ 3 + 14 4 - 12 ö and DE = 6 cm, EF = 4 cm, DF = 5 cm ç ÷ = , and BC = 8 cm è3 3ø D = æ 17 , - 8 ö A ç ÷ è 3 3ø Since, x-coordinate is positive and y-coordinate 6 cm 5 cm is negative. B 8 cm C E F So, the point lies in the IV quadrant. 35. From distance formula, 4 cm D = (x2 - x1 )2 + (y2 - y1 )2 SAMPLE PAPER 10 AB = BC = AC Here, AB = 5 units [given] DE EF DF So, AB = (k - 2 )2 + (3 - 3)2 [Q from Basic Proportionality Theorem] = (k - 2 )2 + 0 Þ AB = 8 = AC Þ (5)2 = (k - 2 )2 645 =(k -2) = ± 5 \\ AB = 2 ´ 6 = 12 cm Þ k =2 ± 5 Þ k = 7, -3 and AC = 5 ´ 2 = 10 cm Now, Perimeter of DABC = AB + BC + CA = 12 + 8 + 10 = 30 cm

CBSE Sample Paper Mathematics Standard Class X (Term I) 185 39. If we have three vertices of triangle (x1 , y1 ), = 95 (x2 , y2 ) and (x3 , y3 ), then \\ 10 p + 5q = 95 [given] …(i) Coordinate of centroid and p + q = 13 …(ii) = æ x1 + x2 + x3 , y1 + y2 + y3 ö Multiply Eq. (ii) by 5 and then subtract Eq. (ii) ç ÷ from Eq. (i) è3 3ø Here, x1 = 3, x2 = - 8 and x3 = 5 5 p = 30 and y1 = - 7, y2 = 6 and y3 = 10 Hence, coordinate of centroid Þ p=6 = æ 3 + (- 8) + 5 , - 7 + 6 + 10 ö \\ q = 13 - 6 = 7 ç ÷ \\Area of each bedroom = 5 ´ 6 = 30 m 2 è3 3ø Area of garden = 5 ´ 7 = 35 m 2 = æ 0, 9 ö 44. Area of living room + Area of each bedroom ç ÷ = 15 ´ 7 = 105 m 2 è 3ø Since, area of each bedroom = 30 m 2 = (0, 3) \\Area of living room = 105 - 30 40. Given, tan q + sin q = m and tan q - sin q = n = 75 m 2 Now, m2 - n2 = (tan q + sin q)2 - (tan q - sin q)2 45. Draw a perpendicular line from centre O to the chord AB. Then, = 4 tan q × sin q [Q(a + b)2 - (a - b)2 = 4ab] = 4 tan2 q × sin2 q DC = 4 sin2 q (sec2 q - 1) O é tan2 A = sin2 A ù AEB êQ cos2 A ú ê ú 1 ëêêand sec2 A = cos2 ú AE = EB = 15 m Aúû 2 =4 sin2 qçæ 1 - 1÷ö Also, BC = 12 m è cos2 ø \\ OE = 1 BC = 12 = 6 m q 22 =4 sin2 q - sin2 q In right DOEB, cos2 q OB = OE2 + EB2 = 4 tan2 q - sin2 q = 4(tan q - sin q)(tan q + sin q) [by pythagoras theorem] = 4 mn = (6)2 + æ 15 ö2 ç ÷ è2 ø Solutions (41-45) 41. From the given figure = 36 + 225 p + 2 + q = 15 4 Þ p + q = 15 - 2 = 13 = 144 + 225 4 42. From the figure SAMPLE PAPER 10 Length = 15 m = 1 369 and Breadth = 2 + 5 + 5 = 12 m 2 Total length of outer boundary = Perimeter of rectangle = 3 41 m2 = 2 (Lenght + Breadth) 2 = 2 (15 + 12 ) = 2 ´ 27 = 54 m Solutions (46-50) 43. Area of two bedrooms = 2 (5 ´ p) = 10 p m2 46. Here, we have to find the HCF of 420 and 130. Area of garden = 5 q m 2 Prime factor of 420 = 7 ´ 3 ´ 2 2 ´ 5 Given, Area of two bedrooms Prime factor of 130 = 13 ´ 2 ´ 5 + Area of two garden \\HCF (420, 130) = 2 ´ 5 = 10 Hence, the maximum capacity of bucket is 10 L. 47. Given, HCF (420, 130) = 420x + 130y

186 CBSE Sample Paper Mathematics Standard Class X (Term I) i.e. 10 = 420x + 130y (a) When x = 3 and y = 1; 420 ´ 3 + 130 ´ 1 = 1260 + 130 = 1390 ¹ 10, so it does not satisfy (b) When x = - 4 and y = 13; 420 ´ (- 4) + 130 ´ (13) = - 1680 + 1690 = 10 Hence, x = - 4 and y = 13 48. Since, smallest composite number = 4 and smallest even number = 2 \\ HCF (4, 2) = 2 49. Subtract the remainders from the given numbers and, then we find the HCF. \\ 285 - 9 = 276 1249 - 7 = 1242 Prime factor of 276 = 2 2 ´ 3 ´ 23 Prime factor of 1242 = 2 ´ 33 ´ 23 \\HCF (276, 1242) = 2 ´ 3 ´ 23 = 138 50. Since, smallest two-digit number = 10 and largest two-digit number = 99 Prime factor of 10 = 2 ´ 5 Prime factor of 99 = 3 ´ 3 ´ 11 \\HCF (10, 99) = 1 SAMPLE PAPER 10