MATHS MCQ SOLVED AND UNSOLVED

94 CBSE Sample Paper Mathematics Standard Class X (Term I) Now, old area of the circle = px2 Since, - 4 is zero of p(x). and new area of the circle = pçæ 9x ö2 = p 81x 2 \\ p(- 4) = 0 ÷ Þ (- 4)2 - (- 4) - (2 k + 2 ) = 0 è 10 ø 100 Þ 16 + 4 - 2 k - 2 = 0 \\Required change in area = px2 - p 81 x2 100 Þ 2 k = 18 = 19 px2 Þ k =9 100 30. Let E be the event passing the test P( E) = x \\Percentage at which area diminished 24 Also, P (not passing the test) P( E) = 7 æ 19 px2 ö ç ÷ 8 ç ÷ Now, P( E) + P( E) = 1 =è 100 ø ´ 100 = 19% px2 x + 7 =1 24 8 27. From figure, we have …(i) Þ x =1-7 Ð1 + Ð4 = 90° …(ii) 24 8 and Ð3 + Ð4 = 90° Þ x =8 -7 C 24 8 Þ x =1 4 24 8 D Þ x = 24 Þ x = 3 1 3B 8 2 31. Let height of vertical pole = AB A Length of wire = AC = 26 m From Eqs. (i) and (ii), we have Distance from the base of the pole to the Ð1 + Ð4 = Ð3 + Ð4 another end of the wire = BC = 10 m Þ Ð1 = Ð3 A Similarly, Ð2 = Ð4 Now, in DABC and DDAC. ÐA = ÐD = 90° SAMPLE PAPER 3ÐC = ÐC [common] 26 m By AA similarity criterion DABC ~ DDAC Again, in DDAC and DDBA C 10 m B Ð1 = Ð3 [proved above] Since, the pole will be perpendicular (vertical) to ground. Ð4 = Ð2 [proved above] \\ ÐABC = 90° By AA similarity criterion, Þ DABC is a right-angled triangle. DDAC ~ DDBA Using pythagoras theorem, Hence, DABC ~ DDAC ~ DDBA. AC2 = AB2 + BC2 28. Given a pack of cards contain card numbered Þ 262 = AB2 + 102 2 to 53. Þ AB2 = 676 - 100 = 576 Prime number less than 20 are 2, 3, 5, 7, 11, 13, 17, 19. Þ AB = 24 m Number of prime numbers = 8 Total number of cards = 53 - 2 + 1 = 52 32. Let the number be 10y + x. \\ Required probability Where x and y are unit’s and ten’s digit. …(i) = Number of favourable outcomes According to the situation …(ii) Total number of outcomes 10y + x = 8(x + y) + 1 \\ Required probability = 8 = 2 Þ 7x - 2y + 1 = 0 52 13 and 10y + x = 13(y - x) + 2 Þ 14x - 3y - 2 = 0 29. Let p(x) = x2 - x - (2 k + 2 ) Multiply Eq. (i) by 2, we get

CBSE Sample Paper Mathematics Standard Class X (Term I) 95 14x - 4y + 2 = 0 …(iii) = cos2 q = 1 [Q sin2 A + cos2 A = 1] cos2 q Now, subtract Eq. (iii) from Eq. (ii), (14x - 3y - 2 ) - (14x - 4y + 2 ) = 0 36. Fencing is done on the circumference of the circular fields. Þ 14x - 3y - 2 - 14x + 4y - 2 = 0 Þ y -4 =0Þ y =4 Let R be the radius of the circular field. Put y = 4 in Eq. (i) Given, area of circular field 7x - 2 ´ 4 + 1 = 0Þ 7x = 7Þ x = 1 = 13.86 hec Hence, the number = 10y + x = 10 ´ 4 + 1 = 41 = 13.86 ´ 10000 m2 33. Let E be the event Sania win the match. Þ (pR2 ) = 138600 So, probability of Sania winning the match Þ R2 = æ 138600 ´ 7 ö ç ÷ è 22 ø = P( E) = 0.68 Þ R2 = 44100 m2 Also, P( E) + P( E) = 1 Þ R = 210 m \\Required probability of Deepika winning the match = P( E) Circumference of the circular field = 1 - P( E) = 2 pR = æ 2 ´ 22 ´ 210 ö ç ÷ è7 ø = 1 - 0.68 = 0.32 = 1320 m 34. In an equilateral triangle, \\ Cost of fencing = ` (1320 ´ 4.40) = ` 5808 AG : GD = 2 : 1 37. Given, tan30° + cot60° Þ AG = 2 x and GD = x tan30° (sin30° + cos60°) A 6 cm 6 cm 1+1 =3 3 G 1 é1 + 1ù 3 êë2 2 ûú 2 BDC =3 6 cm 1 ´1 3 Now, AD = 3 a 2 =2 ´ 3 =2 Again, 3´1 \\ = 3 ´ 6 = 3 3 cm Þ 2 38. Given, join of points A(2 , 6) and B(5, 1) divides \\ and AG + GD = 3 3 by point C in 1 : 3. 2x + x =3 3 x = 3 cm AC : BC = 1 : 3 AG = 2 x = 2 3 cm Thus, k = 3´ 2 + 1´ 5 GD = x = 3 cm 1+3 35. Given, the numerical value = 6 + 5 = 11 44 æ 1 1 öæ 1 1ö ç + ÷ç - ÷ 39. As area of quadrant is equal to the one fourth è cos q cot q ø è cos q cot q ø of the total area of circle. Let r be the radius of the circle. = æ 1 + sin q ö æ 1 - sin q ö \\ Area of quadrant ABDCA = 1 pr2 SAMPLE PAPER 3 ç ÷ç ÷ 4 è cos q cos q ø è cos q cos q ø = 1 ´ 22 ´ (7)2 47 é cot A = cos Aù ëêQ sin A úû [Q r = 7 cm] = 77 cm2 = æ 1 + sin q ö æ 1 - sin q ö ç ÷ ç ÷ 2 è cos q ø è cos q ø = 38.5 cm2 1 - sin2 q In DABE, = cos2 q

96 CBSE Sample Paper Mathematics Standard Class X (Term I) Base = 7 cm, Height = 2 cm 44. The number of vegetables that can be placed in ech stack = 10 Area of DBAE = 1 ´ base × height Prime factor of 10 = 21 ´ 51 2 The exponents are 1 and 1. \\ Sum of exponents = 1 + 1 = 2 = 1 ´ AB´ AE 2 45. Total number of vegetables = 550 = 1 ´ 7 ´ 2 = 7 cm2 2 Hence, area of the shaded portion Number of vegetables that can be placed in = Area of the quadrant ABDCA each stack = 10 - Area of DBAE \\ Number of rows of vegetables = 550 = 55 10 = (38.5 - 7) cm2 Solutions (46-50) = 31.5 cm2 40. Since, 1 is a zeroes of the polynomial 46. The coordinates of P is (4, 6) as the point is 2 4 units on X-axis and 6 unit on Y-axis. x2 + kx - 5, then 47. The coordinates of R is (6, 5) as the point is 4 6 units on X-axis and 5 units on Y-axis. æ 1 ö2 + 1 k - 5 =0 48. The coordinates of Q are (3, 2) and R are (6, 5). ç ÷ \\ The distance between points Q and R is è2 ø 2 4 Þ 1 + 1k - 5 =0 QR = (6 - 3)2 + (5 - 3)2 42 4 = (3)2 + (2 )2 Þ -4 + 1k =0 = 9 + 4 = 13 units 42 49. Since, coordinates of vertices of a DPQR are Þ -1=- 1k P(4, 6), Q(3, 2 ) and R(6, 5). 2 Now, PQ = (3 - 4)2 + (2 - 6)2 Þ k =2 = (-1)2 + (- 4)2 Now, sum of zeroes = - k = - k = - 2 1 Solutions (41-45) = 1 + 16 = 17 units PR = (6 - 4)2 + (5 - 6)2 41. The total number of vegetables = 420 + 130 = 550 42. Since, total number of vegetables is 550 = (2 )2 + (-1)2 Prime factorisation of 550 = 2 ´ 5 ´ 5 ´ 11 = 4 + 1 = 5 units = 21 ´ 52 ´ 111 and QR = 13 The exponents are 1, 2 and 1. Here, PQ ¹ QR ¹ PR \\Product of exponents = 1´ 2 ´ 1 = 2 Hence, DPQR is a scalene triangle. 43. Prime factor of 420 = 2 2 ´ 3 ´ 5 ´ 7 50. The centroid of DPQR is æ 4 + 3 + 6, 6 + 2 + 5ö ç ÷ and 130 = 2 ´ 5 ´ 13 è3 3ø \\ HCF (420, 130) = 2 ´ 5 = 10 So, the number of vegetables that can be i.e. æ 13 , 13 ÷ö. placed in each stack for this purpose is 10. ç è3 3ø SAMPLE PAPER 3

CBSE Sample Paper Mathematics Standard Class X (Term I) 97 SAMPLE PAPER 4 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. What will be the least possible number of the planks, if three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length? (a) 5 (b) 6 (c) 7 (d) None of these 2. The value of k such that the polynomial x 2 - (k + 4)x - 2k + 3 has sum of its zeroes equal to doubled of their product is (a) 1 (b) - 2 (c) 2 (d) 4 5 5 5 5 3. The lengths of the diagonals of a rhombus are 12 cm and 8 cm. The length of each side of the rhombus is (a) 12 cm (b) 2 13 cm (c) 14 cm (d) 17 cm 4. A card is accidently dropped from a pack of 52 playing cards. The probability that it is SAMPLE PAPER 4 a red card, is (a) 1 (b) 1 (c) 1 (d) 12 2 13 52 13 5. The coordinates of the point which is reflection of point (- 3, 5) in X-axis is (a) (3, 5) (b) (3, - 5) (c) (-3, - 5) (d) (-3, 5)

98 CBSE Sample Paper Mathematics Standard Class X (Term I) 6. If 4 tan A = 3, then find the value of the following expression 4 sin 2 A - 2 cos2 A 4 sin 2 A + 3 cos2 A (a) 1 (b) 23 21 41 (c) - 1 (d) Cannot be determined 21 7. Three runners running around a circular track, can complete one revolution in 2, 3 and 4 h respectively. They will meet again at the starting point after (a) 8 h (b) 6 h (c) 12 h (d) 18 h 8. The diameter of a circle, whose area is equal to the areas of two circle of radii 12 cm and 5 cm, is (a) 17 cm (b) 13 cm (c) 26 cm (d) 34 cm 9. The quadratic polynomial whose zeroes are 3 and 2 is 23 (a) x 2 - 13 x + 1 (b) x 2 + 13 x + 2 66 (c) x 2 + 13 x + 1 6 (d) None of these 10. If A = 30°, then the value of 3 cos A - 4 cos3 A is (a) 0 (b) 1 (c) -1 (d) 2 11. In DABC, DE||BC, then the value of x is A 2x x + 2 D E 2x – 3 x–3 B C (a) 4 (b) 5 (c) 6 (d) 8 7 7 7 7 12. In a single throw of a die, the probability of getting a multiple of 2 is (a) 1 (b) 1 (c) 1 (d) 2 2 3 6 3 13. If the point P(k, 0) divides the line segment joining the points A(2, - 2) and B(- 7, 4) SAMPLE PAPER 4 in the ratio 1 : 2, then the value of k is (c) - 2 (d) - 1 (a) 1 (b) 2 14. The decimal expansion of 17 is 125 (a) 1.36 (b) 0.136 (c) 13.6 (d) None of these 15. Sum of zeroes of f (x) = x 2 - 16 is (a) 16 (b) 0 (c) 4 (d) None of these

CBSE Sample Paper Mathematics Standard Class X (Term I) 99 16. The distance between the points (a sin q + b cosq, 0) and (0, a cosq - b sin q) is (a) a2 + b2 (b) a2 - b2 (c) a2 + b2 (d) a2 - b2 17. If x tan 30° cos 30° = sin 60° cot 60°, then x is equal to (a) 1 (b) 3 (c) 1 (d) 1 2 2 18. In the given figure, DABC ~ DAQP. If AB = 6 cm, BC = 8 cm and PQ = 4 cm, then AQ is equal to P Q A B C (a) 2 cm (b) 2.5 cm (c) 3 cm (d) 3.5 cm 19. A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. What is the probability that the drawn ball is white or blue? (a) 1 (b) 2 (c) 5 (d) 7 2 6 12 20. If P is a point on Y-axis, whose ordinate is 3 and Q is a point (-5, 2), then the distance PQ is (a) 26 units (b) 24 units (c) 5 units (d) 65 units Section B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. The decimal expansion of 13 is 8 (a) 1.625 (b) 1.25 (c) 1.675 (d) 1.0625 22. If HCF (306, 657) = 9, then LCM (306, 657) is (a) 22338 (b) 23238 (c) 33228 (d) 32328 23. If x = r sin q cos f, y = r sin q sin f and z = r cosq, then (a) x2 + y2 + z2 = r2 (b) x 2 + y 2 - z2 = r2 SAMPLE PAPER 4 (c) x 2 - y 2 + z2 = r2 (d) z2 + y 2 - x 2 = r2 24. Two numbers ‘a’ and ‘b’ are selected successively without replacement in that order from the integers 1 to 10. The probability that a is an integer, is b (a) 17 (b) 1 (c) 17 (d) 8 45 5 90 45 25. The centre of a circle is (2a, a - 7). If the circle passes through the point (1, - 9) and has diameter 10 2 units, then the value of a is (a) 9 (b) - 3 (c) 3 (d) ± 3

100 CBSE Sample Paper Mathematics Standard Class X (Term I) 26. In the given figure, PB||CF and DP||EF then AD is equal to DE 8cm C F 2cm B P A D (a) 3 (b) 1 E (d) 2 4 3 3 (c) 1 4 27. The LCM of 3x 2 y, 4xz 2 and 5y2 z (where x is a positive integer) is (a) 30xy 2z2 (b) 15xyz2 (c) 30x 3y 2z (d) 60x 2y 2z2 28. The probability of getting a number between 1 and 100 which is divisible by 1 and itself only is (a) 29 (b) 1 (c) 25 (d) 23 98 2 98 98 29. If a and 1 are the zeroes of the quadratic polynomial x 2 - x + 8, then a is 2 (c) - 1 (a) 4 (b)1 (d) 16 44 30. If sin 2 q - 3 sin q + 2 = 1, then q will be cos2 q (a) 0° (b) 30° (c) 45° (d) 60° 31. If in two triangles DABC and DXYZ such that AB = BC = CA, then YZ XZ XY (a) DXYZ ~ DABC (b) DZXY ~ DABC (c) DCBA ~ DXYZ (d) DABC ~ DYZX 32. If tan q = 40, then the value of cosec q is 9 (a) 9 (b) 41 (c) 40 (d) 9 41 40 9 40 33. The ratio in which the segment joining the points (1, -3) and (4, 5) is divided by X-axis is (a) 3 : 5 (b) 5 : 3 (c) 4 : 7 (d) 7 : 4 SAMPLE PAPER 4 34. The sum and product of the zeroes of a quadratic polynomial are - 1 and 1 44 respectively. Then, the quadratic polynomial is (a) 4x2 + x + 1 (b) 3x2 - x + 2 (c) 4x2 - x - 1 (d) 3x2 + x - 2 35. Which of the following cannot be the probability of an event? (a) 2 (b) - 1 (c) 0.1 (d) 0.8 52 36. If a vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow of 28 m long, then the height of the tower is (a) 30 m (b) 35 m (c) 40 m (d) 42 m

CBSE Sample Paper Mathematics Standard Class X (Term I) 101 37. In the given figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm,PC = 2.5 cm, PD = 5 cm, ÐAPB = 50° and ÐCDP = 30°. Then, ÐPBA is equal to A 6 cm D P 5 cm 30° 50° 3 cm 2.5 cm B C (a) 30° (b) 60° (c) 80° (d) 100° 38. 5 ( 11 - 3) is an (a) rational number (b) irrational number (c) undefined (d) None of these 39. 5 yr hence, the age of a man shall be 3 times the age of his son while 5 yr earlier the age of the man was 7 times the age of his son. The present age of the man is (a) 45 yr (b) 50 yr (c) 43 yr (d) 40 yr 40. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(1, - 3) and R(- 3, 0), then the coordinates of P are (a) çèæ 1 , 1 ö÷ø (b) (14, 7) (c) (18, 9) (d) (10, 5) 5 10 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 Major Dhyanchand National Stadium is a very popular multi-purpose sports stadium at Mumbai. It has a capacity to seat 60000 people. The stadium is conducting the annual sports competition soon. The curator of the stadium is asked to figure out the dimensions for carving out some areas allotted for a hockey court and a shooting range, as shown in the given figure. SAMPLE PAPER 4

102 CBSE Sample Paper Mathematics Standard Class X (Term I) A Shooting OC Range B Hockey Court ED The shapes of the hockey court and the shooting range are a square and a triangle respectively. Both of the courts have a common edge that touches the centre of the stadium. The construction of the shooting range is such that the angle at centre is 90°. The radius of the stadium is 180 m. (take p = 3.14) On the basis of above information, answer the following questions. 41. What is the area allotted to shooting range? (a) 12600 m2 (b) 16200 m2 (c) 18660 m2 (d) 16880 m2 42. What is the area alloted to the hockey court? (a) 16200 m2 (b) 22000 m2 (c) 20000 m2 (d) 16880 m2 43. If the team of the curators managing the stadium likes to allot space for some more sports, how much area is available to them? (a) 76980 m2 (b) 95806 m2 (c) 60040 m2 (d) 69336 m2 44. If the boundaries of the hockey court and shooting range are to be fenced, then what is the required length (in m) of the fence? (a) 400(2 + 5 2 ) (b) 180(2 + 3 2 ) (c) 180(2 + 5 2 ) (d) 300(2 + 3 2 ) 45. If the cost of fencing is ` 6 per metre, what is the total cost (in `) of fencing? (a) 1800(2 + 3 2 ) (b) 1080(2 + 5 2 ) (c) 1080(2 + 3 2 ) (d) 2400(2 + 5 2 ) 46-50 are based on Case Study-2. Case Study 2 A tour company provides taxi to their customers for travelling and the taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ` 165 and for a journey of 18 km, the charge paid is ` 277. SAMPLE PAPER 4

CBSE Sample Paper Mathematics Standard Class X (Term I) 103 46. The fixed charges and charges per km respectively are (a) ` 25 and ` 10 (b) ` 10 and ` 25 (c) ` 25 and ` 14 (d) ` 14 and ` 25 47. The taxi fare for a distance of 25 km would be (a) ` 350 (b) ` 375 (c) ` 280 (d) ` 310 48. The pair of linear equations -3x + 4y = 5 and 9 x - 6y + 15 = 0 has 22 (a) unique solution (b) iInfinitely many solutions (c) no solution (d) Cannot be determined 49. The value of k for which the pair of linear equations 2x + 3y = 7 and (k - 1)x + (k + 2)y = 3k have infinitely many solutions is (b) k = 0 (a) k = - 1 (d) k = 7 (c) k = - 7 50. From the graph given below, the area of the triangle formed by the two lines and the Y-axis is Y 7 6 5 P (0, 5) 4 x –2y =0 Q (4, 2) 3 2 B (2, 1) 3x + 4y=20 A (0, 0) 1 X 123456 X¢ –1 0 –1 (a) 10 sq units Y¢ (c) 14 sq units (b) 5 sq units (d) 20 sq units SAMPLE PAPER 4

OMR SHEET SP 4 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) 105 Answers 1. (d) 2. (c) 3. (b) 4. (a) 5. (c) 6. (a) 7. (c) 8. (c) 9. (a) 10. (a) 11. (c) 12. (a) 13. (d) 14. (b) 15. (b) 16. (c) 17. (a) 18. (c) 19. (c) 20. (a) 21. (a) 22. (a) 23. (a) 24. (c) 25. (d) 26. (b) 27. (d) 28. (c) 29. (d) 30. (b) 31. (d) 32. (b) 33. (a) 34. (a) 35. (b) 36. (d) 37. (d) 38. (b) 39. (d) 40. (a) 41. (b) 42. (a) 43. (d) 44. (b) 45. (c) 46. (c) 47. (b) 48. (b) 49. (d) 50. (a) SOLUTIONS 1. Length of the plank will be HCF (42, 49, 63) In DAOB, AB2 = OA2 + OB2 [by pythagoras theorem] The prime factors of 42 = 2 ´ 3 ´ 7 = (6)2 + (4)2 = 36 + 16 = 52 49 = 7 ´ 7 \\ AB = 52 = 2 13 \\ The length of each side of rhombus is and 63 = 3 ´ 3 ´ 7 2 13 cm. \\ HCF (42, 49, 63) = 7 \\ Required number of plank will be 4. Total number of possible outcomes = 52 Number of red cards = 26 = 42 + 49 + 63 So, number of favourable outcomes = 26 7 77 \\ Required probability = 26 = 1 52 2 = 6 + 7 + 9 = 22 2. Let a and b are the zeroes of the polynomial x2 - (k + 4)x - 2 k + 3 \\ Sum of zeroes = a +b = - Coefficient of x 5. The reflection of any point P(x, y) in X-axis is Coefficient of x2 = - [-(k + 4)] = k + 4 given by (x, - y) 1 \\Reflection of (-3, 5) in X-axis will be (-3, - 5). = ab = - Constant term 6. Given, 4 tan A = 3 Coefficient of x2 and product of zeroes tan A = 3 4 = -2 k + 3 = -2 k + 3 1 4 sin2 A - 2 cos2 A 4 sin2 A - 2 4 sin2 A + 3 cos2 A cos2 A Now, it is given that Now, = 4 sin2 A cos2 A a + b = 2ab + 3 a + b = 2ab [divide cos2 A by numerator and denominator] Þ k + 4 = 2(-2 k + 3) 4 tan2 A-2 4 tan2 A+3 Þ k + 4 = -4k + 6 = Þ k + 4k = 6 - 4 Þ 5k = 2 æçè 3 ÷øö 2 9 Þ k=2 4 4 4 ´ - 2 9 -2 9-8 1 5 4 +3 9 + 12 21 = èçæ 3 ø÷ö 2 = = = 4 3. Let ABCD be a rhombus whose diagonals AC 4 ´ + 3 and BD bisect each other at O at right-angled Then, OA = 1 AC = 1 ´ 12 = 6 cm 22 and OB = 1 BD = 1 ´ 8 = 4 cm 22 Also, ÐAOB = 90° DC SAMPLE PAPER 4 4 cm O 7. The prime factors of 2, 3 and 4 are given as 6 cm 2 = 1´2 3 = 1´3 AB 4 =2 ´2 \\ LCM (2, 3, 4) = 2 ´ 2 ´ 3 = 12 So, they will meet at the starting point after 12 h. 8. Let the radius of circle be R and radii of given circles are r1 = 12 cm and r2 = 5 cm, respectively.

106 CBSE Sample Paper Mathematics Standard Class X (Term I) Then, pR2 = pr12 + pr22 13. A( 2, –2) P(k , 0) B(– 7, 4) R2 = r12 + r22 = (12 )2 + (5)2 12 By section formula R = (12 )2 + (5)2 = 144 + 25 P(x, y) = æç m1x2 + m2x1 , m1y2 + m2y1 ö÷ è m1 + m2 m1 + m2 ø = 169 = 13 cm Diameter of a circle = 2 R = 2 ´ 13 P(k , 0) = çèæ 1 ´ (-7) + 2 ´ 2 , 1 ´ 4 +2 ´ (-2 )øö÷ 1+2 1+2 = 26 cm æçè - 7+ 4 4 - 4 øö÷ 9. Given, zeroes of quadratic polynomial are P(k , 0) = 3 , 3 3 and 2 . Then, P(k , 0) = (- 1, 0) 23 \\ k =-1 Sum of zeroes (a + b) = 3 + 2 = 9 + 4 = 13 14. Now, 17 = 17 ´ 23 = 136 = 0.136 23 6 6 125 53 23 (10)3 and product of zeroes (ab) = 3 ´ 2 = 1 Hence, decimal expansion of 17 is 0.136. 23 125 \\ Required polynomial is x2 - (a + b)x + ab i.e. x2 - 13 x + 1 15. We have, f (x) = x2 - 16 6 = x2 + 0 × x - 16 10. Given, Ð A = 30° \\Sum of zeroes = -Coefficient of x = -(0) = 0 \\ 3 cos A = 3 ´ cos30° = 3 ´ 3 = 3 3 Coefficient of x2 1 22 16. Given points are (a sinq + bcosq , 0) and and 4 cos3 A = 4 ´ (cos30° )3 (0, a cosq - bsinq ) \\By distance formula = 4 ´ èçæ 3 øö÷ 3 = 4 ´ 33 = 33 d = (x2 - x1 )2 + (y2 - y1 )2 2 8 2 = (0 - a sinq - bcosq)2 + (a cosq - bsinq - 0)2 \\ 3 cos A - 4 cos3 A = 3 3 - 3 3 = 0 22 = a2 sin2 q + b2 cos2 q + 2 absinq cosq + a2 cos2 q + b2 sin2 q - 2 absinq cosq 11. Given, DE|| BC Then, AD = AE [by BPT theorem] BD CE = a2(sin2 q + cos2 q ) + b2(sin2 q + cos2 q) +2 2x = x -3 = a2 + b2 [Q sin2 x + cos2 x = 1] (2 x - 3) x Þ 2 x(x - 3) = (2 x - 3)(x + 2 ) 17. We have, x tan30° cos30° = sin60° cot60° Þ 2x2 - 6x = 2x2 + 4x - 3x - 6 Þ 6 = 6x + x x´ 1 ´ 3 = 3 ´ 1 \\ x=6 32 2 3 7 é tan30° = 1= cot60° ù Hence, the value of x is 6 êQ 3 ú ê =3 ú 7 êëand sin60° = cos30°úû 2 12. Total number of possible outcomes = 6 SAMPLE PAPER 4 Þ x´ 1 = 1Þx=1 Now, multiple of 2 are 2, 4, 6 22 \\Favourable number of outcomes is 3. 18. Given, DABC ~ DAPQ, BC = 8 cm, PQ = 4 cm and BA = 6 cm Now, Probability Now, DABC ~ DAQP = Number of favourable outcomes \\ BC = AB Total number of outcomes PQ AQ \\ Required probability = 3 = 1 [When two triangles are similar, then the 62 corresponding sides are in proportion]

CBSE Sample Paper Mathematics Standard Class X (Term I) 107 Þ 8= 6 z2 = r2 cos2 q … (iii) 4 AQ Adding Eqs. (i), (ii) and (iii), we get Þ AQ = 6 ´ 4 = 3 cm 8 x2 + y2 + z2 = r2 sin2 q cos2 f + r2 sin2 q sin2 f 19. Total number of balls = 24 + r2 cos2 q Now, x + 2 x + 3x = 24 Þ 6x = 24 x2 + y2 + z2 = r2 sin2 q(cos2 f + sin2 f) + r2 cos2 q Þ x=4 \\ Number of red balls = x = 4 x2 + y2 + z2 = r2 sin2 q + r2 cos2 q Number of white balls = 2 x = 8 Number of blue balls = 3x = 12 x2 + y2 + z2 = r2 [Q sin2 x + cos2 x = 1] Now, Total number of possible outcomes = 24 24. Given, two numbers are selected without replacement from integers 1 to 10. Now, favourable cases for which a will be an b integer are Let E be the event that the ball drawn is white 2 = 2; 3 = 3; 4 = 4; 4 = 2; 5 = 5 or blue 1112 1 6 = 6; 6 = 3; 6 = 2; 7 = 7; 8 = 8, \\ Number of outcomes favourable to E 12 3 11 8 = 4; 8 = 2; 9 = 9; 9 = 3; 10 = 10, = 8 + 12 = 20 2 4 13 1 P( E) = Number of favourable outcomes 10 = 5 and 10 = 2 25 Total number of outcomes = 20 = 5 \\Favourable number of cases = 17 24 6 20. Since, P lies on Y-axis and have ordinate as 3 Total cases are é1 , 1 , ¼1 ù , é2 , 2 ,¼ 2 ùúû, … \\ The point P is (0, 3) and Q is (-5, 2 ). êë 2 3 10 ûú êë 1 3 10 Now, The distance of PQ is é 10 , 10 , 10 , ¼ 10 ù êë 1 2 3 9 úû PQ = (-5 - 0)2 + (2 - 3)2 Total number of cases = 9 ´ 10 = 90 = (-5)2 + (-1)2 \\ Required probability = 17 = 25 + (-1)2 90 PQ = 26 units 25. Let the circle be as given below with centre O(2 a, a - 7) and A(1, - 9) be any point at the 21. As, 13 = 13 circle. 8 23 = 13 ´ 53 = 1625 = 1625 = 1.625 O A(1, – 9) 23 ´ 53 (10)3 1000 (2a, a – 7) Hence, decimal expansion of 13 is 1.625 \\ 2(OA) = 10 2 [Q Diameter = 2 2 given] 8 Þ OA = 5 2 Þ OA2 = (5 2 )2 = 50 22. We know that, Þ (2 a - 1)2 + (a - 7 + 9)2 = 50 SAMPLE PAPER 4 LCM (a, b) ´ HCF (a, b) = a ´ b [using distance formula] \\LCM (306, 657) ´ HCF (306, 657) = 306 ´ 657 (2 a - 1)2 + (a + 2 )2 = 50 Þ LCM (306, 657) = 306 ´ 657 4a2 + 1 - 4a + a2 + 4 + 4a = 50 Þ 5a2 = 50 - 5 HCF (306, 657) Þ 5a2 = 45 Þ a2 = 9 Þ a=±3 = 306 ´ 657 26. In DACF, BP||CF 9 [Q HCF(306, 657) = 9] So, by Basic proportionality theorem AB = AP = 22338 BC PF 23. Given, x = rsinq cos f, y = rsinq sin f and z = rcosq … (i) \\ x2 = r2 sin2 q cos2 f … (ii) y2 = r2 sin2 q sin2 f

108 CBSE Sample Paper Mathematics Standard Class X (Term I) Þ 2 = AP [Q BC = AC - AB] 31. We know that if, two triangles are said to be 8-2 PF similar if their corresponding angles are equal and the corresponding sides are proportion. Þ AP = 2 PF 6 XA Þ AP = 1 PF 3 Again, in DAEF, DP|| EF So, by Basic proportionality theorem AD = AP = 1 Y ZB C DE PF 3 i.e. AB = BC = CA 27. Factors of 3x2y = 3 ´ x2 ´ y, YZ XZ XY 4xz2 = 2 2 ´ x ´ z2 Then, DABC ~ DYZX and 5y2z = 5 ´ y2 ´ z 32. Given, tanq = 40 = AB LCM (3x2y, 4xz2 and 5y2z) 9 BC = 3 ´ 4 ´ 5 ´ x2 ´ y2 ´ z2 = 60x2y2z2 A 28. Number divisible by 1 and itself is a prime. 40 k Number of prime number between 1 and 100 = 25 q B 9k C i.e., (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, Let BC = 9k and AB = 40k 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97) By Pythagoras theorem, \\ Required probability = 25 AC = AB2 + BC2 98 29. We have, a and 1 are the zeroes of the polynomial 2 = (40k )2 + (9k )2 x2 - x + 8, then Constant term = 1600k 2 + 81k 2 Coefficient of x2 Product of zeroes = = 1681k 2 = 41k So, cosec q = AC = 41K = 41 \\ a´ 1 = 8 21 AB 40 K 40 Þ a =8 33. Let P(x, 0) be any point on X-axis, which 2 divides the line segment joining points A(1, - 3) Þ a = 16 and B(4, 5) in the ratio k : 1 sin2 q - 3 sinq +2 = k:1 cos2 q 30. We have, 1 A(1, – 3) P(x, 0) B(4, 5) Þ sin2 q - 3 sinq + 2 = cos2 q \\ (x, 0) = æçè 4k + 1 , 5k -3 ÷øö k +1 k +1 Þ sin2 q + sin2 q - 3 sinq + 2 = cos2 q + sin2 q [Adding sin2q both side] On equating y-coordinate both sides, we get SAMPLE PAPER 4 Þ 2 sin2 q - 3 sinq + 2 = 1[Q sin2 x + cos2 x = 1] 0 = 5k - 3 k +1 Þ 2 sin2 q - 3 sinq + 1 = 0 Þ 2 sin2q - 2 sinq - sinq + 1 = 0 5k - 3 = 0 Þ k = 3 Þ 2 sinq(sinq - 1) - 1(sinq - 1) = 0 5 Þ (2 sinq - 1) (sinq - 1) = 0 \\ The required ratio is 3 : 5. 34. Given, sum of the zeroes = a + b = -1 Þ 2 sinq - 1 = 0 or sinq - 1 = 0 Þ sinq = 1 or sinq = 1 4 2 Product of the zeroes = ab = 1 Þ q = 30° or q = 90° 4 \\ q = 30° \\ Quadratic polynomial [Qq = 90° does not satisfy the given equation]

CBSE Sample Paper Mathematics Standard Class X (Term I) 109 = k [x2 - (a + b)x + ab] 38. We know that, product of a non-zero rational and an irrational number is always irrational = k é x2 - èæç -41ø÷öx + 1ù number. ëê 4 ûú So, 5( 11 - 3 ) is an irrational number. = 4x2 + x + 1 [Q k = 4] 39. Let the present age of man = x yr 35. As, we know that probability of any event is lies from 0 to 1. and present age of son = y yr So, probability can not be negative. After 5 yr, age of man = (x + 5) yr Age of son = (y + 5) yr Hence, option (b) is correct. 36. Let AB be the height of pole and BC be the According to the question, length of its shadow and DE be the height of x + 5 = 3(y + 5) tower and EF be the length of its shadow such that AB = 6 m, BC = 4 m and EF = 28 m. x + 5 = 3y + 15 … (i) x - 3y = 10 Let length of DE be x m. x = 10 + 3y AD and 5 yrs earlier Age of man = x - 5 6m and Age of son = y - 5 B 4 m C E 28 m F According to the question, x - 5 = 7(y - 5) x - 7y = - 30 … (ii) In DABC and DDEF, Subtracting Eq. (i) from Eq. (ii), we get ÐB = ÐE = 90° - 4y = - 40 y = 10 and x = 10 + 3y = 10 + 3 ´ 10 = 40 ÐA = ÐD [Sun’s elevation] Hence, present age of man = 40 yr. DABC ~ DDEF [by AA similarity] \\ AB = BC 40. Let the coordinates of P be (x, y). Since, P is equidistant from Q(1, - 3) and R(- 3, 0). Then, DE EF PQ = PR Þ 6= 4 (1 - x)2 + (- 3 - y)2 = (- 3 - x)2 + (0 - y)2 x 28 Þ x = 6 ´ 7 = 42 m 37. Given, AP = 6 and BP = 3 = 6 [by distance formula] PD 5 PC 2.5 5 On squaring both side, we get Þ AP = BP 1 + x2 - 2x + 9 + y2 + 6y PD PC = 9 + x2 + 6x + y2 Now, in DAPB and DDPC Þ 10 - 2 x + 6y = 9 + 6x AP = BP According to the question, DP CP x = 2y and ÐAPB = ÐDPC = 50°, Þ 10 - 2(2 y) + 6y = 9 + 6(2 y) [vertically opposite angles] Þ 10 - 4y + 6y = 9 + 12 y then DAPB~ DDPC [SAS similarity criterion] Þ 10 - 9 = 12 y - 2 y Þ ÐAPB = 50° and ÐCPD = 50° Þ 10y = 1 SAMPLE PAPER 4 Þ ÐPAB = ÐPDC = 30° [by CPCT] Þ y= 1 In DAPB , 10 ÐAPB + ÐPBA + ÐBAP = 180° Þ x=2´ 1 = 1 Þ 50° + ÐPBA + 30° = 180° 10 5 ÐPBA = 180° - 80° = 100° Hence, coordinates of P are èçæ 1 , 1 ÷øö. 5 10

110 CBSE Sample Paper Mathematics Standard Class X (Term I) Solutions (41-45) Solutions (46-50) 41. The shape of the shooting range is right angle 46. Let the fixed charge be ` x and the charge per triangle km be ` y. O Now, according to the situation …(i) 180 m 90° 180 m x + 10y = 165 …(ii) AB and x + 18y = 277 \\ Area of triangle = 1 ´ Base ´ Height From Eq. (i), we get …(iii) 2 x = 165 - 10y In DOAB, Base = 180 m and Height = 180 m Now, put this value of x in Eq. (ii), we get \\ Area of triangle = 1 ´ 180 ´ 180 165 - 10y + 18y = 277 2 = 90 ´ 180 = 16200 m2 8y = 277 - 165 = 112 Þ y = 14 42. In the given figure Now, put y = 14 in Eq. (iii), we get Let OC = CD = x Using pythagoras theorem, x = 165 - 10y = 165 - 10(14) x2 + x2 = 1802 2 x2 = 32400 = 165 - 140 = 25 x2 = 16200 Þ x = 90 2 Fixed charge = ` 25 As we know, area of square = (side)2 Area of square, OCDE = x2 = (90 2 )2 Charge per km = ` 14 = 8100 ´ 2 = 16200 m2 47. According to the condition, 43. The available area is the difference of the total Fair = ` (x + 25y) area of the circular stadium and the two courts Now, put x = 25 and y = 14 i.e. \\ Fair = 25 + 25 ´ 14 = 25 + 350 = ` 375 \\Required area = pr2 - (Area of hockey court + area of shooting range) 48. On comparing the given equation with = pr2 - (16200 + 16200) = 3.14 ´ 180 ´ 180 - 32400 standard equation = 101736 - 32400 = 69336 m2 a1 = - 3, b1 = 4 and c1 = - 5 9, 15 a2 = 2 b2 = - 6 and c2 = 2 Now, a1 = -3 = - 2, b1 = 4 = - 2 a2 9 3 b2 -6 3 2 -5 -2 and c1 = 15 = 3 c2 2 i.e a1 = b1 = c1 a2 b2 c2 Hence, the given pair of linear equations has infinitely many solutions. 44. From the figure 49. On comparing, In DOAB, AB = OB2 + OA2 a1 = 2, b1 = 3 and c1 = - 7 [by Pythagoras theorem] a2 = k - 1, b2 = k + 2 and c2 = -3k = (180)2 + (180)2 -7 Now, a1 = k 2 , b1 = k 3 2 and c1 = -3k =7 = 180 2 m a2 - 1 b2 + c2 3k The required boundaries be the perimeter of SAMPLE PAPER 4 DOAB and ~ OCDE. For infinitely many solutions, Perimeter of (DOAB + ~OCDE) a1 = b1 = c1 = 180 + 180 + 180 2 + (4 ´ 90 2 ) a2 b2 c2 = 360 + 540 2 = 180(2 + 3 2 ) m 45. Total cost = Total length of fencing Þ k 2 1 = 7 Þ 6k = 7k - 7 Þ k = 7 - 3k ´ Rate per metre = 180(2 + 3 2 ) ´ 6 50. In DAPQ, Base = 5 and Height = 4 = ` 1080(2 + 3 2 ) We know that, Area of triangle = 1 ´ Base ´ height 2 \\ Area of triangle = 1 ´ 5 ´ 4 = 10 sq units. 2

CBSE Sample Paper Mathematics Standard Class X (Term I) 111 SAMPLE PAPER 5 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 ratio between HCF and LCM of 15, 20 and 5 is (a) 1 : 9 (b) 1 : 11 (c) 1 : 12 (d) 3 : 4 2. The graph of y = p(x) are given below, for polynomial p(x). Then, the number of zeroes is Y X¢ X y = p(x) Y¢ (a) 0 (b) 2 (c) 1 (d) 3 3. If tan 3 q - 1 = A sec2 q + B tan q, then A + B is equal to SAMPLE PAPER 5 tan q - 1 (a) 1 (b) -1 (c) 2 (d) 3 4. The condition for which the equations ax + b = 0 and cx + d = 0 are consistent, is (a) ad = bc (b) ad + bc = 0 (c) ab - cd = 0 (d) ab + cd = 0

112 CBSE Sample Paper Mathematics Standard Class X (Term I) 5. The room has its area as 120 m2 and perimeter as 44 m, then the length and breadth of the room are (a) 11 m and 2 m (b) 10 m and 2 m (c) 12 m and 10 m (d) 12 m and 1 m 6. HCF of two number is 23 and their LCM is 1449. If one of the number is 161, then the other number is (a) 207 (b) 307 (c) 1449 (d) None of these 7. Onkar and Neha play a badminton game. If the probability of Onkar winning the match is 0.75, then what is the probability of Neha winning the match? (a) 0.15 (b) 0.25 (c) 0.35 (d) 0.75 8. If 3 chairs and 1 table cost ` 1500 and 6 chairs and 1 table costs ` 2400, then the linear equations to represent this situation is (a) x + 3y = 1500 and 6x - y = 2400 (b) 3x - y = 1500 and 6x - y = 2400 (c) 3x + y = 1500 and 6x + y = 2400 (d) x - y = 1500 and x + y = 2400 9. If p(x) is a polynomial of atleast one degree and p(k) = 0, then k is known as (a) value of p(x) (b) zero of p(x) (c) constant term of p(x) (d) None of these 10. A bag contains 6 green balls and n yellow balls. If probability of drawing a yellow ball is five times that of a green ball, then n is equal to (a) 10 (b) 20 (c) 30 (d) 40 11. If sin q = 1, then the value of sin q(sin q - cosecq) is 2 (a) 3 (b) -3 (c) 3 - 3 4 4 2 (d) 2 12. Which of the following is irrational number? (a) 0.133 (b) 5.329685 ...... (c) 3.5428 (d) 9.265 13. It is proposed to build a single circular park, whose area is equal to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be (a) 10 m (b) 15 m (c) 20 m (d) 24 m 14. If the point P(k, 0) divides the line segment joining the points A(2, - 2) and B(- 7, 4) in the ratio 1 : 2, then the value of k is (a) 1 (b) 2 (c) - 2 (d) - 1 SAMPLE PAPER 5 15. If in two triangles DDEF and DPQR, ÐD = ÐQ and ÐR = ÐE, then which of the following is not true? (a) EF = DF (b) DE = EF (c) DE = DF (d) EF = DE PR PQ PQ RP QR PQ RP QR 16. The area of two similar triangles are respectively 36 cm 2 and 64 cm 2. The ratio of their corresponding sides is (a) 4 : 3 (b) 3 : 4 (c) 3 : 5 (d) 2 : 5

CBSE Sample Paper Mathematics Standard Class X (Term I) 113 17. The ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal, is (a) p : 2 (b) p : 3 (c) 3 : p (d) 2 : p 18. If sin q = cosecq = 1, then the value of sin q + cosecq is (a) 2 5 (b) 2 (c) 3 (d) 9 19. Ramesh buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 9 female fish. Then, the probability that the fish taken out is a male fish, is (a) 5 (b) 5 (c) 6 (d) 7 13 14 13 13 20. The perimeter (in cm) of a square circumscribing a circle of radius ‘a’ cm, is (a) 8a (b) 4a (c) 2a (d) 16a Section B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. The value of k, for which the given pair of linear equation 2x + 3y = 5 and 4x + ky = 10 has infinite solution, is (a) 1 (b) 3 (c) 6 (d) 0 22. If probability of success is 0.9%, then probability of failure is (a) 0.91 (b) 0.091 (c) 0.92 (d) 0.991 23. The number of zeroes lying between -4 and 4 of the polynomial f (x) whose graph is given below is Y f(x) X¢ –4 –2 O 24 X Y¢ (a) 2 (b) 3 (c) 4 (d) 1 24. If tan q = a, then(a sin q - b cosq) is equal to b (a sin q + b cosq) (a2 + b2 ) (a2 - b2 ) (c) (a 2 a2 b2 ) (d) (a2 b2 b2 ) (a) (a2 - b2 ) (b) (a2 + b2 ) + + 25. Denominator of the decimal expression 3.434 is of the form 2m ´ 5n , where m and n are SAMPLE PAPER 5 positive integers. (a) True (b) False (c) Can’t say (d) None of these 26. If the sum and product of zeroes of a quadratic polynomial is 3 and -2 respectively, then the quadratic polynomial is (a) x2 - 3x + 2 (b) x 2 + 3x - 2 (c) x2 - 3x - 2 (d) x2 - 5x - 4

114 CBSE Sample Paper Mathematics Standard Class X (Term I) 27. The system of equations 2x + 3y + 5 = 0 and 4x + ky + 7 = 0 is inconsistent when k is equal to (a) 6 (b) 7 (c) 5 (d) 3 28. In a game, the entry fee is `10. The game consists of tossing of 3 coins. If one or two heads show, Amita win the game and gets entry fee. The probability that she gets the entry fee is (a) 3 (b) 3 (c) 7 (d) 5 4 8 8 8 29. If 3x = secq and 3 = tan q, then the value of x2 - 1 is x x2 (a) 9 (b) 1 (c) 8 (d) 1 9 8 30. The decimal expansion of the rational number 23 will terminate after ………… 22 ´ 5 decimal place(s). (a) 3 (b) 2 (c) 4 (d) doesn’t terminate 31. If one of the zeroes of the quadratic polynomial x 2 - 3x + k is 2, then the value of k is (a) 10 (b) 2 (c) - 7 (d) - 2 32. The pair of equations x = a and y = b graphically represents lines, which are (a) parallel (b) intersecting at (b, a) (c) coincident (d) intersecting at (a, b) 33. In the given figure, three sectors of a circle of radius 7 cm, making angles of 60°, 80°, 40° at the centre are shown. The area (in cm2) of the shaded region is 80º 60º 40º (a) 77 (b) 154 (c) 44 (d) 22 34. The distance between the points (m, - n) and (- m, n) is (a) ( m2 + n2 ) units (b) (m + n) units (c) (2 m2 + n2 ) units (d) ( 2 m2 + n2 ) units SAMPLE PAPER 5 35. In the given figure, AOB is a diameter of a circle with centre O. The value of tan A × tan B will be C 3 cm 2 cm AB O (a) 1 (b) 2 (c) 3 (d) 3

CBSE Sample Paper Mathematics Standard Class X (Term I) 115 36. If a school has two sections A and B in class Xth. Section A has 32 students and section B has 36 students. Then minimum number of books must be in library that can be distributed equally in both section, is (a) 288 (b) 278 (c) 268 (d) 258 37. In DABC and DDEF, ÐB = ÐE, ÐF = ÐC and AB = 3DE. Then, the two triangles are (a) congruent but not similar (b) similar but not congruent (c) neither congruent nor similar (d) congruent as well as similar 38. An arc of a circle is of length 5p cm and the sector bounds an areas of 20pcm2. Then, the radius of the circle is (a) 4 cm (b) 8 cm (c) 12 cm (d) 16 cm 39. The zeroes of the quadratic polynomial x 2 + kx + k, where k > 0 (a) are both positive (b) are both negative (c) are always equal (d) are always unequal 40. The sum of exponents of prime factors of 196 is (a) 3 (b) 4 (c) 5 (d) 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 Bhavna found the picture of Indian flag on the moon hosted by the chandrayan, which was lunar exploration mission by the Indian space research organisation. She wondered, if there is any relation between the height of the flag pole and the shadow of the flag. 41. A flag pole 18 m high casts a shadow 9.6 m long. The distance of the top of the pole SAMPLE PAPER 5 from the far end of the shadow is (a) 18.6 m (b) 20.4 m (c) 17.8 m (d) 15 m

116 CBSE Sample Paper Mathematics Standard Class X (Term I) 42. Which concept is used to find the distance of the top of the pole from the far end of the shadow? (a) Pythagoras theorem (b) Thales theorem (c) Converse of thales theorem (d) Converse of pythagoras theorem 43. Now, if the flag pole 4 m high casts a shadow 3 m long, then the distance of the top of the pole from the far end of the shadow is (a) 5 m (b) 8 m (c) 6 m (d) None of these 44. If the flag pole 15m high and the distance of the top of the pole from the far end of the shadow is 25 m, then the length of shadow is (a) 18 m (b) 14 m (c) 20 m (d) 12 m 45. If the distance of the top of the pole from the far end of the shadow is 13 m and the length of a shadow is 5 m, then the height of the flag pole is (a) 13 m (b) 12 m (c) 15 m (d) 17 m 46-50 are based on Case Study-2. Case Study 2 Shivam went to the hospital near to his home for COVID-19 test along with his family members. The seats in the waiting area were as per the norms of distancing during this pandemic (as shown in the given figure). His family members took their seats surrounded by the circular area 8 Palak Shivam 7 Akshay 6 234 Malika 7 5 56 4 3 2 1 Enter O1 46. Considering O as the origin, what are the coordinates of O? (a) (0,1) (b) (1, 0) (c) (0, 0) (d) (-1, -1) 47. What is the distance between Palak and Shivam? (a) 10 units (b) 2 5 units (c) 10 units (d) 8 units 48. What are the coordinates of seat of Akshay? SAMPLE PAPER 5 (a) (2, 3) (b) (3, 2) (c) (0, 3) (d) (2, 0) 49. What will be the coordinates of a point exactly between Akshay (A) and Malika (M) where a person can be seated? (a) (3.5, 2.5) (b) (2.5, 5) (c) (10, 5) (d) (1.5, 0.5) 50. Determine the shape of the figure while on joining the points where Shivam(S) family members are seated. (a) rectangle (b) square (c) parallelogram (d) rhombus

OMR SHEET SP 5 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)

118 CBSE Sample Paper Mathematics Standard Class X (Term I) Answers 1. (c) 2. (c) 3. (c) 4. (a) 5. (c) 6. (a) 7. (b) 8. (c) 9. (b) 10. (c) 11. (b) 12. (b) 13. (a) 14. (d) 15. (b) 16. (b) 17. (b) 18. (b) 19. (b) 20. (a) 21. (c) 22. (d) 23. (c) 24. (b) 25. (a) 26. (c) 27. (a) 28. (a) 29. (b) 30. (b) 31. (b) 32. (d) 33. (a) 34. (c) 35. (a) 36. (a) 37. (b) 38. (b) 39. (b) 40. (b) 41. (b) 42. (a) 43. (a) 44. (c) 45. (b) 46. (c) 47. (c) 48. (a) 49. (a) 50. (a) SOLUTIONS 1. The prime factors of 15, 20 and 5 are given as Þ b2 -12 b - 10b + 120 = 0 Þ b(b - 12 ) - 10(b -12 ) = 0 15 = 3 ´ 5 Þ (b - 12 )(b -10) = 0 20 = 2 2 ´ 5 Þ b = 10 or 12 \\ l = 22 - 10 or l = 22 - 12 and 5 = 5 l = 12 m or l = 10 m \\ HCF of (15, 20, 5) = 5 LCM of (15, 20, 5) = 2 2 ´ 3 ´ 5 = 60 6. Let the required number be x. \\ Required ratio = HCF = 5 = 1 or 1 : 12 As product of two numbers = (HCF ´ LCM) of two number LCM 60 12 \\ x ´ 161 = 23 ´ 1449 x = 23 ´ 1449 = 207 2. The graph of y = p(x) cuts on X-axis at only one 161 point. So, the number of zeroes is only one. 3. tan3 q - 1 (tan q -1)(tan2 q + 12 + tan q) = 7. Probability of winning the game by Onkar tan q -1 (tan q - 1) = 0.75 [Q a3 - b3 = (a - b) (a2 + b2 + ab)] Total probability = 1 Here, P( A) + P (not A) = 1 = sec2 q + tan q [Q1 + tan2 x = sec2 x] So, probability of winning the game by Neha Þ sec2 q + tan q = Asec2 q + Btan q [given] = 1 - 0.75 = 0.25 On comparing, we get A = 1 and B = 1 8. We have, the cost of 1 chair = ` x and the cost of 1 table = ` y \\ A+ B=1+1=2 Then, according to the given situation 3x + y = 1500 4. Given equations are ax + b = 0 and cx + d = 0 and 6x + y = 2400 On comparision with ax + by + c = 0, we have 9. Let p(x) = ax + b a1 = a, b1 = 0 and c1 = b Put x = k, then a2 = c, b2 = 0 and c2 = d p(k ) = ak + b = 0 [Q p(k ) = 0 (given)] For the system to be consistent, a1 = c1 \\ k is zero of p(x). a2 c2 a=b So, k is known as zero of p(x). cd SAMPLE PAPER 5 10. Given, a bag contains 6 green balls and or ad = bc n yellow balls. 5. Given, area = 120m 2 and perimeter = 44 m Total number of balls = 6 + n Let the length be l and the breadth be b. …(i) So, the possible number of outcomes = 6 + n \\ Area, lb = 120 m2 …(ii) Now, P (drawing a green ball) = 6 and perimeter 2(l + b) = 44 m 6+n and P (drawing a yellow ball) = n Þ l + b = 22 6+n Þ l = 22 - b According to the question, On putting the value of Eq. (ii) in Eq. (i) n æ 6 ö Þ (22 - b) b = 120 + 5çç + ÷ n 30 Þ b2 -22 b + 120 = 0 = è ÷ Þ = ø 6 n 6 n

CBSE Sample Paper Mathematics Standard Class X (Term I) 119 11. We have, sin q = 1 \\ DF = ED = FE 2 QP RQ PR cosec q = 2 é sin x = 1 ù Hence, except option (b), all options are true. cosec êQ x ú 16. Let DABC and DDEF are two similar triangles ë û such that ar (DABC) = 36 cm2 and ar (DDEF) = 64 cm2 Now, sin q (sin q - cosec q) = 1æ 1 -2 ö ç ÷ 2è2 ø = 1 -1 = 1 - 4 = -3 We know that ratio of area of two similar 4 44 triangles is equal to square of ratio of their 12. The number which is non-terminating and corresponding sides. non-repeating are irrational number. ar (DABC) = ( AB)2 Here, only 5.329685 ...... is the required answer. Q ( DE)2 ar (DDEF) 13. Let the radius of new park be r m. 36 æ AB ö2 ç ÷ Given, diameters of other circular parks are Þ = 64 è DE ø d1 = 16 m and d2 = 12 m Þ AB = 6 = 3 DE 8 4 Þ r1 = d1 = 16 =8m 2 2 17. We are given that diameter of circle and side of and r2 = d2 = 12 =6 m an equilateral triangle are equal. 2 2 Let d and a be the diameter and side of circle According to the question, and equilateral triangle respectively. pr2 = pr12 + pr22 \\ d=a \\ pr2 = p(8)2 + p(6)2 Þ pr2 = 64p + 36p We know that, Þ pr2 = 100 p Area of circle = p r2 Þ r = 10 m Area of an equilateral triangle = 3 a2 4 14. When a point P(x, y) divides (internally) the Ratio = Area of circle line segment joining the points A(x1 , y1 ) and Area of equilateral triangle B(x2 , y2 ) in the ratio m : n, then by section formula = pr2 x = mx2 + nx1 and y = my2 + ny1 3 a2 m+ n m+ n 4 (k, 0) pçæ d ö2 1 P2 ÷ = è2 ø A(2, –2) B(–7, 4) 3 a2 4 æ 1´ (- 7) + 2 ´ 2 1 ´ 4 + 2 ´ (- 2 ) ö P(k , 0) = ç 1+2 , 1 + 2 ÷ p ´ a2 ç ÷ è ø =4 [Q d = a] = æ - 7 + 4 , 4 - 4 ö = (- 1, 0) 3 a2 ç ÷ 4 è3 3ø \\ k =-1 p = 15. It is given that in DDEF and DPQR, 3 ÐD = ÐQ andÐR = ÐE \\ Required ratio = p : 3 SAMPLE PAPER 5 Q 18. We have, sin q = cosec q = 1 D \\ sin q + cosec q = 1 + 1 Þ sin q + cosec q = 2 F EP R 19. There are 5 + 9 = 14 fish in the tank. \\ Total number of possible outcomes = 14 \\DDEF - DQRP [by AA similarity criterion] Out of 14 fish, there are 5 male fish. So, total number of favourable outcomes = 5

120 CBSE Sample Paper Mathematics Standard Class X (Term I) \\ Required probability a sin q - b cos q = Number of favourable outcomes = cos q cos q Total outcomes =5 a sin q + b cos q 14 cos q cos q Hence, the probability that the fish taken out [dividing numerator and as a male fish is 5/14. denominator by cos q] 20. Radius of circle = a cm = (a tan q - b) So, diameter of circle = 2 a (a tan q + b) \\Side of square diameter of circle = 2 a Perimeter of square = 4 ´ side a´ a -b a2 - b2 a2 - b2 = 4 ´ 2 a = 8a cm b b = = = PA S a ´ a + b a2 + b2 a2 + b2 bb a 25. Here, 3.434 = 3434 = 3434 = 1717 O 1000 23 ´ 53 22 ´ 53 a Denominator is of the form 2 m ´ 5n QB R m = 2 and n = 3 (positive integers) 21. We know that, 26. Given, sum of zeroes = 3 If a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has infinite solution, then and product of zeroes = -2 a1 = b1 = c1 Quadratic polynomial a2 b2 c2 So, 2 x + 3y - 5 = 0 and 4x + ky - 10 = 0 = x2 - (sum of zeroes)x + (product of zeroes) has infinite solution, then = x2 - 3x + (-2 ) = x2 - 3x - 2 2 = 3 = -5 4 k -10 27. For the given system to be inconsistent Þ 3=1 a1 = b1 ¹ c1 k2 a2 b2 c2 Þ k =6 On comparision the given equation with the 22. Given, probability of success = 0.9% = 0.9 standard equation, we get 100 a1 = 2, b1 = 3 and c1 = 5 Since, probability of failure a2 = 4, b2 = k and c2 = 7 + probability of success = 1 \\ 2 =3 ¹ 5 4k 7 Probability of failure = 1 - probability of success Þ k =6 = 1 - 0.9 = 1 - 9 100 1000 28. In case of tossing three coins. = 1000 - 9 = 991 = 0.991 1000 1000 Total number of possible outcomes = 8 Favourable outcomes are 23. The graph of y = f (x) cuts on X-axis at 4 points. {HHT, HTH, THH, HTT, THT, TTH} i.e 6. \\ Required probability = 6 = 3 So, that number of zeroes lying between -4 and 4 is 4. 84 24. We have, tan q = a 29. Given, 3x = sec q b Now, (a sin q - bcos q) 9x2 = sec2 q …(i) …(ii) SAMPLE PAPER 5 (a sin q + bcos q) and 3 = tan q x 9 = tan2 q x2 Subtracting Eq. (ii) from Eq. (i), we get 9x2 -9 = sec2 q - tan2 q x2 Þ 9æç x2 - 1 ö =1 [Q sec2 x - tan2 x = 1] è x2 ÷ ø Þ x2 - 1 = 1 x2 9

CBSE Sample Paper Mathematics Standard Class X (Term I) 121 30. Since, the denominator is of the form 2 n ´ 5m. 34. Distance between points (m, - n) and (- m, n) = (x2 - x1 )2 + (y2 - y1 )2 So, it is terminating. [by Distance formula] = (- m - m)2 + (n + n)2 Now, 23 = 23 = 1.15 = (-2 m)2 + (2 n)2 2 2 ´ 51 20 = 4m2 + 4n2 = 2 m2 + n2 units So, 23 will terminate after 2 places of 2 2 ´ 51 decimal. 31. Let f (x) = x2 - 3x + k Since, 2 is a zeroes of f (x). \\ f (2 ) = 0 35. In DABC, ÐC = 90° [angle in a semi-circle] Þ 2 2 - 3(2 ) + k = 0 tan A = BC = 2 Þ 4 -6 + k =0 AC 3 Þ k =2 tan B = AC = 3 BC 2 32. We know that, x = a is the equation of a straight line parallel to the Y-axis at a distance \\tan A× tan B = 2 ´ 3 = 1 of ‘a’ units from it. 32 Similarly, y = b is the equation of a straight line 36. Required number of books will be LCM of 32 parallel to the X-axis at a distance of ‘b’ units and 36. Now, from it. 32 = 2 5 and 36 = 2 2 ´ 32 So, the pair of equations x = a and y = b \\ LCM (32, 36) = 2 5 ´ 32 = 288 graphically represents lines which are So, minimum number of books is 288. intersecting at (a, b) as shown below 37. In DABC and DDEF, we have Y (0, b) (a, b) ÐB = ÐE [given] ÐF = ÐC [given] y=b Þ DABC ~ DDEF [by AA similarity criterion] X¢ 0 X D (a, 0) A x=a Y¢ B CE F Hence, the two lines are intersecting at (a, b). Since, AB and DE are corresponding sides. 33. Radius of circle, r = 7 cm But AB = 3 DE [given] Now, Area of shaded region We know that, two triangles are congruent, if they have the same shape and size. = Area of three sectors But here, AB = 3 DE i.e., two triangles are not of same size. = q1 pr2 + q2 pr2 + q3 pr2 SAMPLE PAPER 5 360° 360° 360° \\DABC is not be congruent to DDEF. é = q pr2 ù Hence, the two triangles are similar but not êëQ Area of sector 360° úû congruent. = pr2 (q1 + q2 + q3 ) 38. Let r be the radius of the circle and q be the 360° = 22 ´ 1 ´ 7 ´ 7(60° + 80° + 40°) angle formed by arc of the sector. … (i) 7 360° q … (ii) = 11 ´ 1 ´ 7 ´ 180° Then, length of arc = ´ 2 pr = 5p 180° 360° = 77 cm2 and, area of sector = q ´ pr2 = 20p 360°

122 CBSE Sample Paper Mathematics Standard Class X (Term I) Dividing Eq. (ii) by Eq. (i), we get 43. Let AB be a flag pole of the height 4m and BC be its shadow is 3 m. q pr2 = 20p 360° A q 5p ´ 2 pr 360° Þ r =4 4m 2 Þ r = 8 cm C 3m B 39. Let a and b be the zeroes of x2 + kx + k. Then, a + b = - k and ab = k In right triangle DABC, Since, k > 0, a + b < 0 and ab > 0 ÐB = 90° \\ BC2 + AB2 = AC2 This is possible only when a and b are both negative. [from Pythagoras theorem] 40. Given number is 196. Þ 32 + 42 = AC2 Prime factors of 196 = 2 2 ´ 72 Þ 9 + 16 = AC2 Þ 25 = AC2 So, the exponents are 2 and 2. Þ AC = 25 = 5 m \\ Required sum = 2 + 2 = 4 Solution (41-45) 44. In DPQR, ÐQ = 90° 41. Let AB be a flag pole of height 18 m and BC be P its shadow of 9.6 m long As pole is vertical, so ÐABC = 90° SAMPLE PAPER 5 13 m 25 mA 15 m 18 m RQ C 9.6 m B Let PQ be a flag pole of the height 15 m and the QR be a shadow of the flag pole. The required distance of the top of the pole \\ PQ2 + QR2 = PR2 A from the far end C of the shadow is AC. In right angled DABC, we have [from Pythagoras theorem] 152 + QR2 = 252 AC2 = AB2 + BC2 Þ QR2 = 625 - 225 = 400 [by Pythagoras theorem] QR = 20 m Þ AC2 = (18)2 + (9.6)2 45. In DABC, use pythagoras theorem, Þ AC2 = 324 + 92.16 Þ AC2 = 416.16 A Þ AC = 416.16 B 5m C = 20.4 AC = AB2 - BC2 Hence, the required distance is 20.4 m. AC = (13)2 - (5)2 42. Pythagoras theorem is used to find the = 169 - 25 = 144 = 12 m distance of the top of the pole from the far end of the shadow.

CBSE Sample Paper Mathematics Standard Class X (Term I) 123 Solution (46-50) PS = (3 - 6)2 + (6 - 5)2 46. The coordinates of the origin is O(0, 0). = 9 + 1 = 10 units 47. The coordinates of Palak and Shivam are AM = (2 - 5)2 + (3 - 2 )2 P(3, 6) and S(6, 5) respectively. = 9 + 1 = 10 units The distance between PS = (3 -6)2 + (6 - 5)2 P(3, 6) S(6, 5) = 9+1 = 10 units 48. The coordinates of Akshay are (2, 3). 49. The coordinates of Akshay (A) and Malika (M) A(2, 3) M(5, 2) are (2, 3) and (5, 2) respectively. By using mid-point formula, coordinates of AS = (6 - 2 )2 + (5 - 3)2 point exactly between A and M is = 42 + 2 2 = 16 + 4 æ2 + 5, 3 +2 ö = æ 7 , 5ö = 20 = 2 5 units ç ÷ ç ÷ and PM = (5 - 3)2 + (2 - 6)2 è 2 2 ø è2 2 ø = (2 )2 + (- 4)2 or (3.5, 2.5) = 4 + 16 50. The coordinates of the families member are = 20 = 2 5 units A(2, 3), P(3, 6), M(5, 2) and S(6, 5). Here, AP = PS = MS = AM = 10 units The distance between these points are and diagonals AS = PM = 2 5 units Hence, shape of the figured formed by joining AP = (2 - 3)2 + (3 - 6)2 points is a square. = 1 + 9 = 10 units MS = (5 - 6)2 + (2 - 5)2 = 1 + 9 = 10 units SAMPLE PAPER 5

124 CBSE Sample Paper Mathematics Standard Class X (Term I) SAMPLE PAPER 6 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 square root of all positive integers are (a) rational number (b) irrational number (c) may be rational or irrational number (d) None of these 2. If - 2 is one of the zeroes of the polynomial x 2 - x + k, then the value of k is (a) - 6 (b) - 8 (c) 8 (d) - 7 3. If the pair of lines are coincident, then we say that pair of lines is consistent and it has a (a) unique solution (b) no solution (c) infinite solutions (d) None of these SAMPLE PAPER 6 4. In the given figure, DPMN ~ DPQR and PM = 3 cm, PQ = 4 cm, PN = 6 cm and PR = 8 cm, then relation between MN and QR is P MN (a) parallel QR (c) intersect (b) perpendicular (d) None of these

CBSE Sample Paper Mathematics Standard Class X (Term I) 125 5. If the points A(4, 3) and B(x, 5) are on the circle with centre O(2, 3) , then the value of x is (a) 5 (b) 6 (c) 2 (d) 4 (c) - 1 (d) cos 2 q 6. The value of é sin 2 q - cos2 q ù is ê + cot 2 + tan 2 ú ë1 q 1 qû (a) sin2 q - cos 2 q (b) sin2 q 7. The HCF and LCM of two numbers are 33 and 264 respectively. When the first number is completely divided by 2 the quotient is 33. The other number is (a) 66 (b) 130 (c) 132 (d) 196 8. The parabola representing a quadratic polynomial f (x) = ax 2 + bx + c open downwards when (a) a > 0 (b) a < 0 (c) a = 0 (d) a > 1 9. The values of x and y which satisfy the equations 2x + y + 1 = 0 and 2x - 3y + 8 = 0 are (a) 1 and 2 (b) 11 and 7 (c) - 11 and 7 (d) 2 and 3 84 84 10. If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other, then two triangles are similar, then criteria of similarity is (a) AA similarity (b) SAS similarity (c) ASA similarity (d) None of these 11. If A = (a 2 , 2a) and B = æ 1 , - 2 ö and S = (1, 0), then 1 + 1 is equal to ç a2 a ÷ SA SB è ø (a) 2 (b) 1 (c) 1 (d) 1 2 3 12. If 2 sin 4q = 3 ´ 4, then the value of q is 3 12 (a) 30° (b) 45° (c) 15° (d) 60° 13. If one zero of the quadratic polynomial 2x 2 - 6kx + 6x - 7 is negative of the other, then k is equal to (b) 1 (c) 0 (d) - 1 (a) - 1 2 14. In school, there are two sections, section A and section B of class X. There are 32 students in section A and 36 students in section B. Find the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B. (a) 288 (b) 388 (c) 208 (d) None of these SAMPLE PAPER 6 15. If the pair of linear equations 2x + 3y = 11 and(m + n)x + (2m - n) y - 33 = 0 has infinitely many solutions, then the values of m and n are (a) 5 and 1 (b) 1 and 2 (c) - 1 and 5 (d) 1 and - 5 16. 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

126 CBSE Sample Paper Mathematics Standard Class X (Term I) 17. A girl walks 500 m towards East and then 1200 m towards North, then the travelling distance from the starting point is (a) 1100 m (b) 1200 m (c) 1300 m (d) 1400 m 18. The value of sin q = 4 is 3 (a) possible (b) not possible (c) in special case, if is possible (d) None of these 19. The distance of the point (- 12, 5) from the origin is (a) 12 units (b) 5 units (c) 13 units (d) 169 units 20. DABC is an isosceles triangle in which ÐC = 90°. If BC = 2 cm, then the value of AB is (a) 4 2 cm (b) 2 2 cm (c) 4 cm (d) 2 cm Section B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. On Delhi road, three consecutive traffic lights change after 36, 42 and 72 seconds. If the lights are first switched on at 9.00 am, at what time will they change simultaneously? (a) 9:08:04 (b) 9:08:24 (c) 9:08:44 (d) None of these 22. If the zeroes of the quadratic polynomial ax 2 + x + a are equal, then a 2 is (a) 1 (b) 4 (c) 2 (d) 1 2 4 23. Given that sin q = m, then cot q - tan q is n (a) n m2 - n2 (c) n n2 - 2 m2 m2 - n2 (b) m2 - n2 (d) m m n2 - m2 24. If in two triangles DDEF and DPQR, ÐD = ÐQ and ÐR = ÐE, then which of the following is not true? (a) EF = DF (b) DE = EF PR PQ QR QP (c) DE = DF (d) EF = DE QR PQ RP QR 25. If the distance between the points (x, - 1) and (3, 2) is 5, then the value of x is SAMPLE PAPER 6 (a) - 7 or - 1 (b) - 7 or 1 (c) 7 or 1 (d) 7 or - 1 26. The rational form of 0. 08 is in the form of p, then ( p2 + q 2) is q (a) 1850 (b) 1008 (c) 2041 (d) 3056 27. If both zeroes of the quadratic polynomial x 2 - 2kx + 2 are equal in magnitude but opposite in sign, then value of k is (a) 1 (b) - 1 (c) 0 (d) - 1 2 2

CBSE Sample Paper Mathematics Standard Class X (Term I) 127 28. A number consists of two-digits. The sum of the digits is 12 and the unit’s digit, when divided by the ten’s digit gives the result as 3. The number is (a) 84 (b) 48 (c) 93 (d) 39 29. Given DABC ~ DDEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm. Then, perimeter of DDEF is (a) 20 cm (b) 30 cm (c) 35 cm (d) 47 cm 30. If 5x - 3y + 2k = 0 is a median of the triangle whose vertices are at points A(-1, 3), B(0, 4) and C(-5, 2). The value of k is (a) 19 (b) 9.5 (c) 20.5 (d) 21.5 31. If the HCF of 408 and 1032 is expressible in the form 1032 ´ 2 + 408 ´ p, then the value of p is (a) 5 (b) - 5 (c) 4 (d) - 4 32. If DABC ~ DDEF and DABC is not similar to DDEF then which of the following is not true? (a) BC× DF = AC× EF (b) AB× DF = AE× DE (c) BC× DE = AB× EE (d) BC× DE = AB× FE 33. If x = 3 and y = 1 is the solution of the line 2x + y - q 2 - 3 = 0. The value of q is (a) ± 2 (b) 2 (c) - 2 (d) - 1 34. The line segment joining the points (3, - 1) and (- 6, 5) is trisected. The coordinates of point of trisection are (a) (3, 3) (b) (- 3, 3) (c) (3, - 3) (d) (- 3, - 3) 35. 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 36. If 1 + 1 = k cosec2q, then the value of k is 1 + cosq 1 - cosq (a) 1 (b) - 1 (c) 2 (d) 1/2 37. There is a circular path around a sports field. Priya takes 18 min to drive one round of the field. Harish takes 12 min. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet? (a) 36 min (b) 18 min (c) 6 min (d) They will not meet 38. If D and E are respectively the points on the sides AB and AC of a DABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE||BC. Then, length of DE (in cm) is SAMPLE PAPER 6 (a) 2.5 (b) 3 (c) 5 (d) 6 39. If the sum of the zeroes of the polynomial g(x) = ( p2 - 23) x 2 - 2x - 12 is 1, then p takes the value(s) (a) 23 (b) - 23 (c) 2 (d) ± 5 (d) 65 40. If sin q = 5 , then the value of 2 secq × tan q is 13 72 (a) 65 (b) 45 (c) 65 30 25 75

128 CBSE Sample Paper Mathematics Standard Class X (Term I) 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 Shivani bought a pendulum clock for her living room. The clock contains a small pendulum of length 15 cm. The minute hand and hour hand of the clock are 18 cm and 12 cm long respectively. Based on the above information, answer the following questions. 41. The area swept by the minute hand in 10 min are (a) 22.24 cm 2 (b) 169.71 cm2 (c) 44 cm2 (d) 44.42 cm2 42. If the pendulum covers a distance of 22 cm in the complete oscillation, then the angle described by pendulum at the centre is (a) 40° (b) 42° (c) 45° (d) 48° 43. The angles described by hour hand in 10 min are (a) 5° (b) 10° (c) 15° (d) 20° 44. The area swept by the hour hand in 1 h are SAMPLE PAPER 6 (a) 7.68 cm 2 (b) 8.2 cm2 (c) 8.86 cm2 (d) 37.71 cm2 45. The area swept by the hour hand between 11 am and 5 pm are (a) 452.52 cm2 (b) 62cm2 (c) 70 cm2 (d) 72 cm2

CBSE Sample Paper Mathematics Standard Class X (Term I) 129 46-50 are based on Case Study-2. Case Study 2 On a weekend Rakhi was playing cards with her family. The deck has 52 cards and her brother drew one card. 46. Find the probability of getting a queen of red colour. (a) 1 (b) 1 (c) 1 (d) 1 26 13 52 4 47. Find the probability of getting an ace. (d) 3 13 (a) 1 (b) 1 (c) 2 26 13 13 (d) 3 26 48. Find the probability of getting a jack of diamond. (d) 1 (a) 1 (b) 1 (c) 3 4 26 52 52 (d) 1 49. Find the probability of getting a red face card. 4 (a) 3 (b) 1 (c) 1 26 13 52 50. Find the probability of getting a club. (a) 1 (b) 1 (c) 1 26 13 52 SAMPLE PAPER 6

OMR SHEET SP 6 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) 131 Answers 1. (c) 2. (a) 3. (c) 4. (a) 5. (c) 6. (a) 7. (c) 8. (b) 9. (c) 10. (a) 11. (c) 12. (c) 13. (b) 14. (a) 15. (a) 16. (d) 17. (c) 18. (b) 19. (c) 20. (b) 21. (b) 22. (d) 23. (d) 24. (b) 25. (d) 26. (c) 27. (c) 28. (d) 29. (b) 30. (b) 31. (b) 32. (c) 33. (a) 34. (b) 35. (a) 36. (c) 37. (a) 38. (c) 39. (d) 40. (d) 41. (b) 42. (b) 43. (a) 44. (d) 45. (a) 46. (a) 47. (b) 48. (b) 49. (a) 50. (d) SOLUTIONS 1. Clearly, 9 = ± 3, which is rational number. = sin2 q ´ sin2 q - cos2 q ´ cos2 q Hence, square root of all positive integers may = sin4 q - cos4 q be rational or irrational number. = (sin2 q - cos2 q) (sin2 q + cos2 q) 2. Given, polynomial x2 - x + k [Q a2 - b2 = (a - b) (a + b)] Let p(x) = x2 - x + k = sin2 q - cos2 q [Q sin2 A + cos2 A = 1 ] and - 2 is one of the zeroes of the polynomial. \\ p(- 2 ) = 0 7. Given, HCF = 33 and LCM = 264 Þ 0 = (- 2 )2 - (- 2 ) + k Þ 0=4+2 +k Let the other number be x. First number = 2 ´ 33 = 66 Þ k =-6 Q Product of numbers = HCF ´ LCM \\ x ´ 66 = 33 ´ 264 3. If the pair of lines are coincident, then it has Þ x = 33 ´ 264 infinite number of solutions and hence, consistent. 66 = 132 4. Given, DPMN ~ DPQR 8. If a < 0 in f (x) = ax2 + bx + c, then parabola Therefore, PM = PN Þ 3 = 6 Þ 3 = 3 PQ PR 4 8 4 4 open downwards and a > 0, then parabola open upwards. By converse of Basic proportionality theorem, we can say that MN ||QR. 9. The given equation can be written as 5. Since, A and B lie on the circle having centre O. 2x + y + 1=0 Then, the distance between points A and B Þ 2x + y = -1 …(i) from the centre are same as they are radius of the circle. and 2 x - 3y + 8 = 0 B (x, 5) Þ 2x - 3y = - 8 …(ii) (4, 3) A Subtract Eq. (ii) from Eq. (i), we get O (2, 3) 4y = 7 i.e. OA = OB Þ y=7 From the distance formula, 4 (4 - 2 )2 + (3 - 3)2 = (x - 2 )2 + (5 - 3)2 On putting y = 7 in Eq. (ii), we get 4 2 x - 3æç 7 ö = - 8 SAMPLE PAPER 6 ÷ Þ (2 )2 + 0 = (x - 2 )2 + 4 è4ø On squaring both sides, we get Þ 2 x = 21 - 8 4 = (x - 2 )2 + 4 4 Þ (x - 2 )2 = 0 Þ x - 2 = 0 Þ x = 2 Þ 2 x = 21 - 32 = - 11 44 6. Given, sin2 q - cos2 q + cot2 q + tan2 Þ x = - 11 1 1 q 8 = sin2 q - cos2 q é 1 + cot 2 A = cosec2 Aù \\ x = - 11 and y = 7 cosec2q sec2 q Q 84 êú êë 1 + tan2 A = sec2 A ûú

132 CBSE Sample Paper Mathematics Standard Class X (Term I) 10. Let two right angled triangle be DLMO and 14. It is given that, 32 students in section A and 36 students in section B. DRST. The minimum number of books required is the OT LCM of 32 and 36. As, 32 = 2 5 L MR S 36 = 2 2 ´ 32 \\ LCM (32, 36) = 2 5 ´ 32 = 288 ÐL = ÐR = 90° and ÐM = ÐS [acute angle] Hence, the minimum number of books \\ DLMO ~ DRST [AA similarity] required to distributed equally among students of section A and section B are 288. 11. The distance between points S and A; S and B 15. The pair of linear equation has infinitely many solutions, if are a1 = b1 = c1 SA = (a2 - 1)2 + (2 a)2 a2 b2 c2 = a4 + 1 - 2 a2 + 4a2 Here, on comparison the given equation with standard equations, we get = a4 + 2 a2 + 1 = (a2 + 1)2 a1 = 2 , b1 = 3 and c1 = - 11 = (a2 + 1) units a2 = m + n, b2 = 2 m - n and c2 = - 33 \\ 2 = 3 = -11 and SB = æ1 - 1 ö2 + æ -2 - 0 ö2 ç ÷ ç ÷ m + n 2 m - n -33 è a2 ø èa ø Þ 2 = 3 =1 = 1 +1- 2 + 4 m+ n 2m- n 3 a4 a2 a2 2 = 1 and 3 = 1 = 1 +1+ 2 = æ1 + 1ö÷2 m+ n 3 2m-n 3 a4 a2 ç ø è a2 1 æ 1 + a2 ö Þ m + n = 6 and 2 m - n = 9 …(i) a2 1 ç a2 ÷ units On adding Eqs. (i), (ii) and (iii), we get = + = ç ÷ è 3m = 15 ø Þ m=5 Putting m = 5 in Eq. (i), we get \\ 1+1= 1 + 1 SA SB a2 + 1 1 + a2 5 + n=6 Þ n=1 a2 = 1 + a2 = 1 + a2 =1 16. In a pair of dependent linear equations, second a2 + 1 1 + a2 1 + a2 equation is equivalent to the k times the first equation. Thus, the second equation is 12. Given, 2 sin 4q = 3 33 k(- 5x + 7y - 2 ) = 0 Then, sin 4q = 3 On putting k = - 2, we get 2 é 3ù 10x - 14y + 4 = 0 Þ 4q = 60° êQ sin60° = ë 2 ú or 10x - 14y = - 4 Þ q = 60° = 15° û 4 Hence, 10x - 14y = - 4 is the second equation. SAMPLE PAPER 6 1200 m 13. Given, one zero of the quadratic polynomials is 17. Let a girl starts from point O and walks 500 m negative of the other. towards East, then 1200 m towards North. Let one zero be a. Then, other zero be - a. N N Sum of zeroes = - b WE a S \\ a + (- a) = - (- 6k + 6) = 6k - 6 O 500 m E 22 0 = 6k - 6 6k = 6Þ k = 1

CBSE Sample Paper Mathematics Standard Class X (Term I) 133 By Pythagoras theorem, 22. Given, Quadratic polynomial = ax2 + x + a ON2 = OE2 + NE2 ON2 = (500)2 + (1200)2 Let the zeroes be a and a. …(i) ON2 = 250000 + 1440000 Sum of zeroes = - 1 = 1690000 a or ON = 1690000 = 1300 m Hence, the girl is 1300 m far from the starting a+a=- 1 point. a 18. As we know, sin q = Perpendicular = 4 2a = - 1 Hypotenuse 3 a a Here, sin q is greater than 1, which is not possible. Product of the zeroes = = 1 Hence, sin q = 4 is not possible. a 3 a×a =1 …(ii) a2 = 1 19. The coordinates of origin is O(0, 0). \\The distance of the point (-12 , 5) from Squaring of Eq. (i), then we get O(0, 0) = (0 + 12 )2 + (0 - 5)2 4a2 = 1 = (12 )2 + (- 5)2 a2 = 144 + 25 = 169 = 13 units a2 = 1 4a2 20. Given, DABC is an isosceles triangle a2 = 1 A 4´1 [Q a2 = 1 from Eq. (ii)] a2 = 1 4 23. Given, sin q = m n In right triangle DABC, ÐC = q and ÐB = 90° A C 2 cm B and ÐC = 90° 90º q \\ AC = BC = 2 cm B C In DABC, AB2 = AC2 + BC2 Let AB = mk and AC = nk. Then, Þ [by Pythagoras theorem] BC = AC2 - AB2 = (nk )2 - (mk )2 Þ AB2 = (2 )2 + (2 )2 [by pythagoras theorem] AB2 = 4 + 4 = 8 Þ BC = k n2 - m2 AB = 8 = 2 2 cm cot q = BC = k n2 - m2 21. The factors are AB mk SAMPLE PAPER 6 36 = 2 2 ´ 32 and tan q = AB = mk 42 = 2 ´ 3 ´ 7 BC k n2 - m2 72 = 2 3 ´ 32 \\ LCM (36, 42, 72) = 2 3 ´ 32 ´ 7 = 504 \\ cot q - tan q = n2 - m2 - m m n2 - m2 Here, LCM of 36, 42 and 72 = 504 s n2 - m2 - m2 i.e. 8 min 24 s = The lights are first switched on at 9 am (given) \\The required time will be after 8 min 24 s or m n2 - m2 9:08:24. = n2 - 2 m2 m n2 - m2

134 CBSE Sample Paper Mathematics Standard Class X (Term I) 24. D P Þ a + (- a) = -(-2k) 1 Þ 0 =2k Þ k =0 28. Let two-digits number be xy in expanded form E FQ R of xy = 10x + y In DDEF and DQRP, Given, sum of digits = 12 ÐD = ÐQ i.e. x + y = 12 …(i) …(ii) ÐE = ÐR [given] and unit’s digit divided by ten’s digit, [by CPCT] y =3 From AA-criterion, x DDEF ~ DQRP Thus, y = 3x Þ ÐF = ÐP On putting y = 3x in Eq. (i) \\ ED = FE = DF x + 3x = 12 RQ PR QP Þ 4x = 12 Þ x =3 Hence, option (b) is not true. and y = 9 So, the required number is 3(10) + 9 = 39. 25. Let P(x, - 1) and Q(3, 2 ) by the given point, 29. Given, DABC ~ DDEF D then PQ = 5. (x - 3)2 + (- 1 - 2 )2 = 5 A Þ (x - 3)2 + 9 = 25 4 cm 2.5 cm 7.5 cm Þ x2 - 6x + 9 + 9 = 25 Þ x2 - 6x - 7 = 0 Þ (x - 7) (x + 1) = 0 Þ x = 7 or x = - 1 26. Let x = 0.08, then B CE F 3.5 cm x = 0.08888 .... …(i) \\ Perimeter of DABC = AC Perimeter of DDEF DE On multiplying Eq. (i) by 10, we get Þ AB + BC + AC = AC \\ 10x = 0.8888 …(ii) Perimeter of DDEF DF On subtracting Eq. (i) from Eq. (ii), we get Þ 4 + 3.5 + 2.5 = 2.5 Perimeter of DDEF 7.5 10x - x = 0.888 ...... - 0.0888 ...... Þ 10 = 1 9x = 0.8000 Perimeter of DDEF 3 x = 8000 Þ Perimeter of DDEF = 3 ´ 10 = 30 9 ´ 10000 Hence, perimeter of DDEF is 30 cm. =8 =4 90 45 30. The coordinates of the centroid G of DABC are On comparing with p, we get q p = 4 and q = 45 Gçæ -1 + 0 - 5 , 3 + 4 + 2 ö \\ p2 + q2 = (4)2 + (45)2 ÷ è3 3ø SAMPLE PAPER 6 = 16 + 2025 i.e., Gçæ -6 , 9 ö or G(-2 , 3) p2 + q2 = 2041 ÷ è 3 3ø 27. Given, zeroes of the quadratic polynomial are Since, G lies on the median line equal in magnitude but opposite in sign. 5x -3y + 2k = 0 Let the zeroes of quadratic polynomial be a, \\ 5(-2 ) -3(3) + 2 k = 0 then the other zero be - a Þ -10 -9 +2 k = 0 Sum of zeroes = - Coefficient of x Þ 2 k = 19 Coefficient of x2 Þ k = 19 = 9.5 2

CBSE Sample Paper Mathematics Standard Class X (Term I) 135 31. Prime factors of given numbers are 35. Let the points be P(x, y), A(2 , 1) and B(1, -2 ). 408 = 2 3 ´ 3 ´ 17 Given, P is equidistant from A and B i.e. the distance between A and P, B and P are equal. and 1032 = 2 3 ´ 3 ´ 23 HCF (408, 1032) = 2 3 ´ 31 = 24 The distance AP = (x - 2 )2 + (y - 1)2 But HCF express in the form and BP = (x - 1)2 + (y + 2 )2 1032 ´ 2 + 408 ´ p = 24 \\ AP = BP Þ AP2 = BP2 Þ 2064 + 408 ´ p = 24 Þ (x - 2 )2 + (y - 1)2 = (x - 1)2 + (y + 2 )2 Þ x2 - 4x + 4 + y2 - 2y + 1 Þ 408 p = - 2040 Þ p = - 2040 = - 5 = x2 - 2x + 1 + y2 + 4y + 4 Þ 2x + 6y = 0 Þ x + 3y = 0 408 32. A D 36. 1 + 1 = 1 - cos q + 1 + cos q 1 + cos q 1 - cos q (1 - cos q)(1 + cos q) B CE F =2 1 - cos2 q Given, DABC ~ DDEF [Q sin2 A + cos2 A = 1] \\ AB = BC = AC = 2 = 2 cosec2q ED EF DF sin2 q I II III Þ 2 cosec2q = k cosec2q [given] Taking I and II ratio, we get AB = BC Þ k =2 ED EF 37. Given, Priya takes 18 min whereas Harish Þ AB× EF = ED× BC takes 12 min to drive one round of the field Taking II and III ratio, we get BC = AC Û BC × DF = AC × EF Let us find the LCM of 18 and 12. EF DF We have, factors of 18 and 12 are Taking I and III ratio, we get 18 = 2 ´ 3 ´ 3 = 2 ´ 32 AB = AE Û AB× DF = AE× ED ED DF and 12 = 2 ´ 2 ´ 3 = 2 2 ´ 3 LCM (18, 12) = 2 2 ´ 32 = 36 Hence, (a), (b) and (d) is correct but option (c) is not true. Hence, after 36 min Priya will be at the starting point after completing 2 rounds of the field and Harish also be at same point after 3 rounds. 33. As, x = 3 and y = 1 is the solution of 38. In the given figure, DE|| BC, then by basic 2 x + y - q2 - 3 = 0 proportionality theorem, we have When x = 3 and y = 1, 2 ´ 3 + 1 - q2 - 3 = 0 A Þ 4 - q2 = 0 Þ 4 = q2 2 cm \\ q=±2 D E 3 cm 34. (3, –1) (–6, 5) B 7.5 cm C P ABQ AD = AE = DE SAMPLE PAPER 6 DB EC BC Since, the line segment AB is trisected. Þ AD = DE \\ PB: BQ = 2 : 1 DB CB Þ 2 = DE \\Coordinates of B are 3 7.5 = æ 2( - 6) + 1(3) , 2(5) + 1(- 1) ö Þ DE = 2 ´ 7.5 çç 2 +1 2 +1 ÷÷ è ø 3 = æ - 12 + 3 , 10 -1ö = 2 ´ 2.5 = 5 cm ç ÷ è3 3ø = æ - 9 , 9 ö = (-3 , 3) ç ÷ è 3 3ø

136 CBSE Sample Paper Mathematics Standard Class X (Term I) 39. Let a and b be the zeroes of the polynomial, 44. Angle made by hour hand in 1 h = 360° = 30° 12 g(x) = ( p2 - 23)x2 - 2 x - 12 Also, r = 12 cm (-2 ) 2 Then, (a + b) = - p2 - 23 = p2 - 23 \\ Area swept by hour hand in 1 h = Area of sector having central angle 30° Also given, sum of zeroes = a + b = 1 = pr2 ´ æ 30° ö ç ÷ 2 è 360° ø Þ p2 -23 = 1Þ p2 - 23 =2 Þ p2 = 25 = 22 ´ 12 ´ 12 ´ 1 = 264 = 37.71 cm2 \\ p=± 5 7 12 7 40. Given, sin q = 5 A 45. Number of hours from 11 am to 5 pm = 6 13 B Area swept by hour hand in 1 hour = 37.71 cm2 Let be DABC be right angle, \\ Area swept by hour hand in 6 h triangle, we get = 37.71 ´ 12 = 452.52 cm2 ÐB = 90° and ÐC = q 46. Total number of cards in one deck of cards is 52. \\ AB = 5k and AC = 13k q Total number of outcomes = 52 BC = AC2 - AB2 C Let E1 = Event of getting a queen of red colour = (13k )2 - (5k )2 = 169k 2 -25k 2 \\Number of outcomes favourable of E1 = 2 = k 144 = 12 k [Q there are four queens in a deck of playing cards out of which two So, tan q = AB = 5 and sec q = AC = 13 are red and two are black] BC 12 BC 12 Hence, probability of getting a queen of red \\2 sec q × tan q = 2 ´ 13 ´ 5 = 65 colour, P( E1 ) = 2 =1 12 12 72 52 26 41. Angles made by minutes hand in 60 min = 360° 47. Let E2 = Event of getting an ace. \\ Angle made by minute hand in 10 min \\Number of outcome favourable to E2 = 4 = 360° ´ 10 = 60° [Q in a deck of cards, there are 4 ace cards] 60° Hence, probability of getting an ace, Length of minute hand = 18 cm [given] P( E2 ) = 4 =1 52 13 \\ Area swept by minute hand in 10 min 48. Let E3 = Event of getting a jack of diamond = Area of sector having central angles 60° \\ Number of outcomes favourable to E3 = 1 = pr2çæ 60° ö = 22 ´ 18 ´ 18 ´ 1 [Q there are four jack cards in a deck, namely 1 ÷ of heart, 1 of club, 1 of spade and 1 of diamond] è 360° ø 7 6 Hence, probability of getting a jack of = 1188 = 169.71 cm2 diamond = 1 7 52 42. Let r be the length of the pendulum. Given, r = 15 cm and l = 1 (22 ) = 11 cm 49. Let E4 = Event of getting a red face card. 2 \\ Number of outcomes favourable to E4 = 6 We know that, l = 2 præç q ö ÷ è 360° ø [Q in a deck of cards, there are 12 face cards out of which 6 are red cards] SAMPLE PAPER 6 Þ q = 11 ´ 360° 2 ´ 22 ´ 15 Hence, probability of getting a red face card, 7 P( E4 ) = 6 = 3 52 26 = 90° ´ 7 = 6° ´ 7 = 42 ° 15 50. Let E5 = Event of getting a club. 43. Angle made by hour hand in 12 h = 360° \\ Number of outcomes favourable to E5 = 13 \\ Angle made by hour hand in 10 min or 1 h [Q in a deck of cards, there are 13 spades, 6 13 clubs, 13 hearts and 13 diamonds] æ 360° 1 ö Hence, probability of getting a club, ç ÷ = ´ = 5° P( E5 ) = 13 = 1 52 4 è 12 6 ø

CBSE Sample Paper Mathematics Standard Class X (Term I) 137 SAMPLE PAPER 7 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 k, for which the pair of linear equations kx + y = k 2and x + ky = 1 have infinitely many solution, is (a) ± 1 (b) 1 (c) -1 (d) 2 2. If p is the probability of happening of an event and q is the probability of non-happening of an event, then p + q is equal to (a) 0 (b) 1 (c) - 1 (d) 2 3. In the following figure, O is the centre of the circle. The area of the sector OAPB is 5 12 part of the area of the circle. Find the value of x. O SAMPLE PAPER 7 x (a) 130° (b) 60° AB (d) 150° P (d) sec A (d) BC = PR (c) 45° 4. æ cos A + sin A ö is ç ÷ è cot A ø (a) cot A (b) 2 sin A (c) 2 cos A 5. If DABC is similar to DPRQ, then which of the following is true? (a) AB = PQ (b) AC = PQ (c) AB = QR

138 CBSE Sample Paper Mathematics Standard Class X (Term I) 6. The distance of the point (3, 5) from the X-axis is (a) 3 units (b) 5 units (c) 8 units (d) 4 units 7. The external and internal diameters of a circular path are 12 m and 8 m respectively. The area of the circular path is (a) 9p m 2 (b) 16p m 2 (c) 20p m 2 (d) 36p m 2 8. Which of the following is not an irrational number? (a) 7 5 (b) 2 + 2 2 (c) ( 7 - 3) - 7 (d) 3 + 2 9. An integer is chosen between 0 to 50. What is the probability that it is divisible by 4? (a) 12 (b)13 (c) 1 (d) 4 49 49 7 49 10. If x × tan 45°× cot 60° = sin 30°× cosec 60°, then the value of x is (a) 1 (b) 1 (c) 1 (d) 3 4 2 11. The ratio in which the X-axis divides the line segment joining A(3, 6) and B(12, - 3) is (a) 2 : 1 (b) 1 : 2 (c) - 2 : 1 (d) 1: - 2 12. The values of x and y in x - y + 1 = 0 and 3x + 2y - 12 = 0 are (a) 2, - 3 (b) - 2, 3 (c) - 2, - 3 (d) 2, 3 13. The diameter of a wheel is 1.26 m. How long will it travel in 500 revolutions? (a) 1492 m (b) 2530 m (c) 1980 m (d) 2880 m 14. If cosec 2q(1 + cosq) (1 - cosq) = l, then the value of l is (a) 0 (b) cos 2 q (c) 1 (d) - 1 15. What is the probability of getting a king in a well shuffled pack of 52 cards? (a) 1 (b) 1 (c) 1 (d) 0 13 52 26 16. If the distance between A(k, 3) and B (2, 3) is 5, then the value of k is (a) 5 (b) 6 (c) 7 (d) 8 17. If 4 tan q = 3, then æ 4 sin q - cos q ö is equal to çç 4 sin q + cos q ÷÷ è ø (a) 2 (b) 1 (c) 1 (d) 3 3 3 2 4 SAMPLE PAPER 7 18. In a right-angled triangle, the square of the hypotenuse is equal to the (a) Sum of other two sides (b) Sum of squares of other two sides containing right angle (c) Square of the perpendicular (d) Square of the base 19. For which value(s) of p, will the lines represented by the following pair of linear equations be parallel 3x - y - 5 = 0 6x - 2y - p = 0

CBSE Sample Paper Mathematics Standard Class X (Term I) 139 (a) all real values except 10 (b) 10 (c) 5 (d) 1 2 2 20. If two angles of a triangle are equal to the corresponding two angles of another triangle, then in such case two triangles can be called similar. (a) True (b) False (c) Can’t say (d) None of these Section B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 21. If A(5, 3), B (11, - 5) and P (12, y) are the vertices of a right triangle, right angled at P, then y is equal to (a) - 2 or 4 (b) - 2 or - 4 (c) -4 or 2 (d) 2 or 4 22. In DPQR right angled at Q, QR = 3 cm and PR - PQ = 1 cm. The value of sin 2 R + cosec R is (a) 4 (b) 4 (c) 189 (d) 198 5 9 100 100 23. One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number, which is a multiple of 5, is (a) 1 (b) 3 (c) 4 (d) 1 5 5 5 3 24. In a right-angled DABC, ÐC = 35° and in another right-angled DPQR, ÐR = 35°. Then relation between the two triangles is (a) Congruent (b) Equal (c) Similar (d) No relation 25. The value of k for which the system of equations x + y - 4 = 0 and 2x + ky = 3, has no solution is (a) 2 (b) 8 (c) 3/4 (d) - 2 26. Suppose, O is the centre of a circle of radius 5 cm. The chord AB subtends an angle of 60° at the centre. Area of the shaded portion is equal to (approximately) O 60° AB (a) 50 cm2 (b) 62.78 cm2 SAMPLE PAPER 7 (c) 49.88 cm2 (d) 67.75 cm2 27. 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 28. After how many places, the decimal expansion of the rational number 43 will 24 ´ 53 terminate (a) 1 (b) 2 (c) 3 (d) 4

140 CBSE Sample Paper Mathematics Standard Class X (Term I) 29. Three unbiased coins are tossed together, then the probability of getting exactly 1 head is 2 (c) 3 (d) 1 (a) 1 (b) 8 2 8 8 30. Equations 3x + 4y + 5 = 0 and 6x + 8y + 9 = 0 represents a pair of ............ lines. (a) intersecting (b) coincident (c) parallel (d) None of these 31. In the given figure, if D is mid-point of BC, then value of cot y is cot x A x y C DB (a) 2 (b) 1 (c) 1 (d) 1 4 3 2 32. Name the criteria of similarity by which following triangles are similar R c 4.5 6 3 4 A 5 BP 7.5 Q (a) SSS (b) SAS (c) AAA (d) ASA 33. The ratio in which the line segment joining the points (6,- 8) and (- 3, 10) is divided by (- 1, 6) is (a) 2 : 3 (b) 2 : 5 (c) 7 : 2 (d) 2 : 7 34. A letter is chosen at random from the letter of the word ‘ASSASSINATION’, then the probability that the letter chosen is a vowel is in the form of 6 , then x is equal to 2x + 1 (a) 5 (b) 6 (c) 7 (d) 8 35. The area of the circle that can be inscribed in the square of side 6 cm is SAMPLE PAPER 7 (a) 18 p cm 2 (b) 12 p cm 2 (c) 9 p cm 2 (d) 14 p cm 2 36. Name the criteria of similarity by which following triangles are similar. R c 9.5 6 64 53° BP 53° Q A (d) SAS (a) Not similar (b) ASS (c) SSS

CBSE Sample Paper Mathematics Standard Class X (Term I) 141 37. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively. (a) 4 and 24 (b) 5 and 30 (c) 6 and 36 (d) 3 and 24 38. What is the supplementary angle of the central angle of a semicircle? (a) 0° (b) 90° (c) 180° (d) 360° 39. P(5, - 3) and Q(3, y) are the points of trisection of the line segment joining A(7, - 2) and B(1, - 5), then y equals (a) 2 (b) 4 (c) - 4 (d) - 5 / 2 40. A race track is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Then, width of the track is (a) 7 m (b) 9 m (c) 11 m (d) 17 m 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 Republic day parade is celebrated on 26th January every year in India. The day is celebrated in the form of parades and other military shows in the national capital New Delhi as well as in all headquarters of army. Parade I An Army contingent of 616 members is to march behind an army band of 32 members in parade. The two groups are to march in the same number of columns. Parade II An Army contingent of 1000 members is to march behind an army band of 56 members in parade. The two groups are to march in the same number of columns. Refer to Parade I 41. Number 616 can be expressed as a product of its prime factors as SAMPLE PAPER 7 (a) 2 1 ´ 141 ´ 22 1 (b) 2 2 ´ 111 ´ 141 (c) 2 3 ´ 71 ´ 111 (d) 2 4 ´ 7 2 ´ 111 42. The HCF of 32 and 616 is (c) 18 (d) 12 (a) 8 (b) 16 Refer to Parade II 43. The LCM of 56 and 1000 is (a) 6000 (b) 7000 (c) 8000 (d) 9000

142 CBSE Sample Paper Mathematics Standard Class X (Term I) 44. Number 1000 can be expressed as a product of its prime factors as (a) 2 3 ´ 5 3 (b) 2 2 ´ 5 4 (c) 2 4 ´ 5 2 (d) 2 3 ´ 5 4 45. The maximum number of columns in which army can march is (a) 6 (b) 10 (c) 12 (d) 8 46-50 are based on Case Study-2. Case Study 2 Basketball and football are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a football player uses his feet. Usually, football is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of football and basketball are in the form of parabola representing quadratic polynomial. 46. The shape of the path traced shown in (a) Spiral (b) Ellipse (c) Linear (d) Parabola (d) a ³ 0 47. The graph of parabola opens downwards, if ……… . (a) a = 0 (b) a < 0 (c) a > 0 48. Observe the following graph and answer. Y 6 X¢ –4 –3 –2 2 2 34X –1 –2 –6 Y¢ SAMPLE PAPER 7 In the above graph, how many zeroes are there for the polynomial? (a) 0 (b) 1 (c) 4 (d) 3 49. The four zeroes in the above shown graph are (a) 2, 3, - 1, 4 (b) - 2, 3, 1, 4 (c) - 3, - 1, 2, 4 (d) - 2 , - 3, -1, 4 50. Which will be the expression of the quadratic polynomial? (a) x4 + 2 x2 - 5x - 6 (b) x 3 + 2 x 2 - 5x + 6 (c) 2 x2 + 5x - 6 (d) 5x + 6

OMR SHEET SP 7 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)