maths super 10

Sample Paper-8 SP-61 36. In the given figure, DE || BC and AD : DB = 5 : 4 then ar (DDFE) : ar(DCFB). A DE F (a) 25 : 81 B C (d) 22 : 88 (b) 5 : 81 (c) 81 : 25 37. If x= 4 is a root of the polynomial f (x) = 6x3 – 11x2 + kx – 20, then find the value of k. 3 (a) 10 (b) 19 (c) – 5 (d) 3 38. For what values of k, do the equations 3x – y + 8 = 0 and 6x – ky = –16 represent coincident lines? (a) solution of 3k – 9 = 0 (b) solution of 2k – 8 = 0 (c) 2 (d) 3 39. A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, –5) is the mid point of PQ, then the coordinates of P and Q are respectively (a) (0, –5) and (2, 0) (b) (0, 10) and (–4, 0) (c) (0, 4) and (–10, 0) (d) (4, 0) and (0, 10) 40. The decimal expansion of 21 is : 45 (a) terminating (b) non-terminating and repeating (c) non-terminating and non-repeating (d) none of these SECTION-C Case Study Based Questions: Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted. Q 41. - Q 45 are based on case study-I Case Study-I Two unbiased coins are tossed simultaneously. The word ‘unbiased’ means each outcome is equally likely to occure. 41. The probability of getting two heads is 1 (c) 11 1 1 (c) 3 (d) 4 11 (a) 2 (b) 3 (d) 4 42. The probability of getting one tail is 1 (a) 2 (b)

SP-62 Mathematics 43. The probability of getting no head is 1 11 1 (c) 3 (d) 4 (a) 2 (b) 44. The probability of getting at most one head. (c) 3 (d) 1 11 4 (a) 4 (b) 2 45. The probability of getting at least one head 1 3 (c) 9 (a) 4 (b) 4 2 (d) 1 Q 46 - Q 50 are based on case study-II Case Study-II A horse is tied to a peg at one corner of a square shaped grass field of side 15m. (Use p = 3.14) 46. If rope of horse is 5m long then the area of that part of the field in which the horse can graze is : (a) 19.625m2 (b) 29.625m2 (c) 19 m2 (d) 18.625m2 47. If rope of horse 10 m long then the area of that part of the field in which the horse can graze is: (a) 68.5m2 (b) 78.5m2 (c) 58.5m2 (d) 73.5m2 48. The increase in the grazing area if the rope were 10m long instead of 5m. (a) 58.875m2 (b) 58m2 (c) 57.875m2 (d) 68.87 m2 49. If rope of horse is 5 m long then the area of that part of the field in which the horse can not graze is: (a) 204.37m2 (b) 200.37m2 (c) 205.37m2 (d) 205m2 50. If rope of horse 10m long then the area of that part of the field in which the horse can not graze is : (a) 146.5 m2 (b) 205.37m2 (c) 46.5m2 (d) 146 m2

OMR ANSWER SHEET Sample Paper No –  Use Blue / Black Ball pen only.  Please do not make any atray marks on the answer sheet.  Rough work must not be done on the answer sheet.  Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected. Start time : ____________________ End time ____________________ Time taken ____________________ 1. Name (in Block Letters) 2. Date of Exam SECTION-A 17. a    b    c    d 18. a    b    c    d 3. Candidate’s Signature 9. a    b    c    d 19. a    b    c    d 10. a    b    c    d 20. a    b    c    d 1. a    b    c    d 11. a    b    c    d 2. a    b    c    d 12. a    b    c    d 37. a    b    c    d 3. a    b    c    d 13. a    b    c    d 38. a    b    c    d 4. a    b    c    d 14. a    b    c    d 39. a    b    c    d 5. a    b    c    d 15. a    b    c    d 40. a    b    c    d 6. a    b    c    d 16. a    b    c    d 7. a    b    c    d 49. a    b    c    d 8. a    b    c    d SECTION-B 50. a    b    c    d 21. a    b    c    d 29. a    b    c    d 22. a    b    c    d 30. a    b    c    d 23. a    b    c    d 31. a    b    c    d 24. a    b    c    d 32. a    b    c    d 25. a    b    c    d 33. a    b    c    d 26. a    b    c    d 34. a    b    c    d 27. a    b    c    d 35. a    b    c    d 28. a    b    c    d 36. a    b    c    d 41. a    b    c    d SECTION-C 42. a    b    c    d 43. a    b    c    d 45. a    b    c    d 44. a    b    c    d 46. a    b    c    d 47. a    b    c    d 48. a    b    c    d No. of Qns. Attempted Correct Incorrect Marks

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Sample Paper 9 Time : 90 Minutes Max Marks : 40 General Instructions 1. The question paper contains three parts A, B and C. 2. Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted. 3. Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted. 4. Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions. 5. There is no negative marking. SECTION-A Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted. 1. The height of mountains is found out using the idea of indirect measurements which is based on the (a) principal of congruent figures (b) principal of similarity of figures (c) principal of equality of figures (d) none of these 2. Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively. (a) x2 – 3x – 2 (b) x2 + 3x + 2 (c) x2 – 3x + 2 (d) x2 + 3x – 2 3. The figure given shows a rectangle with a semicircle and 2 identical quadrants inside it. 28 cm 16 cm 23 cm What is the shaded area of the figure? 22 (c) 305 cm2 (d) 216 cm2 (a) 363 cm2 (Use p = 7 ) (b) 259 cm2 4. A lady has 25 p and 50 p coins in her purse. If in all she has 40 coins totalling `12.50, find the number of coins of each type she has. (a) 10, 15 (b) 30, 10 (c) 20, 30 (d) 10, 10 ( )5. The points (a, a) (–a, –a) and − 3a, 3a are the vertices of (a) a scalene triangle (b) a right angled triangle (d) an equilateral triangle (c) an isosceles right angled triangle 6. H.C.F. of pair of co-primes is _________. (a) one (b) product of numbers (c) common factor (d) lowest common factor

SP-66 Mathematics 1 7. What is the maximum value of secθ ? (a) 0 (b) 1 (c) –1 (d) –2 8. If in an isosceles triangle ‘a’ is the length of the base and ‘b’ is the length of one of the equal side, then its area is equal to (a) a2 b2 – 4b2 (b) a2 ( 4b – a2 ) (c) a4 4b2 – a2 (d) 14 a2 + b2 4 9. The zeroes of the polynomial are p(x) = x2 – 10x –75 (a) 5, – 15 (b) 5, 15 (c) 15, – 5 (d) – 5, – 15 10. The points (– 4, 0), (4, 0), (0, 3) are the vertices of a (a) right triangle (b) isosceles triangle (c) equilateral triangle (d) scalene triangle 11. Arjun drew a figure as shown in figure, where a circle is divided into 18 equal parts. He then shaded some of the parts. (Take p = 3.14) 8 cm Find the total area the Arjun shaded. (a) 25.12 cm2 (b) 29.25 cm2 (c) 36.4 cm2 (d) 45.2 cm2 12. L.C.M = ____________ of highest powers of all the factors. (a) product (d) difference (c) sum (d) none of these 13. When two dice are thrown, find the probability of getting same numbers on both dice. (a) 23 (b) 16 (c) 112 (d) 91 14. The points A (9, 0), B (9, 6), C (– 9, 6) and D (– 9, 0) are the vertices of a (a) square (b) rectangle (c) rhombus (d) trapezium 15. A man steadily goes 10 m due east and then 24 m due north. then his distance from the starting point is (a) 28 m (b) 26 m (c) 25 m (d) 18 m 16. The perimeter of a rectangle is 40 cm. The ratio of its sides is 2 : 3. Find its length and breadth. (a) l = 10 cm, b = 8 cm (b) l = 12 cm, b = 18 cm (c) l = 12 m, b=8m (d) l = 40 m, b = 30m 17. If tan A = 3 then, what is the value of sin A? 4 (a) 34 (b) 1 3 (d) 0 (c) 5 18. Which of the following numbers has the terminal decimal representation? (a) 71 (b) 13 (c) 53 (d) 137

Sample Paper-9 SP-67 19. If A(2, 2), B(–4, –4) and C(5, –8) are the vertices of a triangle, then the length of the median through vertex C is (a) 65 (b) 117 (c) 85 (d) 113 20. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, Find the number of blue balls in the bag. (a) x = 10 (b) x = 12 (c) x = 9 (d) x = 8 SECTION-B Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted. 21. Two coins are tossed simultaneously. Find the probability of getting atmost one head. (a) 23 (b) 14 (c) 34 (d) 12 22. which of the following is true if following pair of equations has unique solution? 3x – 2y = – 8 (2m – 5)x + 7y – 6 = 0 (a) m = 11 (b) m = – 141 (c) m ≠ – 141 (d) m ≠ 141 4 23. A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole. (a) 40 cm (b) 24 cm (c) 101 cm (d) 10 cm 24. The graph of y = p(x) is given in fig. below, for a polynomial p(x). The number of zeroes of p(x),is/are (a) 4 (b) 3 (c) no zero (d) 2 25. Given that sin q + 2 cos q = 1, then 2 sin q – cos q = (a) 0 (b) 2 (c) 1 (d) None of these 26. What is the condition that a system of simultaneous equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 must satisfy to have exactly one solution? (a) aa12 = b1 (b) aa12 ≠ bb12 (c) aa12 = cc12 (d) bb12 = cc12 b2 27. The least number which is a perfect square and is divisible by each of 16, 20 and 24 is (a) 240 (b) 1600 (c) 2400 (d) 3600 28. If the end points of a diameter of a circle are A (–2, 3) and B (4, –5), then the coordinates of its centre are (a) (2, –2) (b) (1, –1) (c) (–1, 1) (d) (–2, 2)

SP-68 Mathematics 29. The graph of y = f(x) is shown in the figure. What type of polynomial f(x) is? Y X¢ O X Y¢ (a) cubic (b) quadratic (c) linear (d) none of these 30. If 1 + sin2 q = 3 sin q cos q, then tan q can have values (a) 4, 0 (b) 3 , 1 (c) None of these (d) 1, 1 4 3 2 31. ABC is a right-angled triangle right angled at A. A circle is inscribed in it and the lengths of the two sides containing the right angle are 6 cm and 8 cm. Find the radius of the circle. (a) 1.5 cm (b) 2.2 cm (c) 3 cm (d) 2 cm 32. If (–1)n + (–1)4n = 0, then n is (a) any positive (b) any negative integer (c) any odd natural number (d) any even natural number 33. A chord of a circle of radius 28 cm subtends an angle of 45° at the centre of the circle. Then the area of the minor segment is (a) 30.35 cm2 (b) 30.81 cm2 (c) 30.45 cm2 (d) 30.25cm2 34. In what ratio is the line segment joining the points (3, 5) & (–4, 2) divided by y–axis? (a) 3 : 2 (b) 3 : 4 (c) 2 : 3 (d) 4 : 3 35. What is a system of simultaneous equations called if its graph has intersecting lines? (a) Inconsistent system (b) Consistent system (c) Dependent system (d) Independent system 36. tan θ + secθ −1 = tan θ − secθ +1 (a) 1+cossinθθ (b) cos q + sin q (c) 1+ cos θ (d) cos2q – sin2q sin θ 37. Choose the zeros of the polynomial whose graph is given. Y 3 2 1 X' X –2 –1 0 1 2 3 –1 –2 (b) –2, 1, 3 Y' (d) –2, 2, 3 (a) 1, –1, 2 (c) –2, 0, 3

Sample Paper-9 SP-69 38. In DABC, D is the mid point of BC and AE^ BC. If AC > AB, then (a) AB2 = AD2 – BC2 + BC2 1 (b) AB2 = AD2 – BC ⋅ DE + 4 BC2 1 (c) AD2 = AB2 + 4 BC2 – BC ⋅ DE (d) All of the above 39. Find the H.C.F. of 23 × 32 × 5 × 74, 22 × 35 × 52 × 73, 23 × 53 × 72. (a) 980 (b) 890 (c) 900 (d) 809 40. The perimeter of a sector of a circle with central angle 90° is 25 cm. Then the area of the minor segment of the circle is. (a) 14 cm2 (b) 16 cm2 (c) 18 cm2 (d) 24 cm2 SECTION-C Case Study Based Questions: Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted. Q 41. - Q 45 are based on case study-I Case Study-I A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground . Answer the following questions. 41. The length of her shadow after 4 seconds is : (a) 4.8 m (b) 1.6 m (c) 4 m (d) 2 m (b) 1.6 m 42. Distance travel by girl after 4 second is : (b) 1.8 m (b) SSS (a) 4.8 m (b) ∠B = ∠D (c) 4 m (d) 3 m 43. Distance between their tops is : (a) 4 m (c) 5.4 m (d) 3.2 m 44. Similarity criterion of DABE and DCDE is : (a) AA (c) SAS (d) ASA 45. Which of the following is true ? (a) ∠B = ∠C (c) ∠A = ∠D (d) ∠A = 90°

SP-70 Mathematics Q 46 - Q 50 are based on case study-II Case Study-II A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. Die is rolled once. 46. The probability of obtaining the number 2 is 1 1 (c) 1 (d) None of these (a) 2 (b) 6 3 47. The probability of getting the number 1 or 3 is 11 (c) 1 (d) None of these (a) 3 (b) 6 2 (d) None of these (d) 1 48. The probability of not getting the number 3 is (d) 0 15 (c) 1 (a) 6 (b) 6 2 49. Probability of getting prime number 1 1 (c) 1 (a) 6 (b) 2 3 50. Probability of getting odd number (a) 12 (b) 1 (c) 1 6 3

OMR ANSWER SHEET Sample Paper No –  Use Blue / Black Ball pen only.  Please do not make any atray marks on the answer sheet.  Rough work must not be done on the answer sheet.  Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected. Start time : ____________________ End time ____________________ Time taken ____________________ 1. Name (in Block Letters) 2. Date of Exam SECTION-A 17. a    b    c    d 18. a    b    c    d 3. Candidate’s Signature 9. a    b    c    d 19. a    b    c    d 10. a    b    c    d 20. a    b    c    d 1. a    b    c    d 11. a    b    c    d 2. a    b    c    d 12. a    b    c    d 37. a    b    c    d 3. a    b    c    d 13. a    b    c    d 38. a    b    c    d 4. a    b    c    d 14. a    b    c    d 39. a    b    c    d 5. a    b    c    d 15. a    b    c    d 40. a    b    c    d 6. a    b    c    d 16. a    b    c    d 7. a    b    c    d 49. a    b    c    d 8. a    b    c    d SECTION-B 50. a    b    c    d 21. a    b    c    d 29. a    b    c    d 22. a    b    c    d 30. a    b    c    d 23. a    b    c    d 31. a    b    c    d 24. a    b    c    d 32. a    b    c    d 25. a    b    c    d 33. a    b    c    d 26. a    b    c    d 34. a    b    c    d 27. a    b    c    d 35. a    b    c    d 28. a    b    c    d 36. a    b    c    d 41. a    b    c    d SECTION-C 42. a    b    c    d 43. a    b    c    d 45. a    b    c    d 44. a    b    c    d 46. a    b    c    d 47. a    b    c    d 48. a    b    c    d No. of Qns. Attempted Correct Incorrect Marks

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Sample Paper 10 Time : 90 Minutes Max Marks : 40 General Instructions 1. The question paper contains three parts A, B and C. 2. Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted. 3. Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted. 4. Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions. 5. There is no negative marking. SECTION-A Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted. 1. Find a quadratic polynomial whose zeroes are 8 and 10. (a) k(x2 + 10x + 80) (b) k(x2 – 2x + 1) (c) k(x2 – 18x + 80) (d) k(x2 + 6x + 9) ( )2. What type of a triangle is formed with points (3, –3), (–3, 3) and −3 3, −3 3 as vertices? (a) A scalene triangle (b) An equilateral triangle (c) An isosceles triangle (d) A right triangle 3. The difference between two numbers is 26 and one number is three times the other. Find them. (a) 39, 13 (b) 41, 67 (c) 96, 70 (d) 52, 26 4. A copper wire when bent in the form of an equilateral triangle has area 121 3 cm2. If the same wire is bent into the form of a circle, find the area enclosed by the wire. (a) 345.5 cm2 (b) 346.5 cm2 (c) 342.5 cm2 (d) 340.25 cm2 5. Three wheels can complete respectively 60, 36, 24 revolutions per minute. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again? (a) 3 second (b) 4 second (c) 5 second (d) 7 second 6. If , sin θ = a 2 − b2 then find cosec q + cot q. a 2 + b2 (a) a +a b (b) bb +− aa (c) aa+2b (d) aa +− bb 7. The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2 internally lies in the (a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant 8. An unbiased die is rolled twice. Find the probability of getting the sum of two numbers as a prime (a) 53 (b) 152 (c) 172 (c) 54 9. Given DABC ~ DDEF. If AB = 2DE and area of DABC is 56 cm2 find the area of DDEF. (a) 14 sq.cm (b) 5 sq.cm (c) 18 sq.cm (d) 56 sq.cm

SP-74 Mathematics 10. A sheet is 11 cm long and 2 cm wide. Circular pieces of diameter 0.5 cm are cut from it to prepare discs. Calculate the number of discs that can be prepared. (a) 114 (b) 113 (c) 110 (d) 112 11. If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers, then HCF (a, b) is (a) xy (b) xy2 (c) x3y3 (d) x2y2 12. If tan q + cot q = –k + sec q cosec q 1− cot q 1− tan q 2 Find the value of k. (a) 1 (b) 0 (c) 3 (d) 2 13. Five years ago Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? (a) 50 yrs, 20 yrs (b) 40 yrs, 30 yrs (c) 60 yrs, 40 yrs (d) 45 yrs, 15 yrs 14. ABC is an isosceles triangle in which AB = AC = 10 cm. BC = 12 cm. PQRS is a rectangle inside the isosceles triangle. Given PQ = SR = y cm. and PS = QR = 2x cm. then x = (a) 6 − 3y (b) 6 + 6y (c) 6 + 4y (d) 7x +4 8y 4 3 15. If f(x) = x2 + 5x + p and g(x) = x2 + 3x + q have a common factor, then (p – q)2 = _________ (a) 2(5p – 3q) (b) 2(3p – 5q) (c) 3p – 5q (d) 5p – 3q 16. A month is randomly selected from a year. An event X is defined as ‘the month with 30 days’. Identify the number of outcomes of event X. (a) 1 (b) 6 (c) 3 (d) 4 17. If x2 = 5 , then find whether the variable x is rational or irrational 9 (a) Rational (b) Irrational (c) Composite (d) Integer 18. If P = (2, 5), Q = (x, –7) and PQ = 13, what is the value of ‘x’? (a) 5 (b 3 (c) –3 (d) –5 19. In the figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Then which of the following is true ? (a) PA.PB = PC2 (b) PA.PB = PC.PD (c) (PA)2 = PB.PB (d) PC × PC = PD 20. and tan φ =1−bbsicnoqs q , then 2 a sin φ a If tan q = − a cos b= 1 φ sin q sin q sin φ sin q (a) 1− cos φ (b) 1− cos φ (c) sin q (d) sin φ SECTION-B Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted. 21. xn + yn is divisible by (x + y) when ‘n’ is _________ (a) an even number (b) an odd number (c) a prime number (d) a natural number 22. DABC is an isosceles triangle right angled at B. Similar triangles ACD and aBE are constructed on sides AC and AB. ratio between the areas of DABE and DACD is (a) 1 : 4 (b) 2 : 1 (c) 1 : 2 (d) 4 : 3 23. in the given figure, a circle with centre B overlaps another circle with centre A and a square. The ratio of areas of P and Q 1 is 5 : 4 and the area of Q is 8 the area of circle B. The radii of circle A and circle B are 10 cm and 8 cm respectively.

Sample Paper-10 SP-75 A P B Q 7 cm Find the area of the unshaded part of the figure. (Take p = 3.14) (a) 449.75 cm2 (b) 520.60 cm2 (c) 563.72 cm2 (d) 507.44 cm2 24. The set of real numbers does not satisfy the property of (a) multiplicative inverse (b) additive inverse (c) multiplicative identity (d) none of these 25. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0)is (a) 5 (b) 12 (c) 11 (d) 7 + 5 26. Divide 62 into two parts such that fourth part of the first and two-fifth part of the second are in the ratio 2 : 3. (a) 24, 38 (b) 32, 30 (c) 16, 46 (d) 40, 22 27. For the equations (p + 2)  q − 1  = pq – 5 and (p – 2)  q − 1  = pq – 5, find the solution set (p, q).  2  2  (a)  −10, − 1  (b)  −10, 12  (c) 10, − 12  (d) 10, 12  2  28. If cos q + 3 si=n q 2sin q Then sin q − 3 cos q (a) cos q (b) sin q (c) 2 cos q (d) 2 sin q 29. Which of the following is the quadratic polynomial whose zeros are 1 and −2 ? 3 5 (a) 15x2 + x – 2 (b) 15x2 + 5x – 6 (c) 15x2 – 5x + 6 (d) 15x2 – x + 2 30. Two fair dice are thrown. Find the probability that both dice show different numbers. 1 5 32 29 (a) 6 (b) 6 (c) 36 (d) 36 31. The coordinates of the mid points of the line segment joining the points (3p, 4) and (–2, 2q) are (5, p). Then (a) p = 4, q = 2 (b) q = 6, p = 2 (c) p + q = 8 (d) p – q = –2 32. The sum of a rational and an irrational number is _______. (a) an irrational number (b) a rational number (c) an integer (d) a whole number 33. Solve for q cot cos2 q = 3; (q < 90º ) : 2 q − cos2 q (a) 30° (b) 90° (c) 0° (d) 60° 34. Two poles of heights 6 metres and 11 metres stand vertically on a plane ground. If the distance between their feet is 12 metres, what will be the distance between their tops? (a) 10 m (b) 12 m (c) 13 m (d) 15 m 35. In the given figure, O is the centre of the circle whose diameter is 14 cm. 35 m Find the perimeter of the figure. (Use p = 22 ) 7 (a) 134 cm (b) 124 cm O (c) 112 cm (d) 160 cm 36. Twice the product of the zeroes of the polynomial 23x2 – 26x + 161 is 14p. Then p is 5 (a) 3 (b) 1 (c) 2 (d) (–1)

SP-76 Mathematics 37. In what ratio does the point (–2, 3) divide the line–segment joining the points (–3, 5) and (4, –9) ? (a) 2 : 3 (b) 1 : 6 (c) 6 : 1 (d) 2 : 1 38. The sum of three non-zero prime number is 100. One of them exceeds the other by 36. Find the largest number. (a) 73 (b) 91 (c) 67 (d) 57 39. If DABC ~ DDEF such that BC = 2.1 cm and EF = 2.8 cm. If the area of triangle DEF is 16 cm2, then the area of triangle ABC (in sq. cm) is (a) 9 (b) 12 (c) 8 (d) 13 40. The value of k for which the system of equation kx – y = 2, 6x – 2y = 3 has unique solution is (a) not equal to one (b) equal to three (c) not equal to zero (d) not equal to three SECTION-C Case Study Based Questions: Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted. Q 41. - Q 45 are based on case study-I Case Study-I Situation-1 H.C.F. × L.C.M. = Product of two integers. 41. The H.C.F. of two numbers is 16 and their product is 3072. Find their L.C.M. (a) 182 (b) 121 (c) 192 (d) 3647 42. The sum of two numbers is 135 and their H.C.F. is 27. If their L.C.M. is 162, the numbers are (a) 108, 27 (b) 72, 54 (c) 81, 54 (d) 99, 36 Situation-2 HCF of natural numbers is the largest factor which is common to all the number and LCM of natural numbers is the smallest natural number which is multiple of all the numbers. 43. If p and q are two co-prime natural numbers, then their HCF is equal to (a) p (b) q (c) 1 (d) pq 44. The LCM and HCF of two rational numbers are equal, then the numbers must be (a) prime (b) co-prime (c) composite (d) equal 45. If two positive integers a and b are expressible in the form a = pq2 and b = p3q; p, q being prime number, then LCM (a, b) is (a) pq (b) p3q3 (c) p3q2 (d) p2q2 Q 46 - Q 50 are based on case study-II Case Study-II A chord of a circle of radius 10 cm subtends a right angle at the centre. Q 46. The area of minor sector is (a) 78 cm2 (b) 79 cm2 (c) 78.5 cm2 (d) 77 cm2 47. The area of minor segment is O 90° (a) 28.5 cm2 (b) 27 cm2 AP (c) 26 cm2 (d) 30 cm2 48. The area of major sector is B (a) 236 cm2 (b) 234 cm2 (c) 237 cm2 (d) 235.5 cm2 49. The area of major segment is (a) 285.5 cm2 (b) 286 cm2 (c) 287 cm2 (d) 288 cm2 50. The length of arc APB is (a) 17.15 cm (b) 15.71 cm (c) 25 cm (d) 15 cm

OMR ANSWER SHEET Sample Paper No –  Use Blue / Black Ball pen only.  Please do not make any atray marks on the answer sheet.  Rough work must not be done on the answer sheet.  Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected. Start time : ____________________ End time ____________________ Time taken ____________________ 1. Name (in Block Letters) 2. Date of Exam SECTION-A 17. a    b    c    d 18. a    b    c    d 3. Candidate’s Signature 9. a    b    c    d 19. a    b    c    d 10. a    b    c    d 20. a    b    c    d 1. a    b    c    d 11. a    b    c    d 2. a    b    c    d 12. a    b    c    d 37. a    b    c    d 3. a    b    c    d 13. a    b    c    d 38. a    b    c    d 4. a    b    c    d 14. a    b    c    d 39. a    b    c    d 5. a    b    c    d 15. a    b    c    d 40. a    b    c    d 6. a    b    c    d 16. a    b    c    d 7. a    b    c    d 49. a    b    c    d 8. a    b    c    d SECTION-B 50. a    b    c    d 21. a    b    c    d 29. a    b    c    d 22. a    b    c    d 30. a    b    c    d 23. a    b    c    d 31. a    b    c    d 24. a    b    c    d 32. a    b    c    d 25. a    b    c    d 33. a    b    c    d 26. a    b    c    d 34. a    b    c    d 27. a    b    c    d 35. a    b    c    d 28. a    b    c    d 36. a    b    c    d 41. a    b    c    d SECTION-C 42. a    b    c    d 43. a    b    c    d 45. a    b    c    d 44. a    b    c    d 46. a    b    c    d 47. a    b    c    d 48. a    b    c    d No. of Qns. Attempted Correct Incorrect Marks

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Sample Paper 1 ANSWERKEY 1 (a) 2 (a) 3 (d) 4 (b) 5 (a) 6 (c) 7 (b) 8 (c) 9 (a) 10 (d) 11 (a) 12 (d) 13 (d) 14 (b) 15 (b) 16 (c) 17 (a) 18 (b) 19 (a) 20 (a) 21 (c) 22 (a) 23 (d) 24 (c) 25 (d) 26 (d) 27 (a) 28 (b) 29 (b) 30 (c) 31 (a) 32 (c) 33 (a) 34 (b) 35 (a) 36 (d) 37 (c) 38 (d) 39 (b) 40 (d) 41 (c) 42 (d) 43 (c) 44 (d) 45 (c) 46 (a) 47 (c) 48 (a) 49 (b) 50 (b) 1. (a) Let the required numbers be 15x and 11x. \\ P(E=) nn((ES=)) 13=06 5 Then, their H.C.F. is x. So, x = 13. 18 ∴ The numbers are 5 × 13 and 11 × 13 i.e., 195 and 143. 2. (a) Terminating 8. (c) For reflection of a point with respect to x-axis change 3. (d) (a) is false Q π is the ratio of the circumference of sign of y-coordinate and with respect to y-axis change a circle to the length of the diameter. sign of x-coordinate. It is nearly equal to 22 but not exactly. 9. (a) It is given that AD is the bisector of ∠A. 7 (b) False [Q real numbers can be irrationals also] AB BD AC =6 ×4 3 = 4.5 cm (c) False [Q non-terminating decimal can be AC = DC ⇒ recurring or non-recurring] a b (d) Since 0.2 < 0.21 < 0.3 10. (d) Given, tan θ = 4. (b) Degree of quotient = degree of dividend – degree of divisor ∴ a sin θ − b cos θ = a tan θ − b a2 − b2 a sin θ + b cos θ a tanθ + b = a2 + b2 Degree of quotient = 7 – 4 = 3. 5. (a) 1 is zero of p(x) 11. (a) Let the age of father be ‘x’ years and the age of son ⇒ p (1) = 0 be ‘y’ years ⇒ a(1)2 – 3 (a – 1) (1) –1, = 0 ⇒ –2a + 2 = 0 According to question, x + y = 65 ...(i) ⇒ a = 1 and 2(x – y) = 50 ⇒ x – y = 25 ...(ii) 6. (c) Here, the two triangles are similar. Adding eqs. (i) and (ii), we get, 2x = 90 ⇒ x = 45 Ratio of areas of two similar triangles is equal to the Hence, the age of father = 45 years ratio of squares of their corresponding altitudes.  4 × 3 + 1× 6 3× y +1× 5  h12 12. (d) P(6, 2) =  3 + 1 , 3 +1  h22 25 So, = 36 18 4 Q 6 ≠  (Question is wrong) ∴ h1 = 5 2= 3 y+ 5   ⇒ 3y + 5 = 8 h2 6 4 7. (b) n(S) = 6 × 6 = 36 3y = 3  ⇒  y = 1 E= {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6), (6, 5)} n(E) = 10

S-2 Mathematics 13. (d) We know that sec2θ – tan2θ = 1 and sec θ = x , 20. (a) Let the numbers be 37a and 37b. Then p 37a × 37b = 4107 ⇒ ab = 3 y Now, co-primes with product 3 are (1, 3) tan θ = q So, the required numbers are (37 × 1, 37 × 3) i.e., (37, 111). ∴ x2q2 – p2y2 = p2q2 \\ Greater number = 111 21. (c) We have ABCD is square, 14. (b) Substitute x = 1 in f (x) and x = –2 in g(x), and add 1 f (1) = 2(1) – 6(1) + 4(1) – 5 = –5 ⇒ g(–2) = 3(4) – 9 = 3 AF = BF, BE = 3 BC, f (1) + g(–2) = –2 Area ∆FBE = 108 sq cm. 15. (b) S = {1, 2, 3, ......, 100} n(S) = 100 x x 2 3 E = {11, 22, 33, 44, 55, 66, 77, 88, 99} Let AB = x ⇒ BF = and BE = n(E) = 9 \\ P(E) = 9 DC 100 16. (c) 17. (a) Let the speeds of the cars starting from A and B be x km/hr and y km/hr respectively E x/3 According to problem, 9x – 90 = 9y .... (i) 9 x + 9 y =90 A x F x/2 B 7 7 .... (ii) Solving we get x = 40 km/hr, y = 30 km/hr, Area of DFBE = 1 BF × BE speed of car A = 40 km/hr 2 & speed of car B = 30 km/hr 18. (b) Given (x)2 + (x + 2)2 = 290 = 1  x  x  x2 2 2 3 12 ⇒ x2 + x2 + 4x + 4 = 290 ⇒ 2x2 + 4x – 286 = 0 = 108 sq. cm. (Given) ⇒ x2 = 12 × 108 ⇒ x2 + 2x – 143 = 0 ⇒ x2 = 12 × 12 × 3 × 3 ⇒ x2 + 13x – 11x – 143 = 0 ⇒ (x + 13) (x – 11) = 0 ⇒ x = 12 × 3 = 36 cm ⇒ x = –13, x = 11 In rt. ∆ABC, AC = AB2  BC2 x cannot be negative, discard x = –13, so x = 11 (By Pythagoras Theorem) Hence the two consecutive positive integers are 11, 13 = 362 + 362 = 36 2 cm 19. (a) Given equations are : 7x – y = 5 and 21x – 3y = k 22. (a) Let D be the window at a height of 9m on one side Here a1 =7, b1 =−1, c1 =5 of the street and E be the another window at a height a2 =21, b2 =−3, c2 =k of 12 m on the other side. We know that the equations are consistent with unique In ∆ADC solution AC2 = 152 – 92 a1 b1 = 225 – 81 a2 b2 if ≠ AC = 12 m In ∆ECB Also, the equations are consistent with many solutions CB2 = 152 – 122 if aa=12 bb=12 c1 = 225 – 144 c2 CB = 9 m Width of the street = (12 + 9)m = 21 m ∴ 7 = −1 = 5 ⇒ 1 = 5 ⇒ k = 15 23. (d) x – y = 2 ... (i) 21 −3 k 3 k kx + y = 3 ... (ii) Hence, for k = 15, the system becomes consistent. by adding (i) and (ii) kx + x = 5

Solutions S-3 x(k + 1) = 5 ⇒ – (2b2 + 2a2) y = – 4b2 – 4a2 ⇒ y = 2 5 Substituting y = 2 in eq. (i), we get x = k +1 2ax – 2b × 2 + a + 4b = 0 ⇒ x = – 1/2 ∴ x = – 1/2, y = 2 putting value of x in equation (i) 29. (b) Put x + 1 = 0 or x = – 1 and x + 2 = 0 or 5 x = – 2 in p (x) k +1 − y =2 Then, p (–1) = 0 and p (–2) = 0 ⇒ p (–1) = (−1)3 + 3 (−1)2 − 2α(−1) + β = 0 k 5 − 2 =y +1 ⇒ −1+ 3 + 2α + β = 0 ⇒ β = −2α − 2 .... (i) 5 − 2k − 2 = y k +1 p(−2) = (−2)3 + 3(−2)2 − 2α(−2) + β = 0 y = 3 − 2k ⇒ −8 +12 + 4α + β = 0 ⇒ β = − 4α − 4 .... (ii) k +1 By equalising both of the above equations, we get y should be positive as they intersect in 1st quadrant −2α − 2 = − 4α − 4 ⇒ 2α = −2 ⇒ α = −1 therfore put α = −1 in eq. (i) ⇒ β =−2 (−1) − 2 =2 − 2 =0 y > 0 2k − 3 3 − 2k k +1 k +1 > 0 ⇒ < 0 +. − . + Hence, α = −1, β = 0 − ∞ −1 3 / 2 ∞ ∴ k should lie between – 1 and 3/2 30. (c) Let a, b be two zeroes of 2x2 – 8x – m, where α = 5 . 24. (c) We have, sin x + cosec x 2 ⇒ sin x + 1 \\ α + β = ( −Coefficient of x ) sin Coefficient of x2 x ∴ We know that sum of the number and its reciprocal ⇒ 5 + β = 8 ⇒ β= 8 − 52= 3 . 2 2 2 2 is greater than or equal to 2. 25. (d) We have, sin 5θ = cos 4θ 31. (a) Let f (x) = 2 x 3 – 5x2 + ax + b f (2) = 2 (2)3 – 5 (2)2 + a (2) + b = 0 ⇒ 5θ + 4θ = 90° [ sin α = cos β, than α + β = 90°] ⇒ 16 – 20 + 2a + b = 0 ⇒ 2a + b = 4 f (0) = 2 (0)3 – 5 (0)2 + a (0) + b = 0 ⇒ b = 0 ⇒ 9θ = 90° ⇒ θ = 10° Now, 2 sin 3θ – 3 tan 3θ ⇒ 2a = 4 ⇒ a = 2, b = 0 = 2sin 30° – 3 tan 30° 32. (c) cos A =53 sin A 1− 9 =54 1 3× 1 =1−1= 0 25 = 2 × 2 − 3 ⇒ = consider 26. (d) If two similar triangles have equal area then triangles  9 cot2=A −1 9scinos22=AA −1 9 cos2 A − sin2 A sin2 A are necessarily congruent. 27. (a) Here, the two lines are 2x + 3y = 7 and 2ax + (a + b) y = 28. The above lines are coincident. 9  9 −  16   25  2=5 Two lines aif1xaa+=12 b1ybb=12+ c1c1= 0 and a2x + b2y + c2 = 0 are (81 − 16) ×=1265 65 coincident c2 16 = 16 25 25 So,=22a a=+3 b −7 33. (a) All isosceles triangles are not similar. 28 34. (b) Let α and 6α be roots of equation. ⇒ a = 4, b = 8 14 \\ b = 2a Sum of roots : α + 6α = p 28. (b) 2ax – 2by + a + 4b = 0 ........ (i) and 2bx + 2ay + b – 4a = 0 ........ (ii) ⇒ 7α = 14 ⇒p= 2 p α Multiplying eq. (i) with b and eq. (ii) with a, we get 2abx – 2b2y + ab + 4b2 = 0 ........ (iii) and 2abx + 2a2y + ab – 4a2 = 0 ........ (iv) 8 4 Subtracting (iv) from (iii), we get Product of roots : (α) (6α) = p ⇒p= 3α2 – (2b2 + 2a2) y + 4b2 + 4a2 = 0

S-4 Mathematics ⇒ 2 = 4 cosec θ = a2 + b2 , α 3α2 a2 − b2 ⇒ α =23 co=t θ BA=CC 2ab a2 − b2 Therefore, p = 2 = 3 cosec θ +=cot θ a2 + b2 + 2ab = a +b α a2 − b2 a2 − b2 a −b 35. (a) The equations 3x – (a + 1) y = 2b – 1 37. (c) 3, because it is the exponent of the highest degree 1 5x + (1 – 2a) y = 3b term in the polynomial y3 − 2 y2 − 3y + 2 . The system will have infinite number of solutions if aa=12 bb=12 c1 c2 38. (d) ax + by = c, bx – ay = c Using the cross-multiplication method, −a=cx− bc a=c −y bc 1 Here, a1 = 3, b1 = – (a + 1), c1 = 2b – 1 −a2 − b2 =\\ 53a2 =−=1(5−a, 2+ba12)= 1 – 2a, c2 = 3b 2b −1 −−=aa2c −− bbc2 −−(=ca(2a++bb2)) c(a + b) =⇒ x a2 + b2 3b Taking I and II Taking I and III and 3 −(a + 1) 3 2b −1 5 = 1− 2a 5 = 3b y= ac − bc = c(a − b) = − c(a − b) −a2 − b2 −(a2 + b2) a2 + b2 ⇒ –5a – 5 = 3 – 6a ⇒ 10b – 5 = 9b c(a + b) c(a − b) a2 + b2 a2 + b2 ⇒ –5a + 6a = 3 + 5 ⇒ 10b – 9b = 5 Therefore, x= , y= − a = 8 b = 5 ∴ a = 8, b = 5 (b) =4251 9=2×15 21 32 × 5 . 36. (d) sin θ = a2 − b2 39. a2 + b2 Clearly, 45 is not of the form 2m × 5n. So the decimal expansion of 21 is non-terminating and repeating. 45 40. (d) (sinA + cosecA)2 + (cosA + secA)2 ⇒ sin2A + cosec2A + 2sin A cosec A Since, sin θ = perpendicular base + cos2A + sec2A + 2 secA cosA ⇒ (sin2A + cos2A) + cosec2A + 2 + sec2A + 2 ∴ AC =aa22 −+ b2 ⇒ 1 + 4 + 1 + cot2A + 1 + tan2A AB b2 ⇒ 7 + cot2A + tan2A ∴ (sinA + cosecA)2 + (cosA + secA)2 Now in ∆ ABC, = 7 + cot2A + tan2A ∠B = θ and ∠C = 90° Hence, a = 7 41. (c) 42. (d) 43. (c) 44. (d) 45. (c) 46. (a) 47. (c) 48. (a) 49. (b) 50. (b) (a2 + b2)2 = BC2 + (a2 – b2)2 ∴ BC = 2ab

Sample Paper 2 ANSWERKEY 1 (c) 2 (d) 3 (d) 4 (a) 5 (b) 6 (b) 7 (c) 8 (a) 9 (c) 10 (c) 11 (b) 12 (d) 13 (b) 14 (b) 15 (b) 16 (c) 17 (a) 18 (a) 19 (b) 20 (b) 21 (d) 22 (c) 23 (d) 24 (a) 25 (c) 26 (b) 27 (a) 28 (d) 29 (b) 30 (b) 31 (a) 32 (a) 33 (a) 34 (b) 35 (c) 36 (a) 37 (a) 38 (c) 39 (b) 40 (c) 41 (c) 42 (a) 43 (b) 44 (a) 45 (b) 46 (a) 47 (a) 48 (c) 49 (b) 50 (c) 3. (d) We have, sin 5q = cos 4q 1. (c) cosceocsθθ+ 1 + cos θ = ⇒ 5q + 4q = 90° cosec θ −1 [ sin a = cos b, than a + b = 90°] ⇒ cos θ(cosec θ −1) + cos θ(cosec θ +1) =2 ⇒ 9q = 90° ⇒ q = 10° cosec2θ −1 Now, 2 sin 3q – 3 tan 3θ = 2sin 30° – 3 tan 30° ⇒ cos θ cosec θ − cos θ + cos θ cosec θ + cos θ =2 cot 2θ 2 cos θcosec θ =2 ⇒ coscθot×2 sθin1 θ = 2 × 1 − 3× 1 =1−1= 0 cot2 θ 2 3 ⇒ =1 4. (a) Given equations are : ⇒ cot θ =1 ⇒ cot q = 1 7x – y = 5 and 21x – 3y = k cot2 θ Here a1 = 7 , b1 = −1 , c1 = 5 cot q = cot 45º or q = 45º a2 = 21 , b2 = −3 , c2 = k 2. (d) tan2 θ + cosec2θ We know that the equations are consistent with unique tan2 θ −1 sec2 θ − cosec2θ solution sin2 θ 1 if a1 ≠ b1 a2 b2 cos2 sin 2 = sin2 θ θ + 1 θ Also, the equations are consistent with many solutions cos2θ − cos2 1 1 θ − θ if aa=12 bb=12 c1 sin 2 c2 = sin 2 θ × cos2 θ θ + 1 θ × sin 2 θ cos2 θ ∴ 7 = −1 = 5 ⇒ 1 = 5 ⇒ k = 15 cos2 θ sin2 θ − cos2 sin 2 sin 2 θ − cos2 θ 21 −3 k 3 k = sin 2 sin2 θ θ + sin cos2 θ θ Hence, for k = 15, the system becomes consistent. θ − cos2 2 θ − cos2 5. (b) Let full fare = ` x and reservation charges = ` y = sin2 θ + cos2 θ = 1 sin2 θ − cos2 θ sin2 θ − cos2 θ ∴ x + y = 2125 ....(i )

S-6 Mathematics Also (x + y) + x + y  =3200 from (i), 12. (d) Total number of marbles = 38 – 18 + 1 = 21  2  The multiples of 3 from 18 to 38 are 18, 21, 24, 27, 2125 x y =3200, x + =y 3200 − 2125 30, 33, 36. 2 2 + + These are 7 in numbers ⇒ x + 2y = 1075 ⇒ x + 2y = 2150 ....(ii) ∴ Required probability = 7 = 1 21 3 Solving (i) and (ii), we get – y = – 25 or y = 25 13. (b) Putting the value of y = 25 in (i) 14. (b) 196 = 22 ⋅ 72, sum of exponents = 2 + 2 = 4 x + 25 = 2125 x = 2125 – 25, x = 2100 15. (b) We have, full fare = ` 2100 and reservation charges = ` 25 Area of square metal plate = 40 × 40 = 1600 cm2 6. (b) Diameter of each semi-circle = 42 = 14 cm Area of each hole =πr 2 =272 ×  1 2 =1141 cm2 3  2  Radius of each semi-circle = 7 cm 11 346.5cm2 14 πr 2 ∴ Area of 441 holes = 441× = 2 Area of 6 semi-circle = 6 × = 3pr2 Hence, area of the remaining square plate = 3× 22 × 7× 7 = 462 cm2 = (1600 – 346.5) = 1253.5 cm2 7 Area of cloth piece = 42 × 14 = 588 cm2 16. (c) Area of the coloured portion = 588 – 462 = 126 cm2 17. (a) Given, AB = 2DE and ∆ABC ~ ∆DEF 7. (c) Initial number of workers = 120 Hence, area(∆ABC) = AB2 area(∆DEF) DE2 When 15 male workers are added, then the total number of workers = 120 + 15 = 135 or area(5∆6D=EF) 4=DDEE22 4 [Q AB = 2DE] Number of female workers = 90 \\ Probability of female workers = 90 = 2 area (∆DEF=) 5=46 14sq.cm. 135 3 8. (a) When 2256 is divided by 17 then, 2256 = (24 )64 18. (a) H.C.F. (91, 126) = L.C.9M1.×(91=12,6126) 9=11×81226 13 24 +1 (24 + 1) By remainder theorem when f (x) is divided by x + a the 19. (b) Total number of cards = 52 remainder = f (– a) Total number of diamond cards = 13 Here, f (a) = (24)64 and x = 24 and a = 1 I. P(diamond cards) = 13/52 = 1/4 ∴ Remainder = f (–1) = (–1)64 = 1 9. (c) As PQ is parallel to BC ⇒ DABC ~ DAPQ II. P(an ace of heart) = 1/52 Area of ∆ABC 2 III. P(not a heart) = 1− 1 =43 Area of ∆APQ 1 4 ⇒ = Ratio of sid=es AA=PB 2 IV. P(king or queen) = 4 + 4 = 8 = 2 1 52 52 52 13 \\ AP : AB = 1 : 2 10. (c) 20. (b) Let the required ratio be K : 1 11. (b) Suppose the required ratio is m1 : m2 ∴ The coordinates of the required point on the y-axis is Then, using the section formula, we get x = K(−4) + 3(1) ; y = K(2) + 5(1) m1 ( 4)+ m2 ( –3) K +1 K +1 –2 = m1 + m2 Since, it lies on y-axis ⇒ ––2m1 – 2m2 = 4m1 – 3m2 ∴ Its x-cordinates = 0 ⇒ m2 = 6m1 ⇒ m1 : m2 = 1 : 6

Solutions S-7 ∴ −4K + 3 =0 ⇒ −4K + 3 =0 AP2 = BP2 K +1 i.e. (x – 5)2 + (0 – 4)2 ⇒ K = 3 = (x + 2)2 + (0 – 3)2 4 ⇒x=2 3 25. (c) Given points are A(–2, –19) and B(5, 4) ⇒ Required ratio = 4 : 1 Let P be the point trisection of the line AB, then P divides ∴ ratio = 3 : 4 AB in the ratio 1 : 2 cos θ − sin θ 1− 3 21. (d) We have, cos θ + sin θ = 1+ 3 So coordinate of P is ⇒ (cos θ − sin θ) + (cos θ + sin θ) èæçççç1(5)1++22(-2) ,1(4)+1+2(2-19)øö÷÷÷÷ (cos θ − sin θ) − (cos θ + sin θ) = (1 – 3) + (1+ 3) = æèççç5 - 4 , 4 -338öø÷÷÷ = æèççç13 , -34 öø÷÷÷ (1− 3) − (1+ 3) 3 3 [Applying componendo and dividendo] 26. (b) Since (x, y) is midpoint of (3, 4) and (k, 7) ⇒ −22cs=oisnθθ 2 ⇒ c=ot θ 1 ∴ x = 3 +k and y = 4 + 7 −2 3 3 2 2 ⇒ tan θ = 3 ⇒ tan θ = tan 60° ⇒ θ = 60° Also 2x + 2y + 1 = 0 putting values we get 22. (c) We have, x = a (cosec θ + cot θ) 3+k+4+7+1=0 x ⇒ k + 15 = 0 ⇒ k = – 15 ⇒ a = (cosec θ + cot θ) ...(i) 27. (a) If the lines are parallel, then  1− cos θ  ⇒=by 1 cos θ and y = b  sin θ  sin θ − sin θ aa=12 b1 c1 b2 c2 ≠ ⇒ y = cosec θ – cot θ ...(ii) b Here, a1 = 3, b1 = – 1, c1 = –5, x y ⇒ a × b = (cosec θ + cot θ) (cosec θ – cot θ) a2 = 6, b2 = – 2, c2 = – p ⇒ xy = (cosec2θ – cot2 θ) ∴ xy = ab ⇒ =63 −1 ≠ −5  ... (i) ab −2 −p 23. (d) All the statements given in option (a, b, c) are correct. Taking II and III part of equation (i), we get 24. (a) Since, the required point (say P) is on the x-axis, its ⇒ 1 ≠ −5 ⇒ − p ≠ −10   ⇒  p ≠ 10 ordinate will be zero. Let the abscissa of the point be x. 2 −p Therefore, coordinates of the point P are (x, 0). So, option (a) is correct. P (x, 0) 28. (d) All equilateral triangles are similar ∴ ∆ ABC~ ∆EBD ⇒ Area of ABC  BC2 A Area of BDE BD2 E D is mid-point of BC BD (=2BBDD2)2 4 C 1 \\ BC == 2BD A (5, 4) B (–2, 3) ⇒ Area (∆ABC) : Area (∆BDE) = 4 : 1 Let A and B denote the points (5, 4) and (–2, 3) 29. (b) Coordinates of mid-point are given by respectively.  x1 + x2 , y1 + y2  Given that AP = BP, we have  2 2 

S-8 Mathematics Here, coordinates of mid-point are  a , 4  (2 + 2sin θ) (1− sin θ) 2(1+ sin θ) (1− sin θ)  3  34. (b) (1+ cos θ) (2 − 2 cos θ) = (1+ cos θ) (2) (1 − cos θ) a −6 − 2 So, 3 = 2 csoins22=θθ)) 22csoins22=θθ  15 2 ∴ a = – 12 = 2(1 − cot=2 θ  8  = 225 2(1 − 64 30. (b) [Hint. The outcomes are 1, 2, 3, 4, 5, 6. Out of these, 35. (c) [Hint. The English alphabet has 26 letters in all. The 4 is the only composite number which is less than 5]. word ‘DELHI’ has 5 letter, so the number of favourable outcomes = 5.] 31. (a) In DABC, AB = AC Draw AL ⊥ BC, 36. (a) Required number = H.C.F. {(70 – 5), (125 – 8)} then L is the mid-point of BC = H.C.F. (65, 117) = 13. Using Pythagoras theorem in ∆ABL, we get AL = 8cm 37. (a) In DAFD & DFEB, Also, ∆BPS ≅ ∆CQR, \\ BS = RC ∠1 = ∠2 (V.O.A) SL = LR = x cm \\ BS = CR = 6 – x ∠3 = ∠4 (Alternate angle) In DABL, PS || AL \\ ∆ FBE ~ ∆ FDA PS BS y 6−x So, EF  FB AL BL 8 6 FA DF \\ = ⇒ = 38. (c) PQ = 13 ⇒ PQ2 = 169 or x= 6 − 3 y ⇒ (x – 2)2 + (–7 – 5)2 = 169 4 ⇒ x2 – 4x + 4 + 144 = 169 ⇒ x2 – 4x – 21 = 0 A ⇒ x2 – 7x + 3x – 21 = 0 10 cm 10 cm ⇒ (x – 7) (x + 3) = 0 ⇒ x = 7, –3 P 39. (b) Required number = H.C.F.{(245 – 5), (1029 – 5)} P 2x Q = H.C.F. (240, 1024) = 16. y y 40. (c) BS RC 2x 41. (c) AA'BB=' AC ' ⇒ 155= 3 L A'C A'C ' 32. (a) Since zeroes are reciprocal of each other, so product ⇒ A′C′ = 9 cm of the roots will be 1, so k+2 = 1 , 42. (a) AA'BB=' BC ' ⇒ 155= BC k2 B'C 12 k2 – k – 2 = 0 ⇒ (k – 2)(k + 1) = 0 ⇒ BC = 4 cm 43. (b) Q ∠A = ∠A′ = 80° k = 2, k = –1, Since k > 0 ∴ k = 2 44. (a) Q ∠B = ∠B′ = 60° 45. (b) Q ∠A + ∠B + ∠C = 180° 33. (a) Area of the shaded region 80° + 60° + ∠C = 180° = 40° × 22 × (7)2 − 40° × 22 × (3.5)2 360° 7 360° 7 ∠C = 40° =19 22 =91 22 49 × 7 × (72 − 3.52 ) × 7 ×  49 − 4  46. (a) 47. (a) 48. (c) =19 22 49 =767 cm2 49. (b) 50. (c) 7 4 × × × 3

Sample Paper 3 ANSWERKEY 1 (b) 2 (b) 3 (b) 4 (a) 5 (a) 6 (c) 7 (a) 8 (b) 9 (d) 10 (d) 11 (b) 12 (a) 13 (d) 14 (d) 15 (a) 16 (b) 17 (c) 18 (c) 19 (c) 20 (b) 21 (d) 22 (d) 23 (b) 24 (b) 25 (b) 26 (c) 27 (d) 28 (b) 29 (b) 30 (a) 31 (d) 32 (a) 33 (b) 34 (d) 35 (d) 36 (d) 37 (a) 38 (b) 39 (d) 40 (a) 41 (c) 42 (d) 43 (a) 44 (b) 45 (b) 46 (d) 47 (a) 48 (b) 49 (c) 50 (d) 1. (b) Here, x – y = 3 ...(i) ⇒ AB = 24 cm QR = 9 cm and xy = 54 3. (b) 1 = cos θ and maximum value of cos θ is 1 \\ (x + y)2 = (x – y)2 + 4xy sec θ = (3)2 + 4(54) = 225 ⇒ Maximum value of 1 is 1 sec θ 4. (a) α, 1 are the roots of k2x2 – 17x + (k +2) ⇒ ( x + y) =225 =±15 ...(ii) α Case I : α × 1 = k+2 α k2 If x + y = 15 and x – y = 3 On adding the above two equations ⇒ k2 = k + 2 ⇒ k2 – k – 2 = 0 2x = 18 ⇒ x = 9 ⇒ k = 2 and k = – 1 \\ x + y = 15 ⇒ 9 + y = 15 ⇒ y = 6 But k > 0 \\ k = 2 5. (a) Quadratic polynomial p(x) = k (x + 1)2 Case II If x + y = – 15 and x – y = 3 p(– 2) = k (– 2 + 1)2 = 2 On adding the above two equations k = 2 p(x) = 2 (x + 1)2 2x = – 12 x=–6 p(2) = 2(2 + 1)2 = 2 × 3 × 3 = 18 \\ x + y = – 15 ⇒ – 6 + y = – 15 6. (c) P(raining on both day) = 0.2 × 0.3 = 0.06 ⇒ y = – 15 + 6 ⇒ y = – 9 (Because both independent event) 2. (=b) 196 =APQB 2  BC 2 7. (a) Statement given in option (a) is false.   8. (b) 2πr1 =503 and 2πr2 =437  QR  =⇒ 196  A18B=2 and 196  12 2 r1 503 r2 437    ∴ = 2π and = 2π  QR  =⇒ 34 A18B=and 43 12 Area of ring = π (r1 + r2 ) (r1 − r2 ) QR

S-10 Mathematics π  503 + 437   503 − 437  18. (c) Since, DE || BC ∴ ∆ADE ~ ∆ABC  2π   2π  = AD AE 1.5 1 DB EC 3 EC ∴ = ⇒ = ⇒ EC = 2 cm = 940  66 = 235 × 66 × 7= 235 × 21= 4935 sq. cm. 19. (c) Let the ages of father and son be 7x, 3x 2  2π 22 After 10 years, \\ (7x + 10) : (3x + 10) = 2 : 1 or x = 10 9. (d) \\ Age of the father is 7x i.e. 70 years. 10. (d) L.C.M × H.C.F = First number × second number Hence, required number = 36 × 2 =4. 20. (b) 24 out of the 90 two digit numbers are divisible by ‘3’ 18 and not by ‘5’. 11. (b) 12. (a) The required probability is therefore, 24 = 4 . 13. (d) Sum is 888 ⇒ unit’s digit should add up to 8. This is 90 15 possible only for option (d) as “3” + “5” = “8”. 21. (d) Let x2 = u, y2 = v 14. (d) Let the fraction be x ⇒ 2u + 3v = 35 and u2...(+i)3v =5 y ⇒ 2u + 3v = 35 According to given conditions, ⇒ 3u + 2v = 30 ...(ii) x +1 = 4 ... (i) Multiply (i) by 3 and (ii) by 2 and subtracting (ii) from (i), y +1 we have x −1 7 6u + 9v – 6u + 4v = 60 y −1 and = ... (ii) ⇒ 6u – 6u + 9v – 4v = 105 – 60 ⇒ 5v = 45 ⇒ v = 9 Solving (i) and (ii), we have x = 15, y = 3 substituting v = 9 in (1), we get 2u + 27 = 35 i.e. numbers = 15 ⇒ 2u = 8 ⇒ u = 4 ⇒ x2 = 4, y2 = 9 \\ x =± 2, y =± 3 is the required solution. 15. (a) Let the radii of the two circles be r1 and r2, then 22. (d) Let Area of ∆BEF = x r1 + r2 = 15       (given) ..... (i) ∴ Area of ∆AFE = 3x and πr12 + πr2=2 153π    (given) Let Area of ∆ABF = 3y ⇒ r12 + r22 =153  ..... (ii) ∴ Area of ∆CAF = 2y A On solving, we get r1 = 12, r2 = 3 E Required ratio = 12 : 3 = 4 : 1 1 16. (b) x2 – (m +3)x + mx – m(m + 3) = 0 ⇒ x[x – (m + 3)] + m[x – (m + 3)] = 0 B 3 F 2 C ⇒ (x + m) [x – (m + 3)] = 0 Area ∆ABF = Area ∆BEF + Area ∆AEF \\ x + m = 0   x – (m + 3) = 0 3y = x + 3x 3y = 4x x = –m x=m+3 3  x 4 y 17. (c) We have, sum of zeroes = a + b =− (−24) =2 Area of ∆ABC = Area of ∆ABF + Area of ∆CAF = 3y + 2y = 5y Product of zeroes = ab = 3 Area BEF  x  1  3  3 2 Area ABC 5y 5 4 20 ∴ a2b + ab2 = ab (a + b) = 3 × 2 = 3 23. (b) sin θ + 2 cos θ = 1 ⇒ (sin θ + 2 cos θ)2 = 1 2 ⇒ sin2 θ + 4 cos2 θ + 4 sin θ cos θ = 1 ⇒ 1 – cos2 θ + 4 (1 – sin2 θ + 4 sin θ cos θ =1

Solutions S-11 ⇒ 4 sin2 θ + cos2 θ – 4 sin θ cos θ = 4  2A − 3A × 100  2  ⇒ (2 sin θ – cos θ)2 = 4 ⇒ 2 sin θ – cos θ = 2 ∴ Required percentage increase = 3A [ 2 sin θ – cos θ ≠ –2] 2 24. (b) α + β = 5 ...(i) αβ = k ...(ii) [from (i) & (ii) eqns.] α – β = 1 ...(iii) = 1 × 100 ⇒ 33 1 % 3 3 Solving (i) and (iii), we get α = 3 and β = 2. Putting the value of α and β in (ii), we get 30. (a) Perimeter of sector = 25 cm 25. (b) x + y = 1 & x3 + y3 + 3xy = (x + y)3 – 3xy(x + y) + 3xy = 1 θ ⇒ 2r + 360° × 2pr = 25 26. (c) We have, p(x) = x2 –10x –75 = x2 – 15x + 5x – 75 90° 22 = x (x – 15) + 5 (x –15) = (x – 15) (x + 5) ⇒ 2r + 360° × 2 × 7 × r = 25 ∴ p(x) = (x –15) (x + 5) 11 7 So, p(x) = 0 when x = 15 or x = –5. Therefore required ⇒ 2r + r = 25 ⇒ 25 r = 25 ⇒ r = 7 zeroes are 15 and –5. 7 27. (d) Let cosec x – cot x = 1 Area of minor segment =  πθ − sinθ  r2 3  360° 2  =  22 × 90° − sin 90°  (7)2  7 360° 2  1 cos x 1 ⇒ sin x – sin x = 3  11 1  4  14 2  14 = − × 49 = × 49 = 14 cm2. ⇒ 1 – cos x =13 ⇒ 2 s2insi2xn 2x x =13 Perimeter of ∆∆PAQB=CR PA=QB QB=CR AC sin x 2 2 Perimeter of PR 31. (d) ∵ cos ⇒ tan x = 1 32. (a) The number divisible by 15, 25 and 35 = L.C.M. (15, 2 3 25, 35) = 525 Consider Since, the number is short by 10 for complete division by 15, 25 and 35. tan x = 2 ttaann=22x 2x 2 3 1– 1=–3 19 4 Hence, the required least number = 525 – 10 = 515. 33. (b) [Hint. One digit prime numbers are 2, 3, 5, 7. Out of these numbers, only the number 2 is even.] Thu=s sin x 53=, cos x 4 34. (d) Work ratio of A : B = 100 : 160 or 5 : 8 5 ∴ time ratio = 8 : 5 or 24 : 15 ∴ cos2 x – sin2x = 16 – 9 = 7 25 25 25 If A takes 24 days, B takes 15 days. Hence, B takes 30 days to do double the work. 28. (b) The point satisfy the line 4y = x + 1 35. (d) Out of n and n + 2, one is divisible by 2 and the 29. (b) Let salary of Y be = A and of X is = A other by 4, hence n (n + 2) is divisible by 8. Also n, n + 1, 2 n + 2 are three consecutive numbers, hence one of them is divisible by 3. Hence, n (n + 1) (n + 2) must be divisible by ∴ Total salary of X and Y = 3A ... (i) 24. This will be true for any even number n. 2 36. (d) The L.C.M. of 16, 20 and 24 is 240. The least multiple Let X’ and Y ’ be the new salary after increment, then we of 240 that is a perfect square is 3600 and also we can get easily eliminate choices (a) and (c) since they are not perfect number. Hence, the required least number which X' = 3A and Y' = 5A ⇒ X'+Y' =2A ... (ii) 4 4

S-12 Mathematics is also a perfect square is 3600 which is divisible by each 40. (a) Required probability = 4 = 2 . of 16, 20 and 24. 6 3 37. (a) Since, ∆ABC ~ ∆PQR 41. (c) For getting least number of books, taking LCM of 64, 72 \\ ar(∆PQR) = PR2 = QR2 =9 ∵ QR = 3 =9 8 64, 72 ar(∆ABC) AC 2 BC 2 1 BC 1  8, 9 ⇒ 8 × 8 × 9 = 576 38. (b) Area of rectangle = 28 × 23 = 644 cm2 42. (d) 43. (a) 72 is expressed as prime Radius of semi-circle = 28 ÷ 2 = 14 cm 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 44. (b) 5 × 13 × 17 × 19 + 19 Radius of quadrant = 23 – 16 = 7 cm ⇒ 19 × (5 × 13 × 17 + 1) so given no. is a composite number. Area of unshaded region 45. (b) =  1 × 22 ×14 ×14  +  2 × 1 × 22 × 7 × 7  = 385 cm2 46. (d) parabola  2 7   4 7  47. (a) 2 \\ Shaded area = (644 – 385) = 259 cm2 48. (b) –1, 3 39. (d) 1 = 2 ≠ −3  ∵ For inconsistent 5 k 7   49. (c) x2 – 2x – 3  a1 = b1 ≠ c1   a2 b2 c2  50. (d) 0 ⇒ k = 10

Sample Paper 4 ANSWERKEY 1 (b) 2 (a) 3 (b) 4 (b) 5 (d) 6 (d) 7 (c) 8 (a) 9 (c) 10 (b) 11 (a) 12 (c) 13 (c) 14 (c) 15 (c) 16 (c) 17 (b) 18 (b) 19 (c) 20 (a) 21 (a) 22 (d) 23 (d) 24 (a) 25 (b) 26 (d) 27 (a) 28 (b) 29 (a) 30 (c) 31 (d) 32 (b) 33 (b) 34 (b) 35 (d) 36 (c) 37 (b) 38 (c) 39 (a) 40 (b) 41 (a) 42 (d) 43 (a) 44 (b) 45 (c) 46 (a) 47 (c) 48 (b) 49 (d) 50 (a) 1. (b) x + y = 1 & x3 + y3 + 3xy ∴ x + y = 15 ⇒ 9 + y = 15 ⇒ y = 6 = (x + y)3 – 3xy(x + y) + 3xy = 1 Case II 2. (a) Since, the required point (say P) is on the x-axis, its If x + y = – 15 and x – y = 3 ordinate will be zero. Let the abscissa of the point be On adding the above two equations x. 2x = – 12 Therefore, coordinates of the point P are (x, 0). P (x, 0) x = – 6 ∴ x + y = – 15 ⇒ – 6 + y = – 15 ⇒ y = – 15 + 6 ⇒ y = – 9 4. (b) Let full fare = ` x and reservation charges = ` y ∴ x + y = 2125 ....(i) x Also (x + y) +  2 + y  =3200 from (i),   2125 + x + y =3200, x + =y 3200 − 2125 2 2 A (5, 4) B (–2, 3) ⇒ x + 2y = 1075 ⇒ x + 2y = 2150 . ...(ii) Let A and B denote the points (5, 4) and Solving (i) and (ii), we get – y = – 25 or y = 25 (–2, 3) respectively. Putting the value of y = 25 in (i) Given that AP = BP, we have x + 25 = 2125 AP2 = BP2 x = 2125 – 25, x = 2100 full fare = ` 2100 and reservation charges = ` 25 i.e. (x – 5)2 + (0 – 4)2 = (x + 2)2 + (0 – 3)2 5. (d) We have, tan θ − cot θ sin θ cos θ ⇒x=2 3. (b) Here, x – y = 3 ...(i) = tan θ cot θ and xy = 54 sin θ cos sin θ cos ∴ (x + y)2 = (x – y)2 + 4xy θ − θ = (3)2 + 4(54) = 225 = sin θ − cos θ ⇒ (x + y) =225 =±15 ...(ii) cos θ sin θ cos θ sin θ cos θ cos θ Case I : = 1 θ − sin12=θ sec2 θ − cosec2θ If x + y = 15 and x – y = 3 cos2 On adding the above two equations 2x = 18 ⇒ x = 9 = 1+ tan2 θ −1 − cot=2 θ tan2 θ − cot2 θ

S-14 Mathematics 6. (d) L.C.M × H.C.F = First number × second number = p(5.7 – 4.3) (5.7 + 4.3) = p × 1.4 × 10 sq. cm = 3.1416 × 14sq. cm. = 43.98 sq. cms. Hence, required number = 36 × 2 =4. 18 7. (c) Let BD = x cm 12. (c) Given, area of two similar triangles, Since AC = BC, therefore DABC is an isoscele triangle. A1 = 81cm2 , A2 = 49 cm2 ⇒ ∠B = ∠CAB = 72° Since AD bisects ∠A Ratio of corresponding medians = =AA12 =4891 9 7 \\ ∠DAB = 36° so, In DADB, ∠ADB = 72° ⇒ DADB is an isoscele triangle cos θ cos θ 1− sin θ + sin θ \\ AB = AD = 1cm 13. (c) We have, + 1 =4 ⇒ AB = 1 cm Similarly, DADC is also an isoscele triangle. ⇒ cos θ  1 + sin θ +1− sin θ  =4  1 − sin2 θ  \\ AD = CD ⇒ AD = 1 cm C 2 cos 1 cos2 2 ⇒ θ = 4 ⇒ cos θ= ⇒ θ= 60° θ 36° 1 14. (c) Let the required ratio be k : 1 1 +x D = Then, 2 6=kk−+41(1) or k 3 2 72° 36° x ∴ The required ratio is 3 :1 or 3:2 2 36° 72° A B Now   AC = CD =Also, y 3=(33) ++ 22(3) 3 AB BD ⇒ 1+ x = 1   ⇒  x + x2 – 1 = 0 15. (c) Let unit’s digit : x, tens digit : y 1 x then x = 2y, number = 10y + x Also 10y + x + 36 = 10x + y = ⇒ x – 1 ± (1)2 – 4(1)(–1) –1 ± 5 ∴ 9x – 9y = 36 or x – y = 4 =2 2 Solve, x = 2y, x – y = 4 Substitute x = 2y in x – y = 4 BD = 5 –1 we get, 2y – y = 4 ⇒ y = 4 2 and x = 8 So, the number = 10y + x = 48 8. (a) Since, C (y, – 1) is the mid-point of P (4, x) and Q (–2, 4). 16. (c) Total outcomes = HH, HT, TH, TT We have, 4 − 2 = y and 4 + x = −1 Favourable outcomes = HT, TH, TT 2 2 P(at most one head) = 3 . 4 ∴ y = 1 and x = – 6 17. (b) Given an equilateral triangle ABC in which 9. (c) Required probability = 5 = 1 . AB = BC = CA = 2p A 25 5 and AD ⊥ BC. 10. (b) Value of n = 2. ∴ In ∆ADB, 2p 2p C 11. (a) Let the radii of the outer and inner circles be r1 and r2 AB2 = AD2 + BD2 BD respectively; we have (By Pythagoras theorem) Area = pr12 – pr22 = p(r12 – r22) ⇒ (2p)2 = AD2 + p2 ⇒ AD2 = 3 p. = p(r1 – r2) (r1 + r2)

Solutions S-15 18. (b) x2 + 4x + 2 = (x2 + 4x + 2) – 2 = (x + 2)2 – 2 Alternate method: Lowest value = – 2 when x + 2 = 0  Using identity, sin 2 A = 2 tan A   + tan 2A  19. (c) − 3(1) + 4(2) − 7 = − 4 =94 1 3(−2) + 4(1) − 7 −9 sin 60° = 2 tan 30° = 3 1+ tan2 30° 2  1 22  20. (a) Required area =  72 – 4 × 7 × 72  cm2 27. (a) The largest number of four digits is 9999. Least number divisible by 12, 15, 18, 27 is 540. = (49 – 38.5) cm2 = 10.5 cm2 On dividing 9999 by 540, we get 279 as remainder. 21. (a) Quadratic polynomial p(x) = k (x + 1)2 Required number = (9999 – 279) = 9720. p(– 2) = k (– 2 + 1)2 = 2 28. (b) Let P (x, 0) be a point on X-axis such that AP = BP k = 2 ⇒ AP2 = BP2 p(x) = 2 (x + 1)2 ⇒ (x + 2)2 + (0 – 3)2 = (x – 5)2 + (0 + 4)2 p(2) = 2(2 + 1)2 = 2 × 3 × 3 = 18 22. (d) All the statements given in option (a, b, c) are correct. ⇒ x2 + 4x + 4 + 9 = x2 – 10x + 25 + 16 23. (d) We have, sin 5θ = cos 4θ ⇒ 14x = 28 ⇒ x = 2 ⇒ 5θ + 4θ = 90° Hence, required point is (2, 0). [ sin α = cos β, than α + β = 90°] ⇒ 9θ = 90° ⇒ θ = 10° 29. (a) Let side of a square = x cm ∴ By Pythagoras theorem, x2 + x2 = (16)2 = 256 Now, 2 sin 3θ – 3 tan 3θ = 2sin 30° – 3 tan 30° ⇒ 2x2 = 256 ⇒ x2 = 128 ⇒ x = 8 2 cm. 1 3× 1 =1−1= 0 = 2 × 2 − 3 3x + 4y 9 x + 2y 4 30. (c) = 24. (a) Given equations are : 7x – y = 5 and 21x – 3y = k ⇒ 4(3x + 4y) = 9(x + 2y) Here a1 = 7, b1 = –1, c1 = 5 aW2e=k2n1o,wbt2h=at –3, ecq2 u=atkions the are consistent with unique Hence, 12x + 16y = 9x + 18y or 3x = 2y solution b1 \\ x= 2 y . a1 b2 3 if a2 ≠ 2 Also, the equations are consistent with many solutions Substitute x = 3 y in the required expression. if aa=12 bb=12 c1 i.e. 3x + 5y : 3x − y c2  2 y  2 y ∴ 2=71 −−=31 5 ⇒ 1 = 5 ⇒ k = 15 = 3  3 + 5 y :3  3 − y k 3 k = 2y + 5y : 2y – y = 7y : y = 7 : 1 Hence, for k = 15, the system becomes consistent. 25. (b) α + β = 5 ...(i) 31. (d) αβ = k ...(ii) α – β = 1 ...(iii) 32. (b) 10x = 7.7 or x = 0.7 Solving (i) and (iii), we get α = 3 and β = 2. Putting the value of α and β in (ii), we get Subtracting, 9x = 7  \\  x = 7 9 2 tan 30° 26. (d) We have, 1+ tan2 30° 2=x 19=4 1.555.....=... 1.5 2× 1 2 = 1=3 2 1=+313 2=× 3 3 33. (b) 3  3×4 2 1 +  34. (b) The numbers that can be formed are xy and yx. Hence  (10x + y) + (10y + x) = 11(x + y). If this is a perfect square then x + y = 11.

S-16 Mathematics 35. (d) Since, ∆ABC ~ ∆PQR 200 42. (d) Radius of circle = 2 ∴ aarr((DDPAQBCR))= BC 2 ⇒ 196= (4.5)2 = 100 cm QR2 QR2 16 × (4.5)2 43. (a) Area of each sector 9 ⇒ =QR2 ⇒=QR 6 cm = 60 × pr2 360 36. (c) = 1 × 3.14 × 10000 = 5233.3 cm2 6 37. (b) Centroid is  x1 + x2 + x3 , y1 + y2 + y3   3 3  44. (b) Area of the shaded region  3 + (–8) + 5 –7 + 6 +10   0 9  = Area of DABC – 3 × Area of each sector  3 3   3 3  i.e. , = , = (0, 3) 31400 = 17320.5 – 3 × 6 = 1620.5 cm2 38. (c) Required probability = 1+ 2+1 4 . 45. (c) Perimeter of DABC = 3 × 200 = 600 cm 11 11 = 46. (a) Radius of inner semicircular end 39. (a) Given: The natural number, when divided by 13 leaves = 6=20 30 m remainder 3 The natural number, when divided by 21 leaves remainder 11 47. (c) Radius of outer semicircular end = 30 + 10 = 40 m So, 13 – 3 = 21 – 11 = 10 = k 48. (b) The distance arounded the track along its inner edge Now, LCM (13, 21) = 273 = 106 × 2 + 2 × πr But the number lies between 500 and 600 = 212 + 2 × 22 × 30= 212 +188.57 7 \\ 2 LCM (13, 21) – k = 546 – 10 = 536 = 400.57 m 536 = 19 × 8 + 4 \\ remainder = 4 49. (d) The distance arounded the track along its outer edge 40. (b) Since, ∆ABC ~ ∆APQ = 106 × 2 + 2 × πr ∴ ar(DABC) =PBQC 22 = 212 + 2 × 22 × 40= 212 + 251.43 ar(DAPQ) 7 =PBQC 22 BC  2 = 463.43 m PQ  ⇒ ar(DABC) ⇒  =14 50. (a) The area of the track 4⋅ ar(DABC)  = 2 × Area of ractangle + 2 × Area of BC =12 semicircular ring. PQ ⇒ = 2(10 × 106) + 2 × 1 × 22 × (402 – 302) 2 7 41. (a) Area of DABC = 3 a2 = 2120 + 2200 = 4320 m2 4 17320.5 = 3 a2 4 a2 = 17320.5× 4 = 40000 1.73205 a = 200 cm

Sample Paper 5 ANSWERKEY 1 (c) 2 (d) 3 (c) 4 (b) 5 (c) 6 (d) 7 (d) 8 (c) 9 (b) 10 (b) 11 (c) 12 (a) 13 (b) 14 (a) 15 (c) 16 (b) 17 (a) 18 (c) 19 (c) 20 (c) 21 (b) 22 (c) 23 (a) 24 (b) 25 (a) 26 (c) 27 (c) 28 (a) 29 (b) 30 (d) 31 (a) 32 (c) 33 (d) 34 (c) 35 (a) 36 (d) 37 (a) 38 (b) 39 (a) 40 (d) 41 (a) 42 (d) 43 (c) 44 (d) 45 (b) 46 (b) 47 (a) 48 (c) 49 (a) 50 (d) 1. (c) Let the three points be A(0, 0), B(3, 3 ) and DC C (3, λ). E x/3 ∴ AB = BC = CA [Q ∆ABC an equilateral ∆] A x F x/2 B ⇒ AB2 = BC2 = CA2 Now AB2 = BC2 ( ) ⇒ (0 – 3)2 + 0- 3 2 ( )= (3 – 3)2 + 3 -λ 2 Area of DFBE = 1 BF × BE 2 ⇒ 9 + 3 = 0 + 3 – 2λ 3 + λ2 ⇒ λ2 - 2 3λ -9 = 0 = 1  x  x  x2 2 2 3 12 ⇒ λ2 -3 3λ + 3λ -9 = 0 = 108 sq. cm. (Given) ⇒ λ(λ -3 3)+ 3(λ -3 3) = 0 ⇒ (λ -3 3)(λ + 3) = 0 ⇒ x2 = 12 × 108 ⇒ x2 = 12 × 12 × 3 × 3 ⇒ x = 12 × 3 = 36 cm ⇒ λ = 3 3 or λ = - 3 . In rt. ∆ABC, AC = AB2  BC2 2 (By Pythagoras Theorem) k2 14 2. (d)  = 362 + 362 = 36 2 cm 2  14 ⇒ k2  1 (Given) 4. (b) Let x = 0.31783178...... ...(i) Multiply by 10000 ⇒ k = ± 4 10000 x = 3178.31783178........ ...(ii) 3. (c) We have ABCD is square, Subtracting (i) from (ii) 1 10000 x = 3178.3178........ AF = BF, BE = 3 BC, Area ∆FBE = 108 sq cm. x = 0.3178........ –– xx 999 x = 3178 Let AB = x ⇒ BF = 2 and BE = 3 3178 x = 999

S-18 Mathematics 5. (c) Either a rational number or an irrational number. 9. (b) 6. (d) A 10. (b) Let the number of boys and girls in classroom is x and y. b According to question Q b x −x / 5 =23 ⇒ 4x =23 ⇒ x = 5 ...(i) 2b P y 5y y 6 2b a Also, x− x/5 =52 ⇒ 5 (y4−x 44) =52 a y − 44 B a In DABC C ⇒ 8x = 25y – 1100 ...(ii) From Eqs. (i) and (ii), we get, x = 50, y = 60 AB = AC Let n number of boy leaves the class so number of boys and number of girls become equal. ⇒  ∠C = ∠B ⇒  ∠B = ∠C = a \\ 50 – 10 – n = 60 – 44 By angle sum properly in DABC, n = 40 – 16 = 24 b + a + a = 180 11. (c) Perimeter = 1 × 2πr + 2r 4 ⇒  b + 2a = 180°  ...(i) 1 22 In  DQPB =  2 × 7 × 7 + 2 × 7  cm = 25 cm   ⇒  ∠QPB = 180 – 4b 12. (a) DABC ~ DANM Since ‘APC’ is a straight line \\ Area of ∆ABC = AC 2 ...(i) Area of ∆ANM AM 2 ⇒  180 – 4b + a + b = 180 ⇒  a = 3b  ...(ii) DABC ~ DMPC From equations (i) & (ii) \\ Area of ∆ABC = AC 2 ...(ii) Area of ∆MPC MC 2 180 b + 2(3b) = 180 ⇒ b = 7 From Eqs. (i) and (ii,) we get ∠AQP = 180° – 2  180  = 5 p Area of ∆ANM = AM 2  7  7 Area of ∆MPC MC 2 7. (d) 2πr = 4π ⇒ r = 2 Area of ∆ANM + Area of ∆MPC AM 2 + MC2 Area of ∆MPC MC 2 Area = π(2)2 = 4π = When, 2πr = 8π Now, Area of DANM + Area of DMPC ⇒r=4 = Area of DABC – Area of BNMP Area = 16π Using Area of BNMP = 5 of area of DABC 18 8. (c) (1+ tan θ + sec θ)(1+ cot θ – cosec θ) = 1 + sin θ + 1  × 1 + cos θ – 1 \\ 13 (Area of ∆ABC) = AM 2 + MC2 ...(iii) cos θ cos  sin θ sin θ 18 (Area of ∆MPC) MC 2 θ = {(cos θ + sin θ) +1} ×{(cos θ + sin θ) –1} 13  AC2  AM 2 + MC2 cos θ × sin θ 18  MC2  MC 2 From Eq. (iii), = = (cos θ + sin θ)2 – (1)2 {∵ (a + b)(a – b) =a2 – b2} ⇒ 13 (AM + MC)2 = 18 (AM2 + MC2) cos θ × sin θ AM 1 = 1=+ c2ocsoθs×θ ssiinnθθ –1 2. ⇒ MC = 5, 5 . Hence, option (a) is correct.

Solutions S-19 13. (b) As QAR are collinear 14. (a) Condition for infinite many solutions. \\ ∠QAR = 180° 1p2= 3p= p – 3  aa12= bb12= c1  Q is reflection of P on AB p  c2    \\ ∠QAB = ∠BAP p2 = 36 ; p = {From I and II} R is reflection of P on AC p2 – 3p = 3p {From II and III} p = 6 \\ ∠RAC = ∠CAP \\ p = 6 ∠QAR = 180° 15. (c) Radius of outer concentric circle = (35 + 7) m = 42 m. \\ 2∠BAP + 2∠CAP = 180° Area of path = π (422 – 352) m2 = 22 (422 – 352) m2 ∠BAP + ∠CAP = 90°  ⇒  ∠BAC = 90° 7 21. (b) α + β = –a and αβ = –b 1 1 α +β −a a 16. (b) 9sec2 A – 9 tan2 A = 9(sec2 A – tan2 A) α + β = αβ = −b = b = 9 × 1 = 9. 17. (a) Total three digit number are : 3 × 3 × 2 = 18 1 × 1= 1= 1 Now, numbers divisible by 5 are : α β αβ –β 2 × 3 × 1 + 2 × 2 × 1 = 10 ∴ The required polynomial is a 1 x2  b x  b . So, probability that the slip bears a number divisible 22. (c) H.C.F. of 20 and 15 = 5 10 5 by 5 = 18 = 9 So the 5 students are in each group so 20 +15 35 n = 5 = 5 = 7 18. (c) Let x = 0.235 ...(i) Hence, x = 4, y = 3 and n = 7 1000x = 235.235 ...(ii) 23. (a) Let AB = BC = x units. Subtract (i) from (ii), 999x= 235 ⇒ x= 235 Then Hyp. AC = side 2 = x 2 units 999 x2 E 19. (c) Joining B to O and C to O A3 Let the radius of the outer cirlce be r x 2 x ∴ perimeter = 2πr x2 x2 4 But OQ = BC = r [diagonals of the square BQCO] D1 ∴ Perimeter of ABCD = 4r. x Hence, ratio = 2πr = π Bx C 4r 2 In ∆ABD and ∆CAE 20. (c) Here, ABC is a triangle & P be interior point of a DABC, Q and R be the reflections of P in AB and AC, ∠1 = ∠3 [In equilateral ∆ each angle = 60°] respectively. ∠2 = ∠4 R ∴ ∆ABD ~ ∆CAE (By AA rule) A ∴ area ABD  AB2 area CAE CA2 Q q q θ (By the theorem) φ x2 x2 1   = x2 2  2x2  2 P BC

S-20 Mathematics ∴ area ABD  1 ⇒ 2(x + y) = 10 + y ⇒ x+y=5+ y area CAE 2 2 ⇒ area (∆ABD) = 1 area (∆CAE) Now, (x + y)max when y is maximum & maximum 2 value of y will be 10. (Q y = 10 – 2x) 24. (b) Degree of quotient = degree of dividend – degree of So (x + y)max = 5 + 5 = 10 & (x + y)min when y = 0 divisor \\ minimum value of x + y = 5 Degree of quotient = 7 – 4 = 3. So, sum of (x + y)max & (x + y)min = 15 31. (a) Here, when A = 0° 25. (a) Since, the required point (say P) is on the x-axis, its ordinate will be zero. Let the abscissa of the point be x. LHS = sin 2 A = sin 0° = 0 and RHS = 2 sin A= 2 sin 0° = 2 × 0 = 0 Therefore, coordinates of the point P are (x, 0). In the other options, we will find that P (x, 0) LHS ≠ RHS 32. (c) 1 = 0.142857 7 The second positive integer whose reciprocal have six different repeating decimals is A (5, 4) B (–2, 3) 1 = 0.076923 13 Let A and B denote the points (5, 4) and (–2, 3) respectively. And the third positive integer whose reciprocal have six Given that AP = BP, we have different repeating decimals is AP2 = BP2 1 = 0.047619 21 i.e. (x – 5)2 + (0 – 4)2 Therefore, the values of x are 7, 13, 21 = (x + 2)2 + (0 – 3)2 Hence, the required sum is = 7 + 13 + 21 = 41 ⇒x=2 33. (d) 26. (c) 34. (c) Let distance = d, 2 tan 30° 2  1 Time taken upstream = d = d – tan2 30°  3  15 − 5 10 27. (c) = 1 2 1 3  1 –  d d  Time taken downstream = 15 + 5 = 20 2 Hence, average speed = 3 = 2 × 3 = 3 = tan 60°. = 1d=02+d2d0 2=d3×d 20 40 km/hr 3 2 3 1 – 1 3 28. (a) x = – 1 is the root of the quadratic polynomial p(x) 40 3 So, quadratic polynomial p(x) = k(x + 1)2 Ratio = :15 = 40 : 45 = 8 : 9 p(–2) = k(–2 + 1)2 = 2 ⇒ k = 2 ∴ p(x) = 2(x + 1)2 35. (a) P (E) + P ( ) = 1 Also, p(2) = 2(2 + 1)2 = 2 × 3 × 3 = 18 1 – ttaann=22 4455°° 11=+– ((11))22 1 + 29. (b) 36. (d) 0. 30. (d) Given 2x + y = 10 37. (a) Let the required number be 33a and 33b. on adding y both sides, we get, 2x + y + y = 10 + y Then 33a + 33b = 528 ⇒ a + b =16.

Solutions S-21 Now, co-primes with sum 16 are (1, 15), (3, 13), (5, 11) 2 and (7, 9). = 3 = 2 × 3 = 3 = sin 60°. \\ Required numbers are (33 × 1, 33 × 15), 3 4 2 (33 × 3, 33 × 13), (33 × 5, 33 × 11), (33 × 7, 33 × 9). 1+ 1 3 The number of such pairs is 4. 40. (d) For given numbers, 38. (b) Upstream speed = 4 km/hr and time = x hrs. (55)725, unit digit = 5; (73)5810, unit digit = 9 (22)853, unit digit = 2 Downstream speed = 8 km/hr and time taken = x/2 hrs. 4x +8× x / 2 16 Unit digit in the expression x+x/2 3 55725 + 735810 + 22853 is 6 Hence average speed = = km/hr. 2  1 41. (a) 42. (d) 43. (c)  3  (b) 39. (a) 2 tan 30° = 1 2 44. (d) 45. (b) 46. (a) 1+ tan2 30° 3  1 +  47. (a) 48. (c) 49.  50. (d)

Sample Paper 6 ANSWERKEY 1 (a) 2 (a) 3 (a) 4 (d) 5 (a) 6 (b) 7 (b) 8 (a) 9 (b) 10 (c) 11 (a) 12 (a) 13 (a) 14 (c) 15 (d) 16 (b) 17 (d) 18 (c) 19 (a) 20 (a) 21 (a) 22 (b) 23 (b) 24 (b) 25 (b) 26 (d) 27 (a) 28 (a) 29 (d) 30 (b) 31 (b) 32 (c) 33 (d) 34 (d) 35 (b) 36 (b) 37 (c) 38 (a) 39 (d) 40 (a) 41 (c) 42 (a) 43 (a) 44 (a) 45 (b) 46 (a) 47 (b) 48 (c) 49 (b) 50 (a) 1. (a) P(x) is a polynomial of degree 3. D aC and P(n) = 1 ⇒ n P(n) – 1 = 0 a a  1  n 2 O × d1 × d2 n(P(n)) is a polynomial of degree 4 A aB \\ n P(n) – 1 = k(n – 1)(n – 2)(n – 3)(n – 4) Area of square = 1 × AC × BD −1 2 For n = 0; –1 = 24 k ⇒ k = 24 1 −1 2 For n = 5; 5 × P(5) – 1 = 24 (4)(3)(2)(1) ⇒ 5 ⋅ P(5) – 1 = – 1 ⇒ P(5) = 0 Area of square = × d1 × d2 2. (a) Area of square = 2 cm2 = 1 × 2 5 × 2 5 = 5 cm2 Side of square = 2 cm 2 2 2 OP = 2 cm, OQ = x cm 3. (a) 2 4. (d) Let x & y be the unit and tenth digits respectively of Q a two digit number. Then, x + y = 9 (Q Given) ... (i) 2x and according to given condition, 10x + y = 10 y + x + 27 PO ⇒ 9x – 9y = 27 ⇒ x – y = 3 ... (ii) x On adding (i) & (ii) 2x = 12 ⇒ x = 6 ( ) 2 2  2 2 Hence, from equation (i), ⇒ x2 = +  2  6 + y = 9 ⇒ y = 3 So number will be 10 × 3 + 6 = 36 ⇒ x2= 2 + 2 4 5 5 ⇒ x2 = 2 ⇒ x = 2 cm. AC = 2 5 cm (AC = Diameter) 5. (a) The largest number of four digits is 9999. Least number 2 divisible by 12, 15, 18, 27 is 540. On dividing 9999 by 540, we get 279 as remainder. Required number = (9999 – 279) = 9720.

Solutions S-23 r 6. (b) A (–2, 5) 12. (a) 13. (a) A 35 m 53 m O(x, y) B 66 m C B C Here, a = 66 m, b = 53 m & c = 35 m (–2, 3) (2, –3) a=+ 2b + c 66 + 53 + 35 = s 2 = 77m Let O(x, y) is the centre of the given circle. Area of ∆ = s(s − a)(s − b)(s − c) Join OA, OB & OC. So, area of ∆ = 77(11)(24)(42) = 924 Q OA = OB = OC πr2 = 2(924) \\ OA2 = OB2 ⇒ (x  2)2  ( y  5)2  (x  2)2  ( y  3)2 ⇒ r2 = 2 × 924 + 7 ⇒ r2 = 588 ⇒ x2 + 4 + 4x + y2 + 25 – 10y = x2 + 4 + 4x + y2 + 9 + 6x 22 ⇒ 16y = 16 ⇒ y = 1 ⇒ r = 14 3 m Again: OB2 = OC2 ⇒ (x  2)2  ( y  3)2  (x  2)2  ( y  3)2 14. (c) aa=12 bb=12 cc=12 5 ⇒ x2 + 4 + 4x + (y + 3)2 = x2 + 4 – 4x + (y + 3)2 3 ⇒ 8x = 0 ⇒ x = 0 15. (d) tan 63=00°° 1 \\ centre of the circle is (0, 1). cot =13 1 7. (b) sin 45° + cos 45=° 1 + 1= 2= 2 3 8. (a) 22 2 16. (b) 9. (b) Since, H.C.F. of co-prime number is 1. 17. (d) ∴ Product of two co-prime numbers is equal to their 18. (c) Total number of cards = 25 L.C.M. So, L.C.M. = 117 Prime number are 3, 5, 7, 11, 13, 17, 19, 23, 10. (c) (x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2...(i) ∴ Probability of prime number card = 8 25 Also, (x – 3)2 + (y – 3)2 = (x – 3)2 + (y + 7)2 y2 – 6y + 9 = y2 + 14y + 49 A – 20y = 40 ⇒ y = – 2 Putting y = – 2 in equation (i), we have 19. (a) q (x – 6)2 + (4)2 = (x – 3)2 + (5)2 x2 – 12x + 36 + 16 = x2 – 6x + 9 + 25 90 – q 90 – q B –6x = – 18 ⇒ x = 3 y O 11. (a) Since a, b are co-prime x C ⇒ g. c.d of a, b = 1 ⇒ g. c. d. of a2, b2 = 1 ⇒ a2, b2 are co-prime. ∠AOB = q Q CO ^ OA (b) does not hold. (c) does not hold, (d) does not hold \\ ∠BOC = (90° – q) Q If a = 2, b = 3, then a2 = 4, b2 = 9 \\ a2 is even, b2 is odd.

S-24 Mathematics sin θ = 3 cos θ = 4 ∵ cos θ = 1−sin2 θ  P(h, 5h + 3) y = 5x + 3 5 5 ; sin q = x = 3 M 5 cos q = y = 4 Q (3, –2) 5 \\ point on fourth quadrant is  3 , − 4 Since, M is the mid-point of PQ, therefore by mid-point  5 5  formula, we have M =  h+ 3, 5h + 3−2 .  2 2  20. (a) Required area = π (r + d)2 − r 2   Clearly by observing the options, we can say that M must d lie on the line y = 5x – 7 r 25. (b) Perimeter = 2πr + 2r 2 = πr + 2r ⇒ (π + 2) r = 36 = π[r2 + d2 + 2rd – r2] ⇒  36  – r = 36 rr = π[d2 + 2rd] = πd[d + 2r]  7  21. (a) We know that height of an ⇒ r = 7 cm equilateral triangle 3 a, A Hence, diameter = 7 × 2 = 14 cm. 2 30° 30° 26. (d) f(x) = (x – 1)2 + (x – 2)2 + (x – 3)2 + (x – 4)2 where a is the side of equilateral triangle = 4 x − 5 2 +5 2  \\ AD2 3 a2 3 BC 2 60° = 4 = 4 B 60° 5 2.5 C 2  D f(x) is minimum at x = = …(i) 22. (b) Let speed of boat in still water be x km/hr 27. (a) Let one woman can paint a large mural in W hours and one girl can paint it in G hours and speed of stream be y km/hr 30 According to question, x + y = 3  ⇒  x + y = 10 8 + 12 = 1 ⇒ 2 + 3 = 1 W G 10 W G 40 ...(i) 30 x − y = 5  ⇒  x – y = 6 …(ii) 6 8 1 3 4 1 W G 14 W G 28 Also, + = ⇒ + = ...(ii) From solving equations (i) and (ii) –x + y = 10 On solving equation (i) and (ii), we get –x – y = 6 W = 140 and G = 280   +  – Now, 7 21840 Time1taken 1 (say) 140 t 2y = 4   y = 2 km/hr. and x = 8 km/hr ⇒ 1t 1 1 20 20 N=Too.taolfnfuavmobuerrabolfeoouutctcoommeess 1  ⇒ t = 10 hours 5 =23. (b) Probability 24. (b) Let coordinate of point p be (h, 5h + 3)

Solutions S-25 28. (a) Here, BAC is a right angle triangle then, area = π  r − 10  =πr  81  B    100  AB = 15 & BC = 25 \\ AC= BC2 − AB2= 20 F D 81 100 Thus, area is diminished by 1 −  % = 19%  Area of ∆ABC =12 BC.AD A E C 37. (c) Q ∠BAC = ∠ADC(given) = 1 AB.AC ∠C = ∠C  (common) 2 A ⇒ BC.AD =AB.AC ⇒ 25(AD) =15(20) ⇒  AD = 12 Q AEDF is rectangle then, AD = EF = 12 BD C 29. (d) As (a, 0), (0, b) and (1, 1) are collinear \\ DABC ~ DDAC  (by AA similarity criterion) \\ a(b – 1) + 0(1 – 0) + 1(0 – b) = 0 BC AC AC DC ab – a – b = 0 ⇒  ⇒ BC × DC = AC2 ab = a + b ⇒ BC × DC = (21)2 = area of rectangle with sides BC & DC Now, Area of equilateral triangle = area of rectangle 1= 1 + 1 a b 30. (b) (sin 30° + cos 30°) – (sin 60° + cos 60°) ⇒ 3 (side)2 = (21)2 ⇒ Side = 14 × 33/4 4 1 3  1 3 =  2 + 2  −  2 + 2  =0 38. (a) Since –3 is the zero of (k –1) x2 + kx + 1, 31. (b) Since, 2 is the zero of x2 + 3x + k, \\ (k – 1) (–3)2 + k(–3) + 1 = 0 \\ (2)2 + 3(2) + k = 0  ⇒  k + 10 = 0 ⇒ k = – 10 ⇒ 9k – 9 – 3k + 1 = 0 ⇒ 6k – 8 = 0 ⇒ k= 4 3 32. (c) Possible products are 1, 4, 9, 16, 2, 8, 18, 32, 3, 12, 39. (d) (sec A + tan A) (1 – sin A) 27, 48, 4, 16, 36, 64 =  1 + sin A  × (1 – sin A) So, required probability of getting the product of the two  cos A cos A numbers so obtained is 6 = 3 = (1 + sin A)(1 – sin A) 16 8 cos A 33. (d) 4 P (4, 5) = 1=–csoisnA2 A cos2 A  (∵ cos2 A = 1 – sin2 A) cos A 5 = cosA. 40. (a) x = 1 ⇒ a = 10 and y= 1 ⇒ b = 5 10 5 1+ tan2 A (sec2 A – tan2 A) + tan2 A 34. (d) 1+ cot2 A = (cosec2A – cot2 A) + cot2 A 41. (c) (0, 0) = csoes=ce2c2AA cs=oins22 AA csoi=nsAA 2 42. (a) (4, 6)  tan2 A. 43. (a) (6, 5)  44. (a) (16, 0) 35. (b) (I) Statement I is false. Consistent Linear equations 45. (b) (–12, 6) may have unique or infinite solutions. 46. (a) 47. (b) 48. (c) (II) Statement or is also false 49. (b) 50. (a) Q 132 + 142 = 365 36. (b) Let r be the radius of circle, then area = πr2 When r is diminished by 10%

Sample Paper 7 ANSWERKEY 1 (b) 2 (c) 3 (a) 4 (b) 5 (d) 6 (c) 7 (d) 8 (b) 9 (a) 10 (c) 11 (a) 12 (b) 13 (b) 14 (a) 15 (d) 16 (a) 17 (a) 18 (a) 19 (d) 20 (d) 21 (c) 22 (d) 23 (c) 24 (c) 25 (b) 26 (b) 27 (d) 28 (d) 29 (c) 30 (b) 31 (a) 32 (a) 33 (d) 34 (b) 35 (b) 36 (c) 37 (c) 38 (b) 39 (a) 40 (d) 41 (c) 42 (b) 43 (a) 44 (d) 45 (a) 46 (d) 47 (c) 48 (d) 49 (a) 50 (b) 1. (b) Dividing numerator and denominator by cos2q, 2. (c) (a) x2 + =1x x2 + x–1 is not a polynomial since the 2sin θ cosθ 1=–2ttaann2θθ 2×1123 2 ccooss22cθθos–=2csθoins22 θθ exponent of variable in 2nd term is negative 1–  12  13 1 ∵ tan θ =1123 (b) 2x2 – =3 x +1 2x2 – 3x 2 +1is not a polynomial, since the exponent of variable in 2nd term is a 8. (b) n(S) = [1, 2, 3, ..., 100] = 100 rational number. Q x+ 1 >2 (c) x3 – 3x + 1 is a polynomial. x 3 \\ x2 + 1 > 2x (d) 2x 2 – 5x is also not a polynomial, since the ⇒ x2 – 2x + 1 > 0 exponents of variable in 1st term is a rational ⇒ (x – 1)2 > 0 number x = [2, 3, ... ,100] Hence, (a), (b) and (d) is not a polynomial. n(E) = [2, 3, 4, ..., 100] = 99 3. (a) In ∆ABC, we have DE || BC 99 100 ∴ AD = AE [By Thale’s Theorem] P(E) = = 0.99 DB EC x x+2 9. (a) Let the third side be x cm. Then, by Pythagoras x−2 x −1 ⇒ = theorem, we have p2 = q2 + x2 ⇒ x(x –1) = (x – 2)(x + 2) ⇒ x2 = p2 − q2 =( p − q)( p + q) = p + q [∵ p – q = 1] ⇒ x2 – x = x2 – 4 ⇒ x = 4 ⇒ x= p + q = 2q +1 [∵ p – q = 1 ∴ p = q + 1] 4. (b) No. of sample space = 6 × 6 = 36 Hence, the length of the third side is Sum total of 9 = (3, 6), (4, 5), (5, 4), (6, 3) 2q +1 cm. ∴ P= 4 = 1 10. (c) Given, 5. (d) 36 9 6. (c) Two circle each of radius is 2 and difference between their 7. (d) Given : 13 tan q = 12 ⇒ tan θ=1123 centre is 2 3 2sin θ.cos θ AB = 2 3 1 cos2 θ − sin2 θ ⇒ AC = 2 AB Now given expression is,

Solutions S-27 AC = 3 = CB 17. (a) S and T trisect the side QR. P Let QS = ST = TR = x units Let PQ = y units In right ∆PQS, PS2 = PQ2 + QS2 2 2 qC Ö3 B (By Pythagoras Theorem) A Ö3 = y2 + x2 In right ∆PQT, PT2 = PQ2 + QT2 ...(i) (By Pythagoras Theorem) = y2 + (2x)2 = y2 + 4x2 Q ...(ii) AC 3 In right ∆PQT, PR2 = PQ2 + QR2 AP 2 In ∆APC, cos θ = = (∠C = 90°) (By Pythagoras Theorem) = y2 + (3x)2 = y2 + 9x2 ⇒ θ = 30° R.H.S. = 3PR2 + 5PS2 ...(iii) We know, = 3(y2 + 9x2) + 5(y2 + x2) [From (i) and (iii)] Area of common region = 3y2 + 27x2 + 5y2 + 5x2 = 8y2 + 32x2 = 2 (Area of sector – Area of ∆APQ) = 8(y2 + 4x2) = 8PT2 = L.H.S. [From (ii)] Thus 8PT2 = 3PR2 + 5PS2  60° π(2)2 1 (2)2 60° 18. (a) Unit digit in (795) = Unit digit in [(74)23 × 73] = 2  360° × − 2 × × sin = Unit digit in 73 (as unit digit in 74 = 1)  4π 4 3 2  2 (3.14) (1.73) = Unit digit in 343 2  6 4   3 Unit digit in 358 = Unit digit in (34)4 × 32 = − = − = 2 (2.09 – 1.73) = 2 (0.36) = 0.72. [as unit digit 34 = 1] = Unit digit is 9 So, unit digit in (795 – 358) ∴ Area of region lie between 0.7 and 0.75. 11. (a) = Unit digit in (343 – 9) = Unit digit in 334 = 4 Unit digit in (795 + 358) = Unit digit in (343 + 9) 12. (b) (a) is not true [By def.] (b) holds [Q degree of a zero polynomial is not defined] = Unit digit in 352 = 2 (c) is not true [Q degree of a constant polynomial is ‘0’] So, the product is 4 × 2 = 8 19. (d) In (a) power of x is –1 i.e. negative (d) is not true \\ (a) is not true. [Q a polynomial of degree n has at most n zeroes]. (2 + 2sin θ) (1− sin θ) 2(1+ sin θ) (1 − sin θ) 1 In (b) power of x = 2 , not an integer. 13. (b) (1+ cos θ) (2 − 2 cos θ) = (1+ cos θ) (2) (1− cos θ) \\ (b) is not true In (c) Here also power of x is not an integer = 2(1 − csoins22=θθ)) 22csoins22=θθ cot=2 θ  15 2 = 225 \\ (c) is not true 2(1 −  8  64 (d) holds [Q all the powers of x are non-negative 14. (a) integers.] 20. (d) We have, sin 5θ = cos 4θ 15. (d) All the statements given in option ‘a’, ‘b’ and ‘c’ are ⇒ 5θ + 4θ = 90° [Q sin α = cos β, than α + β = 90°] ⇒ 9θ = 90° ⇒ θ = 10° correct. 16. (a) A Now, 2 sin 3θ – 3 tan 3θ r = 2sin 30° – rO = 2 × 1 − 3 × 1 = 1−1= 0 r 2 3 BC 21. (c) 22. (d) circumference of circle = 2pr ...(i) 23. (c) In ∆PAC and ∆QBC, We have Area of DABC = [ar(DAOB) + ar(DBOC) + ar(DAOC)] ∠PAC = ∠QBC [Each = 90°] = 1 AB × r + 1 × BC × r + AC × r ∠PCA = ∠QCB [Common] 2 2 \\ ∆PAC ~ ∆QBC = 1 r [AB + BC + AC] = 1 r × 7p...(ii) \\ x AC i.e. y = BC ...(i) 2 2 y= BC x AC From (i) and (ii), Similarly z = AC i.e. y = AB ...(ii) y AB z AC CircAmrefaeroefncteriaonfgcliercle 12r2r7 4 7 Adding (i) and (ii), we get BCA+CAB = y + y = y  1 + 1  x z  x z 

S-28 Mathematics AACC = y  1 + 1  ⇒1= y  1 + 1  37. (c)  x z   x z  38. (b) A die is thrown once therefore, total number of ⇒ 1 = 1 + 1 outcomes are {1, 2, 3, 4, 5, 6} y x z (a) P(odd number) = 3/6 = 1/2 (b) P(multiple of 3) = 2/6 = 1/3 24. (c) On adding both the equations, we get x = 3, y = 1 (c) P(prime number) = 3/6 = 1/2 (d) P(greater than 5) = 1/6 25. (b) A(2 – 2), B( – 1, x), AB = 5 39. (a) (By definition of similar triangles). ⇒ AB2 = 25 40. (d) Radius of the circle is 13/4 ⇒ ( – 1–2)2 + (x + 2 )2 = 25 ⇒ 9 + x2 + 4x + 4 = 25 Distance between (0, 0) and  − 3 , 1 is ⇒ x2 + 4x – 12 = 0  4 ⇒ x2 + 6x – 2x – 12 = 0 ⇒ x (x + 6) – 2(x + 6) = 0 2  ⇒ (x – 2) (x + 6) = 0  0 + 3 + (0 − 1)2= 9 +1 4 16 ⇒ x = 2, –6 26. (b) As 1 radian = 1 degree × 180° 1265= 5 13 π 4 4 = < 2π 2π 180° \\ 3 radian =  3 × π  Distance between (0, 0) and  2, 7 is    3  120 2 \\ Time = 6 = 20 min. 7  49 85 ≠so32lut,iosno,toinbfeiniintfeinsiotel,ut−i6=ocn −=21 −2 (2 − 0)2 +  3 − 0 = 4+ 9 = 9 don’t −3  27. (d) For must satisfy. but −1 13 2 exist, for given = 3.073 < 4  −1   2  Distance between (0, 0) and 3, is, equations. 28. (d) All the statements given in option (a, b, c) are correct. (3 − 0)2 +  −1 − 0 2 = 9 + 1 29. (c) Let the coordinate of other end be B(10, y) Given  2  4 point is A(2, –3) 13 AB = 10 ⇒ AB2 = 100 = 3.041 < 4 ⇒ (10 – 2)2 + (y + 3)2 = 100 ⇒ y2 + 6y – 27 = 0 Distance between points (0, 0) and  −6, 5  is ⇒ (y + 9) (y – 3) = 0  2  ⇒ y = –9, 3 30. (b) The probability of an event can never be negative. (−6 − 0)2 +  5 − 0 =2 36 + 245= 169 31. (a) Given, sin A + sin2A = 1  2 4 ⇒ sin A = 1 – sin2A = cos2 A Consider, cos2A + cos4A = sinA + (sin A)2 = 1 32. (a) 12=3 6.5 13 4  7 2 π(49) = 49 cm2. = >  π  = 33. (d) Area of the circle = π π 154 154 × 7 41. (c) AB = (2.4)2 + (1.8)2 = 3m. π 22 Now, consider = = 49 cm2 42. (b) CD = 3.6 – 2.4 = 1.2 m 43. (a) Q DABC ~ DAEF 34. (b) Coefficient of all the terms are positive. So, both AE roots will be negative. \\ AACB = AF 35. (b) Let (x, y) be the point which will be collinear with 1.8 0.9 the points (–3, 4) and (2, –5) ⇒ 3 = AF ⇒ AF = 1.5 m ∴ x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0 44. (d) D ⇒ x (4 + 5) – 3(–5 – y) + 2 (y – 4) = 0 S ⇒ 9x + 15 + 3y + 2y – 8 = 0 45. (a) Time = = 300 = 60 sec = 1 min. ⇒ 9x + 5y = –7 5 By plotting the points given in the options we find 46. (d) 47. (c) 48. (d) that (7, –14) satisfies it. 49. (a) 50. (b) 36. (c) cos θ= 1 − sin2 θ= 1 − a2 = b2 − a2 b2 b

Sample Paper 8 ANSWERKEY 1 (c) 2 (d) 3 (d) 4 (d) 5 (b) 6 (d) 7 (c) 8 (a) 9 (c) 10 (c) 11 (d) 12 (d) 13 (c) 14 (a) 15 (a) 16 (b) 17 (b) 18 (b) 19 (c) 20 (d) 21 (b) 22 (a) 23 (b) 24 (d) 25 (a) 26 (b) 27 (d) 28 (b) 29 (b) 30 (a) 31 (a) 32 (d) 33 (b) 34 (d) 35 (a) 36 (a) 37 (b) 38 (c) 39 (d) 40 (b) 41 (d) 42 (a) 43 (d) 44 (c) 45 (b) 46 (a) 47 (b) 48 (a) 49 (c) 50 (a) 1. (c) Let the speed of the boat in still water be x km/hr and = (-2)2 +(2)2 the speed of the stream be y km/hr then speed of boat in downstream is (x + y) km/hr and the speed of boat = 4 + 4 = 8 = 2 2 upstream is (x – y) km/hr. 3. (d) isosceles and similar Ist case : Distance covered upstream = 12 km 4. (d) Let us first find the H.C.F. of 210 and 55. \\ time = 12 hr Applying Euclid’s division lemma on 210 and 55, we x− y get Distance covered downstream = 40 km 210 = 55 × 3 + 45 ..... (i) 40 \\ time = x+ y hr Since, the remainder 45 ≠ 0. So, we now apply division lemma on the divisor 55 and the remainder 45 to get 12 40 =8 55 = 45 × 1 + 10 ..... (ii) x− y x+ y Total time is 8 hr \\ + ...(i) We consider the divisor 45 and the remainder 10 and apply division lemma to get IInd case : 45 = 4 × 10 + 5 ..... (iii) Distance covered upstream = 16 km We consider the divisor 10 and the remainder 5 and 16 apply division lemma to get \\ time = x − y hr 10 = 5 × 2 + 0 ..... (iv) Distance covered downstream We observe that the remainder at this stage is zero. 32 So, the last divisor i.e., 5 is the H.C.F of 210 and 55. = 32 km \\ time = x + y hr \\ 5 = 210 × 5 + 55y ⇒ y= −1045 = −19 55 Total time taken = 8 hr 5. (b) There are a total of six digits (1, 2, 2, 3, 4, 6) 16 32 \\ x− y + x+ y =8 ...(ii) out of which four are even (2, 2, 4, 6) Solving (i) and (ii), we get, So, required probility = x = speed of boat in still water = 6 km/hr, y = speed of stream = 2 km/hr 6. (d) (2. (d) A 3 +1, 2 -1),B( 3 -1, 2 +1) (a) It is quadratic polynomial [∵ the graph meets the x-axis in two points] ( ) ( ) AB = 3 -1 – 3 -1 2 + 2 +1- 2 +1 2 (b) It is a quadratic polynomial [∵ the graph meets the x-axis in two points]

S-30 Mathematics (c) It is a quadratic polynomial 12. (d) Let the y-axis divides the line segment joining (4, 5) [∵ the graph meets the x-axis in two points] and (– 10, 2) in the ratio k : 1. (d) It is a not quadratic polynomial [∵ the graph meets the x-axis in one point] x coordinate will be zero on y-axis. 7. (c) Let ABCD be a square and two opposite vertices of We know that the coordinates of the point dividing it are A(–1, 2) and C(3, 2). ABCD is square. D C(3, 2) the line segment joining (x1, y1) and (x2, y2) in the mx 2 nx1 m y2 + ny1 ratio m : n are given by  m + n , m +n   +  Here, x coordinate of the point dividing the line segment joining (4, 5) and (– 10, 2) is equal to zero. So, k × (−10) +1× 4 =0 ⇒ – 10k = – 4 ⇒ k = 2 5 k +1 Therefore, the line segment joining (4, 5) and (– 10, 2) is cut by the y-axis in the ratio 2 : 5. A (–1, 2) B(x, y) 13. (c) There are 4 cards of king and 4 cards of Jack n(S) = ⇒ AB = BC 52, n(E) = 4 + 4 = 8 ⇒ AB2 = BC2 ⇒ (x + 1)2 + (y – 2)2 = (x – 3)2 + (y – 2)2 P(E=) nn((ES=)) 58=2 2 ⇒ x2 + 2x + 1 = x2 – 6x + 9 13 ⇒ 2x + 6x = 9 – 1 = 8 14. (a) The graph of y = ax2 + bx + c is a parabola open upward ⇒ 8x = 8 ⇒ x = 1 if a > 0. So, for y = x2 – 6x + 9, a = 1 > 0, the graph ABC is right ∆ at B, then is a parabola open upward. AC2 = AB2 + BC2 (Pythagoras theorem) ⇒ (3 + 1)2 + (2 – 2)2 15. (a) If 1, 1 and 2 are sides of a right triangle then sum of = (x + 1)2 + (y – 2)2 + (x – 3)2 + (y – 2)2 ⇒ 16 = 2(y – 2)2 + (1 + 1)2 + (1– 3) 2 squares of any two sides is equal to square of third ⇒ 16 = 2(y – 2)2 + 4 + 4 ⇒ (y – 2)2 = 4 ⇒ y – 2 = ± 2 side. ⇒ y = 4 and 0 Case 1 (1)2 + (1)2 = 2 ≠ (2)2 i.e., when x = 1 then y = 4 and 0 Case 2 (1)2 + (2)2 = 1 + 4 = 5 ≠ (1)2 Coordinates of the opposite vertices are : Case 3 (2)2 + (1)2 = 5 ≠ (1)2 B(1, 0) or D(1, 4) 16. (b) n(S) = 6 × 6 = 36 8. (a) E = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), 9. (c) In right angled triangle POR PR2 = PO2 + OR2 = (6)2 + (8)2 = 36 + 64 = 100 (5, 4), (5, 6), (6, 5)} ∴ PR = 10 cm n(E) = 10 Again in right angled triangle P(E=) nn((ES=)) 13=06 5 PQR, QR2 = (26)2 = 676 18 PQ2 + PR2 = (24)2 + (10)2 = 576 + 100 = 676 ∴ QR2 = PQ2 + PR2 17. (a) 2 is not a rational number. It can’t be expressed in ∴ ∆PQR is a right angled triangle with right angle the fractional form. 30° 49π 360° 12 at P. 18. (b) Area = θ × πr 2 = × π (7)2 = 360° i.e., ∠QPR = 90° 19. (c) Inconsistent system 20. (d) S = {1, 2, 3, 4......, 25} n(S) = 25 E = {2, 3, 5, 7, 11, 13, 17, 19, 23} n(E) = 9 \\ P(E) = 9 25 10. (c) πr12 = 25 ⇒ r1 = 5 ⇒ 22ππrr12 = 5 5 ×125 = 625 1 +=β13 α(3α+β=)β33 3abc − b3 πr22 16 r2 4 4 = 4 ×125 500 α3 a3 21. (b)  c 3 11. (d) For any rational number p , where prime factorization  a  q =3abcc3− b3 of q is of the form 2n.5m, where n and m are non- ⇒ 1 + 1 negative integers, the decimal representation is α3 β3 terminating.

Solutions S-31 22. (a) S = {S, M, T, W, Th, F, Sa} \\= x 5=1×012 6 n(S) = 7 A non-leap year contains 365 days, Substituting x = 6 in (i) i.e., 52 weeks + 1 day. 3 4 =1 ⇒=4y 1=– 12 1 6 y 2  E = {Sa} ⇒ + \\ y = 8. n(E) = 1 \\ P=(E) nn=((ES)) 1 Hence, x = 6 and y = 8 7 27. (d) 23. (b) Let the given points be A(4, 3) and B(x, 5) 28. (b) n(S) = 6× 6 = 36, E = {(1, 5), (2, 5), (3, 5), (4, 5), (5, Since A and B lies on the circumference of a circle 5), (6, 5), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)} with centre O(2, 3), we have n(E) = 12 OA = OB P(E=) nn((ES=)) 13=26 1 3 ⇒ OA2 = OB2 ⇒ (4 – 2)2 + (3 – 3)2 = (x –2)2 + (5 – 3)2 29. (b) Put x + 1 = 0 or x = – 1 and x + 2 = 0 or x = – 2 in p (x) ⇒ 4 + 0 = (x – 2)2 + 4 Then, p (–1) = 0 and p (–2) = 0 ⇒ p (–1) = ⇒ (x – 2)2 = 0 ⇒ x = 2 ⇒ −1+ 3 + 2α + β = 0 ⇒ β = −2α − 2 .... (i) 24. (d) Since 1123=5 15=33 13 (253)=3 110004=0 0.104 p(−2) = (−2)3 + 3(−2)2 − 2α(−2) + β = 0 (2)3 25. (a) ⇒ −8 +12 + 4α + β = 0 ⇒ β = − 4α − 4 .... (ii) A By equalising both of the above equations, we get −2α − 2 = − 4α − 4 ⇒ 2α = −2   ⇒ α = −1 put a in eq. (i) ⇒ β =−2 (−1) − 2 =2 − 2 =0 Hence, α = −1, β = 0 30. (a) Let D be the window at a height of 9m on one side of the street and E be the another window at a height of 12 m on the other side. BD C In DADC Given : A DABC in which ∠B = 90° and D is the AC2 = 152 – 92 = 225 – 81 midpoint of BC. AC = 12 m Join AD. In DECB In DABC, ∠B = 90°. CB2 = 152 – 122 = 225 – 144 \\ AC2 = AB2 + BC2 ....(i) CB = 9 m  [by Pythagoras’ theorem] Width of the street = (12 + 9)m = 21 m In DABD, ∠B = 90° 31. (a) \\ AD2 = AB2 + BD2 ....(ii) 32. (d) All equilateral triangles are similar  [by Pythagoras’ theorem] \\ D ABC~ D EBD ⇒ AB2 = (AD2 – BD2). ⇒ Area of ABC  BC2 Area of BDE BD2 \\ AC2 = (AD2 – BD2) + BC2 [using (i)] ⇒ AC2 = AD2 – CD2 + (2CD)2 A [∵ BD = CD and BC = 2CD] E   2BD2  4 BD2 1 ⇒ AC2 = AD2 + 3CD2 Hence, AC2 = AD2 + 3CD2 ⇒ Area (DABC) : Area (DBDE) =4: 1 26. (b) 3 + 4 =1 ...(i) 4 + 2 =1121 ...(ii) B DC x y x y 33. (b) As A lies on x-axis and B lies on y-axis, so their Multiplying (ii) by 2 coordinates are (x, 0) and (0, y), respectively. Then, 8 4 =1222 ⇒ x + y ...(iii) x+0 = 4 and 0+y = –3 ⇒ x = 8 and y = – 6 2 2 Subtracting (i) from (iii)  ⇒ 5 = 10 Hence, the points A and B are (8, 0) and (0, –6). x 12

S-32 Mathematics 34. (d) If 6x ends with 5, then 6x would contain the prime 5. ⇒ 6. 64 −11.196 + 4k − 20 =0 But 6x = (2 × 3)x = 2x × 3x. 27 3 ⇒ The only prime numbers in the factorization of 6x ⇒ 128 – 176 + 12k – 180 = 0 are 2 and 3. ⇒ 12k + 128 – 356 = 0 12 k = 228 ⇒ k = 19 38. (c) For coincident lines,=63 ––=k1 8 \\ By uniqueness of fundamental theorem, there are 16 no primes other than 2 & 3 in 6x. So, 6x will never end 12 = 1 with 5. k θ × π=r2 60° × 22 × (=6)2 132 cm2 ⇒ k=2 360° 360° 7 7 35. (a) (a) Area = (b) Area of minor sector = θ × πr 2 39. (d) Let the line x + y =1 meet x-axis at P(a, 0) and 360° a b y-axis at Q(0, b). Since R is mid point at PQ. = 60° × 22 × 14 × 14 = 102.57 cm2 Q (0, b) 360° 7 Area of major sector = Area of circle – Area of minor sector R(2, –5) = 22 (14)2 –102.57 7 = 615.44 – 102.57 = 512.87 cm2 O (0, 0) P (a, 0) (c) CA = 2π ( 5) = 2 a + 0 = 2, 0 + b = -5 π (5)2 5 2 2 \\  θ  \\ a = 4, b = –10 \\ P is (4, 0), Q is (0, –10)  360°  (d) Given, 2πr =22 40. (b) =4251 9=2×15 21 32 × 5  θ  = 36θ0°  πr ∴ Area of sector =  360°  πr 2  2 (2r) Clearly, 45 is not of the form 2m × 5n. So the decimal  θ   r  22 × 6 expansion of 21 is non-terminating and repeating.  360°   2  2 45 = 2πr = = 66 cm2 41. (d) Sample space = {HH, HT, TH, TT} 36. (a) Let AD = 5x cm and DB = 4x cm. Total number of elementary events = 4 Then, Favourable event E = HH AB = (AD + DB) = (5x + 4x) cm = 9x cm. n (E) = 1 In DADE and DABC, we have 1 ∠ADE = ∠ABC (corres. ∠s) P(E) = 4 42. (a) Favourable event E = {TH, HT} ∠AED = ∠ACB (corres, ∠s) n(E) = 2 \\ DADE ~ DABC [by AA-similarity] 21 ⇒ DE = AD = 5x = 5 P(E) = 4 = 2 BC AB 9x 9 43. (d) Favourable event E = {TT} ⇒ DE = 5 ...(i) n(E) = 1 BC 9 1 In DDFE and DCFB, we have P(E) = 4 ∠EDF = ∠BCF (alt. int. ∠s) 44. (c) At most one head = {HT, TH, TT} ∠DEF = ∠CBF (alt. int. ∠s) 3 4 \\ DDFE ~ DCBF P= ⇒ ar (∆DFE) = DE2 = DE2 =  DB=CE 2 =95 2 25 45. (b) At least one head ar (∆CFB) CB2 BC2  81 {HH, HT, TH} ⇒ ar (DDFE) : ar (DCFB) = 25 : 81 3 37. (b) Let f (x) = 6x3 – 11x2 + kx – 20 P= 4 f  34= 6  4 3 −11  4 2 + k  4  − =20 0 46. (a) 47. (b) 48. (a)   3   3   3  49. (c) 50. (a)