EAD Mathematics -10 Sample Paper 1

sample Paper 1 [Easy Concept] Time Allowed : 3 hours] [Maximum Marks : 80 General Instructions: 1. This question paper contains two parts A and B. 2. Both Part A and Part B have internal choices. Part – A: 1. It consists three sections I and II. 2. Section I has 16 questions of 1 mark each. Internal choice is provided in 5 questions. 3. Section II has 4 questions on case study. Each case study has 5 case-based sub-parts. An examinee is to attempt any 4 out of 5 sub-parts. Part – B: 1. Question No 21 to 26 are Very short answer Type questions of 2 mark each 2. Question No 27 to 33 are Short Answer Type questions of 3 marks each 3. Question No 34 to 36 are Long Answer Type questions of 5 marks each. 4. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks. Part – A Section – I Section I has 16 questions of 1 mark each. Internal choice is provided in 5 questions. 1. Express 5005 as a product of its prime factor. OR The product of two numbers is 3691 and their LCM is 3691. Find the HCF. 2. Determine the roots of the equation x2 – 3x – m(m + 3) = 0, where m is any constant. 3. Find whether the following pairs of linear equations 2x – 3y = 1 and 3x – 2y = 4 have unique solution, no solution or infinitely many solution. 4. In an A.P., a = 3.5, d = 0 and n = 101, find an 5. Find the number of zeroes of given polynomial f(x) = (x – 2)2 + 4 OR Determine the zeroes of the polynomial z2 – 27. 6. Which term of the AP 14, 11, 8, ... is – 1. 7. Find the discriminant of the given quadratic equation 3 3 x2 + 10x + 3 = 0 8. In DABC and DDEF, AB = BC . At what possible condition DABC ~ DEDF DE FD 9. Draw a circle and two lines parallel to a given line such that one is a tangent and other is secant to a circle. 10. How many number of tangents to a circle can be drawn which are parallel to a secant of a circle? 11. A line segment AB = 8 cm is divided in the ratio 3 : 2 by a ray at point P lies on AB. Find the distance of point P from A. 12. At sometime, the length of a shadow of a tower is 3 times its height. Find the angle of elevation of the sun at that time. Sample Papers 49

13. For what value of q, cot 3q = cot(30° + q) OR Find the value of q, where 3 sin q – cos q = 0 and 0 < q < 90°. 14. The surface area of a sphere is 616 cm2. Find the radius of the sphere. 15. Find the area of a quadrant of a circle whose circumference is 44 cm. OR Find the perimeter of a square circumscribing a circle of radius x cm. 16. 2000 tickets of a lottery were sold and there are 16 prizes on these tickets. Abhinav purchased one lottery ticket. What is the probability that Abhinav wins a prize? OR A pair of dice is thrown once. What is the probability of getting a same number on each dice. Section – II Case study based questions are compulsory. Attempt any 4 sub parts of each question. Each subpart carries 1 mark. Case Study based - 1 Floor Design 17. A floor designed is made by 2-dimensional geometrical figures as shown on coordinate axes. y (a′, 2y′) A′ A(a, 2y) x′ P(a′, b′) O (a, b)P B(2x, 0) (2x′, 0)B′ y′ Coordinates of points are also represented. Here A′P′B′ is mirror image of APB along y axis. Answer the following questions based on above diagram. (a) Determin coordinates of mid point of PP′. (i) a , b (ii) a2l, bl (iii) (0, 0) (iv) a + b , al + bl 2 2 2 2 2 (b) The distance of point P from y-axis is. (i) a′ (ii) b′ (iii) b (iv) a (c) Distance PP′ is (i) (a – al)2 + (b – bl)2 (ii) a2 + b2 (iii) al2 + bl2 (iv) 0 (ii) (a – al)2 + (2y – 2yl)2 (d) The distance AA′ is (i) (a2 – al2) + (2y)2 (iii) (2y – 2yl)2 + al2 (iv) a2 + al2 + 2y2 + 2yl2 (e) The distance between the points BB′ is. (i) 2x2 + 2xl2 (ii) 4x2 + 4xl2 + 8xxl (iv) (x – xl)2 (iii) (2x – 0)2 + (2xl + y)2 50 Together with®  EAD Mathematics—10

Case Study based - 2 Ratio of the sides of similar triangle 18. A triangle PQR with sides 3 cm, 4 cm and 5 cm is shown. Triangle A, triangle B and triangle C are all similar to triangle DPQR. 5 cm P 8 cm4 cm 5 cm A B C R 3 cm Q 1.5 (a) Which ratio can be used to determine the base of triangle B? (i) 5 : 3 (ii) 3 : 4 (iii) 4 : 5 (iv) 5 : 5 (iv) 4 : 4 (b) Which ratio can be used to determine the hypotenuse of triangle B? (i) 5 : 4 (ii) 5 : 3 (iii) 3 : 4 (c) Which ratio can be used to determine the hypotenuse of triangle A? (i) 5 : 4 (ii) 4 : 3 (iii) 5 : 3 (iv) 4 : 5 (d) The unknown lengths of triangle A are (i) 2 cm, 2 cm (ii) 2.5 cm, 2 cm (iii) 2.5 cm, 2.5 cm (iv) 1 cm, 2 cm (e) The unknown length of triangle C are (i) 6.4 cm, 4.8 cm (ii) 6.8 cm, 4.4 cm (iv) 4 cm, 2 cm (iii) 6 cm, 2.4 cm Case Study based - 3 application of linear equations 19. A cab company charges ` 200 boarding rate in addition to this they also charges ` 10 for travelling every kilometre (a) What linear equation represents the rate of this company? (i) y = 200 + 10 (ii) y = 200x + 10 (iii) y = 10x + 200 (iv) y = 12x + 1200 where ‘x’ is distance travelled in km and ` y is total charges. (b) What are the charges, if a person travels 12 km in day (i) ` 320 (ii) ` 372 (iii) ` 720 (iv) ` 1272 (c) The graphical representation of given situation is. (i) y (ii) y 600 600 ` 400 ` 400 200 200 x x 10 20 30 10 20 30 Distance Distance Sample Papers 51

(iii) y (iv) y 600 600 ` 400 ` 400 200 200 x x 10 20 30 10 20 30 Distance Distance (d) For the given linear equation y = 10x + 200, which of the following linear equations in two variables represents intersecting lines. (i) x – 10y + 200 = 0 (ii) 10x – y + 200 = 0 (iii) y – 200 = 10 x (iv) 10x – y = – 200 (e) For the given linear equation y = 10x + 200; which of the following linear equation in two variables represents parallel lines (i) x – 10y + 200 = 0 (ii) – x + 10y – 200 = 0 (iii) 10x – y + 200 = 0 (iv) 10x + 10y + 200 = 0 Case Study based - 4 SpINNING aN ARROW 20. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. 8 1 7 2 6 3 5 4 (a) What is the probability that it will point at 8? (i) 8 (ii) 1 (iii) 1 (iv) 7 8 8 (b) What is the probability that it will point at an odd number? (i) 2 (ii) 3 (iii) 1 (iv) 1 4 4 2 (c) What is the probability that it will point at a number greater than 8? (i) 0 (ii) 1 (iii) sure event (iv) 1 2 (d) What is the probability that it will point at a number less than 9? (i) 0 (ii) 1 (iii) impossible event (iv) 1 2 (e) What is the probability that it will point at 0? (i) 1 (ii) 0 (iii) 1 (iv) 1 2 3 Part – B All questions are compulsory. In case of internal choices, attempt any one. 21. Find the LCM and HCF of 840 and 144 by applying the Fundamental Theorem of Arithmetic. OR Find the prime factors of 15300 using the factor tree. 52 Together with®  EAD Mathematics—10

22. Let P(x, y) be any point on the line joining the points A(a, 0) and B(0, b). Show that x + y = 1. a b 23. If tan θ = 1 , then evaluate cosec2θ − sec2θ 3 cosec2θ + sec2θ OR If sin (A – B) = 1 and cos(A + B) = 1 , Find A and B. 2 2 24. 5 books and 7 pens together cost ` 79, whereas 7 books and 5 pens together cost ` 77. Represent this situation in the form of linear equation in two variables. Find the cost of one book and one pen. 25. In the given figure, O is the centre of the circle. The radius of the circle is 3.1 cm and PA is a tangent drawn to the circle from point P. If OP = x cm and AP = 6.2 cm, then find the value of x. A 6.2 cm P x cm 3.1 cm O 26. AB is a tangent drawn from a point A to a circle with centre O and BOC is a diameter of the circle such that ∠AOC = 110°. Find ∠OAB. 27. Two sets of Maths and Science books containing 1680 and 1056 books respectively in a library have to be stacked in such a way that all the books are stored subjectwise and the height of each stack is the same. Assuming that the books are of the same thickness, determine the total number of stacks. 28. Find the 11th term from the end in the AP 56, 63, 70, ….., 329. OR If the 3rd and the 6th terms of an AP are 7 and 13 respectively, find its 10th term. 29. A solid metallic sphere of radius 8 cm is melted and recast into eight equal solid spherical balls. Determine the diameter of the balls. 30. The lengths of tangents drawn from an external point (point outside the circle) to a circle are equal. Prove it. OR ABC is an isosceles triangle, in which AB = AC, circumscribed about a circle. Show that BC is bisected at the point of contact. 31. A quiz was conducted by a group of students as a part of their environment awareness programme. The marks obtained by them is represented by the data shown below. Construct the frequency distribution table for the data represented. Find the modal class of the data. Marks Number of students Less than 10 12 Less than 20 24 Less than 30 37 Less than 40 56 Less than 50 69 Less than 60 76 Sample Papers 53

32. Simplify: (1 + tan q + sec q)(1 + cot q – cosec q) 33. The following table shows the distribution of mineral water bottles according to the number of   boxes. The boxes contains varying number of mineral water bottles. Number of mineral water bottles Number of boxes 50 – 52 20 53 – 55 120 56 – 58 105 59 – 61 125 62 – 64 30 Find the mean number of mineral water bottles kept in a packing box. 34. Two men standing on either side of a cliff 90 m high, observes the angles of elevation of the top of the cliff to be 30° and 60° respectively. Find the distance between the two men. 35. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. How many litres of water is left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. (Use p = 22 ) 7 OR The cost of fencing a circular field at the rate of ` 24 per m is ` 5280. The field is to be ploughed at the rate of ` 0.50 per m2. Find the cost of ploughing the field. (Use p = 22 ) 36. Solve the following pair of linear equations graphically. 7 x + 3y = 6 and 2x – 3y = 12 Solutions 5 5005 7 1001 1. 5005 in prime factors expressed as 13 143 5005 = 5 × 7 × 11 × 13 11 OR Product of two numbers = 3691 LCM of these numbers = 3691 We know HCF = Product of two numbers LCM = 33669911 = 1 2. x2 + mx – (m + 3)x – m(m+3) = 0 x(x + m) – (m + 3)(x + m) = 0 (x + m)[x – (m + 3)] = 0 x + m = 0 ⇒ x = – m or x – (m + 3) = 0 or x = m + 3 Roots are – m and (m + 3) 54 Together with®  EAD Mathematics—10

3. (a) The given equations are 2x – 3y = 1 and 3x – 2y = 4 Here a1 = 2 , b1 = –3 = 3 ⇒ aa21 ≠ 3 b2 –2 2 a2 b1 b2 The given pair of linear equations has unique solution. 4. In given AP an = a + (n – 1)d Where a is 1st term, d is common difference and n is number of terms So, an = 3.5 + (101 – 1)0 = 3.5 + 0 = 3.5 5. The given polynomial is f(x) = (x – 2)2 + 4 for zeroes, put f(x) = 0 ⇒ (x – 2)2 + 4 = 0 ⇒ (x – 2)2 = –4 Which is not possible, Hence this polynomial has no zeroes. OR We have z2 – 27 For zeros, put z2 – 27 = 0 ⇒ z2 – _3 3 i2 = 0 ⇒ _ z – 3 3 i_ z + 3 3 i = 0 ⇒ z – 3 3 = 0 or z + 3 3 = 0 ⇒ z = 3 3 or z = – 3 3 Hence zeroes of the polynomial are 3 3 and –3 3 6. AP is 14, 11, 8 ... Here a = 14, d = 11 – 14 = – 3 Let an = – 1 an = a + (n – 1)d \\ –1 = 14 + (n – 1)(– 3) –1 = 14 – 3n + 3 – 1 = 17 – 3n ⇒ n = 6 Hence 6th term is – 1 7. The given quadratic equation is 3 3 x2 + 10x + 3 = 0 For D = (10)2 – 4_3 3 i_ 3 i = 100 – 36 = 64 8. In DABC and DDEF A D AB = BC Also DABC ~ DEDF DE DF This is possible when ∠B = ∠D. B CE F Sample Papers 55

9. l′ (tangent) l (secant) diameter l′ || l and both are parallel to diameter. 10. Two tangents can be drawn which are parallel to secant of a circle. 11. Here AB = 8 cm P divides AB in the ratio 3 : 2 AP = 3 AB = 3 (8) = 24 = 4.8 cm \\ 5 5 5 Hence AP = 4.8 cm Distance of point P from A is 4.8 cm 12. Here AB is tower of height h m. Its shadow BC = 3 h m A q be the angle of elevation \\ In DABC, tan q = AB = h =1 hm BC 3h 3 tan q = tan 30° ⇒ q = 30° q C 13. We have cot 3q = cot (30° + q) 3 hm B ⇒ 3q = 30° + q ⇒ 2q = 30° q = 15° OR 3 sin q – cos q = 0 ⇒ 3 sin q = cos q sin q = 1 cos q 3 tan q = 1 3 = tan 30° ⇒ q = 30° 14. Let radius of sphere be a cm \\ Surface area of sphere = 4pa2 \\ \\ 4 × pa2 = 616 4 × 22 a2 = 616 ⇒ 7 616 × 7 a2 = 22 × 4 = 49 a = 7 cm 56 Together with®  EAD Mathematics—10

15. Circumference of the circle = 44 cm ⇒ 2pr = 44 cm ⇒ r = 44 × 7 = 7 cm \\ 2 × 22 Area of the quadrant of a circle = 14 πr2 = 1 × 22 × 7 × 7 = 77 cm2 4 7 2 OR A square circumscribing a circle of radius x cm \\ side of square = diameter of circle Side of square = 2(x) = 2x O x Perimeter square = 4(2x) = 8x 16. Number of lottery tickets = 2000 Total number of prizes = 16 \\ Probability that Abhinav wins a prize = 16 = 1 = 0.008 2000 125 OR A pair of dice thrown once, possible outcomes for same number on dice. (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6) Number of favourable cases = 6 Total possible outcomes = 36 \\ Required probability = 6 = 1 36 6 17. (a) (iii) coordinates of P′(a′, b′) and P(a, b) P′ is mirror image of P along y-axis Also, P′ and P lies on x-axis. \\ Origin O is mid point of PP′ \\ Coordinates of origin are (0, 0) Option (iii) is correct. (b) (iv) distance of point P from y-axis is x-coordinate i.e. a Option (iv) is correct. (c) (i) Distance PP′ = (a – al)2 + (b – bl)2 Option (i) is correct. (d) (ii) Distance AA′ = (a – al)2 + (2y – 2yl)2 Option (ii) is correct. (e) (ii) Distance between BB′ is BB′ = (2x – 2xl)2 = 4x2 + 4xl + 8xxl Option (ii) is correct. 18. (a) (ii) 3 : 4 As triangle B is similar to DPQR (b) (i) 5 : 4 (c) (iii) 5 : 3 Sample Papers 57

(d) (ii) Let unknown length of triangle A are x and y DPQR is similar to triangle A P 4 cm 5 = x 3 1.5 R 3 cm Q 5 cm x y x = 7.5 = 2.5 cm 8 cm A b 3 1.5 4 = y 3 1.5 y = 6.0 = 2 cm 3 Option (ii) is correct. (e) (i) Let unknown lengths of triangle C are a and b. P 4 cm DPQR is similar to triangle C. Q 3 = a 5 8 5 cm a = 24 = 4.8 cm 3 cm 5 C 4 = b ⇒ b = 32 = 6.4 cm a 5 8 5 R 19. (a) (iii) Given that distance travelled is x km and total charges are ` y Total charges = boarding rate + (rate per km × distance) y = 200 + 10x Possible equation is y = 10x + 200 (b) (i) Linear equation representing the situation is y = 10x + 200 Here x = 12 km y = 10(12) + 200 = 120 + 200 = ` 320 (c) (i) as fixed charges are ` 200 there far at zero distance charges are 200 at 10 km cahrges are = 200 + 10(10) = 300 at 30 km charges are = 200 + 10(30) = 500 Table is x 0 10 30 y 200 300 500 Graph (i) represents the situation (d) (i) x – 10y + 200 = 0 and y = 10x + 200 Represents intersecting lines because 1 ≠ –10 ≠ 200 10 –1 200 1 10 10 ≠ 1 ≠1 58 Together with®  EAD Mathematics—10

(e) y = 10x + 200 and 10x – y + 200 = 0 Represents parallel lines because 10 = 1 = 200 10 1 200 1 = 1 = 1 20. (a) (iii) P(8) = Favourable cases = 1 Total cases 8 (b) (iv) P(odd number) = Total odd numbers Total numbers = 4 = 1 8 2 (c) (i) it is impossible event \\ probability is zero (d) (ii) It is sure event, probability is one. (e) (ii) 0 It is impossible events there is no zero. 21. The prime factors of 840 = 2 × 2 × 2 × 3 × 5 × 7 = 23 × 3 × 5 × 7 and 144 = 2 × 2 × 2 × 2 × 3 × 3 = 24 × 32 \\ LCM (840, 144) = 24 × 32 × 5 × 7 = 5040 and HCF (840, 144) = 23 × 3 = 24 OR 15300 2 7650 2 3825 3 1275 3 425 5 85 5 17 22. Let the point P(x, y) be any point on the line segment AB. \\  P, A, B are collinear. P(x, y) \\  Area of DPAB = 0 A(a, 0) B(0, b) ⇒ 1 |x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)| = 0 2 1 ⇒ 2 |ay – ab + bx| = 0 ⇒ |ay – ab + bx | = 0 ⇒ bx + ay = ab Dividing both sides by ab, we get y x + b = 1  Hence proved. a Sample Papers 59

23. Given, tan q = 1 3 As we know that tan 30° = 1 3 ⇒ q = 30° Putting q = 30° in cosec2 θ – sec2 θ , we get cosec2 θ + sec2 θ ccoosseecc22 30° – sec2 30° (2)2 – f 22 4 – 4 8 1 30° + sec2 30° (2)2 + f 3p 3 16 2 = 22 = = = 3p 4 4 + 3 OR sin(A – B) = 21 ⇒ sin(A – B) = sin 30° A – B = 30° …(i) …(ii) cos(A + B) = 21 cos(A + B) = cos 60° ... (i) ... (ii) A + B = 60° Adding (i) and (ii), we get 2A = 90° ⇒ A = 45° Putting in (i), 45° – B = 30° B = 15° 24. Let the cost of 1 book be ` x and the cost of 1 pen be ` y According to question, 5x + 7y = 79 and 7x + 5y = 77 Solving (i) and (ii) for x and y Adding (i) and (ii) 12x + 12y = 156 x + y = 13 x = 13 – y Put in (i) 5(13 – y) + 7y = 79 65 – 5y + 7y = 79 2y = 14 y = 7 ⇒ x = 6 25. In right-angled ∆OAP, (Using Pythagoras Theorem) A 6.2 cm OP2 = OA2 + AP2 x cm 3.1 cm ⇒ x2 = (3.1)2 + (6.2)2 O ⇒ x2 = 9.61 + 38.44 P ⇒ x2 = 48.05 ⇒ x = 6.93 cm 60 Together with®  EAD Mathematics—10

26. ∠AOB + ∠AOC = 180° (linear pair) C \\ ∠AOB = 180° – ∠AOC = 180° – 110° = 70° 110° O In DAOB, ∠OBA + ∠OAB + ∠AOB = 180° B \\ 90° + ∠OAB + 70°= 180° ∠OAB = 180° – 160° A ∠OAB = 20° 27. Height of each stack is the HCF of 1680 and 1056. Prime factors of 1680 = 2 × 2 × 2 × 2 × 3 × 5 × 7 = 24 × 3 × 5 × 7 Prime factors of 1056 = 2 × 2 × 2 × 2 × 2 × 3 × 11 = 25 × 3 × 11 \\  HCF of 1680 and 1056 = 24 × 3 = 48 So, we make stacks of 48 books each. \\  Total number of stacks = 1680 + 1056 = 35 + 22 = 57 48 48 28. We know that 11th term from the end of AP 56, 63, 70, ... 322, 329 is same as 11th term from the beginning of the AP 329, 322, ... 56 Here, a = 329, d = 322 – 329 = –7 \\ a11 = a + 10d = 329 + 10 × (–7) = 329 – 70= 259 OR Here, a3 = a + 2d = 7  ...(i) and a6 = a + 5d = 13 ...(ii) Subtracting equation (i) from (ii), we get a + 5d = 13 a + 2d = 7 – – – 3d = 6  ⇒  d = 2 Putting d = 2 in (i), we get a + 2(2) = 7 ⇒ a + 4 = 7 ⇒ a = 3 Therefore, 10th term = a10 = a + 9d = 3 + 9 × 2 = 21 29. Radius of metallic sphere = 8 cm Volume of the sphere = 4 pr3 = 4 p(8)3 cm3 3 3 Let the radius of each spherical ball be x cm \\ Volume of one spherical ball = 4 p(x)3 cm3 3 4 32 \\ Volume of 8 spherical balls = 8 × 3 p(x)3 cm3 = 3 p(x)3 cm3 According to question, 32 p(x)3 = 4 p(8)3 3 3 ⇒ 32 (x)3 = 4 (8)3 3 3 ⇒ x3 = 4 × 83 × 3 3 32 ⇒ x3 = 64   ⇒  x = 4 cm \\ Diameter of spherical ball = 2x = 2 × 4 = 8 cm Sample Papers 61

30. Given: A circle C(O, r). P is a point outside the circle and PA and PB are tangents to a circle. To prove: PA = PB A Construction: Join OA, OB and OP. Proof: In DOAP and DOBP, OP ∠OAP = ∠OBP = 90° (Radius is perpendicular to the tangent at the point of contact) B OA = OB (Radii of the same circle) OP = OP (Common) \\ DOAP @ DOBP (RHS congruence rule) ⇒ PA = PB (CPCT) Hence proved. OR Given: In an isosceles DABC, AB = AC, circumscribed a circle. To prove: BD = DC Proof: Here, AB = AC  (Given) ...(i)  A AF = AE  (Tangents from an external point A to a circle are equal) ...(ii) Subtracting (ii) from (i), we get AB – AF = AC – AE F E ⇒ BF = CE ...(iii) Now, BF = BD  (Tangents from an external point B to a circle are equal) Also, CE = CD  (Tangents from an external point C to a circle are equal) B D C ⇒ BD = CD \\ BC is bisected at the point of contact. Hence proved. 31. Frequency distribution table: Class intervals Frequency 0 – 10 12 Modal class of the data is 30 – 40 10 – 20 24 – 12 = 12 20 – 30 37 – 24 = 13 30 – 40 56 – 37 = 19 40 – 50 69 – 56 = 13 50 – 60 76 – 69 = 7 32. We have (1 + tan q + sec q)(1 + cot q – cosec q) = e1 + csoinsθθ + 1 θ oe1 + cos θ – 1 θo cos sin θ sin = e cos θ + sin θ + 1 oe sin θ + cos θ – 1 o cos θ sin θ = >(sin θ + cos θ)2 – 1 sin θ cos θ H = sin2 θ + cos2 θ + 2 sin θ cos θ – 1 sin θ cos θ = 1 + 2ssininθθccoossθθ – 1 = 2ssininqqccoossqq = 2 62 Together with®  EAD Mathematics—10

33. Number of mineral Number of boxes (fi) Class marks (xi) di = xi – A fidi water bottles 49.5 – 52.5 20 51 –6 – 120 52.5 – 55.5 120 54 –3 – 360 55.5 – 58.5 105 57 = A 0 0 58.5 – 61.5 125 60 3 375 61.5 – 64.5 30 63 6 180 Total n = 400 Sfidi = 75 Let A = 57, n = 400 and Sfidi = 75 By direct method, Mean, x = A + Sfi di Sfi = 57 + 75 = 57 + 0.1875 = 57.1875 ≈ 57.19 (approx.) 400 34. Let AB be the cliff and two men are standing at P and Q respectively. A Now, AB = 90 m 90 m 60° ∠APB = 30° and ∠AQB = 60° BQ In right-angled DABP, AB BP = tan 30° ⇒ 90 = 1 ⇒ BP = 90 3 m BP 3 In right-angled DABQ, 30° AB 90 P BQ BQ = tan 60° ⇒ = 3 ⇒ BQ = 90 m 3 Now, PQ = BP + BQ = 90 3 + 90 = 270 + 90 = 360 = 120 3 m 333 35. Radius of cone = 60 cm Height of cone = 120 cm \\ Volume of cone = 1 pr2h 3 = 1 p × (60)2 × 120 3 = 144000p cm3 120 cm 180 cm Radius of hemisphere = 60 cm \\ Volume of hemisphere = 2 pr3 = 2 p × (60)3 = 144000p cm3 3 3 60 cm \\ Volume of solid = Volume of cone + Volume of hemisphere 60 cm = 144000p + 144000p = 288000p cm3 Now, Volume of cylinder = pr2h = p × (60)2 × 180 = 648000p cm3 Sample Papers 63

Volume of water left in the cylinder= Volume of cylinder – Volume of solid = 648000p – 288000p = 360000p cm3 = 360000 × 22 cm3 7 = 1131428.57 cm3 = 1131428.57 l = 1131.42 l 1000 OR ` 24, is the cost for fencing 1 m of circular field. ` 5280, is the cost for fencing = 1 × 5280 = 220 m of circular field. 24 Circumference of the field = 220 m ⇒      2pr = 220 ⇒   2× 22 × r = 220 ⇒ 7 220 × 7       r = 44 = 35 m \\ Area of the field = pr2 = p(35)2 = 1225p m2 Cost of ploughing = ` 0.50 per m2 Total cost of ploughing the field = ` 1225p × 0.50 = ` 1225× 22 ×1 = ` 175 × 11 = ` 1925 7×2 36. 1st equations is x + 3y = 6 x = 6 – 3y Y Table 5 x 6 09 4 y 0 2 –1 3 2nd equation is 2x – 3y = 12 (0, 2) 2 x = 12 + 3y 2 1 Table X′ (6, 0) – 4 –3 –2 –1 O 1 23456 X –1 x 6 30 1st equation y 0 –2 –4 Solution of equations –2 is (6, 0). –3 – 4 –5 Y′ 64 Together with®  EAD Mathematics—10