SET-II DESIGN OF THE QUESTION PAPER Mathematics Class X Time : 3 Hours Maximum Marks : 80Weightage and the distribution of marks over different dimensions of the question shall be as follows: (A)Weightage to Content/ Subject Units : (B)Weightage to Forms of Questions : (C)Scheme of Options S.No. Content Unit Marks 1. 2. 3. 4. 5. 6. 7. Number Systems Algebra Trigonometry Coordinate Geometry Geometry Mensuration Statistics and Probability 04 20 12 08 16 10 10 Total : 80 S.No. Form of Questions Marks for each Question Number of Questions Total Marks 1. 2. 3. 4. MCQ SAR SA LA 01 02 03 06 10 05 10 05 10 10 30 30 Total 30 80 All questions are compulsory, i.e., there is no overall choice. However, internal choices are provided in one question of 2 marks, three questions of 3 marks each and two questions of 6 marks each. (D) Weightage to Difficulty level of QuestionsS.No. Estimated Difficulty Level of Questions Percentage of Marks 1. 2. 3. Easy Average Difficult 206020 Note : A question may vary in difficulty level from individual to individual. As such, the assessment in respect of each will be made by the paper setter/ teacher on the basis of general anticipation from the groups as whole taking the examination. This provision is only to make the paper balanced in its weight, rather to determine the pattern of marking at any stage. BLUE PRINT MATHEMATICS CLASS X Form of Question Units MCQ SAR SA LA Total Number Systems 2(2) 2(1) --4(3) Algebra 3(3) 2(1) 9(3) 6(1) 20(8) Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmatic Progressions Trigonometry 1(1) 2(1) 3(1) 6(1) 12(4) Introduction to Trigonometry, Some Applications of Trigonometry Coordinate Geometry 1(1) 4(2) 3(1) -8(4) Geometry 1(1) -9(3) 6(1) 16(5) Triangles, Circles, Constructions Mensuration 1(1) -3(1) 6(1) 10(3) Areas related to Circles, Surface Areas and Volumes Statistics & Probability 1(1) -3(1) 6(1) 10(3) Total 10(10) 10(5) 30(10) 30(5) 80(30) SUMMARY Multiple Choice Questions (MCQ) Number of Questions : 10 Marks : 10 Short Answer Questions with Resasoning (SAR) Number of Questions : 05 Marks : 10 Short Answer Questions (SA) Number of Questions : 10 Marks : 30 Long Answer Qustions (LA) Number of Questions : 05 Marks : 30Total 30Mathematics Class X Maximum Marks : 80 Time : 3 Hours General Instructions 1. All questions are compulsory. 2. The question paper consists of 30 questions divided into four sections A, B, C, and D.Section A contains 10 questions of 1 mark each, Section B contains 5 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 5 questions of 6 marks each. 3. There is no overall choice. However, an internal choice has been provided in one question of 2 marks, three questions of 3 marks and two questions of 6 marks each. 4. In questions on construction, the drawing should be neat and exactly as per given measurements. 5. Use of calculators is not allowed. Section A 1. The largest number which divides 318 and 739 leaving remainders 3 and 4, respectively is (A) 110 (B) 7 (C) 35 (D) 105 2. The number of zeroes lying between –2 to 2 of the polynomial f (x), whose graph is given below, is (A) 2 (B) 3 (C) 4 (D) 1 3. (A) 8 (B) 64 (C) − 1 (D)33 64. If , a, 4 are in AP, the value of a is5 13 26 (A) 1 (B) 13 (C) (D)55 5. If in the following figure, Δ ABC � Δ QPR, then the measure of ∠ Q is (A) 60° (B) 90° (C) 70° (D) 50° 6. In the adjoining figure, Δ ABC is circumscribing a circle. Then, the length of BC is (A) 7 cm (B) 8 cm (C) 9 cm (D) 10 cm 7. If sinθ = 1 3 , then the value of (9 cot2 θ + 9) is (A) 1 (B) 81 (C) 9 (D) 1 81 8. The radii of the ends of a frustum of a cone 40 cm high are 20 cm and 11 cm. Its slant height is (A) 41 cm (B) 20 5 cm (C) 49 cm (D) 521 cm 9. A bag contains 40 balls out of which some are red, some are blue and remaining 11 1are black. If the probability of drawing a red ball is and that of blue ball is ,20 5 then the number of black balls is (A) 5 (B) 25 (C) 10 (D) 30 10. Two coins are tossed simultaneously. The probability of getting at most one head is 1 13 (A) (B) (C) (D) 14 24 SECTION B 11. Which of the following can be the nth term of an AP? 3n + 1, 2n2 + 3, n3 + n. Give reasons. 12. Are the points (–3, –3), (–3, 2) and (–3, 5) collinear? Give reasons. 13. ABC and BDE are two equilateral triangles such that D is the mid point of BC. What is the ratio of the areas of triangles ABC and BDE? Justify your answer. 1 114. cos (A + B) = and sin (A–B)= , 0° < A + B < 90° and A – B > 0°. What are 2 2 the values of ∠ A and ∠ B? Justify your answer. 15. A coin is tossed twice and the outcome is noted every time. Can you say that head must come once in two tosses? Justify your answer. OR 2 A die is thrown once. The probability of getting a prime number is . Is it true?3 Justify your answer. SECTION C 16. Show that square of an odd positive integer is of the form 8q + 1, for some positive integer q. OR 357 Write the denominator of the rational number in the form of 2m 5n , m, n are5000non-negative integers and hence write its decimal expansion, without actual division. 17. If (x – 2) is a factor of x3+ax2+bx+16 and b = 4a, then find the values of a and b. 18. The sum of reciprocals of a child’s age (in years) 3 years ago and 5 years from 1now is . Find his present age.3 OR Solve for x : 6 a2x2 – 7abx – 3b2 = 0, a ≠ 0, using the quadratic formula. 19. Find the sum of all two digit natural numbers which are divisible by 7. 20. Find the ratio in which the line x + 3y – 14 = 0 divides the line segment joining the points A (–2, 4) and B (3, 7). 21. Find the area of the quadrilateral whose vertices in the same order are (–4, –2), (–3, –5), (3, –2) and (2, 3). 22. Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that ∠ APB = 2 ∠ OAB. (see the following figure). 23. Construct a triangle with sides 3 cm, 5 cm and 7 cm and then construct another 5 triangle whose sides are of the corresponding sides of the first triangle.3secθ cosecθ 24. Prove the identity (1 + cot θ+ tanθ) (sin θ– cos θ) = 2–2 cosec θ sec θ OR Find the value of 22cos32° + cos58° 2 2 – 4 tan13° tan37° tan53° tan77°sec 50°– cot 40°25. The area of an equilateral triangle is 49 3 cm2. Taking each vertex as centre, circles are described with radius equal to half the length of the side of the triangle. Find the area of the part of the triangle not included in the circles. [Take 31.73, = π=22 ]7 SECTION D 26. In a bag containing white and red balls, half the number of white balls is equal to the one third the number of red balls. Twice the total number of balls exceeds three times the number of red balls by 8. How many balls of each type does the bag contain? 27. Prove that in a right triangle, the square of the hypotenuse is equal to sum of squares of the other two sides. Using the above theorem, prove that in a triangle ABC, if AD is perpendicular to BC, then AB2 + CD2 = AC2 + BD2. 28. A pole 5m high is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point A on the ground is 60° and the angle of depression of point A from the top of the tower is 45°. Find the height of the tower.(Take 3 = 1.73) 29. The interior of a building is in the form of a cylinder of diameter 4 m and height 3.5 m, surmounted by a cone of the same base with vertical angle as a right angle. Find the surface area (curved) and volume of the interior of the building. OR A vessel in the form of an open inverted cone of height 8 cm and radius of its top is 5 cm. It is filled with water up to the brim. When lead shots, each of radius 0.5 cm are dropped into the vessel, one fourth of the water flows out. Find the number of lead shots dropped in the vessel. 30. Find the mean, median and mode of the following frequency distribution: Class 0-10 10-20 20-30 30-40 40-50 50-60 60–70 Frequency 4 5 7 10 12 8 4 OR The following distribution gives the daily income of 50 workers of a factory: Daily income (in Rs) 100-120 120-140 140-160 160-180 180-200 Number of workers 12 14 8 6 10 Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive. Find the median from this ogive. MARKING SCHEME SECTION A MARKS 1. (D) 2. (A) 3. (B) 4. (C) 5. (A) 6. (D) 7. (B) 8. (A) 9. (C) 10. (C) (1 × 10 = 10) SECTION B 1 11. nth term is 3n +1, ()2 1because, nth term of an AP can only be a linear relation in n. (1)2 112. Yes, ()2 Since all the three points are on the line x = –3. (11 )2 1 13. 4 : 1 ()2 ar ABC BC2 BC2 4 1 = 2 = 2= (1)ar BDEBD1 2⎡1 ⎤(BC)⎢⎥⎣ 2 ⎦ 114. ∠ A = 45°, ∠ B = 15° ( )2 1A + B = 60° and A – B = 30°, solving, we get ∠ A = 45°, ∠ B = 15° (1 )2 115. No. ()2 Head may come and head may not come. In every toss, there are two equally likely outcomes. (11 )2 OR 1 No. ()2 31 1P (a pirme number) = P (2, 3, 5) = = (1 )62 2 SECTION C 16. An odd positive integer can be of the form, 4n+1 or 4n+3 (1) Therefore, (4n + 1)2 = 16n2 + 8n + 1 = 8 (2n2 + n) + 1 = 8q + 1. (1)(4n + 3)2 = 16n2 + 24n + 9 = 8 (2n2 + 3n +1)+1 = 8q + 1. (1) OR 357 357 = 34 (1)5000 25× 357 × 2 714 = 4 4 = (10)4 (1)25×= 0.0714 (1) 17. (x–2) is a factor of x3 + ax2 + bx + 16 Therefore, (2)3 + a(2)2 + b(2) + 16 = 0 (1) 4a + 2b + 24 = 0 or 2a + b + 12 = 0 (1) Given b = 4a, so a = –2 (1) and b = –8 18. Let the present age be x years. (1) 1 11Therefore, + = x −3 x +53 or 3 [(x + 5) + (x – 3)] = (x – 3) (x + 5) or 6x + 6 = x2 + 2x – 15. or x2 – 4x –21 = 0 or (x – 7) (x + 3) = 0 (1) i.e., x = 7, x = –3 (rejected) Therefore, present age = 7 years (1) OR 6a2x2 – 7abx – 3b2 = 0 B2 – 4AC = [(–7ab)2 – 4 (6a2) (–3b2)]= 49a2b2 + 72a2b2 = 121a2 b2 (1) –(–7 ab) ±11 abTherefore, x = 2 (1)12a 18ab – 4ab = or12a2 12a2 3bb = or − (1)2a3a 19. Numbers are 14, 21, ..., 98 (1) 98 = 14 + (n – 1) 7, i.e., n = 13 (1) 13 S13 = [14 + 98] = 728. (1) 2 20. Let C (x, y) be the point where the line x + 3y – 14 = 0 divides the line segment in the ratio k:1.3 – 2k 7 + 4k So, x = , y = (1)k + 1 k + 1 3 –2 7 + 4 1 and, k +3.k –14 = 0 ( )k + 1 k +1 2 i.e., 3k – 2 + 21k + 12 – 14k – 14 = 0, i.e., 10k – 4 = 0 42 i.e., k = = 105 Therefore, ratio is 2 : 5 21. Area of Δ ABC 1 = [– 4 (–5 + 2) – 3 (–2 + 2) + 3 (–2 + 5)]2 1 21 = [12 + 9] = sq.units (1)221 area of Δ ACD = [–4 (–2 –3) + 3(3 + 2) + 2(–2 + 2)]21 35 = [20 + 15] = sq. units (1) 2221 +35 56Therefore, area of quadrilateral ABCD = = = 28 sq. units (1)22 1 22. AP = PB. So, ∠ PAB = ∠ PBA = [180º – ∠ APB]2 1= 90º – ∠ APB (1) 2 ∠ OAB = 90° – ∠ PAB (1) 1 1 = 90° – [90° – ∠ APB] = ∠ APB22 i.e., 2 ∠ OAB = ∠ APB (1) 23. Correct construction of Δ with sides 3, 5 and 7 cm (1) Correct construction of similar triangle (2) ⎛ cosθ sinθ ⎞ 124. LHS = ⎜1+ + ⎟ (sinθ – cosθ) ()⎝ sinθ cosθ⎠2 33(sinθ cosθ + cos 2θ + sin 2θ) (sinθ – cosθ) sin θ – cos θ 1 = = (1)sinθ cosθsinθ cosθ 2 sin2 θ cos 2θ secθ cosecθ = – = – (1) cosθsinθcosec 2θ sec 2 θOR 22cos 58° = sin 32°, tan53° =cot37° 2 2 (2) sec 50° =cosec 40, tan77° =cot13°Given expression 22cos 32° + sin 32° = 2 2 – 4 tan13° tan 37° cot 37° cot 13° (1) cosec 40° – cot 40°= 1 – 4 = – 3 2 25. Area of ΔABC = 49 3cm2 = 3 a 4 So, a = 14 cm (1) 60 49πArea of one sector = π × 72 = (1) 3606349 × 22 ⎛⎞ 49 3– ×Therefore, required area = ⎜⎟67⎝⎠ = 49 3– 77 = 84.77 – 77 = 7.77 cm2 (1) SECTION D 26. Let the number of white balls be x and number of red balls be y 11 1 Therefore, x = y , i.e., 3x – 2y =0 (I) (1 )23 2 and 2 (x + y) = 3y + 8 1i.e., 2x – y =8 (II) (1)2 Solving (I) and (II), we get x = 16, y = 24 (2) Therefore, number of white balls = 16 Number of red balls = 24 (1) 1 ×=)27. For correct given, to prove, construction and proof ( 42 2 For correct proof (2) 1 DESIGN OF THE QUESTION PAPER, SET-II 233 AD2 = AB2 – BD2 ( 1 2 ) and AD2 = AC2 – CD2 ()2 129. For correct figure ( )2 Here, ∠ Q = 45° , i.e., height of cone = radius = 2m (1) Therefore, surface area = πrl + 2 πrh = πr ( l+ 2h) (1) 1 = π × 2 × (2 2+7) (2) = (14 + 4 2) π m2 (1) 12 1 Volume = πrh1 + πr2h ()32 ⎡h ⎤ = π r2 1 + h⎢⎥⎣ 3 ⎦ 2 2 10.5 1π 4 3.5 4π= ()332 50π = m3(1) 3 OR 11Volume of water = π × (5)2 × 8 ( )32 200π = cm3 (1)31 50πth volume = cm3 (1)43 4 3 0.5π 1Volume of one lead shot = π(0.5) = cm3 (1)33 2 Let number of shots be n. 0.5π 50π Therefore, × n = (1) 33i.e., n = 100. (1) 30. 1 (1 )2 CI 0–10 10–20 20–30 30–40 40–50 50–60 60–70 Total fi 4 5 7 10 12 8 4 50 xi 5 15 25 35 45 55 65 ui –3 –2 –1 0 1 2 3 fi ui –12 –10 –7 0 12 16 12 11 cf 4 9 16 26 38 46 50 ∑ fi = 50 Σfiui = 11 11Mean = 35 + ×10 = 35 + 2.2 = 37.2 (1) 50 ⎛ n ⎞– cf ⎜⎟21l + ⎜ ⎟× hMedian = ⎜ f ⎟ ()2⎝⎠ 25 – 16= 30 + ×10 = 30 + 9 = 39 (1) 10 f – f 1Mode = l+ 10× h ()2f – f – f 2 12 – 10 1 10 2= 40 + ×10 ()24 – 10–8 2 20= 40 + = 43.33 (1) 6 OR Writing as (1) Daily income (in Rs) cf Less than 120 12 Less than 140 26 Less than 160 34 Less than 180 40 Less than 200 50 (5) Note: Full credit should be given for alternative correct solution.