SET-I 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 Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmatic Progressions 3(3) 2(1) 9(3) 6(1) 20(8) Trigonometry Introduction to Trigonometry, Some Applications of Trigonometry 1(1) 2(1) 3(1) 6(1) 12(4) Coordinate Geometry 1(1) 4(2) 3(1) - 8(4) Geometry Triangles, Circles, Constructions 1(1) - 9(3) 6(1) 16(5) Mensuration Areas related to Circles, Surface Areas and Volumes 1(1) - 3(1) 6(1) 10(3) 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 : 30 Total 30 80 Mathematics 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. After how many decimal places will the decimal expansion of the number 47 32 terminate?25(A) 5 (B) 2 (C) 3 (D) 1 2. Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where (A) 0 ≤ r ≤ a (B) 0 < r < b (C) 0 ≤ r ≤ b (D) 0 ≤ r < b 3. The number of zeroes, the polynomial p (x) = (x – 2)2 + 4 can have, is (A) 1 (B) 2 (C) 0 (D) 3 4. A pair of linear equations ax + by + c = 0; ax + by + c= 0 is said to be 111222 inconsistent, if ab abc abc ac11 111 111 11≠ ≠= =≠≠(A) (B) (C) (D) ab abcabcac22 222 222 22 5. The smallest value of k for which the equation x2 + kx + 9 = 0 has real roots, is (A) – 6 (B) 6 (C) 36 (D) –3 6. The coordinates of the points P and Q are (4, –3) and (–1, 7). Then the abscissa of PR 3 =a point R on the line segment PQ such that isPQ 5 18 1717(A) (B) (C) (D) 15 587. In the adjoining figure, PA and PB are tangents from a point P to a circle with centre O. Then the quadrilateral OAPB must be a (A) square (B) rhombus (C) cyclic quadrilateral (D) parallelogram 1 8. If for some angle θ, cot 2θ = 3 , then the value of sin3θ, where 2θ < 90º is 1 3(A) (B) 1 (C) 0 (D)2 2 9. From each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in figure. The area of the remaining (shaded) portion is (A) (16 – 2π) cm2 (B) (16 – 5π) cm2 (C) 2π cm2 (D) 5π cm2 10. A letter of English alphabets is chosen at random. The probability that it is a letter of the word ‘MATHEMATICS’ is 11. Is there any natural number n for which 4n ends with the digit 0? Give reasons in support of your answer. 12. Without using the formula for the nth term, find which term of the AP : 5, 17, 29, 41, ... will be 120 more than its 15th term? Justify your answer. OR Is 144 a term of the AP : 3, 7, 11, ... ? Justify your answer. 13. The coordinates of the points P, Q and R are (3, 4), (3, –4) and (–3, 4), respectively. Is the area of ΔPQR 24 sq. units? Justify your answer. 14. The length of a line segment is 10 units. If one end is (2, –3) and the abscissa of the other end is 10, then its ordinate is either 3 or –9. Give justification for the two answers. 15. What is the maximum value of ? Justify your answer. (A) 11 26 (B) 5 13 (C) 9 26 (D) 4 13 SECTION B 3 cosec SECTION C 16. Find the zeroes of the polynomial p (x) = 43 x 2–2 3 x– 2 3 and verify the relationship between the zeroes and the coefficients. OR On dividing the polynomial f (x) = x3 – 5x2 + 6x – 4 by a polynomial g(x), the quotient q (x) and remainder r (x) are x – 3, –3x + 5, respectively. Find the polynomial g (x). 17. Solve the equations 5x – y = 5 and 3x – y = 3 graphically. 18. If the sum of the first n terms of an APis4n – n2, what is the10th term and the nth term? OR How many terms of the AP : 9, 17, 25, ... must be taken to give a sum 636? 19. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find the values of x and y. 20. The sides AB, BC and median AD of a ΔABC are respectively propotional to the sides PQ, QR and the median PM of ΔPQR. Show that ΔABC ~ ΔPQR. 21. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 7 cm, respectively. Find the sides AB and AC. 22. Construct an isosceles triangle whose base is 6 cm and altitude 5 cm and then 7another triangle whose sides are of the corresponding sides of the isosceles5triangle. cos –sin 1 1 23. Prove that sin cos –1 cosec – cot . OR Evaluate: 3cos 43 2 cos 37 cosec53 – sin47 tan5 tan25 tan45 tan65 tan85 24. In the figure, ABC is a triangle right angled at A. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region. 25. A bag contains white, black and red balls only. A ball is drawn at random from the 32bag. The probability of getting a white ball is and that of a black ball is . Find105 the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag. SECTION D 26. If the price of a book is reduced by Rs 5, a person can buy 5 more books for Rs 300. Find the original list price of the book. OR The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is this situation possible? If so, determine their present ages. 27. Prove that the lengths of the tangents drawn from an external point to a circle are equal. Using the above theorem, prove that: If quadrilateral ABCD is circumscribing a circle, then AB + CD = AD + BC. OR Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides. Using the above theorem, do the following : ABC is an iscosceles triangle right angled at B. Two equilateral triangles ACD and ABE are constructed on the sides AC and AB, respectively. Find the ratio of the areas of Δ ΑΒΕ and Δ ACD. 28. The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distances between the building and the tower. 29. A well of diameter 3 m and 14 m deep is dug. The earth, taken out of it, has been evenly spread all around it in the shape of a circular ring of width 4 m to form an embankment . Find the height of the embankment. 30. The following table shows the ages of the patients admitted in a hospital during a month: Age (in years) : Number of patients : 5 -15 6 15 - 25 11 25 - 35 21 35 - 45 23 45 - 55 14 55 - 65 5 Find the mode and the mean of the data given above. MARKING SCHEME SECTION A MARKS 1. (C) 2. (D) 3. (C) 4. (C) 5. (A) 6. (D) 7. (C) 8. (B) 9. (A) 10. (D) (1 × 10 = 10) SECTION B 1 11. No ()2 4n = 22n Therefore, 2 is the only prime number in its prime facorisation, so it cannot end 1 with zero. (1 )2 1 12. 25th term (2) 120 will be added in 10 terms (since d = 12) 1 Therefore, 15 + 10 = 25 (1 )2 OR 1No ()2 Here, a = 3 (odd), d = 4 (even) 1Sum of (odd + even) = odd but 144 is even (1 )2113. Yes ()2 Here, PQ = 8, 11PR = 6, therefore, area = 2 . 8.6 = 24 sq. units. (12 ) 14. Let ordinate of the point be y. Then (10 – 2)2 + (y + 3)2 = 102 , i.e.,y + 3 = ± 6, i.e., y = 3 or –9 (1+1) 115. Maximum value = 3 ( )2 31 Since cosec θ = 3 sinθ, and sinθ≤ 1, therefore, 3 sinθ ≤ 3 (1)2 SECTION C 16. p(x) = 43 x2 –23 x – 23 = 23 (2x2 – x – 1) = 23(2x + 1) (x – 1) 1Therefore, two zeroes are – , 1 (1) 2 Here a = 43, b = 23, c = –23 11 b 23 1 b –– = –Therefore, α + β = + 1 = 2, = , i.e., α + β = (1)2 a 43 2 a ⎛ 1 ⎞ 1 c –23 1 c – – = –αβ = ⎜⎟ 1 = 2 , = 2 , i.e., αβ = (1)⎝ 2⎠ a43 a OR f(x)= g(x) q(x) + r(x) Therefore, x3 – 5x2 + 6x – 4 = g(x) (x – 3) + (–3x + 5) (1) 32 32x –5x + 6x –4+ 3x –5 x –5x +9x –4 Therefore, g(x) = = (1)x –3 x –3 = x2 – 2x + 3 (1) 17. 5x – y = 5 3x – y = 3 x 1 2 3 y 0 5 10 x 1 2 3 y 0 3 6 For correct graph (2) Solution is x = 1, y = 0 (1) 118. S = 4n – n2. Therefore, t= S– S = (40 – 100) – (36 –81) ( )n 10 10 92 = –60 + 45 = –15 (1) 1 t = S – S = (4n – n2) – [4 (n – 1) – (n – 1)2] (2)nn n – 1= 4n – n2 – 4n + 4 + n2 + 1 – 2n = 5 – 2n (1) OR a = 9, d = 8, Sn = 636 nn 1Using S= [2a + (n – 1) d], we have 636 = [18 + (n – 1) 8] (1 )n 2 22 1Solving to get n = 12 (1 )2 19. Let A(1, 2), B (4, y) and C (x, 6) and D (3, 5) be the vertices. ⎛ x +1⎞ 1The mid-point of AC is ⎜ ,4 ⎟ ()⎝ 2 ⎠ 2 ⎛ 7 y + 5 ⎞ 1and mid-point of BD is ⎜⎟ ()⎝ 2, 2 ⎠ 2 x+ 17 =ABCD is a parellologram. Therefore, 2 2 ,i.e., x = 6 (1) y+ 5 = 4 , i.e., y = 3 (1)2 ABBC BD ADGiven == = PQ QRQM PM Therefore, Δ ABD ∼ ΔPQMAB BC =Therefore, ∠ Β = ∠Q. Also, since PQ QR 1 Δ ABC ~ Δ PQR [SAS] (1 )2 21. Let AE (=AF) = x cm. 1 Area ΔABC = . 4 . (AB + BC + AC)2 = s (sa)( – b)(sc)– s – i.e., 4 s= s(–)( – )(–) sas bsc 16 s= (s– a) (s– b) (s– c) i.e., 16 (15 + x) = x. 8 . 7, i.e., x= 6 Therefore, AB = 14 cm and AC = 13 cm 22. Construction of isosceles Δ with base 6 cm and altitude 5 cm 7Construction of similar Δ with scale factor 5 cos θ– sin θ+ 1 cot θ –1 + c osec θ 23. LHS = = sin θ+cos θ–1 1+cot θ – cosec θ cot θ –1 + cosec θ cosec θ+cot θ–1 = = 221– (cosec θ– cot θ) (cosec θ– cot θ) – (cosec θ– cot θ) cosec θ+cot θ–1 1 = = (cosec θ– cot θ) (cosec θ+ cot θ–1) cosec θ– cot θ OR 3cos43 2 cos 37 cosec 53 – sin47 tan5 tan25 tan45 tan65 tan85 ⎡3cos43 °⎤2 cos37 °.sec 37 ° = ⎢ – cos43 °⎥ tan 5 °tan 25 °(1) cot 25 cot 5 °° ⎣⎦ 1= (3)2 – = 9 – 1 = 8124. ⎧area of semicircle withdiameter AB + ⎪⎪area of semicircle withdiameter AC + Required area = ⎨area of right triangle ABC – ⎪ ⎩area of semicircle withdiameter BC ⎪2151()2 1 ()2 (1) 1 ()2 (1) (2) (1) (1) (1) (2) (1) (1) 1 21 21 12Required area = .(3) (4) 6 8– (5) sq. units (1)22221 = 24 + π (9 + 16 – 25) = 24 sq. units (1) 2 25. P(Red ball) = 1 – {P(White ball) + P(Black ball )} (1)⎧ 32⎫ 31 = 1 – ⎨ +⎬ = ()⎩10 5⎭ 10 2 Let the total number of balls be y. 202 1,i.e., y 50 Therefore, (1 )y 52 SECTION D 26. Let the list price of a book be Rs x 300 1Therefore, number of books, for Rs 300 = x (2) 300 1No. of books, when price is (x – 5) = ()x–52 300 300 Therefore, – = 5 (2)x–5 x300 (x – x + 5) = 5x (x – 5) 300 = x (x – 5), i.e., x2 – 5x – 300 = 0 (1) i.e., x = 20, x = –15 (rejected) (1) Therefore, list price of a book = Rs 20 (1) OR Let the present age of one of them be x years, so the age of the other = (20 – x) years Therefore, 4 years ago, their ages were, x – 4, 16 – x years (1) Therefore, (x – 4) (16 – x) = 48 (1 21) i.e., – x2 + 16 x + 4x – 64 – 48 = 0 x2 – 20x + 112x = 0 Here B2 – 4 AC = (20)2–4(112) = – 48 Thus, the equation has no real solution Hence, the given situation is not possible 27. For correct, given, to prove, contruction and figure For correct proof AP =AS ⎫ (tangents to a circle from external⎪BP=BQ ⎪point are equal)⎬DR = DS ⎪ CR =CQ⎪⎭ Adding to get (AP + BP) + (DR + CR) = (AS + DS) + (BQ + CQ) i.e., AB + CD = AD + BC OR For correct,given, to prove, construction and figure For correct proof Let AB = BC = a, i.e., AC = a2 +a2 = 2 a area ΔABC AB2 a21 = = = area Δ ACD AC2 2a22 28. For correct figure ABIn Δ ABD, = tan 60° = 3BD Therefore, AB = 3 BD (I) AE AE 1 =In Δ ACE, = tan 30° =EC BD 3i.e., (AB –50) 1 = , i.e., 3 (AB – 50) = BDBD 3 217 (1) 1 ()2 (1) (1) (2) (2) (1) (1) (2) (2) 1 ()2 1 (1 )2 (1) (1) (1) (1) Therefore, from (I) AB = 3 . 3 (AB – 50) ,i.e., AB = 3AB – 150 ,i.e., AB = 75 m BD = 3 (75 – 50) = 25 3 m29. Volume of earth dug out = πr2h = π (1.5)2 × 14 = 31.5 π m3 Area of circular ring = π[R2 – r2] = π[(5.5)2 – (1.5)2] = π(7) (4) = 28π m2Let height of embankment be h metres Therefore, 28π × h = 31.5 π 31.5 h = = 1.125 m 28 30. (1) (1) (2) (1) (1) (1) (1) 1()2 (1) fxi 2830∑ iMean = = = 35.375 years (1)∑ fi 80 1Modal class is (35 – 45) ( )2 f– f Therefore, Mode = l + 10 × h (1)2– f –ff10 2 Putting l = 35, f1 = 23, f0 = 21, f2 = 14 and h = 10, we get (1) Mode = 35 + 2 ×10 = 36.81 years (1) 11Note: Full credit should be given for alternative correct solution.