# DESIGN OF THE QUESTION PAPER Mathematics Class X NCERT. Time : 3 Hours Maximum Marks : 80

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1 DESIGN OF THE QUESTION PAPER Mathematics Class X Weightage and the distribution of marks over different dimensions of the question shall be as follows: (A) Weightage to Content/ Subject Units : S.No. Content Unit Marks. Number Systems 04. Algebra 0 3. Trigonometry 4. Coordinate Geometry Geometry 6 6. Mensuration 0 7. Statistics and Probability 0 (B) Weightage to Forms of Questions : Total : 80 SET-II Time : 3 Hours Maximum Marks : 80 S.No. Form of Marks for each Number of Total Marks Questions Question Questions. MCQ SAR SA LA Total 30 80

2 0 EXEMPLAR PROBLEMS (C) Scheme of Options All questions are compulsory, i.e., there is no overall choice. However, internal choices are provided in one question of marks, three questions of 3 marks each and two questions of 6 marks each. (D) Weightage to Difficulty level of Questions S.No. Estimated Difficulty Percentage of Marks Level of Questions. Easy 0. Average Difficult 0 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.

3 DESIGN OF THE QUESTION PAPER, SET-II BLUE PRINT MATHEMATICS CLASS X Form of Question Units MCQ SAR SA LA Total Number Systems () () - - 4(3) Algebra 3(3) () 9(3) 6() 0(8) Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmatic Progressions Trigonometry () () 3() 6() (4) Introduction to Trigonometry, Some Applications of Trigonometry Coordinate Geometry () 4() 3() - 8(4) Geometry () - 9(3) 6() 6(5) Triangles, Circles, Constructions Mensuration () - 3() 6() 0(3) Areas related to Circles, Surface Areas and Volumes Statistics & Probability () - 3() 6() 0(3) Total 0(0) 0(5) 30(0) 30(5) 80(30) SUMMARY Multiple Choice Questions (MCQ) Number of Questions : 0 Marks : 0 Short Answer Questions with Resasoning (SAR) Number of Questions : 05 Marks : 0 Short Answer Questions (SA) Number of Questions : 0 Marks : 30 Long Answer Qustions (LA) Number of Questions : 05 Marks : 30 Total 30 80

4 EXEMPLAR PROBLEMS Maximum Marks : 80 General Instructions Mathematics Class X Time : 3 Hours. All questions are compulsory.. The question paper consists of 30 questions divided into four sections A, B, C, and D.Section A contains 0 questions of mark each, Section B contains 5 questions of marks each, Section C contains 0 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 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. The largest number which divides 38 and 739 leaving remainders 3 and 4, respectively is (A) 0 (B) 7 (C) 35 (D) 05. The number of zeroes lying between to of the polynomial f (x), whose graph is given below, is (A) (B) 3 (C) 4 (D) 3. The discriminant of the quadratic equation 3 3 x + 0x + 3 = 0 is (A) 8 (B) 64 (C) (D) 3 3 3

5 DESIGN OF THE QUESTION PAPER, SET-II 3 4. If 6, a, 4 are in AP, the value of a is 5 (A) (B) 3 (C) 3 5 (D) If in the following figure, Δ ABC Δ QPR, then the measure of Q is (A) 60 (B) 90 (C) 70 (D) 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) 0 cm 7. If sinθ = 3, then the value of (9 cot θ + 9) is (A) (B) 8 (C) 9 (D) 8

7 DESIGN OF THE QUESTION PAPER, SET-II 5 SECTION C 6. Show that square of an odd positive integer is of the form 8q +, for some positive integer q. 357 Write the denominator of the rational number 5000 in the form of m 5 n, m, n are non-negative integers and hence write its decimal expansion, without actual division. 7. If (x ) is a factor of x 3 +ax +bx+6 and b = 4a, then find the values of a and b. 8. The sum of reciprocals of a child s age (in years) 3 years ago and 5 years from now is. Find his present age. 3 Solve for x : 6 a x 7abx 3b = 0, a 0, using the quadratic formula. 9. Find the sum of all two digit natural numbers which are divisible by Find the ratio in which the line x + 3y 4 = 0 divides the line segment joining the points A (, 4) and B (3, 7).. Find the area of the quadrilateral whose vertices in the same order are ( 4, ), ( 3, 5), (3, ) and (, 3).. Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that APB = OAB. (see the following figure).

8 6 EXEMPLAR PROBLEMS 3. Construct a triangle with sides 3 cm, 5 cm and 7 cm and then construct another triangle whose sides are 5 of the corresponding sides of the first triangle. 3 secθ 4. Prove the identity ( + cot θ + tanθ ) (sin θ cos θ ) = cosec θ cosecθ sec θ Find the value of cos3 + cos58 sec 50 cot 40 4 tan3 tan37 tan53 tan77 5. The area of an equilateral triangle is 49 3 cm. 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 3=.73, π = ] 7 SECTION D 6. 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? 7. 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 AB + CD = AC + BD. 8. 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 =.73) 9. 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.

9 DESIGN OF THE QUESTION PAPER, SET-II 7 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 Frequency The following distribution gives the daily income of 50 workers of a factory: Daily income (in Rs) Number of workers Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive. Find the median from this ogive.

10 8 EXEMPLAR PROBLEMS MARKING SCHEME SECTION A MARKS. (D). (A) 3. (B) 4. (C) 5. (A) 6. (D) 7. (B) 8. (A) 9. (C) 0. (C) ( 0 = 0) SECTION B. n th term is 3n +, ( ) because, n th term of an AP can only be a linear relation in n. ( ). Yes, ( ) Since all the three points are on the line x = 3. ( ) 3. 4 : ( ) ar ABC ar BDE = BC BD = BC (BC) = 4 ( ) 4. A = 45, B = 5 ( ) A + B = 60 and A B = 30, solving, we get A = 45, B = 5 ( ) 5. No. ( ) Head may come and head may not come. In every toss, there are two equally likely outcomes. ( ) No. ( ) P (a pirme number) = P (, 3, 5) = 3 6 = ( )

11 DESIGN OF THE QUESTION PAPER, SET-II 9 SECTION C 6. An odd positive integer can be of the form, 4n+ or 4n+3 () Therefore, (4n + ) = 6n + 8n + = 8 (n + n) + = 8q +. () (4n + 3) = 6n + 4n + 9 = 8 (n + 3n +)+ = 8q +. () 357 = = (0) = 4 = () 7. (x ) is a factor of x 3 + ax + bx + 6 Therefore, () 3 + a() + b() + 6 = 0 () 4a + b + 4 = 0 or a + b + = 0 () Given b = 4a, so a = () and b = 8 8. Let the present age be x years. () Therefore, x 3 + x + 5 = 3 or 3 [(x + 5) + (x 3)] = (x 3) (x + 5) or 6x + 6 = x + x 5. or x 4x = 0 or (x 7) (x + 3) = 0 () i.e., x = 7, x = 3 (rejected) Therefore, present age = 7 years () 6a x 7abx 3b = 0 B 4AC = [( 7ab) 4 (6a ) ( 3b )] = 49a b + 7a b = a b () ( 7 ab) ± ab Therefore, x = a () () ()

12 30 EXEMPLAR PROBLEMS 8ab 4ab or a a = = 3 b a b 3a () 9. Numbers are 4,,..., 98 () 98 = 4 + (n ) 7, i.e., n = 3 () S 3 = 3 [4 + 98] = 78. () 0. Let C (x, y) be the point where the line x + 3y 4 = 0 divides the line segment in the ratio k:. So, x = 3 k, y = 7 k + 4 k + k + and, 3 k 7 k = 0 k + k + ( ) i.e., 3k + k + 4k 4 = 0, i.e., 0k 4 = 0 i.e., k = 4 0 = 5 Therefore, ratio is : 5 ( ). Area of Δ ABC = [ 4 ( 5 + ) 3 ( + ) + 3 ( + 5)] = [ + 9] = sq.units () area of Δ ACD = [ 4 ( 3) + 3(3 + ) + ( + )] = 35 [0 + 5] = sq. units () () ()

13 DESIGN OF THE QUESTION PAPER, SET-II Therefore, area of quadrilateral ABCD = = 56 = 8 sq. units (). AP = PB. So, PAB = PBA = [80º APB] = 90º APB () OAB = 90 PAB () = 90 [90 APB] = APB i.e., OAB = APB () 3. Correct construction of Δ with sides 3, 5 and 7 cm () Correct construction of similar triangle () 4. LHS = = = cosθ sinθ + + (sinθ cosθ) sinθ cosθ (sinθ cosθ + cos θ + sin θ) (sinθ cosθ) sin θ sinθ cosθ cos θ secθ cosecθ cosθ sinθ = cosec θ sec θ cos 58 = sin 3, tan53 = cot 37 sec 50 = cosec 40, Given expression = tan77 = cot3 = 3 3 sin θ cos θ sinθ cosθ cos 3 + sin 3 4 tan3 tan 37 cot 37 cot 3 ( ) ( ) cosec 40 cot 40 = 4 = 3 () () ()

14 3 EXEMPLAR PROBLEMS a 5. Area of Δ ABC = 49 3 cm = 3 4 So, a = 4 cm () 60 Area of one sector = π = 49π 6 Therefore, required area = = = = 7.77 cm () SECTION D 6. Let the number of white balls be x and number of red balls be y Therefore, x = 3 y, i.e., 3x y = 0 (I) ( ) and (x + y) = 3y + 8 i.e., x y = 8 (II) ( ) Solving (I) and (II), we get x = 6, y = 4 () Therefore, number of white balls = 6 Number of red balls = 4 () 7. For correct given, to prove, construction and proof ( 4 = ) For correct proof () ()

15 DESIGN OF THE QUESTION PAPER, SET-II 33 AD = AB BD ( ) and AD = AC CD ( ) i.e., AB BD = AC CD ( ) or AB + CD = AC + BD ( ) 8. For correct figure () Let height of tower be h metres and AB = x metres. ( ) Therefore, x h = cot 45 = () i.e., x = h. ( ) Also, h + 5 = tan60 = 3 () x i.e., h + 5 = 3x = 3h ( ) i.e., ( 3 )h = 5 h = = ( 3 + ) 5(.73) = ( ) = 3.65 = 6.85 m ()

16 34 EXEMPLAR PROBLEMS 9. For correct figure ( ) Here, Q = 45, i.e., height of cone = radius = m () Therefore, surface area = π rl + π rh = π r ( l+ h) () = π ( +7) ( ) = (4 + 4 ) π m () Volume = π 3 rh + π r h ( ) = π r h 3 + h 0.5 = π π 3 3 ( ) = 50π 3 m3 () Volume of water = 3 π (5) 8 ( ) = 00π cm 3 3 () 4 th volume = 50 π cm 3 () 3 Volume of one lead shot = Let number of shots be n π π(0.5) = cm 3 ( ) 3 3 Therefore, 0.5π n = 50π 3 3 () i.e., n = 00. ()

17 DESIGN OF THE QUESTION PAPER, SET-II CI Total f i x i u i f i u i cf = 50 Σ i f i f u = i Mean = = = 37. () 50 Median = n cf l + h f = = = 39 () 0 f Mode = f0 l+ h f f f = = = () 6 Writing as () Daily income (in Rs) cf Less than 0 Less than 40 6 Less than Less than Less than ( ) ( ) ( ) ( )

18 36 EXEMPLAR PROBLEMS Note: Full credit should be given for alternative correct solution. (5)

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