KENDRIYA VIDYALAYA GACHIBOWLI, GPRA CAMPUS, HYD-32 SAMPLE PAPER TEST 06 FOR BOARD EXAM 2023 (ANSWERS) SUBJECT: MATHEMATICS CLASS : X MAX. MARKS : 80 DURATION : 3 HRS General Instruction: 1. 2. 3. 4. 5. 6. This Question Paper has 5 Sections A-E. Section A has 20 MCQs carrying 1 mark each. Section B has 5 questions carrying 02 marks each. Section C has 6 questions carrying 03 marks each. Section D has 4 questions carrying 05 marks each. Section E has 3 case based integrated units of assessment (04 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively. 7. All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2marks questions of Section E 8. Draw neat figures wherever required. Take π =22/7 wherever required if not stated. SECTION – A Questions 1 to 20 carry 1 mark each. 1. If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a = (a) 1 (b) 2 (c) 3 Ans. (d) 4 (d) 4 2. The sum of exponents of prime factors in the prime-factorisation of 196 is: (a) 3 (b) 4 (c) 5 (d) 6 Ans. (b) 4 3. If the zeroes of the quadratic polynomial x2 + (a + 1)x + b are 2 and -3, then (a) a = -7, b = -1 (b) a = 5, b = -1 (c) a = 2, b = –6 (d) a = 0, b = –6 Ans. (d) a = 0, b = –6 4. Three cubes each of side 15 cm are joined end to end. The total surface area of the cuboid is: (a) 3150 cm2 (b) 1575 cm2 (c) 1012.5 cm2 (d) 576.4 cm2 Ans. (a) 3150 cm2 5. The point which lies on the perpendicular bisector of the line segment joining point A (–2, –5) and B (2, 5) is: (a) (0, 0) (b) (0, –1) (c) (–1, 0) (d) (1, 0) Ans. (a) (0, 0) 6. The point on the x-axis which is equidistant from (– 4, 0) and (10, 0) is: (a) (7, 0) (b) (5, 0) (c) (0, 0) (d) (3, 0) Ans. (d) (3, 0) 7. If x = 2sin2θ and y = 2cos2θ + 1 then x + y is: (a) 3 (b) 2 Ans: (a) 3 (c) 1 8. If cos θ + cos2 θ = 1, the value of sin2 θ + sin4 θ is : (a) –1 (b) 0 (c) 1 Prepared by: M. S. KumarSwamy, TGT(Maths) (d) 1/2 (d) 2 Page - 1 - Ans: (c) 1 9. In the figure given below, AD = 4 cm, BD = 3 cm and CB = 12 cm, then cot θ equals : (a) 3/4 Ans: (d) 12/5 (b) 5/12 (c) 4/3 (d) 12/5 10. The perimeters of two similar triangles are 26 cm and 39 cm. The ratio of their areas will be : (a) 2 : 3 (b) 6 : 9 (c) 4 : 6 (d) 4 : 9 Ans: (d) 4 : 9 11. If ∆ABC ~ ∆EDF and ∆ABC is not similar to ∆DEF, then which of the following is not true? (a) BC.EF = AC.FD (b) AB.EF = AC.DE (c) BC.DE = AB.EF (d) BC.DE = AB.FD Ans. (c) BC.DE = AB.EF 12. In the given figure, from an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O. If ∠QPR = 90°, then length of PQ is (a) 3 cm Ans: (b) 4 cm (b) 4 cm (c) 2 cm (d) 2.2 cm 13. In a circle of diameter 42cm, if an arc subtends an angle of 60º at the centre, then the length of the arc is: (a) 22/7 cm (b) 11cm (c) 22 cm (d) 44 cm Ans: (c) 22 cm 14. If the circumference of a circle increases from 2π to 4π then its area ......the original area : (a) Half (b) Double (c) Three times (d) Four times Ans: (d) Four times 15. The radii of 2 cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Then, the ratio of their volumes is: (a) 19 : 20 (b) 20 : 27 (c) 18:25 (d) 17:23 Ans: (b) 20 : 27 16. Consider the following frequency distribution Class 0–5 6 – 11 Frequency 13 10 The upper limit of the median class is (a) 7 (b) 17.5 (c) 18 Prepared by: M. S. KumarSwamy, TGT(Maths) 12 – 17 15 18 – 23 8 24 – 29 11 (d) 18.5 Page - 2 - Ans: (b) 17.5 17. For the following distribution: Marks Below 10 No. of students 3 the modal class is (a) 10-20 (b) 20-30 Ans: (c) 30-40 Below 20 12 Below 30 27 (c) 30-40 Below 40 57 Below 50 75 Below 60 80 (d) 50-60 18. A box contains cards numbered 6 to 50. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square is : (a) 1/45 (b) 2/15 (c) 4/45 (d) 1/9 Ans. (d) 1/9 P(perfect Square)= 5/45 = 1/9 Direction : In the question number 19 & 20 , A statement of Assertion (A) is followed by a statement of Reason(R) . Choose the correct option 19. Assertion (A): The mid-point of the line segment joining the points A (3, 4) and B (k, 6) is P (x, y) and x + y – 10 = 0, the value of k is 7 x x y y2 Reason (R): Midpoint of line segment is 1 2 , 1 2 2 (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true and R is false (d) A is false and R is true Ans: (b) Both A and R are true but R is not the correct explanation of A 20. Assertion (A): The largest number that divide 70 and125 which leaves remainder 5 and 8 is 13 Reason (R): HCF (65,117) =13 (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true and R is false (d) A is false and R is true Ans: (b) Both A and R are true but R is not the correct explanation of A SECTION-B Questions 21 to 25 carry 2M each 21. The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 2 days Ans: In 2 days, the short hand will complete 4 rounds. ∴ Distance moved by its tip = 4(circumference of a circle of radius 4 cm) 704 22 4 2 4 cm cm 7 7 In 2 days, the long hand will complete 48 rounds. ∴ Distance moved by its tip = 48(circumference of a circle of radius 6 cm) 12.672 22 48 2 6 cm cm 7 7 704 12672 Hence, sum of distances moved by the tips of two hands of the clock cm 7 7 = 1910.85 cm Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 - OR A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle of 120°. Find the total area cleaned at each sweep of the blades Ans: Here, r = 21 cm, θ = 120° 1200 22 2 Area of a sector = r 21 21 3600 3600 7 = 462 cm2 ∴ Total area cleaned by two wipers = 2 × 462 = 924 cm2 22. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. Ans: We know that the lengths of the tangents drawn from an external point to the circle are equal. DR = DS ...... (i) BP = BQ ...... (ii) AP = AS ...... (iii) CR = CQ ...... (iv) Adding (i), (ii), (iii), (iv), we get DR + BP + AP + CR = DS + BQ + AS + CQ By rearranging the terms we get, (DR + CR) + (BP + AP) = (CQ + BQ) + (DS + AS) ⇒ CD + AB = BC + AD Hence it is proved AB + CD = AD + BC. 1 1 , cos(A + B) = , 00< A + B ≤900 , A > B. Find A and B. 2 2 1 1 Ans: sin( A B) sin( A B) 30 sin 30 2 2 On equating both sides A B 30 (1) 1 1 cos( A B) cos( A B ) cos(60 ) cos(60 ) 2 2 On equating both sides A B 60..(2) Adding (1) and (2),we get 2A = 900 ⇒ A = 450 Putting value of A in (i) 450 + B = 600 ⇒ B = 150 23. If sin(A – B) = 24. Find the value of m for which the pair of linear equations: 2x + 3y – 7 = 0 and (m – 1) x + (m + 1) y = (3m – 1) has infinitely many solutions Ans: For infinitely many solutions the condition is a1 b1 c1 2 3 7 a2 b2 c2 m 1 m 1 3m 1 Now, 2(m + 1) = 3(m – 1) ⇒ m = 5 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 - and 3(3m – 1) = 7(m + 1) ⇒ m = 5 Hence, for m = 5, the system has infinitely many solutions. 25. In the figure, QR QT and ∠1 = ∠2, Show that ∆PQS ∼ ∆TQR. QS PR Ans: In △PQR, Since, ∠1=∠2 ∴ PR=PQ (Opposite sides of equal angles are equal) .....(1) QR QT . (Given) In △PQS and △TQR, QS PR QR QT .( From 1) i.e., QS PQ Also, ∠Q is common ∴ By SAS criterion of similarity, △PQS∼△TQR. OR ABCD is a trapezium in which AB || CD and its diagonals intersect each other at the point O. Using a similarity criterion of two triangles, show that = Ans: ABCD is a trapezium with AB∥CD and diagonals AB and CD intersecting at O. In △OAB and △OCD ∠AOB =∠DOC [ Vertically opposite angles ] ∠ABO = ∠CDO [ Alternate angles ] ∠BAO = ∠OCD [ Alternate angles ] ∴ △OAB ∼ △OCD [ AAA similarity ] We know that if triangles are similar, their corresponding sides are in proportional OA OB OC OD SECTION-C Questions 26 to 31 carry 3 marks each 26. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. Find the minimum distance each should walk so that each can cover the same distance in complete steps. Ans: The minimum distance each should walk so that each can cover the same distance in complete steps is the LCM of 40 cm, 42 cm and 45 cm. Prime factorisation of 40, 42 and 45 gives Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 - 40 = 23 × 5, 42 = 2 × 3 × 7, 45 = 32 × 5 LCM (40, 42, 45) = Product of the greatest power of each prime factor involved in the numbers = 23 × 32 × 5 × 7 = 8 × 9 × 35 = 72 × 35 = 2520 cm. 27. If a, b are the zeroes of the polynomial 2x2 – 5x + 7, then find a polynomial whose zeroes are 2a + 3b, 3a + 2b Ans: Since a, b are the zeroes of 2x2 – 5x + 7 (5) 5 7 ∴a+b= = and ab = 2 2 2 The given zeroes of required polynomial are 2a + 3b and 3a + 2b 5 25 Sum of the zeroes = 2a + 3b + 3a + 2b = 5a + 5b = 5(a + b) = 5 × = 2 2 Again, product of the zeroes = (2a + 3b) (3a + 2b) = 6 (a2 + b2) + 13ab = 6 [(a + b)2 – 2ab)] + 13 ab = 6(a + b)2 + ab = 6(5/2)2 + 7/2 = 75/2 + 7/2 = 82/2 = 41 Now, required polynomial is given by 25 k [x2 – (Sum of the zeroes) x + Product of the zeroes] = k [ x2 – x + 41] 2 k = [2 x 2 25 x 82] , where k is any non-zero real number. 2 Hence the required polynomial is 2 x 2 25 x 82 . 28. Prove that (sinA + cosecA)2 + (cosA + secA)2 = 7 + tan2A + cot2A Ans: L.H.S. =(sinA + cosecA)2 + (cosA + secA)2 = sin2A + cosec2A + 2sinAcosecA + cos2A + sec2A + 2cosAsecA = 1 + 2 + 2 + 1 + tan2A + 1 + cot2A = 7 + tan2A + cot2A = R.H.S. OR cos A 1 sin A 2sec A 1 sin A cos A cos A 1 sin A Ans: LHS 1 sin A cos A 2 2 cos A (1 sin A) cos 2 A 1 sin 2 A 2sin A cos A(1 sin A) cos A(1 sin A) 2 2sin A 2 2sec A RHS cos A(1 sin A) cos A Prove that 29. Find the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7) Ans: Let P(x, y) be the point on the line 2x + y – 4 = 0 dividing the line segment joining the points A(2, –2) and B(3, 7) in the ratio k : 1. 3k 2 7 k 2 ∴ The coordinate of P are , k 1 k 1 Since, point (x, y) lies on the line 2x + y = 4. 6k 4 7k 2 3k 2 7k 2 2 4 4 k 1 k 1 k 1 ⇒ 13k + 2 = 4k + 4 ⇒ 9k = 2 ⇒ k = 2/9 Thus, required ratio is 2 : 9. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 - 30. In the given figure, OP is equal to diameter of the circle. Prove that ABP is an equilateral triangle. Ans: Join OP and let it meets the circle at point Q. Since OP = 2r (Diameter of the circle) ⇒ OQ = QP = r Consider ΔAOP in which OA ⊥ AP and OP is the hypotenuse. ∴ OQ = AQ = OA (Mid-point of the hypotenuse is equidistant from the vertices) ⇒ OAQ is an equilateral triangle. ⇒ ∠AOQ = 60° (Each angle of an equilateral triangle is 60°) Consider right-angled triangle OAP. ∠AOQ = 60° (Proved above) ∠OAP = 90° ⇒ ∠APO = 30° ∠APB = 2∠APO = 2 × 30° = 60° Also PA = PB (Tangents to a circle from an external point are equal.) ⇒ ∠PAB = ∠PBA (Angles opposite to equal sides in ΔPAB) In ΔABP, ∠APB = 60° 1800 600 ⇒ ∠PAB = ∠PBA = 600 2 ⇒ Each angle of DPAB is 60° ⇒ PAB is an equilateral triangle. OR A circle is inscribed in a ΔABC having sides 8 cm, 10 cm and 12 cm as shown in the following figure. Find AD, BE and CF. Ans: Let AD = x1, BE = x2 and CF = x3; then AF = AD = x1, BD = BE = x2 and CE = CF = x3. ∴ x1 + x2 = 12; x2 + x3 = 8; x1 + x3 = 10 (1) Adding, Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 - 2(x1 + x2 + x3) = 30 ⇒ x1 + x2 + x3 = 15 Solve for x1, x2 and x3 to get AD = 7 cm, BE = 5 cm, CF = 3 cm 31. From a pack of 52 playing cards, jacks, queens, kings and aces of red colour are removed. From the remaining a card is drawn at random. Find the probability that the card drawn is (i) a black queen (ii) a red card (iii) a face card. Ans: From the total playing 52 cards, red coloured jacks, queen, kings and aces are removed(i.e., 2 jacks, 2 queens, 2 kings, 2 aces) ∴ Remaining cards = 52 – 8 = 44 (i) Favourable cases for a black queen are 2 (i.e., queen of club or spade) ∴ Probability of drawing a black queen = 2/44 = 1/22 (ii) Favourable cases for red cards are 26 – 8 = 18 (as 8 cards have been removed) (i.e.9 diamonds + 9 hearts) ∴ Probability of drawing a red card = 18/44 = 9/22 (iii) Favourable cases for a face card are 6 (i.e. 2 black jacks, queens and kings each) ∴ Probability of drawing a face card = 6/44 = 3/22 SECTION-D Questions 32 to 35 carry 5M each 32. A survey regarding the heights (in cm) of 50 girls of class Xth of a school was conducted and the following data was obtained. Find the mean, median and mode of the given data. Heights (in cm) 120 – 130 130 – 140 140 – 150 150 – 160 160 – 170 No. of Girls 2 8 12 20 8 Ans: 33. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see below figure). In how may rows are the 200 logs placed and how many logs are in the top row? Ans: Here, a is the first term, d is a common difference and n is the number of terms. It can be observed that the number of logs in rows are forming an A.P. 20, 19, 18, ... n We know that sum of n terms of AP is given by the formula Sₙ = [2a + (n - 1) d] 2 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 - n [2 × 20 + (n - 1)(- 1)] ⇒ 400 = n [40 - n + 1] ⇒ 400 = n [41 - n] 2 ⇒ 400 = 41n - n² ⇒ n² - 41n + 400 = 0 ⇒ n² - 16n - 25n + 400 = 0 ⇒ n(n - 16) - 25(n - 16) = 0 ⇒ (n - 16)(n - 25) = 0 ⇒ Either (n -16) = 0 or (n - 25) = 0 ∴ n = 16 or n = 25 The number of logs in nth row will be aₙ = a + (n - 1) d ⇒ a₁₆ = a + 15d ⇒ a₁₆ = 20 + 15 × (- 1) ⇒ a₁₆ = 20 – 15 ⇒ a₁₆ = 5 Similarly, a₂₅ = 20 + 24 × (- 1) ⇒ a₂₅ = 20 – 24 ⇒ a₂₅ = - 4 Clearly, the number of logs in the 16th row is 5. However, the number of logs in the 25th row is negative 4, which is not possible. Therefore, 200 logs can be placed in 16 rows. The number of logs in the top (16th) row is 5. OR The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP. Ans: Here, a is the first term, d is the common difference and n is the number of terms. Given: a₃ + a₇ = 6 ----- (1) a₃ × a₇ = 8 ----- (2) We know that nth term of AP is aₙ = a + (n - 1)d Third term, a₃ = a + 2d ----- (3) Seventh term, a₇ = a + 6d ----- (4) Using equation (3) and equation (4) in equation (1) to find the sum of the terms, (a + 2d) + (a + 6d) = 6 ⇒ 2a + 8d = 6 ⇒ a + 4d = 3 ⇒ a = 3 - 4d ----- (5) Using equation (3) and equation (4) in equation (2) to find the product of the terms, (a + 2d ) × (a + 6d ) = 8 Substituting the value of a from equation (5) above, (3 - 4d + 2d) × (3 - 4d + 6d) = 8 ⇒ (3 - 2d) × (3 + 2d) = 8 ⇒ (3)² - (2d)² = 8 [Since (a + b)(a - b) = a² - b² ] ⇒ 9 - 4d² = 8 ⇒ 4d² = 1 ⇒ d² = ¼ ⇒ d = ½, -½ Case 1: When d = ½ a = 3 - 4d = 3 - 4 × ½ = 3 - 2 = 1 n 16 Sₙ = [2a + (n - 1) d] ⇒ S₁₆ = [ 2 × 1 + (16 - 1) × ½ ] = 8 × 19/2 = 76 2 2 Case 2: When d = - ½ a = 3 - 4d = 3 - 4 × (- ½) = 3 + 2 = 5 n 16 Sₙ = [2a + (n – 1) d] ⇒ S₁₆ = [2 × 5 + (16 - 1) × (- ½)] = 8 [10 - 15/2] = 8 × 5/2 = 20 2 2 ⇒ 200 = 34. Prove that if a line is a drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. Using the above theorem. Prove that = if LM II CB and LNII CD as shown in the figure. Ans: Given, To Prove, Figure and Construction – 1 ½ marks Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 - Proof - 1 ½ mark In △ABC, LM ∥ BC ∴ By proportionality theorem, AM AL .(1) AB AC Similarly, In △ADC, LN ∥ CD ∴ By proportionality theorem, ∴ from (1) and (2), AN AL (2) AD AC AM AN AB AD 1 hours. If one pipe takes 3 hours more than 13 the other to fill it, find the time in which each pipe would fill the cistern. Ans: Let time taken by faster pipe to fill the cistern be x hrs. Therefore, time taken by slower pipe to fill the cistern = (x + 3) hrs Since the faster pipe takes x minutes to fill the cistern. 1 ∴ Portion of the cistern filled by the faster pipe in one hour = x 1 Portion of the cistern filled by the slower pipe in one hour = x3 1 13 Portion of the cistern filled by the two pipes together in one hour = 40 40 13 1 1 13 x 3 x 13 According to question, x x 3 40 x( x 3) 40 ⇒ 40 (2x + 3) = 13x (x + 3) ⇒ 80x + 120 = 13x2 + 39x ⇒ 13x2 – 41x – 120 = 0 ⇒ 13x2 – 65x + 24x – 120 = 0 ⇒ 13x (x – 5) + 24 (x – 5) = 0 ⇒ (x – 5) (13x + 24) = 0 Either x – 5 = 0 or 13x + 24 = 0 ⇒ x = 5 as x = −24/13 not possible. Hence, the time taken by the two pipes is 5 hours and 8 hours respectively. OR If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now? [NCERT Exemplar] Ans: Let the present age of Zeba be x years. Age before 5 years = (x – 5) years According to given condition, (x – 5)2 = 5x + 11 ⇒ x2 + 25 – 10x = 5x + 11 ⇒ x2 – 10x – 5x + 25 – 11 = 0 ⇒ x2 – 15x + 14 = 0 ⇒ x2 – 14x – x + 14 = 0 ⇒ x (x – 14) – 1 (x – 14) = 0 ⇒ (x – 1) (x – 14) = 0 ⇒ x – 1 = 0 or x – 14 = 0 ⇒ x = 1 or x = 14 But present age cannot be 1 year. Hence, Present age of Zeba is 14 years. 35. Two pipes running together can fill a cistern in 3 SECTION-E (Case Study Based Questions) Questions 36 to 38 carry 4M each 36. On the roadway, Points A and B, which stand in for Chandigarh and Kurukshetra, respectively, are located nearly 90 kilometres apart. At the same time, a car departs from Kurukshetra and one from Chandigarh. These cars will collide in 9 hours if they are travelling in the same direction, Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 10 - and in 9/7 hours if they are travelling in the other direction. Let X and Y be two cars that are travelling at x and y kilometres per hour from places A and B, respectively. On the basis of the above information, answer the following questions: (a) When both cars move in the same direction, then find the situation can be represented algebraically. [2] OR (a) When both cars move in the opposite direction, then find the situation can be represented algebraically. [2] (b) Find the speed of car x. [1] (c) Find the speed of car y. [1] Ans: (a) Suppose two cars meet at point Q. Then, Distance travelled by car X = AQ, Distance travelled by car Y = BQ. It is given that two cars meet in 9 hours. ∴ Distance travelled by car X in 9 hours = 9x km = AQ = 9x Distance travelled by car Y in 9 hours = 9y km = BQ = 9y Clearly, AQ - BQ = AB = 9x - 9y = 90 = x - y = 10 OR Suppose two cars meet at point P. Then Distance travelled by car X = AP and Distance travelled by car Y = BP. 9 9 In this case, two cars meet in hours. Distance travelled by car X in hours 7 7 9 9 = x km ⇒ AP = x 7 7 9 Distance travelled by car Y in hours 7 9 9 = y km ⇒ BP = y 7 7 Clearly, AP + BP = AB 9 9 9 ⇒ x + y = 90 ⇒ (x + y) = 90 ⇒ x + y = 70 7 7 7 (b) We have x - y = 10 and x + y = 70 Adding equations (i) and (ii), we get 2x = 80 ⇒ x = 40 Hence, speed of car X is 40 km/hr. (c) We have x - y = 10 ⇒ 40 - y = 10 ⇒ y = 30 Hence, speed of car y is 30 km/hr 37. In a toys manufacturing company, wooden parts are assembled and painted to prepare a toy. One specific toy is in the shape of a cone mounted on a cylinder. For the wood processing activity center, the wood is taken out of storage to be sawed, after which it undergoes rough polishing, then is cut, drilled and has holes punched in it. It is then fine polished using sandpaper. For the retail packaging and delivery activity center, the polished wood sub-parts are assembled together, then decorated using paint. The total height of the toy is 26 cm and the height of its conical part is 6 cm. The diameters of the base of the conical part is 5 cm and that of the cylindrical part is 3 cm. On the basis of the above information, answer the following questions: Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 11 - (a) If its cylindrical part is to be painted yellow, find the surface area need to be painted. [1] (b) If its conical part is to be painted green, find the surface area need to be painted. [2] OR (b) Find the volume of the wood used in making this toy. [2] (c) If the cost of painting the toy is 3 paise per sq cm, then find the cost of painting the toy. (Use π = 3.14) [1] Ans: Let the radius of cone be r, slant height of cone be l, height of cone be h, radius of cylinder be r′ and height of cylinder be h′. Then r = 2.5 cm, h = 6 cm, r′ = 1.5 cm, h′ = 26 – 6 = 20 cm and Slant height, l r 2 h 2 2.52 62 6.25 36 42.25 6.5cm (a) Area to be painted yellow = CSA of the cylinder + area of one base of the cylinder = 2πr′h′ + π(r′)2 = πr′ (2h′ + r′) = (3.14 × 1.5) (2 × 20 + 1.5) cm2 = 4.71 × 41.5 cm2 = 195.465 cm2 (b) Area to be painted green = CSA of the cone + base area of the cone – base area of the cylinder = πrl + πr2 – π(r′)2 = π[(2.5 × 6.5) + (2.5)2 – (1.5)2] cm2 = π[20.25] cm2 = 3.14 × 20.25 cm2 = 63.585 cm2 OR Volume of wood used in making the toy = Volume of cone + Volume of cylinder 1 1 1 = r 2 h r '2 h ' r 2 h r '2 h ' 3.14 2.5 2.5 6 1.5 1.5 20 3 3 3 3 3.14(12.5 4.5) 53.38cm (c) Total area of painting = 195.465 + 63.585 = 259.05 cm2 Cost of painting 1 cm2 = Re. 0.03 Total cost of painting = Rs. 0.03 x 256.05 = Rs. 7.77 38. Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure, including microwave dishes. They are among the tallest human-made structures. There are 2 main types: guyed and self-supporting structures. On a similar concept, a radio station tower was built in two sections A and B. Tower is supported by wires from a point O. Distance between the base of the tower and point O is 36 m. From point O, the angle of elevation of the top of section B is 30° and the angle of elevation of the top of section A is 45°. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 12 - (i) What is the height of the section B? (1) (ii) What is the height of the section A? (1) (iii) What is the length of the wire structure from the point O to the top of section A? (2) OR (iii) What is the length of the wire structure from the point O to the top of section B? (2) Ans: Given, that the distance between the base of the tower and point O = 36 m BC (i)Consider ΔOCB, tan 300 = ⇒ = OC Hence, BC= 12√3=20.78 m AB BC (ii)In ΔOAC, tan 450 = ⇒ = 1 ⇒ AC = 36m OC (iii) length of the wire structure from the point O to the top of the section cos 450 = ⇒ OA = 36√2M OR length of the wire structure from the point O to the top of the section cos 300 = ⇒ = ⇒ OB = 72 / √3 = 24√3m Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 13 -