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CBSE Class 10 Board Exam 2025 : Mathematics

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Umesh Mukati
D.A.V. College (DAV), Jalandhar

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Class X Session 2024-25 Subject - Mathematics (Standard) Sample Question Paper - 7 Time Allowed: 3 hours Maximum Marks: 80 General Instructions: 1. This Question Paper has 5 Sections A, B, C, D and E. 2. Section A has 20 MCQs carrying 1 mark each 3. Section B has 5 questions carrying 02 marks each. 4. Section C has 6 questions carrying 03 marks each. 5. Section D has 4 questions carrying 05 marks each. 6. Section E has 3 case based integrated units of assessment carrying 04 marks each. 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 2 marks questions of Section E 8. Draw neat figures wherever required. Take = 22 7 wherever required if not stated. Section A 1. 2. The number 5+ 2 5 2 [1] is a) an irrational number b) an integer c) not a real number d) a rational number The graph of y = f(x) is shown in the figure for some polynomial f(x). [1] The number of zeroes of f(x) is 3. a) 3 b) 4 c) 0 d) 2 The number of solutions of two linear equations representing intersecting lines is/are Page 1 of 19 [1] 4. a) 1 b) 2 c) 0 d) The product of two consecutive even integers is 528. The quadratic equation corresponding to the above [1] statement, is 5. 6. 7. 8. 9. a) x (x + 2) = 528 b) 2x (2x + 1) = 528 c) (1 + x) 2x = 528 d) 2x (x + 4) = 528 If the sum of n terms of an A.P. is 3n2 + 5n then which of its term is 164? a) 27th b) none of these. c) 28th d) 26th The distance of a point from the x-axis is called [1] a) due point b) origin c) abscissa d) ordinate If the point C(k, 4) divides the join of the points .4(2, 6) and B(5,1) in the ratio 2:3 then the value of k is a) 16 b) 28 c) d) 16 8 5 [1] [1] 5 5 [1] Find the value of a given figure. a) 12 cm. b) 6 cm. c) 10 cm. d) 15 cm. 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 = 90o, then length of PQ is: Page 2 of 19 [1] 10. 11. 12. 13. a) 3 cm b) 4 cm c) 2 cm d) 2 2 cm How many tangents can be drawn to a circle from a point on it? a) Two b) Zero c) Infinite d) One [1] If x = a cos and y = b sin , then the value of b2x2 + a2y2 is a) a + b b) a2b2 c) a - b d) ab sin + 1 cot cos 1 tan [1] is equal to [1] a) sin + cos b) sin - cos c) 0 d) 1 The top of a broken tree has its top end touching the ground at a distance 15 m from the bottom, the angle made [1] by the broken end with the ground is 30o. Then length of broken part is a) 10 m c) 3 m 14. 15. b) 10 3 m d) 5 3 m Find the area of the sector of a circle having radius 6 cm and of angle 30o. a) 8.42 cm2 b) 9 cm2 c) 9.42 cm2 d) 9.52 cm2 A piece of wire 20cm long is bent into the form of an arc of a circle subtending an angle of 60o at its centre. The [1] [1] radius of the circle is a) c) 16. 17. 18. 20 6+ 60 cm cm b) d) 30 6+ 15 6+ cm cm In a family of two children, the probability of having at least one girl is: a) 1 c) 3 2 4 b) 2 d) 1 5 4 If a letter of English alphabet is chosen at random, then the probability of this letter to be a consonant is: a) 11 c) 21 13 26 b) d) [1] [1] 10 13 5 26 The marks obtained by 9 students in Mathematics are 59, 46, 31, 23, 27, 40, 52, 35 and 29. The mean of the data [1] Page 3 of 19 is 19. a) 30 b) 41 c) 23 d) 38 Assertion (A): In a solid hemisphere of radius 10 cm, a right cone of same radius is removed out. The surface area of the remaining solid is 570.74 cm2 [Take = 3.14 and 2 [1] = 1.4] Reason (R): Reason (R): Expression used here to calculate Surface area of remaining solid = Curved surface area of hemisphere + Curved surface area of cone a) Both A and R are true and R is the correct b) Both A and R are true but R is not the explanation of A. correct explanation of A. c) A is true but R is false. 20. d) A is false but R is true. Assertion (A): The sum of the first n terms of an AP is given by Sn = 3n2 - 4n. Then its nth term an = 6n - 7 [1] Reason (R): nth term of an AP, whose sum to n terms is Sn, is given by an = Sn - Sn - 1 a) Both A and R are true and R is the correct b) Both A and R are true but R is not the explanation of A. correct explanation of A. c) A is true but R is false. d) A is false but R is true. Section B 5 2 21. Prove that is irrational. 22. ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersects AE at F, prove that AF F B = EF F D 23. [2] [2] . In the given figure, common tangents AB and CD to the two circles with centres O1 and O2 intersect at E. Prove [2] that AB = CD. 24. If sin (A + B) = 1 and cos (A B) = 1, find A and B. [2] OR Prove the trigonometric identity: sec4 - sec2 = tan4 + tan2 . 25. In the given figure, two concentric circles with centre O are shown. Radii of the circles are 2 cm and 5 cm respectively. Find the area of the shaded region. Page 4 of 19 [2] OR Reeti prepares a Rakhi for her brother Ronit. The Rakhi consists of a rectangle of length 8 cm and breadth 6 cm inscribed in a circle as shown in the figure. Find the area of the shaded region. (Use = 3 14) Section C 26. On morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. [3] What is the minimum distance each should walk so that each can cover the same distance in complete steps? 27. The angle of elevation of a jet plane from a point A on the ground is 60 . After a flight of 30 seconds, the angle 3 of elevation changes to 30 . If the jet plane is flying at a constant height of 3600 [3] m, find the speed of the jet plane. 28. Solve: 1 2x 3 + 1 x 5 = 1 1 9 ,x 3 2 [3] ,5 OR At t minutes past 2 pm the time needed by the minutes hand of a clock to show 3 pm was found to be 3 minutes less than 29. 2 t 4 minutes. Find t. If radii of the two circles are equal, prove that AB = CD where AB and CD are common tangents. [3] OR In the given figure, PT and PS are tangents to a circle with centre O, from a point P, such that PT = 4 cm and TPS = 60o. Find the length of the chord TS. Also, find the radius of the circle. 2 p 1 [3] 30. If sec + tan = p, show that 31. Find the mean of the following frequency distribution: 2 = sin p +1 Class [3] Frequency Page 5 of 19 0-10 12 10-20 18 20-30 27 30-40 20 40-50 17 50-60 6 Section D 32. Points A and B are 70 km. apart on a highway. A car starts from A and another car starts from B simultaneously. [5] If they travel in the same direction, they meet in 7 hours, but if they travel towards each other, they meet in one hour. Find the speed of the two cars. OR Ved travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and the rest by car. He takes 20 minutes longer if he travels 200 km by train and the rest by car. Find the speed of the train and the car. 33. In a trapezium ABCD, AB | | DC and DC = 2AB. EF | | AB, where E and F lie on BC and AD respectively such that 34. BE EC = 4 3 [5] . Diagonal DB intersects EF at G. Prove that, 7EF = 11AB. The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the [5] product of two middle terms is 7 : 15. Find the number. OR If the ratio of the sum of the first n terms of two APs is (7n +1): (4n + 27) then find the ratio of their 9th terms. 35. The following table shows the ages of the patients admitted in a hospital during a month: [5] Age(in years) 6 - 15 16 - 25 26 - 35 36 - 45 46 - 55 56 - 65 Number of patients 6 11 21 23 14 5 Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Section E 36. Read the following text carefully and answer the questions that follow: Aashish and his family went for a vacation to Manali. There they had a stay in tent for a night. Aashish found that the tent in which they stayed is in the form of a cone surmounted on a cylinder. The total height of the tent is 42 m, diameter of the base is 42 m and height of the cylinder is 22 m. i. What is curved surface area of cone? (1) ii. If each person needs 126 m2 of floor, then how many persons can be accommodated in the tent? (1) iii. What is the curved surface area of cylinder? (2) OR Page 6 of 19 [4] How much canvas required to make a tent? (2) 37. Read the following text carefully and answer the questions that follow: [4] Two friends Govind and Pawan decided to go for a trekking. During summer vacation, they went to Panchmarhi. While trekking they observed that the trekking path is in the shape of a parabola. The mathematical representation of the track is shown in the graph. i. What are the zeroes of the polynomial whose graph is given? (1) ii. What will be the expression of the given polynomial p(x)? (1) iii. What is the product of the zeroes of the polynomial which represents the parabola? (2) OR In the standard form of quadratic polynomial, ax2 + bx + c, whar are a, b, and c? (2) 38. Read the following text carefully and answer the questions that follow: Some students were asked to list their favourite colour. The measure of each colour is shown by the central angle of a pie chart given below: i. If a student is chosen at random, then find the probability of his/her favourite colour being white? (1) ii. What is the probability of his/her favourite colour being blue or green? (1) iii. If 15 students liked the colour yellow, how many students participated in the survey? (2) OR What is the probability of the favourite colour being red or blue? (2) Page 7 of 19 [4] Solution Section A 1. (a) an irrational number 5+ 2 Explanation: 5 2 5+ 2 = 5+ 2 5 2 5+ 2 2 ( 5+ 2) = 2 2 ( 5) ( 2) 2 2 = ( 5) + ( 2) +2 5 2 = 5+2+2 10 = 7+2 10 5 2 3 3 Here 10 = 2 5 Since 2 and Therefore, 5 5+ 2 5 2 both are an irrational number is an irrational number. 2. (c) 0 Explanation: Here y = f(x) is not intersecting or touching the X-axis. Number of zeroes of f(x) = 0 3. (a) 1 Explanation: The number of solutions of two linear equations representing intersecting lines is 1 because two linear equations representing intersecting lines has a unique solution. 4. (a) x (x + 2) = 528 Explanation: Let the first number = x Second number = x + 2 According to question x(x + 2) = 528 5. (a) 27th Explanation: Sum of n terms of an A.P = 3n2+5n Let a be the first term and d be the common difference S = 3n + 5n S = 3(1) + 5 1 = 3 + 5 = 8 S = 3(2) + 5 2 = 12 + 10 = 22 First term (a)=8 2 n 2 1 2 2 a2 = S2 S1 = 22 8 = 14 d = a2 a1 = 14 8 = 6 Now an = a + (n - 1)d 164 = 8 + (n 1) 6 6n - 6 = 164 - 8 = 156 6n = 156 + 6 = 162 n = = 27 162 6 164 is 27th term 6. (d) ordinate Explanation: The distance of a point from the x-axis is the y (vertical) coordinate of the point and is called ordinate. 7. (d) 16 5 Page 8 of 19 Explanation: By Section Formula, The X-coordinate of C = K = 16 5 2(5)+3(2) 2+3 8. (c) 10 cm. Explanation: Here, CAD = 180 (130 + 25 ) = 25o Now, since CAD = DAB, therefore, the AD is the bisector of BAC. Since the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. BD AB = DC AC 15 6 x = 9 x 6 = 15 9 = 10 cm 9. (b) 4 cm Explanation: Join OR & OQ Now PQOR becomes a quadrilateral QPR = 90o (given) PQ = PR ( tangent from external point) OQ = OR = (radius of same circle) OQP = ORP = 90o ( tangents and radius are perpendicular) QOR = 360o - OQP - QPR - ORP QOR = 360o - 90 - 90 - 90 QOR = 90o PQOR becomes a square sides of a square are same 10. PQ = 4 cm (a) Two Explanation: Two 11. (b) a2b2 Explanation: Given: x = a cos and y = b sin b2x2 + a2y2 = b2(a cos )2 + a2(b sin )2 = b2a2 cos2 + a2b2sin2 b2x2 + a2y2 = a2b2(cos2 + sin2 ) = a2b2 [ sin2 + cos2 = 1] 12. (a) sin + cos Explanation: We have, = sin sin sin cos 2 = sin sin cos 2 = = sin cos 2 sin cos + sin 1 cot cos cos cos sin cos 2 + cos 1 tan = sin 1 c os sin + cos 1 sin c os sin cos (sin +cos )(sin cos ) sin cos = sin + cos 13. (b) 10 3 m Explanation: Let the length of broken part (BC) = x Page 9 of 19 In ABC, cos 30o = x x = 15 x 3 2 = 15 5 3 2 = 10 3 m 14. (c) 9.42 cm2 Explanation: Radius of a circle = r = 6 cm Central angle = = 30o Area of the sector = 3.14 6 6 30 = ( 360 ) 2 r 360 cm2 = 9.42 cm2 15. (c) 60 cm Explanation: Given: Length of arc = 20 cm 2 r= 20 360 360 r 3 = 20 r( r( 2 r= 20 60 3 r = 3 ) = 20 ) = 20 60 cm 16. (c) 3 4 Explanation: 3 4 17. (c) 21 26 Explanation: We have, Number of vowels = 5 ( a, e, i, o, u) Number of consonants = 21 ( 26 - 5 = 21 ) Number of possible outcomes = 21 Number of total outcomes = 26 Required Probability = 21 26 18. (d) 38 Explanation: Given: 59, 46, 31, 23, 27, 40, 52, 35 and 29 Mean = Sum of all observations Number of observations = 59+46+31+23+27+40+52+35+29 = 342 9 9 = 38 19. (d) A is false but R is true. Page 10 of 19 Explanation: A is false but R is true. 20. (a) Both A and R are true and R is the correct explanation of A. Explanation: nth term of an AP be an = Sn - Sn - 1 an = 3n2 - 4n - 3(n - 1)2 + 4(n - 1) an = 6n - 7 So, both A and R are true and R is the correct explanation of A. Section B 21. Let us assume that 5 2 is rational. Then, there exist positive co-primes a and b such that 5 2 = a b 2 = a 2 = a 5b 5 b b As a-5b and b are integers . So, a 5b is rational number . b 2 But is not rational number . Since a rational number cannot be equal to an irrational number. Our assumption that 5 2 is rational wrong. Hence, 5 2 is irrational. 22. Given: ABCD is a parallelogram and E is a point on BC. The diagonal BD intersects AE at F. To prove: AF F B = EF F D Proof: Since ABCD is a parallelogram, then its opposite sides must be parallel. In ADF and EBF FDA = EBF and F AD = F EB [Alternate interior angles] [vertically opposite angles] Therefore,by AAA criteria of similar triangles,we have, ADF = EBF Since the corresponding sides of similar triangles are proportional. Therefore,we have, AF D = BF E AF FD = EF FB AF F B = EF F D 23. We know that tangent segments to a circle from the same external point are congruent. So, EA = EC for the circle having centre O1 And, ED = EB for the circle having centre O2 Now, Adding ED on both sides in EA = EC , we get EA + ED = EC + ED EA + EB = EC + ED AB = C D 24. sin(A + B) = 1 = sin 90 A + B = 90 (1) cos(A B) = 1 = cos 0 Solving (1) and (2) we get A B = 0 (2) A = 45 , B = 45 OR L.H.S = sec4 - sec2 (sec2 -1) = sec2 = sec2 (tan2 ) [ 1 +tan 2 tan2 ) tan2 = (1+ Hence proved. = tan2 25. Area of sector OABC = Area of sector OED = = sec 2 + tan4 2 5 60 360 2 2 60 Area of shaded region = 360 25 6 - = 4 6 or tan 2 2 = sec 1] = R.H.S = 4 6 = 25 6 cm2 cm2 21 6 22 7 = 11 cm2 Page 11 of 19 OR Diagonal of rectangle = 6 + 8 = 10 Radius of circle r = = 5 2 2 10 2 Area of circle = 3.14 5 5 = 78.5 Area of rectangle = 6 8 = 48 Area of shaded region = 78.5 - 48 = 30.5 cm2 Area of shaded region is 30.5 cm2 Section C 26. Since, the three persons start walking together. The minimum distance covered by each of them in complete steps = LCM of the measures of their steps 40 = 8 5 = 2 5 3 42 = 6 7 = 2 3 7 Hence LCM (40, 42, 45) 2 45 = 9 5 = 3 5 The minimum distance each should walk so that each can cover the same distance 3 = 2 2 3 5 7 = 8 9 5 7 = 2520 = 2520 cm = 25.20 meters. 27. Let P and Q be the two positions of the plane and let A be the point of observation. Let ABC be the horizontal line through A. It is given that angles of elevation of the plane in two positions P and Q from point A are 60 and 30 respectively. PAB = 60o, QAB = 30o. It is also given that PB = 3600 3 metres In ABP, we have tan 60o = 3 = BP AB 3600 3 AB AB = 3600 m In ACQ, we have tan 30o = 1 3 = CQ AC 3600 3 AC AC = 3600 3 = 10800 m PQ = BC = AC - AB = 10800 - 3600 = 7200 m Thus, the plane travels 7200 m in 30 seconds. Hence, Speed of plane = = 240 m / sec = 60 60 = 864 km / hr 7200 240 30 1000 28. The given equation is: 1 2x 3 + 1 x 5 = x 5+2x 3 (2x 3)(x 5) 3x 8 (2x 3)(x 5) 10 9 = = 10 9 10 9 27x - 72 = 10[(2x - 3) ( x - 5 )] ( By cross multiplication method) 27x - 72 = 10[2x2 - 10x - 3x + 15] 27x - 72 = 10[2x2 - 13x + 15] Page 12 of 19 27x - 72 = 20x2 - 130x + 150 20x2 - 157x + 222 = 0 Here , a = 20, b = -157, c = 222 Therefore, by quadratic formula we have: 2 b b 4ac x = 2a 2 157 ( 157) 4(20)(222) x = x = 40 157 24649 17760 40 157 6889 x = x = x = 40 157 83 40 157+83 40 x = 6 or x = or x = 157 83 40 37 20 OR Since, there are 60 minutes gap between 2 PM & 3 PM. Time needed by minutes hand after t minutes past 2 PM to show 3 PM = (60 - t) minutes According to the question ; 60 - t = 2 - 3 t 4 2 63 = t 63 = t +4t 4 + t 2 4 t2 + 252 = 4t t2 + 4t - 252 = 0 t2 + 18t - 14t - 252 = 0 t(t + 18) - 14(t + 8) = 0 (t + 18)(t - 14) = 0 t + 18 = 0 or t - 14 = 0 t = -18 or t = 14 Since time cannot be negative, t -18 Hence, t = 14 minutes. 29. Given: AB and CD are two common tangents to two circles of equal radii. To prove Construction: OA, OC, O'B and O'D proof Now, OAB = 90 and OCD = 90 as OA AB and OC CD A tangent at any point of a circle is perpendicular to radius through the point of contact Thus, AC is a straight line. Also, O BA = O DC = 90 A tangent at a point on the circle is perpendicular to the radius through point of contact so ABCD is a quadrilateral with four sides as AB, BC, CD and AD But as A = B = C = D = 90 so, ABCD is a rectangle. Hence, AB = CD opposite sides of the rectangle are equal. OR PT = PS (tangents from an external point P) PTS = PST Page 13 of 19 Using Angle Sum Property in PTS PTS + PST + TPS = 180o 2 PTS = 180 - 60 = 120o PTS = 60o PTS Is a equilateral triangle So, TS = 4 cm Now, In PTO As PO is angle bisector of TPS, OTP = 90o tan 30o = 1 3 = OT = OT = OT 4 4 3 OT TP 4 3 cm 3 4 3 radius of circle = cm 3 30. We have, 2 2 LHS = p 1 2 p +1 (sec +tan ) 1 = LHS = 2 sec LHS = 2 + tan sec 2 2 LHS = tan sec 2 1)+ tan 2 2 2 + tan LHS = 2 tan 2 sec 31. +2 sec tan +2 sec tan + sec 2 2 +2 sec tan +2 sec tan +(1+ tan 2 +2 sec tan 1 sec 2 + tan2 +2 sec tan +1 (sec (sec +tan ) +1 2 +2 tan sec +2 sec tan 2 = ) 2 tan (tan +sec ) 2 sec (sec +tan ) = tan sec = sin cos sec = sin =RHS Class interval Frequency Midpoint fixi 0-10 12 5 60 10-20 18 15 270 20-30 27 25 675 30-40 20 35 700 40-50 17 45 765 50-60 6 55 330 Total fi =100 fi xi = 2800 fi xi Mean= = fi 2800 100 = 28 Section D 32. Suppose, P and Q be the cars starting from A and B respectively and let their speeds be x km/hr and y km/hr respectively. Case I: When the cars P and Q move in the same direction. Distance covered by the car P in 7 hours = 7x km Distance covered by the car Q in 7 hours = 7y km Let the cars meet at point M. AM = 7x km and BM = 7y km AM BM = AB 7x 7y = 70 x y = 10 ..........(i) Case II: 7(x y) = 70 Page 14 of 19 When the cars P and Q move in the opposite directions. Distance covered by the car P in 1 hour = x km Distance covered by the car Q in 1 hour = y km In this case, let the cars meet at the point N AN = x km and BN = y km x + y = 70 ........(ii) Adding (i) and (ii), we get AN + BN = AB 2x = 80 Putting x = 40 in (i), we get 40 y = 10 y = (40 10) = 30 x = 40, y = 30 x = 40 Speed of car starting from point A = 40km/hr. Speed of car Starting from point B = 30km/hr. OR Suppose the speed of the train be x km/hr and the speed of the car be y km/hr. CASE I Distance covered by car is (600 120)km = 480km. Now, Time taken to cover 480 km by train hrs [ Time = Time taken to cover 480 km by car = 120 + x 8( 480 15 15 y 60 60 Speed hrs ....................(i) 1 = 0 y y Distance x ) = 8 = 1 y + x 60 + x + x = 8 y 15 480 120 CASE II Distance travelled by car is (600 200)km = 400km Now, Time taken to cover 200km by train = hrs 200 x Time taken to cover 400km by train = 400 y hrs In this case the total time of journey is 8 hour 20 minutes 200 200 200 400 400 8 16 + x 1 25 and 60 hrs = 8 1 3 hrs] 3 1 = 0 y = u x 3 = 1 y 48 + x Putting 48 20 ) = 1 = y + x 24 y [ 8 hrs 20 minutes = 8 3 3 16 + x 1 25 = y 25 ( 24 = 8 y + 8 = 8 hrs 20 minutes y + x x 400 + x 1 y ....................(ii) = v in equations (i) and (ii), we get 15 u + 60v -1=0 .............................(iii) 24u + 48v -1=0 ..............................(iv) By using cross-multiplication, we have u 60 1 48 1 u u = 12 u = Now, u = and, v = 1 y = 9 60 1 80 720 60 1 1 1 = 720 1 = 15+24 v 12 x 15 1 24 1 v = 60+48 v = = 1 y 1 x 15 48 24 60 1 720 1440 and v = = 1 = 9 720 1 80 x = 60 y = 80 = Page 15 of 19 ] Speed of train = 60km/hr Speed of car=80km/hr. 33. In a trapezium ABCD, AB|| DC ,. EF || AB and CD=2AB BE and also 4 = EC -------------(1) 3 AB || CD and AB || EF = = AF BE 4 FD EC 3 In BGE and BDC BEG = BC D ( corresponding angles) GBE = DBC (Common) BGE BDC [ By AA similarity] EG N ow, BE = CD ...........(2) BC from (1) EC EC EC+BE BC BE BE 3 = 4 + 1 4 7 = 7 So EG = 4 or BE 4 CD 7 FG DA AB = = EC 7 ......(3) DBA (by AA similarity) 7 3 AB 7 4 ...(4) BE = 7 3 = BC 4 AB = AD = CD 3 [ 7 EG FG FG = AF 4 = BC Similarly, DGF 3 4 + 1 = from equation (2), DF 4 = 3 = BE BE BE EC BC DE = 7 ] DA Adding equations (3) and (4),we get, 4 EG + F G = EF = = 8 7 4 3 CD + 7 (2AB) + 7 3 AB + 7 AB = 7 11 7 AB 3 7 AB AB 7EF = 11AB 34. Let the four parts be (a-3d), (a-d), (a+d) and (a+3d). Then, (a-3d)+(a-d)+(a+d)+(a+3d)=32 4a=32 a=8 It is given that (a 3d)(a+3d) (a d)(a+d) 2 a 9d 2 a d 64 d 2 2 64 9d = = 7 15 2 2 = 7 15 7 15 960-135d2=448-7d2 128d2=512 d2=4 d= 2 When d=2, a-3d=8-3(2)=2 Page 16 of 19 a-d=8-2=6 a+d=8+2=10 a+3d=8+3(2)=14 when d=-2, a-3d=8-3(-2)=14 a-d=8-(-2)=10 a+d=8+(-2)=6 a+3d=8+3(-2)=2 Thus, the four parts are 2,6,10,14 or 14,10,6,2 OR Let a1 and a2 be the first terms and d1 and d2 be the common difference of the two APs respectively. Let Sn and S'n be the sums of the first n terms of the two APs and Tn and T'n be their nth terms respectively. n Then, Sn S 7n+1 = 4n+27 n 2 n 2 2a1 +(n 1) d1 2a2 +(n 1) d2 [2a1 +(n 1) d1 ] 7n+1 = 7n+1 = 4n+27 [2a2 +(n 1) d2 ] ........(i) 4n+27 To find the ratio of mth terms, we replace n by (2m -1) in the above expression. Replacing n by (2 9 -1), i.e., 17 on both sides in (i), we get 2a1 +(17 1) d1 = 2a2 +(17 1) d2 a1 +8d1 24 19 a1 +(9 1) d1 a1 +(9 1) d1 T9 T = 24 19 24 19 24 = a2 +(9 1) d2 2a1 +16d1 2a2 +16d2 = 120 95 = a2 +(9 1) d2 9 = a2 +8d2 7 17+1 4 17+27 19 required ratio = 24:19. Frequency fi Class Interval Class marks xi ui = = 35. ( xi A) h fiui ( xi 40.5) 10 5.5 - 15.5 6 10.5 -3 -18 15.5 - 25.5 11 20.5 -2 -22 25.5 - 35.5 21 30.5 -1 -21 35.5 - 45.5 23 40.5 = A 0 0 45.5 - 55.5 14 50.5 1 14 55.5 - 65.5 5 60.5 2 10 Total we know that, Mean, x = A + {h ( 37) = 40.5 + {10 = 40.5 8 A = 40.5, h = 10 37 fi = 80 80 } f ui i fi } The modal class of the given data is 35.5 - 45.5. xk = 35.5, h = 10, fk = 23, fk-1 = 21, fk+1 = 14 = 40.5 4.63 = 35.87 we know that, Mode, Mo = x k = 35.5 + {10 = 35.5 + ( 10 2 11 + {h (23 21) (2 23 21 14) ) (fk fk 1 ) } (2fk fk 1 fk+1 ) } = 35.5 + 1.82 = 37.32 Clearly, mode > mean. Page 17 of 19 fiui = -37 Section E 36. i. Curved surface area of cone = rl = 22 7 21 x29 = 1914 m2 [ l = r 2 2 + h 1 2 2 = (21) + (20) = 841 = 29 m] ii. Area of floor = r2 = 22 7 21 21 = 1386 m2 Number of persons that can be accommodated in the tent = 1386 126 = 11 iii. Curved surface area of cylinder = 2 rh = 22 7 21 x 44 = 2904 m2 OR Required area of canvas = Curved surface area of cone + Curved surface area of cylinder = rl + 2 rh = r (l + 2h) = 22 7 21 (29 + 44) = 4818 m2 37. i. Point of intersection of graph of polynomial, gives the zeroes of the polynomial. zeroes = -4 and 7 ii. Since, zero's are = -4, = 7 = -4 + 7 = 3 = -4 7 = -28 + P(x) = x2 - (Sum of zeroes)x + product of zeroes P(x) = x2 - 3x + (-28) P(x) = x2 - 3x - 28 iii. Product of zeroes = -4 7 = -28 Page 18 of 19 OR a is a non-zero real number, b and c are any real numbers c. 38. i. Since, Total angle in a pie chart is 360o So, Total no. of Sample Space = 360 Let 'E' be the event of having 'White' as favourite colour. P(E) = = 120 = 1 360 f avourable outcome T otal Outcome 3 ii. P(Blue or Green) = = 120 360 = 60+60 360 1 3 iii. Since, Yellow represent 90o in the Pie Chart 90o = 15 Students 360o = 15 90 360 = 60 students Hence, 60 Students participated in the survey. OR P(Red or Blue) = = 90 360 = 30+60 360 1 4 Page 19 of 19

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