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ICSE Class X Question Bank 2025 : Commercial Studies

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ICSE 2025 EXAMINATION Sample Question Paper - 1 Mathematics Time Allowed: 2 hours and 30 minutes Maximum Marks: 80 General Instructions: Answers to this Paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent reading the question paper. The time given at the head of this Paper is the time allowed for writing the answers. Attempt all questions from Section A and any four questions from Section B. All work, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answers. Omission of essential work will result in a loss of marks. The intended marks for questions or parts of questions are given in brackets [ ] Mathematical tables are provided. Section A 1. Question 1 Choose the correct answers to the questions from the given options: (a) A retailer purchases a fan for 1500 from a wholesaler and sells it to a consumer at 10% profit. If the [15] [1] sales are intra-state and the rate of GST is 12%, the cost of the fan to the consumer inclusive of tax is: (b) a) 1848 b) 1830 c) 1650 d) 1800 A factory kept increasing its output by the same percentage every year. Then, the percentage, if it is [1] known that the output is doubled in the last two years, will be (c) a) 44.4% b) 14.4% c) 41.4% d) 44.1% When ax3 + 6x2 + 4x + 5 is divided by (x + 3), the remainder is -7. [1] The value of constant a is (d) a) 2 b) -2 c) -3 d) 3 If A = a) 3 1 1 2 [ ] 87 149 [ ] 149 [1] , then the value of matrix A5 is 62 b) 87 149 149 62 [ ] c) 62 ] 149 (e) d) 149 [ 62 149 149 87 [ 87 ] An AP starts with a positive fraction and every alternate term is an integer. If the sum of the first 11 [1] terms is 33, then the fourth term is (f) (g) a) 3 b) 6 c) 5 d) 2 If (4, 3) and (-4, -3) are opposite two vertices of a rectangle, then other two vertices are a) (4, -3) and (-4, 3) b) (-4, -3) and (-4, -3) c) (-4, 4) and (-3, 4) d) (4, -3) and (-3, 4) [1] Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting [1] AC at L and AD produced at E. The values of EL and ar ( AEL) are respectively (h) a) ar ( CBL) and BL b) 2BL and 4 ar ( CBL) c) 4 ar ( CBL) and 2BL d) BL and ar ( CBL) A sphere of radius a units is immersed completely in water contained in a right circular cone of semi- [1] vertical angle 30 and water is drained off from the cone till its surface touches the sphere. Then, the volume of water remaining in the cone will be a) c) (i) 5 3 b) 2 a a3 5 3 3 d) 5 a3 a 3 Graph the range of the inequation 2 2 3 x+ 1 3 3 1 3 , x R on the number line. If the solution [1] set is consider as a diagonal of a square on the number line, then the area of obtained figure, is (j) (k) a) 11 sq units b) 14 sq units c) 17 sq units d) 18 sq units The probability that the minute hand lies from 5 to 15 min in the wall clock, is a) 1 c) 1 b) 6 d) 5 If A = 3 0 0 3 [ ] a) [ 3n 0 0 (l) 5 6 1 10 , then An (where, n is a natural number) is equal to ] b) n 3 [ 3n c) [ 3 0 0 3 ] [1] 1 0 0 1 [1] ] d) l2 2 The sum of the squares of the distances of a moving point (x, y) from two fixed points (a, 0) and (- a, [1] 0) is equal to a constant quantity 2b2. The value of x2 + y2 + a2 is equal to (m) a) b2 b) -a2 c) ab d) -b2 If P, Q, S and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. Then, the perimeter of the quadrilateral PQSR will be [1] (n) a) 2( 3 + 1)r b) 2 3 + r c) 2r d) 2 3r Observe the data given in three sets [1] P: 3, 5, 9, 12, x, 7, 2 Q: 8, 2, 1, 5, 7, 9, 3 R: 5, 9, 8, 3, 2, 7, 1 If the ratio between P's and Q's means is 7 : 5, then the ratio between P s and R s means is (o) a) 7 : 5 b) 5 : 7 c) 6 : 7 d) 7 : 6 Assertion (A): Sum of first 10 terms of the arithmetic progression -0.5, -1.0, -1.5, ... is 27.5 Reason (R): Sum of n terms of an A.P. is given as Sn = n 2 [2a + (n 1)d] [1] where a = first term, d = common difference. a) Both A and R are true and R is the correct explanation of A. c) A is true but R is false. 2. b) Both A and R are true but R is not the correct explanation of A. d) A is false but R is true. Question 2 (a) [12] Mrs. chopra deposits 1600 per month in a Recurring Deposit Account at 9% per annum simple [4] interest. If she gets 65592 at the time of maturity, then find the total time for which the account was held. 3. (b) Find the mean proportional of (a4 - b4)2 and [(a2 - b2)(a - b)]-2. (c) If cosec = x + 1 4x , then prove that cosec + cot = 2x or 1 2x [4] . [4] Question 3 (a) [13] The given solid figure is cylinder surmounted by a cone. The diameter of the base of the cylinder is 6 cm. The height of the cone is 4 cm and the total height of the solid is 25 cm. Take = 22 7 [4] . Find the: i. Volume of the solid ii. Curved surface area of the solid Give your answer correct to the nearest whole number. (b) The equation of a line is y = 3x - 5. Write down the slope of this line and the intercept made by its on [4] the Y-axis. Hence or otherwise, write down the equation of a line, which is parallel to the line and which passes through the point (0, 5). (c) Use graph paper for this question (Take 2 cm = 1 unit along both x and y axis). ABCD is a quadrilateral whose vertices are A(2, 2), B(2, - 2), C (0, -1) and D (0, 1) [5] i. Reflect quadrilateral ABCD on the y-axis and name it as A'B'CD. ii. Write down the coordinates of A' and B' iii. Name two points which are invariant under the above reflection. iv. Name the polygon A'B'CD. Section B Attempt any 4 questions 4. Question 4 (a) [10] The price of a Barbie Doll is 3136 inclusive tax (under GST) at the rate of 12% on its listed price. A [3] buyer asks for a discount on the listed price, so that after charging GST, the selling price becomes equal to the listed price. Find the amount of discount which the seller has to allow for the deal. 5. (b) Find the values of k, for which the equation x2 + 5kx + 16 = 0 has no real roots. [3] (c) The mean of the following distribution is 49. Find the missing frequency a. [4] Class Interval 0-20 20-40 40-60 60-80 80 -100 Frequency 15 20 30 a 10 Question 5 [10] (a) Find the values of x, y, a and b, when [ (b) Two chords AB and CD of a circle intersect each other at a point E inside the circle. If AB = 9 cm, x+ y a b a + b 2x 3y ] = [ 5 3 1 5 ] . [3] [3] AE = 4 cm and ED = 6 cm, then find CE. (c) 6. Determine, whether the polynomial g(x) = x - 7 is a factor of f(x) = x3 - 6x2 - 19x + 84 or not. Question 6 [4] [10] (a) Find the points of trisection of the line segment joining the points (5, -6) and (-7, 5). [3] (b) Prove the following identities. [3] i. sin4 + cos4 = 1 - 2 sin2 cos2 ii. (c) 1 cosec cot 1 sin = 1 sin 1 cosec +cot 150 workers were engaged to finish a job in a certain number of days, 4 workers dropped out on [4] second day, 4 more workers dropped out an third day and so on. It took 8 more days of finish the work. Find the number of days in which the work was completed. 7. Question 7 (a) [10] A two-digit positive number, such that the product of its digits is 6. If 9 is added to the number, then [5] the digits interchange their places. Find the number. (b) The marks obtained by 120 students in a test are given below: Marks Number of Students 0 - 10 5 10 - 20 9 20 - 30 16 30 - 40 22 40 - 50 26 50 - 60 18 [5] 60 - 70 11 70 - 80 6 80 - 90 4 90 - 100 3 Draw an ogive for the given distribution on a graph sheet. (Use suitable scale for ogive to estimate the following) i. the median. ii. the number of students who obtained more than 75% marks in the test. iii. the number of students who did not pass the test, if minimum marks required to pass is 40. 8. Question 8 (a) [10] Two players Niharika and Shreya play a tennis match. It is known that the probability of Niharika [3] winning the match is 0.62. What is the probability of Shreya winning the match? (b) A conical military tent is 5 m high and the diameter of the base is 24 m. Find the cost of canvas used [3] in making this tent at the rate of 14 per sq m. (c) In the given figure CE is a tangent to the circle at point C. ABCD is a cyclic quadrilateral. If ABC = [4] 93o and DCE = 35o find: i. ADC ii. C AD iii. AC D 9. Question 9 (a) [10] Given: A = {x : 3 < 2x - 1 < 9, x R}, B = {x : 11 3x + 2 23, x R} where R is the set of real [3] number. i. Represents A and B on number lines ii. On the number line also mark A B. (b) Find the missing frequency for the given frequency distribution table, if the mean, of the distribution [3] is 18. (c) Class interval 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Frequency 3 6 9 13 f 5 4 In the given figure, M = N = 46o. Express x in terms of a, b and c, where a, b and c are the lengths [4] of LM, MN and NK, respectively. 10. Question 10 (a) The ages of A and B are in the ratio 7 : 8. Six years ago, their ages were in the ratio 5 : 6. Find their [10] [3] present ages. (b) Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents [3] to the circle and measure their lengths. (c) The angle of elevation from a point P of the top of a tower QR, 50 m high is 60o and that the tower PT from a point Q is 30o. Find the height to the tower PT, correct to the nearest metre. [4] Solution Section A 1. Question 1 Choose the correct answers to the questions from the given options: (i) (a) 1848 Explanation: { Here, selling price of fan = 1650 GST on fan = 12% of 1650 = 1650 12 100 = 198 Thus, cost of a fan to the consumer inclusive of tax = (1650 + 198) = 1848 (ii) (c) 41.4% Explanation: { Let P be the initial production (2 yr ago) and the increase in production every year be x%. Then, production at the end of first year = P + PX = P (1 + 100 x 100 ) Production at the end of second year = P = (1 + = P (1 + x 100 x ) + Px x ) (1 + 100 [(1 + 100 100 ) x )] 100 = P(1 + 2 x 100 ) Since, the production is doubled in last two years. P(1 + x 100 2 ) = 2P (1 + 2 x 100 ) = 2 (100 + x)2 = 2 (100)2 10000 + x2 + 200x = 20000 On comparing it with ax2 + bx + c = 0, we get a = 1, b = 200 and c = -10000 2 b b 4ac By quadratic formula, x = 2a 2 x= 200 (200) +40000 2 2 = -100 100 = 100(-1 4 2) = 100 (-1 + 1.414) [ percentage cannot be negative] = 100(0.414) = 41.4 Hence, the required percentage is 41.4%. (iii) (a) 2 Explanation: { Let f(x) = ax3 + 6x2 + 4x + 5 By remainder theorem, f(-3) = -7 a(-3)3 + 6(-3)2 + 4(-3) + 5 = -7 -27a + 54 - 12 + 5 = -7 -27a = -54 a = 2 (iv) (c) [ 62 149 149 87 ] Explanation: { Given, A = [ 3 1 1 2 ] Now, A2 = A A = [ 3 1 1 2 ][ 3 1 1 2 ] 9 1 3 + 2 = [ 3 2 1 + 4 8 5 5 3 A4 = A2 A2 = [ 39 55 55 16 = [ 5 5 3 ] ] 8 5 5 3 ][ ] 64 25 40 + 15 40 15 25 + 9 =[ ] Now, A5 = A4 A = [ 39 55 3 1 1 2 ][ 55 117 55 39 + 110 165 + 16 55 32 = [ (v) 8 ]= [ 16 ] 62 149 149 87 ]= [ ] (d) 2 Explanation: { Given, S11 = 33 11 2 (2a + 10d) = 33 [ Sn = n 2 [2a + (n - 1)d] a + 5d = 3 i.e. a6 = 3 a4 = 2 [ alternate terms are integers and the given sum is possible] (vi) (a) (4, -3) and (-4, 3) Explanation: { Since, the image of point (4, 3) under X-axis is (4, - 3) and the image of point (4, 3) under Y-axis is (-4, 3). Other two vertices of the rectangle are (4, -3) and (-4, 3). (vii) (b) 2BL and 4 ar ( CBL) Explanation: { In BMC and EMD, we have BMC = EMD [vertically opposite angles] MC = MD [ M is the mid-point of CD] MCB = MDE [alternate angles] So, by AAS congruence criterion, we have BC = ED [ corresponding parts of congruent triangles are equal] In AEL and CBL, we have ALE = CLB [vertically opposite angles] and EAL = BCL [alternate angles] So, by AA criterion of similarity, we have BM C EM D AEL C BL AE BC = EL BL = AL CL [ if two triangles are similar, then their corresponding sides are proportional] On taking first two terms, we get EL AE = BL BC+BC = BC = 2 [ AD = SC as sides opposite to parallelogram and DE = BC, proved above] 2BC = BC AD+DE = BC BC EL = 2BL ...(i) ar( AEL) Now, ar( CBL) EL = ( BL 2 [ ratio of areas of two similar triangles is equal to the square of the ratio of their ) corresponding sides] = ( 2 2BL = (2)2 [from Eq. (i)] ) BL ar( AEL) = 4 ar( CBL) ar ( AEL) = 4 ar ( CBL) (viii) (b) 5 3 a3 Explanation: { Let radius of sphere be a, i.e. OK = OA = a. Then, the centre O of a sphere will be centroid of the BCD 1 OA = AB AB = 3(OA) 3 In right angled OKB, sin 30o = 1 2 = OK = OB a OB a OB OB = 2a Now, AB = OA + OB = a + 2a = 3a Now, in right angled BAC, = tan 30o AC AB AC = AC = = AB 3 3a = AC AB 1 3 3a 3 units Now, volume of a cone BCD = = 1 3 2 (a 3) 3a 3 (AC)2 AB = 3 a 3 Volume of water remaining in the cone = Volume of the cone BCD - Volume of a sphere = 3 a3 (ix) 1 4 3 a3 = 5 3 (d) 18 sq units Explanation: { Given, 2 x + 8 3 8 3 a3 cu units 2 1 3 3 x + 1 3 3 (x + 1 3 10 3 3 1 3 )3 10 3 3 [multiplying by 3 in each term] -8 3x + 1 10 -8 - 1 3x + 1 - 1 10 - 1 [subtracting 1 from each term] -9 3 x 9 9 3 3x 3 9 3 [dividing by 3 each term] -3 x 3 Since, x R. Range of x is [-3, 3]. Representation of range of x on the number line is given as Now, consider the following figure Here, AC = 6 units, which is a diagonal of square. Let side of a square ABCD be a. In right angled ABC, AC2 = AB2 + BC2 62 = a2 + a2 36 = 2a2 a2 = 18 Now, area of a square ABCD= (Side)2 = a2 = 18 sq units. (x) (a) 1 6 Explanation: { In a wall clock, the minute hand cover the 60 min in on complete round. Total number of possible outcomes = 60 The minute hand cover the time from 5 to 15 min, Number of outcomes favourable to E = Distance from 5 to 15 min = 10 Required probability = = (xi) (b) 3 n 1 0 0 1 [ ] 10 1 60 6 Explanation: { We have, A = [ 3 0 0 3 1 0 0 1 ]= 3[ ] = 3l An = (3l)n = 3nln = 3nl [ ln = l, for all natural numbers n] n = 3 1 0 0 1 [ ] (xii) (a) b2 Explanation: { Let P (x, y) be the moving point. Let given two fixed points be A (a, 0) and B (-a, 0). According to the given condition, PA2 + PB2 = 2b2 (x - a)2 + (y - 0)2 + (x + a)2 + (y - 0)2 = 2b2 [by distance formula] x2 - 2ax + a2 + y2 + x2 + 2ax + a2 + y2 = 2b2 2x2 + 2y2 + 2a2 = 2b2 x2 + y2 + a2 = b2 [dividing both sides by 2] (xiii) (a) 2( 3 + 1)r Explanation: { As PQR is an equilateral triangle, hence PS will be perpendicular to QP and will divide it into 2 equal parts. Since, P and S will be supplementary, so S = 120o and QSA = RSA = 60o Now, PA = PQ cos 30o and OA = OQ sin 30o = AS = OA = Hence, PQ = r 2 and PA = PO + OA = r + r+ PA = cos 30 r 2 r 2 r = 3r 2 3 2 r In QAS, AS = QS cos 60o QS = 2 1 = r 2 Since, AQ = AR, AS is common and QAS = RAS = 90o So, QS = RS. Perimeter of PQSP = 2(PQ + QS) = 2( 3 + 1)r (xiv) (a) 7 : 5 Explanation: { Mean of the observations of sets P= 3+5+9+12+x+7+2 Q= 8+2+1+5+7+9+3 7 7 and R = 38+x = = 7 35 7 5+9+8+3+2+7+1 7 = 5 = 35 7 = 5 Ratio of means of sets P and Q = 7 : 5 [given] Let P s mean = 7y and Q s mean = 5y. 5y = 5 y = 1 Q's mean = 5 Now, P's mean = 7y 38+x 7 = 7 1 38 + x = 48 x = 11 Mean of P : Mean of R = 7y : 5 = 7 1 : 5 = 7 : 5 (xv) (a) Both A and R are true and R is the correct explanation of A. Explanation: { Both are correct. Reason is the correct reasoning for Assertion. Assertion, S10 = 10 2 [2(-0.5) + (10 - 1)(-0.5)] = 5[-1 - 4.5] = 5(-5.5) = 27.5 2. Question 2 (i) p = 1600/month r = 9% p.a. m.v. = 65,592 n = ? m.v. = pn + p r n(n+1) 2400 1600 9 n(n+1) 65,592 = 1600n + 65,592 = 1600n + 6n(n + 1) 65,592 = 1600n + 6n2 + 6n 6n2 + 1606n - 65,592 = 0 2400 2 1606 (1606) 4(6)( 65,592) n= 2(6) n= 1606 4153444 n= 1606 2038 n= 1606+2038 12 12 12 432 n= 12 n = 36 months n = 3 years or n = 1606 2038 12 or n = 3649 12 or n = -303.66 months rejected. As 'n' is no. of months here. So can't be -ve. (ii) Let the mean proportional between (a4 - b4)2 and [(a2 - b2)(a - b)]-2 be x. (a4 - b4)2, x and [(a2 - b2)(a - b)]-2 are in continued proportion. (a4 - b4)2 : x = x : [(a2 - b2)(a - b)]-2 x2 = (a4 - b4)2 [(a2 - b2)(a - b)]-2 4 2 4 (a b ) 2 x = 2 2 2 [( a b )(a b)] 4 4 a b x = 2 2 2 2 ( a b )(a b) 2 2 ( a + b )( a b ) x = 2 ( a2 b )(a b) 2 x = 2 a +b a b 1 (iii)Given, cosec = x + 4x ...(i) We know that, cot2 = cosec2 - 1 cot2 = (x + cot2 = x 2 1 4x 2 ) 1 + 2 - 1 [from Eq. (i)] + 2x 16x =x 2 =x 2 + + 1 2 16x 1 16x2 + 1 1 2 2x cot = x 1 4x or cot = (x =x 1 4x 2 + 1 1 4x 1 2 16x = (x 1 4x [ (a + b)2 = a2 + b2 + 2ab] 1 2 ) 2 [ a2 + b2 - 2ab = (a - b)2] ...(ii) 1 4x ) ...(iii) On adding Eqs. (i) and (ii), we get cosec + cot = 2x Now, adding Eqs. (i) and (iii), we get cosec + cot = 1 2x Hence, cosec + cot = 2x or 1 2x . 3. Question 3 (i) Given total height of the solid = 25 cm Height of the cone (h2) = 4 cm Diameter of the cylinder = 6 cm Height of the cylinder (h1) = 25 - 4 = 21 cm Radius of the cone = Radius of the cylinder = (r) = = 3 cm 6 2 Slant height of cone = h 2 2 = 2 2 4 + 3 = 16 + 9 25 = = 5 cm. 2 + r i. Volume of the solid = Volume of cylinder + Volume of cone = r h + r h 1 2 1 = r 2 = 22 = 22 2 2 3 1 (h1 + 3 h2 ) 3 3 (21 + 7 9 7 67 3 4 3 ) = 631.71 cm3 632 cm3. (Approx.) ii. Curved surface area of the solid = C.S.A of cylinder + C.S.A. of cone = 2 rh1 + rl = r(2h1 + l) = 22 = 22 7 7 3(2 21 + 5) 3 47 = 443.14 cm2 Curved surface area = 443 cm2 (Approx.). (ii) Given eqn of line y = 3x - 5 Comprare with y = mx + c we get. Slope (m) = 3 and y-intercept (c) = -5 Now slope of the line parellel to the given line will be 3 and it passes through (0, 5). Thus eqn of line will be y - y1 = m(x - x1) y - 5 = 3(x - 0) y - 5 = 3x Y= 3x + 5 (iii) i. ii. Co-ordinates of A' (-2, 2) Co-ordinates of B' (-2, - 2) iii. Two invariant points are C (0, -1) and D (0,1). iv. A'B'CD is an isosceles Trapezium polygon. Section B 4. Question 4 (i) Let the List price of doll be x. Total Amount = x + 12% of x =x+ x 12 100 = 112 100 ATQ x= x 112 100 x = 3136 3136 100 112 x = 2800 List price of doll 2800. Now the reduced price of the doll = (2800 - y) amount of GST on (2800 - y) 12% of (2800 - y) = (2800 - y) 12 100 Now the selling price of doll = (2800 - y) + (2800 y) 12 100 = (2800 - y)(1 + 112 = (2800 - y) 100 12 100 ) According to given condition, selling price of doll = list price of doll (i.e 2800) i.e (2800 y) = 2800 112 100 2800 - y = 2800 100 112 2800 - y = 2500 2800 - 2500 = y y = 300 Hence, amount of discount is 300. (ii) Given equation is x2 + 5kx + 16 = 0 On comparing it with ax2 + bx + c = 0, we get a = 1, b = 5k and c = 16 Now, discriminant, D = b2 - 4ac = (5k)2 - 4 1 16 = 25k2 - 64 Since, the given equation has no real roots. D < 0 25k2 - 64 < 0 2 25 (k k2 < (iii) 64 25 64 25 ) 8 5 <0 k 2 <k< Class 64 25 < 0 8 5 Frequency (f) x f 0-20 15 10 150 20-40 20 30 600 40-60 30 50 1500 60-80 a 70 70a 80-100 10 90 900 f = 75 + a fx = 3150 + 70a Given mean = 49 fx or, f = 49 3150+70a 75+a = 49 3150 + 70a = 3675 + 49a 70a - 49a = 3675 - 3150 21a = 525 a = 25 5. Question 5 (i) We know that two matrices are said to be equal if each matrix has the same number of rows and same number of columns. Corresponding elements within each matrix are equal. Given: [ x + y a b a+ b 2x 3y 5 3 1 5 ]= [ x + y = 5 ...(i), 2x - 3y = -5 ...(ii) ] a - b = 3 ...(iii) a + b = -1 ...(iv) Solving eqn (i) and (ii) y = 3 Putting the value of y in eqn (i) x + 3 = 5 x = 2 Again solving eqn (iii) and (iv) a = 1 Putting the value of a in eqn (iv) 1 + b = -1 b = -2 x = 2, y = 3, a = 1, b = -2 (ii) Given, two chords AB and CD are intersect each other at point E. AB = 9 cm AE = 4 cm ED = 6 cm So, BE = AB - AE = 9 - 4 = 5 So, AE EB = DE CE 4 5 = 6 CE CE = = 3.34 4 5 6 (iii)g(x) = 0 x - 7 = 0, x = 7 By factor theorem, g(x) will be a factor of f(x) if f(7) = 0 Now, f(7) = (7) 6 (7) 19 7 + 84 3 2 = 343 - 294 - 133 + 84 = 427 - 427 = 0 f(7) = 0, So, g(x) is a factor of f(x). 6. Question 6 (i) Let P and Q be the points of trisection of AB. Given points be A(5, -6) and B(-7, 5) P divides AB in the ratio 1 : 2 [By section formula, m x2 +n x1 m+n m y +n y , 2 1 m+n ] the coordinate P are 1 ( 7)+2 5 ( , 1+2 P (1, 7 3 ) 1 5+2 ( 6) 1+2 ) =( 7+10 3 , 5 12 3 ) = (1, 7 3 ) Q divides AB in the ratio 2 : 1 then the coordinates of Q are 2 ( 7)+4 5 ( Q ( 3, 2 5+1 ( 6) , 2+1 4 ) 3 ) 2+1 =( 14+5 3 10 6 , 3 ) = ( 3, 4 3 ) Hence, the points of trisection of AB are P (1, 7 3 ) and Q ( 3, 4 3 ) . (ii) i. LHS = sin4 + cos4 = 1 - 2 sin2 cos2 = (sin2 + cos2 )2 - 2 sin2 cos2 [ a2 + b2 = (a + b)2 - 2ab] = 12 - 2 sin2 cos2 = 1 - 2sin2 cos2 [ sin2 A + cos2 A = 1] = RHS Hence proved. ii. 1 cosec cot if 1 1 cosec cot sin = 1 sin 1 + cosec +cot cosec +cot = 1 (cosec cot )(cosec +cot ) i.e. if 2 cosec cosec i.e. if 2 cot 2 cosec 1 2 + sin (cosec +cot )+(cosec cot ) i.e. if is true 1 = 2 sin is true 1 sin is true = 2 cosec is true = 2 cosec is true. [ cosec2 - cot2 = 1] which is true. Hence proved. (iii)Let total work be 1 and let total work completed in days. work dream in 1 day = Total work Number of days to complete work 1 n This is the work done by 150 workers work done by 1 worker in one day = 1 150n Number of workers 150 146 142 work done per worker in 1 day Total work done in 1 day 1 150 150n 150n 1 146 150n 150n 1 142 150n 150n Given that, In this manner, it took 8 more days to finish the work i.e. work finished in (n + 8) days. 150 150n 1 150n + 146 150n + 142 150n + ... + (n + 8) terms = 1 [150 + 146 + 142 + ... + (n + 8) terms] = 1 150 + 146 + 142 + ... +(n + 8) terms = 150n Now a = 150 d = 146 - 150 = -4 diff. is equal It forms A.P. Sn = [2a + (n - 1)d] n 2 150 + 146 + 142 +.. (n = 8) terms = 150n becomes. n+8 2 [2(150) + (n + 8 - 1)] = 150n (n+8) 2 2 [150 - 2(n + 7)] = 150n (n + 8)(150 - 2n - 14) = 150n (n + 8)(136 - 2n) = 150n 136n - 2n2 + 1088 - 16n = 150n 2n2 - 120n - 1088 + 150n = 0 2n2 + 30n - 1088 = 0 n2 + 15n - 544 = 0 n2 + 32n - 17n - 544 = 0 n(n + 32) - 17(n + 32) = 0 (n + 32)(n - 17) = 0 n + 32 = 0 or n = 17 n = -32 n = 17 Reject n = -32 as n should be natural no. n = 17 work was complete in 17 + 8 = 25 days. 7. Question 7 (i) Let the two digit no. be 10x + y product of their digits i.e., xy = 6 y = ...(i) 6 x According to the question 10x + y + 9 = 10y + x 9x - 9y + 9 = 0 x - y = -1 ...(ii) Substituting the value of y from equal (i), = -1 x 6 1 x 2 x 6 x = -1 x2 - 6 = -x x2 + x - 6 = 0 x2 + 3x - 2x - 6 = 0 x(x + 3) - 2(x + 3) = 0 (x - 2)(x + 3) = 0 x = 2, x = - 3 (according to question rejected as digits are never negative) put the value of x in eqn. (i) y= 6 x = 6 2 = 3 Thus, x = 2 and y = 3 Hence, the required no. = 10 2 + 3 = 23 (ii) C.I. f c.f 0 - 10 5 5 10 - 20 9 14 20 - 30 16 30 30 - 40 22 52 40 - 50 26 78 50 - 60 18 96 60 - 70 11 107 70 - 80 6 113 80 - 90 4 117 90 - 100 3 120 N = 120 i. Given, n = 120 which is even Median = th term 120 2 = 60th term Median = 43 ii. The number of students who obtained more than 75% marks in test = 120 - 110 = 10. 8. Question 8 (i) Let E and F denote the events that Niharika and Shreya win the match, respectively. It is clear that, if Niharika wins the match, then Shreya losses the match and if Shreya wins the match, then Niharika losses the match. Thus, E and F are complementary events. P(E) + P(F) = 1 Since, probability of Niharika s winning the match, i.e. P(E) = 0.62 Probability of Shreya s winning the match, P(F) = P (Niharika losses the match) = 1 - P(E) [ P (E) + P (F ) = 1] = 1 - 0.62 = 038 (ii) Given: h = 5m, d = 24 m, r = 12 m 2 2 l = h + r 2 2 = 5 + 12 l = 13 m Convas Required = C.S.A of conical tent = rl = 22 7 12 13 Convas Required = 490.28 m2 Total cost = 490.28 14 = 6814 Hence, total cost of convas used = 6814 (iii) i. ABC + ADC = 180 (Opposite angles of cyclic quadrilateral) 93 + ADC = 180 ADC = 180 93 ADC = 180 93 87o = ii. C AD = EC D (Alternate segment theorem) C AD = 35 iii. In ADC , AC D + C AD + ADC = 180o (Sum of internal angles of a triangle = 180 ) AC D + 35 + 87 = 180 AC D = 180 (35 + 87 ) AC D = 180 122 AC D = 58 9. Question 9 (i) i. A = {x : 3 < 2x - 1 < 9, x R} 3 < 2x - 1 < 9 3 + 1 < 2x - 1 + 1 < 9 + 1 4 < 2x < 10 < < 4 2x 10 2 2 2 2 < x < 5 A = (2, 5) R B = {x : 11 3x + 2 23, x R} 11 3x + 2 23 11 - 2 3x + 2 - 2 23 - 2 9 3x 21 9 3x 21 3 3 3 3 x 7 B = [3, 7] R ii. A B = (2, 5) [3, 7] = [3, 5) (ii) Class Interval Required mid value 11 - 13 3 12 36 13 - 15 6 14 84 15 - 17 9 16 144 17 - 19 13 18 234 19 - 21 f 20 20f 21 - 23 5 22 110 23 - 25 4 24 96 fi = 40 + f ) = mean (x 18 = 704+20f 40+f ( fi xi ) ( fi ) 720 + 18f = 704 + 20f f = 8 (iii) Given: In the given figure. fi xi LMN = PNK = 46o (fi xi ) = 704 + 20f LM || PN (as corresponding angles are equal) Now consider LMK and PNK LMK = PNK (corresponding angles are equal) LKM = PKN (common) LMK PNK (AA similarity) = ML MK NP a x NK = x= b+c c b+c c Hence we get the result x = ac b+c 10. Question 10 (i) Let the present age of A and B are 7x and 8x respectively. Then, 6 years ago their ages are 7x - 6 and 8x - 6 So, 7x 6 8x 6 5 = 6 6(7x - 6) = 5(8x - 6) 42x - 36 = 40x - 30 42x - 40x = -30 + 36 2x = 6 x = = 3 6 2 Hence, the present age of A and B are 21 and 24. (ii) i. Draw a circle with radius 6 cm and centre C. ii. Take a point P at 10 cm from centre and join CP. iii. Draw perpendicular bisector of CP which cuts CP at O. iv. Take O as centre and OC as radius draw a circle which cuts the previous circle at A and B. v. Join PA and PB. vi. PA and PB are required tangents. (iii)We have, RPQ = 60o and PQT = 30o and QR = 50 m Let PT = x m and PQ = y m In PQR, tan 60o = QR PQ 3 = 50 or y = ...(i) 50 3 y In PQT, tan 30o = 1 3 or x = = y 3 PT PQ x y ...(ii) From eq. (i) and (ii), we get = x= = 16.66 50 50 3 3 3 = 17 m (correct to the nearest meter)

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