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ICSE Class X Sample / Model Paper 2026 : Mathematics

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ICSE 2026 EXAMINATION Sample Question Paper - 5 Mathematics Time Allowed: 3 hours 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) [15] A hollow cone of radius 6 cm and height 8 cm is vertical standing at the origin, such that the vertex of [1] the cone is at the origin. Some pipes are hanging around the circular base of the cone, such that they touch the surface of the graph paper. Then, the total surface area of the formed by the figure will be (b) (c) (d) a) 948.84 cm2 b) 494.68 cm2 c) 484.98 cm2 d) 489.84 cm2 The sum of a number as its square is 20, then, the number is a) -5 only b) 5 or -4 c) 2 or 3 d) -5, or 4 When x3 - 3x2 + 5x - 7 is divided by x - 2, then the remainder is a) 2 b) 0 c) -1 d) 1 Solve for x : 8 [ 12 48 32 8 28 ] = x[ 2 a) 4 8 7 [1] [1] ] b) 0.5 c) 0.25 [1] d) 2 Page 1 of 19 (e) (f) The sum of the series 452 - 432 + 442 - 422 + 432 - 412 + 422 - 402 + ... upto 30 terms is a) 2220 b) 1110 c) 4440 d) 3330 A Recurring Deposit Account of 1200 per month has a maturity value of 12440. If the rate of [1] [1] interest is 8% and the interest is calculated at the end of every month, then find the time of this Recurring Deposit Account. (g) a) 20 months b) 50 months c) 30 months d) 10 months 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) BL and ar ( CBL) b) 4 ar ( CBL) and 2BL c) ar ( CBL) and BL d) 2BL and 4 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) 3 c) (i) 3 3 The solution set of the inequation c) ( , (k) (l) 5 3 2 a d) 5 a3 a a) ( , (j) b) a3 5 5 3 1 5+3x 0 is [1] ) 5 3 ) b) ( d) ( 5 3 , ) 5 3 , ) The probability that a non-leap year selected at random will have 53 Mondays is a) 3 c) 1 b) 0 7 d) 7 If A = [1] [ 5 5 0 0 ][ 0 0 5 5 200 ] and A n = [ 5 0 2 7 200 5 ] , then the value of n is [1] 0 a) 50 b) 100 c) 75 d) 25 Suppose PQ be a pole, whose coordinates are P(1, 3) and 0(3, 3) and A be the position of a man [1] whose coordinates are (1, 1). i. If a pole makes an angle of elevation to the point A, then the angle is ii. Also, if we shift the origin at (1, 1), then the angle is (m) a) 75, 45 b) 45 , 60 c) 45 , 90 d) 45 , 45 From a point O, 13 cm away from the centre of a circle, the length of tangent P Q to the circle is 12 cm. The radius of the circle (in cm) is Page 2 of 19 [1] (n) 313 a) 25 b) c) 5 d) 1 For an ogive, the cumulative frequencies are plotted against which of the following? a) Mid-point of the class interval [1] b) Lower limits of class intervals or Midpoint of the class interval c) Upper limits of class intervals (o) d) Lower limits of class intervals Assertion (A): an - an - 1 is not independent of n then the given sequence is an AP. [1] Reason (R): Common difference d = an - an - 1 is constant or independent of n. a) Both A and R are true and R is the b) Both A and R are true but R is not the correct explanation of A. correct explanation of A. c) A is true but R is false. 2. 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. (b) Show that (2x + 7) is factor of 2x3 + 5x2 - 11x - 14. Hence, factorise the given expression completely, [4] using factor theorem. (c) 3. Prove that: sin 1 cot + cos 1 tan = cos + sin Question 3 (a) [4] [13] In a solid hemisphere of radius 10 cm, a maximum volume of sphere is cut out. Find the surface area [4] and volume of the remaining solid. [Take = 3.14] (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) In the given figure, O is the centre of the circle, chord PQ is parallel and equal to the chord RS and [4] QR is the diameter. Prove that i. arc PR = arc QS ii. arc PQ = arc RS. Section B Attempt any 4 questions 4. Question 4 (a) [10] A dealer in Mumbai sold a telescope to an end-user in Bangalore. The marked price of the telescope was 25000 and the dealer offered a discount of 20%. If the rate of GST is 28%, calculate the IGST, CGST and SGST charged from the end-user. Also determine the total amount of bill. Page 3 of 19 [3] (b) A number consists of two digits, whose product is 18. When 27 is subtracted from the number, the [3] digits interchange their places. Find the number. (c) Find both the quartiles for the following distribution. i. ii. 5. [4] Marks 30 40 50 60 70 80 90 No. of students 4 6 11 19 20 26 14 Size 4 5 6 7 8 Frequency 2 5 8 9 6 Question 5 (a) If A = [10] 8 ] 2 (b) 6 [ and B = 3 5 1 0 [ 4 ] , then solve for 2 2 matrix X, such that X - B = A. In the given figure PQ is a tangent to the circle at A. AB and AD are bisectors of CAQ and PAC. [3] [3] If BAQ = 30 prove that: i. BD is a diameter of the circle. ii. ABC is an isosceles triangle. (c) If one zero of the polynomial 2x2 - 5x - (2k + 1) is twice the other, then find both the zeroes of the [4] polynomial and the value of k. 6. 7. Question 6 [10] (a) If the vertices of ABC are A(5, -1), 5 (-3, -2), C(-1, 8), find the length of median through A. [3] (b) If a cos - b sin = x and a sin + b cos = y, then prove that a2 + b2 = x2 + y2. [3] (c) Solve the equation - 4 + (-1) + 2 + ... + x = 437 [4] Question 7 (a) [10] Find the nature of roots of the following quadratic equations. If the real roots exist, then also find [5] them. 2x2 - 3x + 5 = 0 3x2 - 4 3x + 4 = 0 (b) The following table show the heights of a group of students: Height (in cm) [5] Number of Students 140 - 145 8 145 - 150 12 150 - 155 18 155 - 160 22 160 - 165 26 165 - 170 10 170 - 175 4 Page 4 of 19 Use a graph sheet to draw an Ogive for the distribution. Use the Ogive to find: i. the inter quartile range ii. the number of students whose height is more than 168 cm. iii. the number of students whose height is less than 148 cm. 8. Question 8 (a) [10] A box consists of 4 red, 5 black and 6 white balls. One ball is drawn out at random. Find the [3] probability that the ball drawn is: i. Black ii. Red or white (b) Find the volume of a hemisphere, whose diameter is 14 cm. [Taken = (c) Draw a circle of radius 2 cm and construct a tangent to it from an external point without using the 22 7 ] [3] [4] centre. 9. Question 9 (a) [10] Solve the following inequation and write down the solution set: [3] 11x - 4 < 15x + 4 13x + 14, x W Represent the solution on a real number line. (b) From the following cumulative frequency table, draw ogive and then use it to find [3] Marks (Less than) 10 20 30 40 50 60 70 80 90 100 Cumulative Frequency 5 24 37 40 42 48 70 77 79 80 i. median ii. lower quartile iii. upper quartile. (c) 10. In the given figure, if DE || BC and AD : DB = 5 : 4, then find ar( DF E) ar( C F B) . Question 10 [4] [10] (a) If (x + 2y ) : (2x - y) is equal to the duplicate ratio of 3 : 2, then find x : y. [3] (b) Using a ruler and a pair of compasses only, construct: [3] i. A triangle ABC, given AB = 4 cm, BC = 6 cm and ABC = 90o. ii. A circle when passes through the points A, B and C and mark its centre as O. (c) From the top of a light house 100 m high, the angles of depression of two ships on opposite sides of it, [4] are 48o and 36o, respectively. Find the distance between the two ships to the nearest metre. Page 5 of 19 Solution Section A 1. Question 1 Choose the correct answers to the questions from the given options: (i) (d) 489.84 cm2 Explanation: According to the given information, a shape of figure is shown below When the hanging pipes touches the surface paper, a circular shape ABCD is formed on the graph paper. The size of circle ABCD is equal to the size of circular base of the cone. Radius of the circle ABCD is 6 cm. Hence, the coordinates of A, B, C and Dare (6, 0), (0, 6), (-6, 0) and (0, -6), respectively. The figure formed in the given information is cylindrical in outer surface and conical in the inner surface. Now, total surface area of the figure = Curved surface area of the cylinder + Curved surface area of the cone = 2 rh + rl = r (2h + l) 2 2 + h ) = r(2h + r 2 2 + 8 ) = 3.14 6(2 8 + 6 36 + 64 = 18.84(16 + ) = 18.84(16 + 100) = 18.84(16 + 10) = 18.84 26 = 489.84 cm2 (ii) (d) -5, or 4 Explanation: Let the required number be Then x + x = 20 x + x 20 = 0 x + 5x 4x 20 = 0 x(x + 5) 4(x + 5) = 0 2 2 2 (x + 5)(x 4) = 0 either x + 5 = 0 or x 4 = 0 x = 5 or x = 4 x = 5, 4 (iii) (c) -1 Explanation: f(x) = x3 - 3x2 + 5x - 7 g(x) = x - 2, if x - 2 = 0, then x = 2 Remainder will be f(2) = (2)3 - 3(2)2 + 5 2 - 7 Page 6 of 19 = 8 - 12 + 10 - 7 = 18 - 19 = -1 Remainder = -1 (iv) (d) 2 Explanation: Given 8 [ 12 12 8 2 7 (v) 48 32 8 28 ]= x[ 2 8[ 8 ] 7 12 8 2 7 ] = 4x [ ] 8 = 4x x = 2 (a) 2220 Explanation: Let S = (452 - 432) + (442 - 422) + (432 - 412) + (422 - 402) + ... upto 15 terms = (45 + 43) (45 - 43) + (44 + 42) (44 - 42) + (43 + 41) (43 - 41) + (42 + 40) (42 - 40) + ... upto 15 terms [ a2 - b2 = (a - b) (a + b)] = (45 + 43)2 + (44 + 42)2 + (43 + 41)2 + (42 + 40)2 + ... upto 15 terms = 2[{45 + 44 + 43 + ... upto 15 terms} + {43 + 42 + 41 + ...upto 15 terms}] = 2 [ {2 45 + (15 - 1) (- 1)} + (2 43 + (15 - 1) (- 1)}] [ S = {2a + (n 1)d} ] = 2[ 15 2 15 15 2 2 (76) + 15 2 (72)] n n = 15(76 + 72) = 15 148 = 2220 Hence, the sum of the given series is 2220. (vi) (d) 10 months Explanation: r(n+1) We have, MV = Pn [1 + 12440 = 1200 n [1 + ] 2400 8(n+1) 2400 2400+8n+8 12440 = 1200n ( 24880 = 8n2 + 2408 n n2 + 301n - 3110 = 0 2400 ] ) n = -311 or 10 n = 10 months (vii) (d) 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 BM C EM D 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 Page 7 of 19 2 AEL C BL AE BC EL = BL [ if two triangles are similar, then their corresponding sides are proportional] AL = CL 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 ) BL = (2)2 [from Eq. (i)] ar( AEL) = 4 ar( CBL) ar ( AEL) = 4 ar ( CBL) (viii) (a) 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 OA = AB AB = 3(OA) 1 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 = 3 3a AB 3a = AB AC 3 = 1 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 (c) ( , a3 = 5 3 Explanation: We have, ) 1 5+3x 5 3 a3 cu units 0 This is possible, when 5 + 3x < 0 3x < -5 x < 5 3 x ( , (x) 5 3 ) 1 (c) 7 Explanation: Page 8 of 19 A non-leap year contains 365 days. Since 365 = 52 7 + 1 . Thus non-leap year contains 52 Mondays and we left with 1 day which can be any day of the week. Total no. of outcomes = 7 Now a non-leap year contains 53 Mondays, if the left day must be Monday Total no. of favourable outcomes = 1 Thus P (getting 53 Mondays) Total no. of favourable outcomes = (xi) 1 = Total no. of outcomes 7 (b) 100 Explanation: We have, A = [ 5 5 0 A2 = A A = [ 4 4 5 5 0 An = [ ] 0 ] 0 2 5 0 = [ 0 ][ 5 2 ][ 0 0 2 5 5 0 2 25 5 2 5 ]= [ 5 2 5 25 =[ 0 ] 0 0 ] 0 0 2n 5 2n 5 0 2n Thus, [ 5 ] 0 2n 5 0 200 ]= [ 5 0 0 200 5 ] 0 52n = 5200 2n = 200 n = 100 (xii) (d) 45 , 45 Explanation: Given, coordinates of pole be P(1, 3) and Q (3, 3) and A(1, 1) be the position of man 2 2 + (3 1) i. Now, AP = (1 1) 2 2 + 2 = 0 [ distance = (x 2 2 PQ AP 2 2 + (y2 y1 ) ] = 2 units 2 2 + 0 and PO = (3 1) + (3 3) = 2 Now, in APQ, we have tan = x1 ) tan = 2 2 2 = 2 units = 1 = 45o [ tan 45o = 1] ii. When we shift the origin at (1, 1), then the angle will remain same, i.e. = 45 . (xiii) (c) 5 Explanation: Since OP P Q [ radius of circle is to the tangent to circle at that point] Page 9 of 19 By pythagoras theorem for OP Q , OQ = OP + PQ 2 2 OP 2 2 2 2 = 13 12 = 169 144 = 25 radius of circle be 5 cm. OP = 5 (xiv) (c) Upper limits of class intervals Explanation: Upper limits of class intervals (xv) (d) A is false but R is true. Explanation: We have, common difference of an AP d = an - an - 1 is independent of n or constant. So, A is false but R is true. 2. Question 2 (i) p = 1600/month r = 9% p.a. m.v. = 65,592 n = ? p r n(n+1) m.v. = pn + 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 n= 432 12 n = 36 months n = 3 years 1606 2038 or n = or n = 12 3649 12 or n = -303.66 months rejected. As 'n' is no. of months here. So can't be -ve. (ii) 2x + 7 = 0 x = 7 2 Remainder = Value of 2x3 + 5x2 - 11x - 14 at x = 7 2 = 2( = 4 2 3 ) + + 5( 245 4 + 4 7 2 77 2 343+245+154 56 = 343 7 2 ) 11 ( 14 = 0 7 2 ) 14 (2x + 7) is a factor of 2x3 + 5x - 11x - 4 2x3 + 5x2 - 11x - 14 = (2x + 7)(x2 - x - 2) = (2x + 7)(x2 - 2x + x - 2) = (2x + 7)[x(x - 2) + 1(x - 2)] Page 10 of 19 = (2x + 7)(x - 2)(x + 1) (iii) sin + 1 cot LHS = cos 1 tan sin 1 = cos + sin cos + c os 1 sin = = 2 sin sin cos 2 sin sin cos 2 cos 2 cos + sin 2 cos sin cos 2 sin cos = sin = (sin +cos )(sin cos ) sin cos c os sin cos [Using, (a - b)(a - b) - (a2 - b2)] = sin + cos = RHS Hence proved 3. Question 3 (i) Given, radius of hemisphere, r1 = 10 cm Since, the maximum sphere is cut out of diameter 10 cm. Since, the maximum sphere is cut out of diameter 10 cm. Therefore, radius of the sphere, r2 = = 5 cm 10 2 Now, surface area of the hemisphere = 3 r 2 1 = 3 3.14 (10)2 = 942 cm2 and surface area of the sphere = 4 r = 4 3.14 (5)2 2 2 = 314 cm2 Surface area of the remaining solid = Surface area of the hemisphere + Surface area of the sphere = 942 + 314 = 1256 cm2 Now, volume of the hemisphere = 2 3 3 r 1 = 2 3 3 3.14 (10) = 6280 = 1570 3 = 2093.33 cm3 and volume of the sphere = 4 3 3 r 2 = 4 3 3 3.14 (5) 3 = 523.33 cm3 Volume of the remaining solid = Volume of hemisphere - Volume of sphere = 2093.33 - 523.33 = 1570 cm3 (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 Page 11 of 19 (iii) Construct: Join PS. i. Proof: POR = 2 PQR (angle at the centre is twice the angle at the remaining circumference) SOQ = 2 SRQ (angle at the anthe is twice the angle at the remaining circumference) But PQR = SRQ (Alt. angles PQ || RS) POR = SOQ Arc PR = Arc QS (In the same circle arcs subtending equal angles at the centre are equal) PQR = SRQ arc PR = arc QS ii. arc PQ = arc RS Consider POQ and SOR OR = OQ (radii of same circle) OS = OP (radii of same circle) PQQ = ROS (vertically OPP. angle) POQ SOR (SAS cong. rule) PQ = RS (CPCT) arc PQ = arc RS arc subtented on equal chords are equal. Hence proved. Section B 4. Question 4 (i) Since, it is a case of inter-state transaction of goods and services IGST = GST and CGST = SGST = Nil Given, Marked price of telescope = 25000 Discount % = 20% Discount amount = 20% of 25000 = 25000 = 5000 20 100 S.P. of telescope = M.P. - discount amount = (25000 - 5000) = 20000 Also, given rate of GST = 28% I GST = GST = 28% IGST = 28% of 20000 = 20000 = 5600 28 100 Now, total bill amount = S.P. telescope + IGST = 20000 + 5600 = 25600 (ii) Let the 1st no. be x i.e. unit digit then 2nd no. = i.e. ten's digit 18 x 18 Required no. = + x According to question: x 180 x + x 27 = 10x + x - 27 - 10 = -9x - 27 = -9(x + 3) = x2 + 3x - 18 = 0 18 x 18 x 180 x 18 180 x 162 x x2 + 6x - 3x - 18 = 0 x(x + 6) - 3(x + 6) = 0 (x - 3) (x + 6) = 0 x = 3 { digit can never be -ve} Required no. = + x 180 x = 180 3 + 3 Page 12 of 19 = 60 + 3 = 63 (iii) i. Marks No. of Students (f ) c.f. 30 4 4 40 6 10 50 11 21 60 19 40 70 20 60 80 26 86 90 14 100 Total 100 which is odd Here n = 100 n+1 Q1 = 101 = = 25.25 4 100+1 = 4 4 th term th term Which lies in 40 c.f. Q = 60 1 n+1 Q3 = 303 = 4 3 = 4 100+1 3 4 101 3 = 4 th term th = 75.75 th term which lies in 86 c.f. Q3 = 80 ii. Size Frequency (f ) c.f. 4 2 2 5 5 7 6 8 15 7 9 24 8 6 30 Total 30 Here n = 30 Q1 th term = n+1 4 = 30+1 th 4 = 31 4 th = 7.75 th term Which in 15 c.f. Q1 = 6 n+1 th term = Q3 4 30+1 3 = 3 = 4 31 3 4 = 93 4 = 23.25 Q3 = 7 5. Question 5 (i) 8 6 A= [ ] 2 and B = [ 4 3 5 ] 1 Let x = [ 0 a b c d ] Given: X - B = A a b [ 3 5 ] [ c d ] 1 a+ 3 =[ 0 b 5 [ 8 6 2 4 ] 8 6 2 4 ]= [ c 1 d 0 ] a + 3 = 8, b - 5 = 6, c - 1 = -2, d - 0 = 4 a = 5, b = 11, c = -1, d = 4 x=[ 5 11 1 4 ] (ii) i. AB is the bisector of CAQ. C AB = BAQ = 30 Page 13 of 19 th term which lies in 24th c.f. C AQ = 2 30 = 60 AD is the bisector of PAC = C AD = P AD 180 60 2 = 120 = 60 2 BAD = BAC + CAD = 30 + 60 = 90 BD is a diameter of the circle. ii. BAQ = BC A = 30 (Alternate segment theorem) and ACB = ADB (Angles in same segment are equal) In AEB and CEB AE = EC [ DB is a diameter so it bisects the chord perpendicularly] AEB = C EB = 90 and EB = EB (common) AEB C EB (R.H.S. Congruency) [byCPCT] i,e, AB = BC (Sides opposite to equal angle are equal) Hence, ABC is an isosceles triangle. BAC = AC B = 30 (iii)Let and 2 are the zeroes of the polynomial 2x2 - 5x - (2k + 1). Then, 2 2 - 5 - (2k + 1) = 0 and 2 (2 )2 - 5(2 ) - (2k + 1) = 0 2 2 - 5 = 2k + 1 ...(i) and 8 - 10 = 2k + 1 ...(ii) From Eqs. (i) and (ii), we get 2 - 5a = 8 - 10 6 = 5 = 2 2 2 2 = 5 2 2 = 6 5 3 Now, substituting = 2 36 25 6 2k + 1 = 2k = k=- 68 36 [ 0] 6 Thus, the zeroes of the polynomial are 25 5 5 6 5 6 and 5 3 in Eq. (i), we get = 2k + 1 50 150 36 100 36 = 2k + 1 1 2k = = 136 36 100 36 17 9 6. Question 6 (i) Let the three vertices of triangle ABC be A(5, -1), B(-3, -2) and C(-1, 8) AD is the median, So, D is the mid-point of BC. x= 3 1 2 x = -2 y= = 3 2+8 2 Coordinate of D are (-2, 3) Distance between A(5, -1) and (-2, 3) Page 14 of 19 2 2 = (5 + 2) + ( 1 3) = 49 + 16 = 65 unit Hence, the length of median AD is 65 unit. (ii) Given, a cos - b sin = x ...(i) and a sin + b cos = y ...(ii) On adding the squares of Eqs. (i) and (ii), we get x2 + y2 = (a cos - b sin )2 + (a sin + b cos )2 x2 + y2 = a2 cos2 + b2 sin2 - 2ab cos sin + a2 sin2 + b2 cos2 + 2ab sin cos [ (a b)2 = a2 + b2 2ab] x2 + y2 = a2 cos2 + b2 sin2 + a2 sin2 + b2 cos2 x2 + y2 = a2 (cos2 + sin2 )+ b2 (sin2 + cos2 ) x2 + y2 = a2 + b2 [ sin2 A + cos2 A = 1] Hence proved. (iii)a = -4, d = (-1) -(-4) = -1 + 4 d = 3 Sn = 437 Sn = n 2 [2a + (n - 1)d] 437 = n 2 437 2 [2( 4) + (n 1)(3)] = n(-8 + 3n - 3) 874 = -n(3n - 11) 874 = 3n2 - 11n 3n2 - 11n - 874 = 0 3n2 - 57n + 46n - 874 = 0 3n(n - 19) + 46(n - 19) = 0 (3n + 46)(n - 19) = 0 3n + 46 = 0 or n - 19 = 0 3n = 46 or n = 19 n= 46 3 Reject n = 46 3 as n should be ration no. n = 19 So. x = a + (n - 1)d = (-4) + (19 - 1) (3) x = -4 + 18 3 = -4 + 54 = 50 x = 50 7. Question 7 (i) i. Given equation is 2x2- 3x + 5 = 0 On comparing it with ax2 + bx + c = 0. we get a = 2, b = - 3 and c = 5 Now, discriminant, D - b2 - 4ac = (-3)2 - 4 (2) (5) = 9 - 40 = - 31< 0 The given equation has no real roots. ii. Given equation is 3x2 - 3x + 4 = 0...(i) On comparing it with ax2 + bx + c = 0, we get a = 3, b = - 4 3 and c = 4 Now, discriminant, D = b2 - 4ac = (- 4 3)2 - 4(3)(4) = 48 - 48 = 0 Page 15 of 19 The given equation has two equal real roots. Now, Eq. (i) can be written as ( 3x)2 - 2 ( 3x) (2) + (2)2 = 0 ( 3x - 2)2 = 0 [ a2 - 2ab + b2 = (a - b)2] ( 3x - 2) ( 3x - 2) = 0 3x - 2 = 0 and 3x - 2 = 0 x = , 2 2 3 3 Hence, the equal roots are (ii) 2 3 and 2 3 . C.I. f. C.f. 140 - 145 8 8 145 - 150 12 20 150 - 155 18 38 155 - 160 22 60 160 - 165 26 86 165 - 170 10 96 170 - 175 4 100 i. Here n = 100 Median = ( =( 100 2 th ) n 2 ) th term term = 50th term = 157.5 cm Q1 (Lower quartile = 1 4 100th term = 25th term = 152 Q3 (Upper quartile) = 3 4 100th term = 75th term = 163 Inter quartile range = Q3 - Q1 = 163 - 152 = 11 ii. Number of students whose height is more than 168 cm = 100 - 96 = 4 iii. The number of students whose height is less than 148 cm = 18. 8. Question 8 (i) Given, number of red balls in the bag (R) = 4 Number of black balls in the bag (B) = 5 Number of white balls in the bag (W) = 6 Total number of possible outcomes = 4 + 5 + 6 = 15 Page 16 of 19 i. Probability of getting a black ball P (B) = = 5 15 = Number of favourable outcomes Total number of possible outcome 1 3 ii. Probability of getting a red or white ball P (R or W) = = = 4+6 10 2 15 15 3 (ii) Given, diameter of a hemisphere, d = 14 cm Then, radius of a hemisphere, r = = = 7 cm d 14 2 Now, volume of a hemisphere = = 2 = 2156 3 3 22 7 3 (7) = 718.67 = 44 3 2 7 2 3 2 r3 cm3 (iii)Steps of construction: i. Draw a circle with centre O and radius 2 cm. ii. Take a point P outside the circle. iii. From P draw a straight line which intersects the circle at A and B. iv. With BP as diameter draw a semicircle. v. At A, draw a perpendicular which meets the semicircle at C. vi. With centre P and radius PC, draw an arc which intersects the given circle at T and S . vii. Join PT. PT is the required tangent. 9. Question 9 (i) 11x - 4 < 15x + 4 11x - 15x < 4 + 4 -4x < 8 x > -2 -2 < x And 15x + 4 13x + 14 15x - 13x 14 - 4 2x 10 x 5 Solution Set : {x: -2 < x 5, x R} Or {0, 1, 2, 3, 4, 5 x W} Page 17 of 19 (ii) i. n = 80 Median Term , + 1 80 80 2 2 40+41 = 2 = 40.5 ii. for lower quantile n 4 = 80 4 = 20 lower quantile (Q1) = 19 iii. for upper quantile 3n 4 = 3 80 4 = 60 upper quantile (Q3) = 66.5 (iii)Given, DE || BC and Proof: DB AD AD DB DB+AD AD AB AD 4 4 + 1 = = 5 4 + 1 5 4+5 = = 5 = AD DB 5 ...(i) 9 5 Now consider, ADE and ABC A = A (common) 1 = 2 (corresponding angles as DE || BC) ABC ADE (By AA similarity criteria) AB AD BC DE DE BC AC = = AE = = AB BC DE = AD 9 5 (from (i)) ...(ii) 5 9 Now again, consider DFE and CFB DFE = C F B (Vertically opp. angles) DEF = FBC (Alternate angles) DF E C F B (By AA similarly criteria) We know that, the ratio of area of similar is equal to square of the ratio of their corresponding sides. ar( DFE) ar( CFB) = ( ar( DFE) ar( CFB) = ( ar( DFE) ar( CFB) = 2 DE ) BC 5 9 25 81 2 ) [from (ii)] Hence ar( DFE) : ar( CFB) = 25 : 81. 10. Question 10 (i) Given, = (x+2y) 3 (2x y) x+2y 2x y = 9 4 2 2 2 4(x + 2y) = 9(2x - y) 4x + 8y = 18x - 9y 18x - 4x = 8y + 9y Page 18 of 19 14x = 17y = x 17 y 14 Hence, x : y = 17 : 14 (ii) Steps of construction: i. Draw a line BC = 6 cm. ii. Construct an angle 90o, at B. iii. Cut BA = 4 cm and join AC to get ABC. iv. Draw the bisectors of AC and AB which intersect at O. v. Draw the circle taking O as centre and OA as radius, which passes through A, B and C. (iii)Let the distance between two ships be a and b. In ABD, tan 48o = 1.11 = a= 100 a 100 1.11 AD BD = 90.09 m a = 90.09 m In ADC, tan 36o = 0.7265 = b= 100 0.7265 100 b AD DC = 137.64 m b = 137.64 Distance between two sides = a + b = 90.09 + 137.64 (nearest metre) = 227.73 m = 228 m. (round off) Page 19 of 19

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