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ICSE Class X Board Exam 2015 : Mathematics

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Delhi Public School (DPS), Newtown, Kolkata

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ICSE QUESTION PAPER Class X MATHS (2015) (Two and a half hours) 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 in 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 working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. The intended marks for questions or parts of questions are given in brackets [ ]. Mathematical tables are provided. SECTION A (40 Marks) Attempt all questions from this Section. Question 1 (a) A shopkeeper bought an article for Rs. 3,450. He marks the price of the article 16% above the cost price. The rate of sales tax charged on the article is 10%. Find the: (i) marked price of the article. (ii) price paid by a customer who buys the article. (b) Solve the following inequation and write the solution set: 13x 5 < 15x + 4 < 7x + 12, x R Represent the solution on a real number line. (c) Without using trigonometric tables evaluate: sin65 cos32 sin28 .sec62 cosec230 cos25 sin58 [3] [3] [4] Question 2 3 x 9 16 , B , find x and y where A2 B. (a) If A 0 1 0 y [3] (b) The present population of a town is 2,00,000. The population is increased by 10% in the first year and 15% in the second year. Find the population of the town at the end of two years. [3] (c) Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2) (i) the coordinate of the fourth vertex D (ii) length of diagonal BD (iii) equation of the side AD of the parallelogram ABCD [4] Question 3 (a) In the given figure, ABCD is a square of side 21 cm. AC and BD are two diagonals of the square. Two semicircles are drawn with AD and BC as diameters. Find the area of the shaded region. Take 22 7 [3] (b) The marks obtained by 30 students in a class assignment of 5 marks are given below. [3] Marks 0 1 2 3 4 5 No. of Students 1 3 6 10 5 5 Calculate the mean, median and mode of the above distribution. (c) In the figure given below, O is the centre of the circle and SP is a tangent. If SRT = 65 , find the value of x, y and z. [4] Question 4 (a) Katrina opened a recurring deposit account with a Nationalised Bank for a period of 2 years. If the bank pays interest at the rate 6% per annum and the monthly instalment is Rs. 1,000, find the: (i) Interest earned in 2 years. (ii) Matured value [3] (b) Find the value of K for which x = 3 is a solution of the quadratic equation, (K + 2)x2 Kx + 6 = 0. Thus find the other root of the equation. [3] (c) Construct a regular hexagon of side 5 cm. Construct a circle circumscribing the hexagon. All traces of construction must be clearly shown. [4] SECTION B (40 Marks) Attempt any four questions from this Section Question 5 (a) Use a graph paper for this question taking 1 cm = 1 unit along both the x and y axis : (i) Plot the points A(0, 5), B(2, 5), C(5, 2), D(5, -2), E(2, -5) and F(0, -5). (ii) Reflect the points B, C, D and E on the y-axis and name them respectively as B , C , D and E . (iii) Write the coordinates of B , C , D and E . (iv) Name the figure formed by B C D E E D C B . (v) Name a line of symmetry for the figure formed. [5] (b) Virat opened a Savings Bank account in a bank on 16th April 2010. His pass book shows the following entries : Date Particulars Withdrawal (Rs.) Deposit (Rs.) Balance (Rs.) April 16, 2010 By cash - 2500 2500 April 28th By cheque - 3000 5500 May 9th To cheque 850 - 4650 May 15th By cash - 1600 6250 May 24th To cash 1000 - 5250 June 4th To cash 500 - 4750 June 30th To cheque - 2400 7150 July 3rd By cash - 1800 8950 Calculate the interest Virat earned at the end of 31st July, 2010 at 4% per annum interest. What sum of money will he receive if he closed the account on 1st August, 2010? [5] Question 6 (a) If a, b, c are in continued proportion, prove that (a + b + c) (a b + c) = a2 + b2 + c2. [3] (b) In the given figure ABC is a triangle and BC is parallel to the y axis. AB and AC intersect the y axis at P and Q respectively. (i) (ii) (iii) (iv) Write the coordinates of A. Find the length of AB and AC. Find the ratio in which Q divides AC. Find the equation of the line AC [4] (c) Calculate the mean of the following distribution : [3] Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 8 5 12 35 24 16 Question 7 (a) Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed. [3] (b) Find 'a' of the two polynomials ax3 + 3x2 9 and 2x3 + 4x + a, leaves the same remainder when divided by x + 3. [3] (c) Prove that sin cos cos sin 1 cot 1 tan [4] Question 8 (a) AB and CD are two chords of a circle intersecting at P. Prove that AP PB = CP PD [3] (b) A bag contains 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is: (i) a green ball (ii) a white or a red ball (iii) is neither a green ball nor a white ball. [3] (c) Rohit invested Rs. 9,600 on Rs. 100 shares at Rs. 20 premium paying 8% dividend. Rohit sold the shares when the price rose to Rs. 160. He invested the proceeds (excluding dividend) in 10% Rs. 50 shares at Rs. 40. Find the: (i) original number of shares (ii) sale proceeds (iii) new number of shares. (iv) change in the two dividends. [4] Question 9 (a) The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is 30 and 24 respectively. [4] Find the height of the two towers. Give your answer correct to 3 significant figures. (b) The weight of 50 workers is given below : [6] Weight in Kg 50-60 60-70 70-80 80-90 90-100 100-110 110-120 No. of Workers 4 7 11 14 6 5 3 Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis and 2 cm = 5 workers along the other axis. Use a graph to estimate the following: (i) The upper and lower quartiles. (ii) If weighing 95 kg and above is considered overweight, find the number of workers who are overweight. Question 10 (a) A wholesaler buys a TV from the manufacturer for Rs. 25,000. He marks the price of TV 20% above his cost price and sells it to a retailer at a 10% discount on the market price. If the rate of VAT is 8%, find the : (i) Market price (ii) Retailer s cost price inclusive of tax. (iii) VAT paid by the wholesaler. 3 7 0 2 1 5 ,B and C (b) If A 2 4 5 3 4 6 Find AB 5C. [3] [3] (c) ABC is a right angled triangle with ABC = 90 . D is any point on AB and DE is perpendicular to AC. Prove that: (i) ADE ACB (ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD. (iii) Find. Area of ADE : Area of quadrilateral BCED. [4] Question 11 (a) Sum of two natural numbers is 8 and the difference of their reciprocal is 2 . 15 Find the numbers. (b) Given x3 12x y 3 27y . Using componendo and dividendo find x : y. 6x2 8 9y 2 27 [3] [3] (c) Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and BAC = 105 . Hence: (i) Construct the locus of points equidistant from BA and BC. (ii) Construct the locus of points equidistant from B and C. (iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC. [4] Class X Mathematics Board Paper 2015 (Solution) SECTION A 1. (a) Given, b is the mean proportion between a and c. a b k(say) b c a bk, b ck a (ck)k ck 2 ,b ck L.H.S. a 4 a2b2 b4 b4 b2c2 c4 ck ck ck ck 4 2 2 2 2 ck ck 4 2 c2 c 4 c 4 k 8 c2k 4 c2k 2 c 4 k 4 c k c k c c4 4 4 2 2 2 c4k 8 c4k 6 c4k 4 c4k 4 c4k 2 c4 c4k 4 k 4 k 2 1 c k k 1 4 4 2 k4 R.H.S. a2 c2 ck 2 2 c2 c2 k 4 2 c k4 Hence, L.H.S. R.H.S. 4 (b) Given equation is 4x 2 5x 3 0. Comparing with ax 2 bx c 0,we get a 4, b 5 and c 3 x b b2 4ac 2a ( 5) ( 5)2 4(4)( 3) 2 4 5 25 48 8 5 73 8 5 8.54 8 3.54 13.54 or 8 8 1.6925 or 0.4425 1.69 or 0.44 (c) Join OA and OC. Since the perpendicular from the centre of the circle to a chord bisects the chord. Therefore, N and M are the mid-points of AB and CD respectively. Consequently, 1 1 AN NB AB 24 12 cm and 2 2 1 1 CM MD CD 10 5 cm 2 2 In right-angled triangles ANO and CMO, we have OA2 ON2 AN2 and OC2 OM2 CM2 132 ON2 122 and 132 OM2 52 ON2 132 122 and OM2 132 52 ON2 169 144 and OM2 169 25 ON2 25 and OM2 144 ON 5 and OM 12 Now, NM ON OM 5 12 17cm Hence, the distance between the two chords is 17 cm. 2. (a) 1 sin2 28 sin2 62 tan2 38 cot 2 52 sec2 30 4 1 sin2 28 sin2(90 28 ) tan2 38 cot 2(90 38 ) sec2 30 4 1 2 sin 28 cos 28 tan 38 tan 38 4 3 1 4 1 0 4 3 1 1 3 4 3 2 2 2 2 2 (b) 1 3 2 1 Given : A , B and A 2 5B2 5C 3 4 3 2 1 3 1 3 Now, A 2 A A 3 4 3 4 1 1 3 3 1 3 3 4 3 1 4 3 3 3 4 4 1 9 3 12 3 12 9 16 10 15 15 25 2 1 2 1 And, B2 B B 3 2 3 2 2 ( 2) 1 ( 3) 2 1 1 2 3 ( 2) 2 ( 3) 3 1 2 2 4 3 2 2 6 6 3 4 1 0 0 1 10 15 1 0 10 15 5 0 5 15 1 3 Now, A 2 5B2 5 5 5C 15 25 0 1 15 25 0 5 15 20 3 4 1 3 Hence, C 3 4 (c) For 1st year: P Rs. 50,000; R 12% and T 1 year 50,000 12 1 Rs. 6,000 100 And, Amount Rs. 50,000 Rs. 6,000 Rs. 56,000 Interest Rs. Since Money repaid Rs. 33,000 Balance Rs. 56,000 Rs. 33,000 Rs. 23,000 For 2nd year: P Rs. 23, 000; R 15% and T 1 year 23,000 15 1 Rs. 3,450 100 And, Amount Rs. 23,000 Rs. 3,450 Rs. 26,450 Interest Rs. Thus, Jaya must pay Rs. 26,450 at the end of 2nd year to clear her debt. 3. (a) List price Rs.42,000 Discount 10% of Rs. 42,000 10 Rs. 42,000 100 Rs. 4,200 Discounted price Rs. 42,000 Rs. 4,200 Rs. 37,800 Off-season discount 5% of Rs. 37,800 5 Rs. 37,800 100 Rs. 1,890 Sale-price Rs. 37,800 Rs. 1,890 Rs. 35,910 (i) The amount of sales tax a customer has to pay 8% of Rs. 35,910 8 Rs. 35,910 100 Rs. 2872.80 (ii) The total price, a customer has to pay for the computer Sale-price Sales Tax Rs. 35,910 Rs. 2872.80 Rs. 38782.80 (b) Given, P(1, 2), A(3, 6) and B(x,y) AP : PB 2:3 2 x 3 3 2 y 3 ( 6) 2x 9 2y 18 Hence, coordinates of P , 5 , 2 3 5 2 3 But, the coordinates of P are (1, 2). 2x 9 2y 18 1 and 2 5 5 2x 9 5 and 2y 18 10 2x 4 and 2y 8 x 2 and y 4 Hence, the coordinates of B are ( 2,4). (c) Data in ascending order: 13, 35, 43, 46, x, x 4, 55, 61, 71, 80 Median 48 Number of observations n 10 (even) n 2 Median 10 2 48 th th n term 1 term 2 2 th th 10 term 1 term 2 2 th th 5 term 6 term 48 2 x x 4 48 2 2x 4 48 2 48 x 2 x 46 x 4 46 4 50 Thus, the observations are 13, 35, 43, 46, 46, 50, 55, 61, 71, 80 Observation 46 is appearing twice. Hence, the mode of the data is 46. 4. (a) Let the number to be subtracted from the given polynomial be k. Let f(y) 16x3 8x 2 4x 7 k It is given that (2x 1) is a factor of f(y). 1 f 0 2 3 2 1 1 1 16 8 4 7 k 0 2 2 2 1 1 16 8 2 7 k 0 4 8 2 2 2 7 k 0 1 k 0 k 1 Thus, 1 should be subtracted from the given polynomial. (b) Length of a rectangle Radius of two semi-circles Diameter of a circle 5 5 10 20 cm Breadth of a rectangle Diameter of a circle 2 5 10 cm Area of a rectangle Length Breadth 20 10 200 sq. cm Area of a circle 22 5 5 78.571 sq. cm 7 1 And, area of two semi-circles each of radius 5 cm 2 78.571 78.571 sq. cm 2 Now, Area of shaded region Area of a rectangle Area of a circle Area of two semi-circles 200 78.571 78.571 200 157.142 42.858 sq. cm (c) 1 1 1 8 4x 7 , x I 2 2 2 17 1 15 4x , x I 2 2 2 17 1 Take 4x 2 2 17 1 4x 2 2 16 4x 2 8 4x 2 x 1 15 4x 2 2 15 1 4x 2 2 16 4x 2 4x 8 x 2 Thus, on simplifying, the given inequation reduces to 2 x 2. Since x I, the solution set is { 2, 1, 0, 1}. The required graph on number line is as follows: SECTION B (40 Marks) Attempt any four questions from this section 5. (a) 1 1 2 Given : B and X B 4B 8 3 2 Now, B B B 1 1 1 1 8 3 8 3 1 1 1 8 1 1 1 3 8 1 3 8 8 1 3 3 1 8 1 3 8 24 8 9 9 4 32 17 9 4 1 1 9 4 4 4 5 0 X B2 4B 4 32 17 8 3 32 17 32 12 0 5 a 5 Now, X b 50 5 0 a 5 0 5 b 50 5a 0b 5 0a 5b 50 5a 5 5b 50 5a 5 and 5b 50 a 1 and b 10 (b) 10 Rs. 50 Rs. 5 100 Total dividend Rs. 450 Number of shares bought 90 Dividend on 1 share Rs. 5 Since Dividend on 1 share 10% of Rs. 50 Since market value of each share Rs. 60 Sum invested by the man 90 Rs. 60 Rs. 5,400 Percentage return Total return Rs. 450 100% 100% 8.33% 8% Sum invested Rs. 5400 (c) Outcomes: a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p Total number of all possible outcomes 16 (i) When the selected card has a vowel, the possible outcomes are a, e, i, o. Number of favourable outcomes 4 4 1 16 4 (ii) When the selected card has a consonant, Required probability Number of favourable outcomes 16 4 12 12 3 Required probability 16 4 (iii)When the selected card has none of the letters from the word median, the possible outcomes are b, c, f , g, h, j, k, l, o, p. Number of favourable outcomes 10 Required probability 10 5 16 8 6. (a) Steps of construction: (i) Draw line AC = 5 cm and CAB = 60o. Cut off AB = 7 cm. Join BC, ABC is the required triangle. (ii) Draw angle bisectors of A and B. (iii) Bisector of B meets AC at M and bisector of A meets BC at N. (iv) Similarly, draw the angle bisector of C which meets AB at D. (v) P is the point which is equidistant from AB, BC and AC. (vi) With DP as the radius, draw a circle touching the three sides of the triangle (incircle.) (b) Let h bethe height and r be the radius of the base of the conical tent. According to the given information, 1 77 16 = r2h 3 1 22 77 16 = 7 7 h 3 7 1 77 16 = 22 7 h 3 77 16 3 h h = 24 m 22 7 Now, l2 = r2 + h2 l2 = 72 + 242 625 l = 25 m 22 7 25 = 550 m2 7 Hence, the height of the tent is 24 m and the curved surface area of the tent is 550 m2 . Curved surface area = rl= (c) (i) 7m 2n 5 7m 2n 3 By Componendo Divinendo, we get 7m 2n (7m 2n) 5 3 7m 2n (7m 2n) 5 3 14m 8 4n 2 7m 4 2n 1 m 8 n 7 m : n 8 :7 m 8 m2 82 (ii) 2 2 n 7 n 7 ApplyingComponendo Divinendo, we get m2 n2 82 72 2 m n2 82 72 m2 n2 64 49 2 m n2 64 49 m2 n2 113 2 m n2 15 7. (a) Principal for the month of Jan = Rs. 5600 Principal for the month of Feb = Rs. 4100 Principal for the month of Mar = Rs. 4100 Principal for the month of Apr = Rs. 2000 Principal for the month of May = Rs. 8500 Principal for the month of June = Rs. 10000 Total Principal for one month = Rs. 34300 Rate of interest = 6% pa PRT 34300 6 1 = Rs.171.50 100 100 12 (ii) Totalamount Rs.10000 Rs.171.50 Rs.10171.50 (i) Simple interest = (b) The image of point (x, y) on Y-axis has the coordinates ( x, y). Thus, we have Coordinates of B = ( 2, 3) Coordinates of C = ( 1, 1) Coordinates of D = ( 2, 0) Since, Y-axis is the line of symmetry of the figure formed, the equation of the line of symmetry is x = 0. 8. (a) Let the assumed mean A = 25 Mid-value Marks f x d x A t x A x 25 i 10 ft 0 10 5 10 20 2 20 10 20 15 9 10 1 9 20 30 25 25 0 0 0 30 40 35 30 10 1 30 40 50 45 16 20 2 32 50 60 55 10 30 3 30 f 100 Mean A ft 63 63 i 25 10 25 25 6.3 31.3 f 100 10 (b) (i) BAQ 30 Since AB is the bisector of CAQ CAB BAQ 30 AD is the bisector of CAP andP A Q, DAP CAD CAQ 180 CAD CAD 60 180 CAD 60 So, CAD CAB 60 30 90 Since angle in a semi-circle 90 Angle made by diameter to any point on the circle is 90 So,BD is the diameter of the circle. ft 63 (ii) SinceBD is the diameter of the circle, so it will pass through the centre. By Alternate segment theorem, ABD DAP 60 So, in BMA, AMB 90 ....(Use AngleSumProperty) We know that perpendicular drawn from the centre to a chord of a circle bisects the chord. BMA BMC 90 In BMA and BMC, BMA BMC 90 BM BM (common side) AM CM (perpendicular drawn from the centre to a chord of a circle bisects the chord.) BMA BMC AB BC (SAS congruence criterion) ABC is an isosceles triangle. (c) (i) Printed price of an air conditioner = Rs. 45000 Discount = 10% 45000 (100 10) 100 45000 90 Rs. 100 Rs.40500 C.P. f the air conditioner = Rs. 12 Rs.4860 100 So, the shopkeeper paid VAT of Rs. 4860 to the government. VAT (12%)=40500 (ii) Discount = 5% of the marked price 45000 (100 5) 100 45000 95 Rs. 100 Rs. 42750 C.P. f the air conditioner= Rs. 12 Rs.5130 100 So, the total amount paid by the customer inclusive of tax VAT (12%)=42750 = Rs. 42750 Rs.5130 Rs.47880 9. (a) (i) DAE 70 ....(given) BAD DAE 180 ....(linear pair) BAD 70 180 BAD 110 Since ABCD is a cyclic quadrilateral, sum of the measures of the opposite angles are supplementary. So, BCD BAD 180 BCD 110 180 BCD 70 (ii) BOD 2 BCD (Inscribed angle theorem) BOD 2(70 ) 140 (iii) In OBD, OB = OD ....(radii of same circle) OBD ODB By Angle Sum property, OBD ODB BOD 180 2 OBD BOD 180 2 OBD 140 180 2 OBD 40 OBD 20 (b) Given vertices: A( 1, 3), B(4, 2) and C(3, 2) (i) Coordinates of the centroid G of ABC are given by 1 4 3 3 2 2 6 3 G , , (2, 1) 3 3 3 3 (ii) Since the line through G is parallel to AC, the slope of the lines are the same. m y2 y1 2 3 5 x2 x1 3 ( 1) 4 So, equation of the line passing throughG(2, 1) and with slo pe y y 1 m(x x1 ) 5 (x 2) 4 4y 4 5x 10 y 1 5x 4y 14 is the required equation. 5 is given by, 4 (c) sin 2sin3 L.H.S. 2cos3 cos sin (1 2sin2 ) cos (2cos2 1) sin (1 2sin2 ) cos [2(1 sin2 ) 1] sin (1 2sin2 ) cos (2 2sin2 1) sin (1 2sin2 ) cos (1 2sin2 ) tan R.H.S. (proved) 10. (a) Let Vivek's age be x years and Amit's age be (47-x) years. According to the given information, x(47 x) 550 47x x2 550 x2 47x 550 0 (x 25)(x 22) 0 x 25 or x = 22 So, Vivek's age is 25 years and Amit's age is 22 years. (b) The cumulative frequency table of the given distribution is as follows: Wages in Rs. Upper Limit No. of workers Cumulative frequency 400-450 450-500 500-550 550-600 600-650 650-700 700-750 450 500 550 600 650 700 750 2 6 12 18 24 13 5 2 8 20 38 62 75 80 The ogive is as follows: Number of workers = n = 80 th n (i) Median = term 40th term 2 Through mark 40 on the Y-axis, draw a horizontal line which meets the curve at point A. Through point A, on the curve draw a vertical line which meets the X-axis at point B. The value of point B on the X-axis is the median, which is 605. th 80 (ii) Lower quartile (Q1 ) term 20th term 550 4 (ii) Through mark of 625 on X-axis, draw a verticle line which meets the graph at point C. Then through point C, draw a horizontal line which meets the Y-axis at the mark of 50. Thus, number of workers that earn more than Rs. 625 daily 80 50 30 11. (a) Let PQ be the lighthouse. PQ 60 In PQA, PQ AQ 60 3 AQ 60 AQ 3 20 3 AQ 3 tan 60 AQ 20 3 3 3 AQ 20 3 m In PQB, tan 45 PQ QB 60 QB QB 60 m 1 Now, AB AQ QB 20 3 60 20 1.732 60 94.64 95 m (b) (i) In PQR and SPR, we have QPR PSR ....(given) PRQ PRS ....(common) So, by AA-axiom similarity, we have PQR SPR ....(proved) (ii) Since PQR SPR ....(proved) PQ QR PR SP PR SR QR PR Consider ....[From (1)] PR SR QR 6 6 3 6 6 QR 12 cm 3 PQ PR Also, SP SR 8 6 SP 3 8 2 SP 8 SP 4 cm 2 Area of PQR PQ 2 82 64 (iii) 4 Area of SPR SP2 42 16 (c) (i) Let the deposit per month Rs. P Number of months (n) 36 Rate of interest (r) 7.5% p.a. n(n 1) r 2 12 100 36 37 7.5 8325 P 2 12 100 3 37 7.5 8325 P 2 100 8325 2 100 P Rs. 2000 3 37 7.5 (ii) Maturity value P n S.I. Rs.(2000 36 8325) Rs. 80,325 S.I. P

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