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MATHEMATICS 10th ICSE 2011 (Two hours and a half) Instructions 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. 1. (a) Find the value of k if (x 2) is a factor of x3+2x2 kx+10 Hence determine whether (x+5) is also a factor. [3] (b) 3 If A= 4 5 2 and B= , is the product AB possible? Give a reason. If yes, find AB.[3] 2 4 (c) Mr. Kumar borrowed Rs. 15,000 for two years. The rate of interest for the two successive years are 8% and 10% respectively. If he repays Rs. 6,200 at the end of the first year, find the outstanding amount at the end of the second year. Sol: (a) f(x) = x3+2x2 kx+10 [4] x 2 is a factor of f(x) f(2)=0 (2)3+2(2)2 k(2)+10=0 8+2(4) 2k+10=0 26=2k k=13 To check x+5 is a factor of f(x) put x= 5 in f(x) f( 5)=( 5)3+2( 5)2 k( 5)+10 = 125+2(25)+5k+10 = 125+50+5k+10 = 125+50+5(13)+10 = 125+125 =0 Hence x+5 is a factor of f(x). 3 5 (b) A= 4 2 2 2 3 5 AB= 4 2 2 B= 4 2 1 2 6 + 20 = 4 8 8 26 0 Yes, the product AB is possible as order of A is 2 2 & order of B is 2 1 Order of AB should be 2 1 (c) Principle for the first year = Rs. 15000 Interest for the first year=Rs. 15000 8 1 100 =1200 Amount after the first year =15000+1200 = 16200 Money refunded at the end of first year=Rs. 6200 Principal for the second year=Rs. 16200 6200=Rs. 10,000 Rate = 10% = 10,000 10 = Rs. 1000 100 Amount at the end of second year = Rs. 10,000+1000=11000 2. (a) From a pack of 52 playing cards all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is (i) a face card (King, Jack or Queen) (ii) an even numbered red card? (b) x [3] Solve the following equation: 18 =6. Give your answer correct to two significant figures. x (c) [3] In the given figure O is the centre of the circle. Tangents at A and B meet at C. If ACO=300, find (i) BCO (ii) [4] AOB (iii) APB Sol: (a) Total Number of cards = 52 Cards whose numbers are multiples of 3 = 3, 6, 9 No of cards removed=3 4=12 Hence Cards left=52 12=40. (i) Number of face cards=3 4=12 P(card drown is a face card)= 12 3 = = 0.3 40 10 (ii) Red cards = 26 Even numbered red cards=5 2=10 P(card drawn is even numbered red card)= (b) x 10 1 = 40 4 18 =6 x x2 18 6x=0 x2 6x 18=0 x= 6 36 4( 18) 6 36 + 72 = 2 2 x= 3 3 3 x= 3+3 3 , x=3 3 3 = 3+3 1.73 = 3 3 1.73 = 3+5.19 = 3 5.19 = 8.19 = 2.19 (c) ACO= 30 6 108 6 6 3 2 2 (i) In ACO & BCO, OC=OC (common) CA=CB(Length of tangents from an external point of a circle are equal) OAC = OBC= 900 OAC OBC OCA= OCB (c.p.c.t) BCO=300 (ii) ACBO is a cyclic quadrilateral. Sum of opposite angles of a cyclic quadrilateral is 1800 600 + AOB =1800 AOB = 1800 600=1200 (iii) AOB=1200 APB= 120 =600 2 ( Angle subtended by an arc at the centre is double the angle subtended by it on the remaining part of circle) 3. (a) Ahmed has a recurring deposit account in a bank. He deposits Rs. 2,500 per month for 2 years. If he gets Rs. 66,250 at the time of maturity, find (i) The interest paid by the bank (ii) The rate of interest. (b) Calculate the area of the shaded region, if the diameter of the semi circle is equal to 14 [3] cm. Take = (c) 22 . 7 ABC is a triangle and G (4, 3) is the centroid of the triangle. If A = (1, 3), B= (4, b) and C= (a, 1) find a and b . Find the length of side BC. [4] Sol: (a) Deposit per month=Rs. 2500 Total deposit=Rs.2500 24=Rs. 60,000 Amount of maturity=Rs. 66,250 Interest paid by the bank=S.I. =66250 60,000=6250 n(n + 1) (ii) Equivalent principal for 1 month= .p 2 Rs. [3] 2500 24 25 2 =750000 Let the rate of interest =r% Rate of interest = = S.I. 100 P T 6250 100 12 750000 1 Rate of interest =10% (b) Diameter of semicircle = 14cm. Radius of semicircle = 7cm. r2 Area of semicircle = 2 = 1 22 7 7 = 77 cm2 2 7 Area of upper shaded region= Area of ABCD Area of 2 quadrants r2 Area of 2 quadrants = 2 4 1 22 = 2 7 7 = 77cm2 4 7 Area of red rect. ABCD= (14) 7 = 98cm2 Area of shaded region = 98 77=21cm2 Total shaded portion = 77+21 cm =98cm2 (c) Given: ABC is a triangle with G as the centroid To find: Length of BC & a, b. Proof: The Coordinates of centroid G is 1+ 4+ a 3+ b+1 5+a 4+ b , , = 3 3 3 3 But Coordinates of G is (4, 3) 5+ a =4 5+a=12 a=7 , 3 4+b =3 4+b=9 b=5 3 Coordinates of B is (4, 5) Coordinates of C is (7, 1) Length of BC = (7 4)2 + (1 5)2 = 9 + 16 = 5 units. 4. (a) 2x 5 (b) Solve the following in equation and represent the solution set on the number line 5x+4 < 11, where x I Evaluate without using trigonometric tables. 2 2 tan350 cot550 sec 400 + 3 2 0 0 0 cot55 tan35 cosec50 (c) [3] [3] A Mathematics aptitude test of 50 students was recorded as follows: Marks 50 60 4 No. of Students 60 70 70 80 8 14 80 90 90 100 19 5 Draw a histogram for the above data using a graph paper and locate the mode. Sol: (a) 2x 5 5x+4 <11 2x 5 5x+4 & 5x+4<11 2x 5+( 5x+5) 5x+4+( 5x+5) & 5x+4+( 4)<11+( 4) 3x 9 & 5x<7 x 3 &x< 3 x 7 5 (2) (1) From (1) & (2) 3 x < 7 ; x R 5 7 The Solution set = x : x R, 3 x < 5 2 2 tan350 cot550 sec 400 + (b) 2 3 0 0 0 cot55 tan35 cosec50 2 2 tan(900 550 ) cot(900 350 ) 1 1 / 2 + 3 0 0 0 0 cot 55 tan35 cos40 sin50 [4] 2 2 cot 550 tan350 sin500 + 3 =2 0 0 0 cot 55 tan35 cos40 = 2+1 3 [sin 500 sec 400] =3 3[sin (900 400) sec 400] = 3 3 [cos 400 sec 400] =3 3 1 = 3 3=0 (c) SECTION B 5. (a) A manufacturer sells a washing machine to a wholesaler for Rs. 15,000. The wholesaler sells it to a trader at a profit of Rs. 1,200 and the trader in turns sells it to a consumer at a profit of Rs. 18,00. If the rate of VAT is 8 % find: (i) The amount of VAT received by the State Government on the sale of this machine from the manufacturer and the wholesaler. (ii) The amount that the consumer pays for the machine. (b) A solid cone of radius 5 cm and height 8 cm is melted and made into small spheres of radius 0.5 cm. Find the number of spheres formed. [3] (c) ABCD is a parallelogram where A(x, y), B (5,8), C(4,7) and D(2, 4). Find (i) Coordinates of A (ii) Equation of diagonal BD. [ 4] Sol: (a) Rate of VAT=8% VAT from manufacturers= VAT from Traders = 8 15000 =1200 8 1200 = Rs. 96 100 (i) VAT received By (Govt.) = 600+96 =Rs. 696 (ii) VAT paid by Trader= 1800 = Rs. 144 Total amount that customer pay =15000 1200(VAT) 1200 (Profit By Wholesaler) 96 (VAT by trader to wholesaler) 1800 (Profit by traders) 144 (VAT by Trader to Govt. ) [3] Total = Rs. 19440 (b) Vol. of cone= Volume of spheres 1 2 4 r h = r3 n 3 3 4 5 5 5 5 5 8= n 3 10 10 10 n= 2 10 10 2=4 100 = 400 (c) ABCD is a parallelogram. AB=CD AB=Diagonals of a parallelogram bisect each other. 4+x 2 5+2 2 4 + x = 7 x=3 7 + y 4 + 8 = 7+y=4 y= 3 2 2 x=3, y = 3 Equation of BD : y y1 = m (x x1) y y1= y2 y1 (x x1) x 2 x1 y 8= 4 8 (x 5) 2 5 y 8 = 12 (x 5) 3 y 8 = 4(x 5) 4x y 20+8 = 0 4x y 12=0 6. (a) Use a graph to answer the following questions. (Take 1cm=1 unit on both axes) (i) Plot A (4, 4), B(4, 6) and C(8,0) the vertices of a triangle ABC. (ii) Reflect ABC on the y axis and name it as A B C . (iii) Write the coordinates of the images A , B and C . (iv) Give a geometrical name for the figure AA C B BC (v) Identify the line of symmetry of AA C B BC. (b) Mr. Choudhury opened a Saving s Bank Account at State Bank of India on 1st April [5] 2007. The entries of one year as shown in his pass book are given below. Date Particulars 12th 2007 By Cash April To self Deposits Balance (in Rs.) 1st April 2007 Withdrawals (in Rs.) (In Rs.) 8550.00 8550.00 1200.00 7350.00 24th April 2007 By Cash 4550.00 11900.00 8th July 2007 By Cheque 1500.00 13400.00 10th Sept. 2007 By Cheque 3500.00 16900.00 17th Sept. 2007 To Cheque 2500.00 14400.00 11th Oct. 2007 By Cash 800.00 15200.00 6th Jan. 2008 To Self 2000.00 13200.00 950.00 14150.00 9th March 2008 By Cheque If the bank pays interest at the rate of 5% per annum, find the interest paid on 1st April, 2008, Give your answer correct to the nearest rupee Sol: (a) (iv) Figure AA C B BC is a hexagon (v) From the figure it is observed that the line of symmetry are y axis and x axis. (b) Since the interest is earned on the minimum balance between 10th day and the last day of month, as per entries of the passbook, we have: Minimum balance for April = Rs.7350/ Minimum balance for May = Rs.11900/ Minimum balance for June = Rs.11900/ Minimum balance for July = Rs.13400/ Minimum balance for August = Rs.13400/ Minimum balance for September = Rs. 14400/ Minimum balance for October= Rs.14400/ Minimum balance for November = Rs.15200 Minimum balance for December = Rs.15200/ Minimum balance for January = Rs.13200/ Minimum balance for February = Rs.13200/ Minimum balance for March= Rs.14150/ Rate of interest = 5% Interest = Pr incipal Rate 1 12 100 = 157700 5 1 12 100 = 788500 1200 = Rs. 657.083 7. (a) Using componendo and dividendo, find the value of x 3x + 4 + 3x 5 3x + 4 3x 5 =9 [3] 2 5 4 2 t If A= , B= and I is the identity matrix of the same order and A is the 1 3 1 3 (b) transpose of matrix A, find At. B +BI (c) [3] In the adjoining figure ABC is a right angled triangle with BAC=900 (i) Prove ADB CDA. (ii) If BD=18cm, CD=8 cm find AD. (iii) Find the ratio of the area of ADB is to area of CDA. Sol: (a) 3x + 4 + 3x 5 3x + 4 3x 5 By componendo & dividendo 3x + 4 + 3x 5 + 3x + 4 3x 5 3x + 4 + 3x 5 3x + 4 3x 5 2 3x + 4 10 = 8 2 3x 5 Squaring both sides, 3x + 4 25 = 3x 5 16 3x + 4 5 = 3x 5 4 = 9+1 9 1 [4] 16(3x+4)=25(3x 5) 48 x+64=75x 125 64+125=75x 48x 189=27x x= 189 21 = =7 27 3 x=7 2 5 4 2 (b) A= , b= 1 3 1 3 At B+BI = ? 2 5 4 2 8 1 4 + 3 At B= 1 3 = 20 3 10 + 9 = 1 3 7 1 17 1 4 2 1 0 4 + 0 0 2 4 2 BI = 0 1 = 1 + 0 0 + 3 = 1 3 1 3 7 1 4 2 11 3 At B+BI = + = 17 1 1 3 16 2 (c) (i) In ADB & CDA AD=AD (common) ADB= ADC =900 ADB ~ CDA (ii) BD=18 cm, CD= 8cm , AD=? In ADC, (AC) =82+(AD)2 (i) In ADB, (AB)2=(AD)2+(18)2 (AB)2=(AD)2+324 (ii) Subtracting (i) & (ii), (AC)2 (AB)2=64 324 (AC)2 (AB)2= 260 (AB)2 (AC)2=260 (iii) A = 900 (given) (BC)2 =(AC)2+(AB)2 (26)2= (AC)2+ (AB)2 (iv) From (iii) & (iv) (AB)2+(AC)2=676 (AB)2 (AC)2=260 2 (AB)2 = 936 (AB)2= 936 = 468 2 Put in (AB)2 in (ii), 468 = (AD)2 +324 468 324=(AD)2 144=(AD)2 AD=12 cm. (iii) Area of ADB= AD BD 1 = 12 18= 108 cm 2 Area of ADC= = 1 1 8 12=48 cm 2 Ratio = =9:4 8. (a) 108 9 = 4 48 DC AD (i) Using step deviation method, calculate the mean marks of the following distribution. (ii) State the modal class. [5] Class interval 50 55 55 60 60 65 65 70 70 75 75 80 80 85 85 90 Frequency 20 10 10 6 12 8 (b) 5 9 Marks obtained by 200 students in an examination are given below: Marks 0 10 10 20 20 30 30 40 40 50 50 60 60 70 70 80 80 90 90 100 No. of 5 11 10 20 28 37 40 29 14 6 students Draw an give for the given distribution taking 2 cm =10 marks on one axis and 2 cm=20 students on the other axis. Using the graph, determ (i) The median marks. (ii) The number of students who failed if minimum marks required to pass is 40. (iii) If scoring 85 and more marks is considered as grade one. Find the number of students who secured grade one in the examination. Sol: (a) [5] Classes Class Mark u i= yi yi A Frequency fi ui c fi 50 55 52.5 4 5 20 55 60 57.5 3 20 60 60 65 62.5 2 10 20 65 70 67.5 1 10 10 70 75 72.5 0 9 0 75 80 77.5 1 6 6 80 85 82.5 2 12 24 85 90 87.5 3 8 24 80 56 Assumed Mean (A) = 72.5 C=5 Mean=A+C = 72.5+5 f u f i i ( 56) 80 = 72.5 3.5=69 marks (ii) Modal Class = 55 60 (As the number of frequency is more as compare to others) (b) Marks No. of students Cumulative freq. 0 10 5 5 10 20 11 16 20 30 10 26 30 40 20 46 40 50 28 74 50 60 37 111 60 70 40 151 70 80 29 180 80 90 14 194 90 100 6 200 To Find the Median Here n (no. of Students)=200 = which is even (i) Let A be the point on y Axis representing frequency 1 n n 1 = + + 1 = [100+101]=100.5 2 2 2 2 The point M represents the Median marks in the graph . Median marks=57 (ii) If Minimum marks required to pass =40 Then 43 students are failed (iii) The number of students who get grade one =200 190=10 (approximately) 9. (a) Mr. Parekh invested Rs. 52,000 on Rs. 100 shares at a discount of Rs. 20 paying 8% dividend. At the end of one year he sells the shares at a premium of Rs. 20. Find (i) The annual dividend. (ii) The profit earned including his dividend. [3] (b) Draw a circle of radius 3.5 cm. Mark a point P outside the circle at a distance of 6 cm from the centre. Construct two tangents from P to the given circle. Measure and write down the length of one tangent. (c) [3] Prove that (cosec A sinA) (secA cosA) sec2A=tanA. Sol: (a) Face value = Rs 100 Market Value= 100 20= Rs 80 Dividend=8% number of shares= 52000 = 650 80 dividend of one share= 8 100 = Rs. 8 100 (i) Total dividend=5200 (650 Market value2=(100+20) Income from selling shares = 650 120 Total dividend=5200 (ii) Profit earned =78000+5200= Rs. 83200 (b) Hence it is clear from the figure that the length of tangent is 3.8 cm (approx) (C) (Cosec A sinA)(secA CosA)sec2A=tanA L.H.S : (Cosec A sinA)(secA CosA)sec2A 1 1 sin A cos A sec 2 A = sin A cos A 1 sin2 A 1 cos2 A = sec 2 A sin A cos A cos2 A sin2 A 1 = sin A cos A cos a = 2 2 sin A + cos A = 1 sin A = tan A = R.H.S. cos A 10. (a) 6 is the mean proportion between two numbers x and y and 48 is the third proportional of x and y. Find the numbers. [3] (b) In what period of time will Rs. 12,000 yield Rs. 3,972 as compound interest at 10% per annum, if compounded on an yearly basis? [ 3 ] (c) A man observes the angle of elevation of the top of a building to be 300. He walks towards it in a horizontal line through its base. On covering 60m the angle of elevation changes to 600. Find the height of the building correct to the nearest metre. Sol: (a) 6 is the mean proportion between x &y (i) 48 is 3rd proportional of x & y (ii) from (i), 6 = from (ii), xy (iii) x y = (iv) y 48 from (iii), 36 = xy y= 36 x put the value of y in (iv) , x x 36 = 36 48x x2 36 = 36 48x 48x3 = (36)2 x 3= 36 36 =27 48 x3 = 27 x=3 Hence y= 36 =12 3 x=3 , y = 12 [4] (b) Principal=Rs.12000 Amount=Rs. 12000+ Rs. 3972 R Amount= P 1 + 100 T 10 15972=12000 1 + 100 15972 10 = 1 + 12000 100 15972 110 = 12000 100 15972 11 = 12000 10 T 1331 11 = 1000 10 3 11 11 = 10 10 T = 3 years (c) T T T T T In ADB, tan 600= 3= h x h x In ABC, tan 300 = 1 h = 3 60 + x 60+x = h 3 60+ h =h 3 3 1 60=h 3 3 3 1 60=h 60= 2h 3 30 3 = h 30 3 =h. h =30 1.73 h 60 + x = 51.9 m. 11. (a) ABC is a triangle with AB=10cm, BC=8cm and AC=6cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles. (b) [3] Rs. 480 is divided equally among x children. If the number of children were 20 more than each would have got Rs. 12 less. Find x . [ 3 ] (c) Given equation of line L 1 is y=4. (i) Write the slope of line L2 if L is the bisector of angle O. (ii) Write the co ordinates of point P. (iii) Find the equation of L2. Sol: (a) CR=CO=radius = x BR=BP=radius = y AO=AP=radius = z CR+RB=8 x+y=8 (i) BP+PA=10 y+ z=10 (ii) CO+OA=6 x+ z=6 (iii) Subtract (ii) & (iii), y x = 4 .(iv) Adding x+y = 8 x+ y=4 2y=12 y =6 & (iv) [4] x+6 =8 (x=2) put the value of x in (iii) 2+z=6 z=4 Hence Radii of the three circles are 2, 6, 4 cm (b) Total Money distributed =Rs. 480 Total no. of children = x Money given to each child = 480 x If no. of children were 20 more then their no. =x+20 Money given to each child = A.T.Q, 480 x + 20 480 480 = 12 x + 20 x 480 12x 1 = 480 x x + 20 480x = (480 12x)(x+20) 480x = 480x +9600 12x 240x 480x = 12 x2+240x+9600 12x2+480x 240x 9600=0 12x2+240x 9600=0 x2+20x 800=0 x2+40x 20x 800=0 x(x+40) 20(x+40)=0 x= 40 , 20 x= 40 (rejected) x=20 (c) Equation of L1 : y=4 XOY = 900 Since L2 is bisector of angle O L2 0Y = L2 0X=450 (i) Slope of line L2 (m) = tan 450 =1 (ii) Coordinates of point P is (4, 4) (iii) Equation of L2 y 0 = 4 0 (x 0) 4 0 y = x

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