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ICSE Prelims 2018 : Mathematics (Shishuvan English Medium School, Mumbai)

13 pages, 65 questions, 3 questions with responses, 3 total responses,

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Shishuvan English Medium School, Mumbai

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MATHEMATICS (X) (Two hours and a half) ----------------------------------------------------------------------------------------------------------------SECTION A (40 Marks) Attempt all questions from this section QUESTION 1: A Factorise the expression f (x)=2 x 3 7 x2 +14 x 8 . Hence, find all possible values of x for which f(x) = 0. [3 ] B Mrs. Geeta deposited ` 350 per month in a bank for 1 year and 3 months under the Recurring Deposit Scheme. If the maturity value of her deposits is ` 5565, find the rate of interest per annum. [4 ] Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 Page 1 C Find the sum of 32 terms of an A.P. whose third term is 1 and sixth term is 11. Given a + 2d = 1 and a + 5d = 11 Solving it simultaneously, we get a = 9 and d = 4 n S n= [2 a+ ( n 1 ) d ) 2 32 S 32= [ 2 9+ ( 32 1 ) ( 4 ) ] = 1696 2 [3 ] QUESTION 2: A Solve and graph the solution set of: 2 1 1 2 x+ <3 ; x R 3 3 3 [3 ] B The printed price of an air conditioner is ` 45000. The wholesaler allows a discount of 10% to the shopkeeper. The shopkeeper sells the article to the customer at a discount of 5% of the marked price. Sales tax (under VAT) is charged at the rate of 12% at every stage. Find : (i) VAT paid by the shopkeeper to the government. (ii) The total amount paid by the customer inclusive of tax. [4 ] CP for shopkeeper = 45000 10% of 45000 = 40500 Tax paid by the shopkeeper = 12% of 40500 = 4860 SP for shopkeeper = 45000 5% of 45000 = 42750 Tax charged by the shopkeeper = 12% of 42750 = 5130 VAT paid by the shopkeeper = Tax charged tax paid = 5130 4860 = 270 The total amount paid by the customer = 42750 + 5130 = 47880 C AB is a diameter of the circle with centre C. APBR is as shown in the figure. APQ and RBQ are straight lines. Find : (i) PRB = PAB = 35 (angles in the same segment) (ii) PBR = PQB + BPQ = 90 + 25 = 115 Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 [3 ] Page 2 ( APB = 90 angle in the semicircle BPQ = 90 PBR is exterior angle of triangle PBQ) BPR = 180 ( PRB + PBR) = 180 (35 + 115 ) = 30 (Sum of all angles of a triangle is 180 ) (iii) QUESTION 3: A B C Solve sin 65 cos 32 + sin28 . sec 62 +cose c 2 30 cos 25 sin 58 If [ A= 1 1 a b ] and A2 = I; find a and b. If quadratic equation x2 (m + 1)x + 6 = 0, has one root as x = 3, find the value of m and the other root of the equation. Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 [3 ] [4 ] [3 ] Page 3 QUESTION 4: A A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box. Let the number of black balls be n N(black balls) = n N(S) = n + 30 N(white balls) = 30 Given P(black ball) = 2/5 of P(white ball) n 2 30 = n=12 n+30 5 n+30 [3 ] Compute all the quartiles from the following data Weekl y 5 5 6 6 6 6 6 Incom 8 9 0 1 2 3 4 e No of 1 1 worker 2 3 6 5 4 5 0 s 1 2 3 4 4 CF 2 5 1 6 6 1 5 [3 ] ( B ) 6 5 6 6 3 1 4 8 4 9 ( n+1 ) 50 = =12.5 th observation=61 4 4 3 ( n+1 ) 150 Q3= = =37.5 th observation=63 4 4 ( n+1 ) 50 Median= = =25 thobservation=61 2 2 Q1= C An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cubic cm of iron weight is 7.8 gms. V ( cylinder )= r 2 h= ( 8 )2 240=15360 cubic cm 1 V ( cone ) = r 2 h= ( 8 )2 36=768 cubic cm 3 3 22 total volume=15360 +768 =16128 =50688 cubic cm 7 weight of the pillar=50688 7.8=395366.4 gm=395.3664 kg [4 ] SECTION B (40 Marks) Attempt any four questions from this section QUESTION 5: Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 Page 4 A Fourth term of an AP is equal to 3 times of its first term and seventh term exceeds twice the third term by 1. Find the first term and the common difference. [3] B The difference between the outer and the inner curved surface area of a hollow cylinder, 14 cm long, is 88 sq. cm. Find the outer and the inner radii of the cylinder, given that the volume of metal used is 176 cu. cm. 2 ( R r ) 14=88 R r =1 (1) [3] ( R 2 r 2 ) 14=176 R2 r 2=4 ( R+r ) ( R r )=4 R+r=4 ( 2) solving simultaneously ( 1 ) (2 ) , we get R=2.5 cm r=1.5 cm C Points A Find (i) (ii) (iii) (iv) and B have the coordinates (-2, 4) and (-4, 1) respectively. [4] The coordinate of A`, the image of A in the line x = 0 The coordinate of B`, the image of B in y-axis The coordinate of A``, the image of A in the line BB` Hence, write the angle between the lines A`A`` and BB`. Assign a special name to the figure B`A`BA``. QUESTION 6: A In the given figure, ABC is a triangle with EDB = ACB. Prove that ABC EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of BED = 9 cm 2, calculate the (i) Length of AB (ii) Area of ABC Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 [4] Page 5 B M is the mid-point of a line segment joining the points A(-3, 7) and B(9, -1). Find the coordinates of point M. Further, if R(2, 2) divides the line segment joining M and the origin in the ratio p : q, find the ratio p : q. 3+9 7 1 M= , = (3, 3 ) 2 2 ( [3] ) q 0 p+3 q 3q 3q , =( , ( 0 p+3 ) p+q p+q p+ q p +q ) ( 2,2 )= 3q p 1 =2 = p :q=1: 2 p+ q q 2 C Given A = [ ] 3 7 2 4 ,B= [ ] 0 2 5 3 and C = [ 1 5 4 6 ] , find AB [3] 5C. AB = 5C = 5 [ ][ ] [ ] [ ][ ] [ ][ ][ 3 7 0 2 = 35 27 2 4 5 3 20 16 1 5 = 5 25 4 6 20 30 AB 5C = 35 27 5 25 = 30 52 20 16 20 30 40 14 ] QUESTION 7: Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 Page 6 A B In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If BCG = 108 and O is the centre of the circle, find (i) Angle BCT (ii) Angle DOC BCG + BCD = 180 (linear pair of angles) BCD = 180 108 = 72 Draw segment BD. given CD = CB triangle BCD is an isosceles triangle BDC = DBC = 54 BCT = BDC = 54 (angles in the alternate segment) DOC = 2 DBC = 108 (central angle is twice the angle subtended by it at the circumference) How many terms of the geometric progression 1 + 4 + 16 + 64 + . must be added to get sum equal to 5461? a = 1, r = 4, Sn = 5461 a ( r n 1 ) 1 ( 4n 1 ) 4n 1 5461= = r 1 4 1 3 n n 5461 3+1=4 16384=4 n=7 Using step deviation method, calculate the mean marks of the following distribution. Class 50 55 60 65 70 75 80 85 interva 55 60 65 70 75 80 85 90 l Frequen 5 20 10 10 9 6 12 8 cy Also state the modal class. [3 ] [3 ] S n= C CI X f 50-55 55-60 60-65 65-70 52.5 57.5 62.5 67.5= A 72.5 77.5 82.5 87.5 70-75 75-80 80-85 85-90 t=d/i ft 5 20 10 10 d= x A -15 -10 -5 0 -3 -2 -1 0 -15 -40 -10 0 9 6 12 8 5 10 15 20 1 2 3 4 9 12 36 32 [4 ] ft 24 Mean=A + i=67.5+ 5=67.5+1.5=69 f 80 Modal class = 55 60 QUESTION 8: A Prove that [3] 2 2 co s A co s B =cosec 2 A cose c 2 B 2 2 sin A si n B 2 2 co s A co s B co t 2 A co t 2 B= 2 2 sin A si n B co t 2 A co t 2 B= Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 Page 7 cos2 A cos2 B cos 2 A sin 2 B cos 2 B sin2 A LHS=co t A co t B= 2 2 = 2 2 sin A sin B sin A sin B cos 2 A (1 cos 2 B) cos 2 B(1 cos2 A ) sin 2 A sin2 B 2 2 co s A co s B =RHS (1) sin 2 A sin 2 B co s 2 A co s2 B =cose c2 A cosec 2 B 2 2 si n A sin B 1 1 sin2 B sin2 A RHS=cose c 2 A cose c2 B= 2 2 = 2 2 sin A sin B sin A sin B 2 2 2 2 1 cos B 1+cos A co s A co s B = =LHS ( 2 ) sin2 A sin 2 B si n2 A si n2 B co s2 A co s 2 B 1 2, co t 2 A co t 2 B= =cosec 2 A cose c 2 B 2 2 si n A si n B Hence proved 2 B 2 The line 5x + 4y + 20 = 0 meets the x-axis at point A and the y-axis at point B. Find (i) The co-ordinates of A and B (ii) The length of segment AB (iii) The slope of AB and slope of perpendicular to AB, by using the ordinates of A and B only. Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 [4] Page 8 C If ax = by = cz, prove that x2 y 2 z 2 bc ca ab + + = + + yz zx xy a2 b2 c 2 Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 [3] Page 9 QUESTION 9: A A trader bought a number of articles for ` 1200. Ten were damaged and he sold each of the remaining articles at ` 2 more than what he paid for it, thus getting a profit of ` 60 on the whole transaction. Taking the number of articles he bought as x, form an equation in x and solve it. Total CP = `1200 1200 CP of each article = x 1200 SP of each article = +2 x 1200 Total SP = ( + 2) (x 10) x Profit = SP CP = 60 1200 +2 ( x 10 ) 1200=60 x x 2 40 x 6000=0 ( x 100 ) ( x+60 )=0 x=100 ( B [4] ) The daily wages of 80 workers in a building project are given below [6] CF Coordinates Wages (`) No of workers 6 (30, 0), (40, 30 40 6 6) 16 (50, 16) 40 50 10 31 (60, 31) 50 60 15 50 (70, 50) 60 70 19 62 (80, 62) 70 80 12 70 (90, 70) 80 90 8 76 (100, 76) 90 100 6 80 (110, 80) 100 110 4 Using graph paper, draw an ogive for the above distribution. Use your ogive to estimate (i) The median wages of workers (ii) The percentage of workers who earn more than ` 75 a day (iii) The upper quartile wage of the workers (iv) The lower quartile wage of the workers (v) Inter quartile range QUESTION 10: A In the given diagram given, AC is the diameter of the circle with centre O, CD and BE are parallel to each other. AOB = 80 , ACE = 10 . Calculate (i) BEC, BCD, CED COB + BOA = 180 (Linear pair of angles) COB = 180 80 = 100 BEC = COB (central angle is twice the angle subtended by it at the circumference) BEC = of 100 = 50 BCA = BOC = 40 (central angle is twice the angle subtended by it at the circumference) ECD = BEC = 50 (alternate angles) Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 [4] Page 10 BCD = BCA + ACE + ECD = 40 + 10 + 50 = 100 (angle addition property) BCD + BED = 180 (opposite angles of a cyclic quadrilateral is supplementary) 100 + BEC + CED = 180 CED = 180 100 50 = 30 B Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1. Does the line 3x = y + 1 bisect the line segment joining the given two points? Slope of line 3x = y + 1 is m1 = 3 Slope of line passing through (-2, 3) and (4, 1) is y 1 y 2 3 1 1 m2= = = x 1 x 2 2 4 3 1 m1 m 2=3 = 1 3 The two lines are perpendicular to each other. Midpoint of (-2, 3) and (4, 1) is x 1+ x 2 y 1+ y 2 2+ 4 3+ 1 , = , =( 1,2) 2 2 2 2 Substituting (1, 2) in 3x = y + 1 LHS = 3x = 3(1) = 3 RHS = y + 1 = 2 + 1 = 3 Since LHS = RHS, point (1, 2) lie on the line 3x = y + 1 line 3x = y + 1 bisects the given line [3] A manufacturer sells an article to a wholesaler for ` 20000. The wholesaler sells it to a shopkeeper at 10% profit and the shopkeeper in turn sells to a consumer at a profit of ` 2000. If the rate of VAT is 10%, find (i) The VAT paid by the wholesaler (ii) The VAT paid by the shopkeeper (iii) The price paid by the consumer. Profit for wholesaler = 10% of 20000 = 2000 SP for wholesaler = 20000 + 2000 = 22000 VAT paid by the wholesaler = 10% of 2000 = 200 VAT paid by the shopkeeper = 10% of 2000 = 200 SP for shopkeeper = 22000 + 2000 = 24000 Price paid by the customer = 24000 + 10% of 24000 = 24000 + 2400 = 26400 [3] ( C )( ) QUESTION 11: A A man invests ` 20020 in buying shares of nominal value ` 26 at 10% premium. The dividend on the shares is 15% per annum. Calculate (i) The number of shares he buys. (ii) The dividend he receives annually. (iii) The rate of interest he gets on his money. Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 [3] Page 11 B If a : b = c : d, show that ( a c ) b 2 : ( b d ) cd=( a2 b 2 ab ) :(c 2 d 2 cd) [3] C Construct a triangle ABC in which base BC = 5.5 cm, AB = 6 cm and ABC = 120 . (i) Construct a circle circumscribing the triangle ABC (ii) Draw a cyclic quadrilateral ABCD, so that D is equidistant from B and C. [4] Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 Page 12 Std 10 Pre Preliminary Examination 2017-18 Thursday 4th December 2017 Page 13

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