Mathematics Class X Past Year Paper Time: 2½ hour Total Marks: 80

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1 Pas Year Paper Mathematics Class X Past Year Paper Time: ½ hour Total Marks: 80 Solution SECTION A (40 marks) Sol. 1 (a) A + X B + C X X X X X X (b) P 4000, C.I. 134, n 3 years Amount, A P + C.I r A P r r n 3 3 r r r r 10% The observations in ascending order are: 11, 1, 14, (x - ), (x + 4), (x + 9), 3, 38, 47

2 Number of observations 9 (odd) n 1 Median th observation 5 th observation x x 0 Thus, the observations are: 11, 1, 14, 18, 4, 9, 3, 38, Mean Sol. (a) Let the number added be x. (6 + x) : (15 + x) :: (0 + x) (43 + x) 6 x 0 x 15 x 43 x (6 x)(43 x) (0 x)(15 x) 58 6x 43x x 300 0x 15x x 49x 35x x 4 x 3 Thus, the required number which should be added is 3. (b) Let p(x) x 3 + ax + bx - 14 Given, (x - ) is a factor of p(x) Remainder p() 0 () 3 + a() + b() a + b a + b + 0 a + b (1) Given, when p(x) is divided by (x - 3), it leaves a remainder 5. p(3) 5 (3) 3 + a(3) + b(3) a + 3b a + 3b a + b () Subtracting (1) from (), we get, a a 5 From (1), 10 + b b -11

ICSE X Mathematics Past Year Paper Steps for calculation of mode. (i) Mark the end points of the upper corner of rectangle with maximum frequency as A and B.

3 ICSE X Mathematics Past Year Paper Steps for calculation of mode. (i) Mark the end points of the upper corner of rectangle with maximum frequency as A and B. (ii) Mark the inner corner of adjacent rectangles as C and D. (iii) Join AC and BD to intersect at K. From K, draw KL perpendicular to x-axis. (iv) The value of L on x- axis represents the mode. Mode 13 Sol. 3 (a) 3 cos80o. cosec10o + sin59o sec 31o 3cos(90o - 10o) cosec10o + sin (90o - 31o) sec 31o 3 sin 10o cosec10o + cos31o, sec31o [ sin (90o - q) cos q, cos (90o - q) sin q ] [ sin q.cosec q 1, cos q.sec q 1] 3+5 (b) (i) In D ABD, Ð DAB + Ð ABD + Ð ADB 180o 65o + 70o + Ð ADB 180o

4 Ð ADB 180 o - 70 o - 65 o 45 o Now, Ð ADC Ð ADB + Ð BDC 45 o + 45 o 90 o Ð ADC is the angle of semi-circle so AC is a diameter of the circle. (ii) Ð ACB Ð ADB (angle subtended by the same segment) Ð ACB 45 o (i) Radius AC ( 3 + ) + (- -7 5) ( 5) ( 1) units (ii) Let the coordinates of B be (x, y) Using mid-point formula, we have x - 7+ y x y x - 7 y 17 Thus, the coordinates of points B are (-7, 17). Sol. 4 (a) x²-5x-10 0 Use quadratic formula: a 1, b -5, c -10 x x a b b 4ac (1) ( 5) ( 5) 4(1)( 10) x 5 65 x x x, x 6.53, 1.53 (b) (i) In DABC and D DEC,

$ICSE X Matheatics Past Year Paper Ð ABC Ð DEC 90 (perpendiculars to BC) Ð ACB Ð DCE (Common) \ D ABC D DEC (AA criterion) (ii) Since DABC D DEC, AB AC DE CD 6 15 4 CD 6 CD 60 60 CD 10 cm 6 (iii) It$

5 ICSE X Matheatics Past Year Paper Ð ABC Ð DEC 90 (perpendiculars to BC) Ð ACB Ð DCE (Common) \ D ABC D DEC (AA criterion) (ii) Since DABC D DEC, AB AC DE CD CD 6 CD CD 10 cm 6 (iii) It is known that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides. ar( ABC): ar( DEC) AB : DE 6 : 4 36 : 16 9 : 4 (i) (ii) Co-ordinates of A' (-6, -4) Co-ordinates of B' (0, -4) (iii) ABA'B' is a parallelogram.

(iv) AB A'B' 6 units In D OBO', OO' 3 units OB 4 units BO' OB + OO' 4 + 3 5 5 units Since BO' 5 units BA' 10 units AB' Perimeter of ABA'B' (6 + 10 + 6 + 10) units 3 units SECTION B (40 marks) Sol.

6 (iv) AB A'B' 6 units In D OBO', OO' 3 units OB 4 units BO' OB + OO' units Since BO' 5 units BA' 10 units AB' Perimeter of ABA'B' ( ) units 3 units SECTION B (40 marks) Sol. 5 (a) The given inequation is - x x < Î, x R x x x 1 1 x x 4-1 < x 1 4 x + 3x 4 < + ³ x x 4 < ³ and x 9 5x ³ 8 < 6 8 x ³ 18 5 x < 6 x ³ 1.6 x < 3 The solution set is {x: x R and 1.6 x < 3} It can be represented on a number line as follows: (b) Maturity amount 8088 Period (n) 3 yrs 36 months Rate 8% p.a.

7 Let x be the monthly deposit. ( + ) S.I. P n n 1 r P x Total amount of maturity 36x+ 4.44x 40.44x Now, 40.44x 8088 x 00 Thus, the value of monthly installment is 00. (i) No. of shares 50 Market value of one share 13 Salman's investment (13 x 50) 6600 (ii) Dividend on one share 7.5% of His annual income (iii) Salman wants to increase his income by 150. Income on one share No. of extra shares he buys Sol. 6 (a) LHS 1 - CosA 1 + CosA (1 - CosA)(1+ CosA) (1 + CosA)(1+ CosA) 1 - Cos A (1 + CosA) Sin A (1 + CosA) SinA RHS 1 + CosA (b) In cyclic quadrilateral ABCD, Ð B + Ð D 180 (opp. angles of cyclic quad are supplementary) Ð ADC 180 Ð ADC 80

8 Now, in D ACD, Ð ACD + Ð CAD+ Ð ADC Ð CAD Ð CAD Now Ð DCT Ð CAD (angle s in the alternate segment) \ Ð DCT 60 On completing the given table, we get: Date Particulars Withdrawls Deposit Balance Feb8 Feb 18 April 1 June 15 July 8 B/F To self By cash To self By cash Principal for the month of Feb 4500 Principal for the month of March 4500 Principal for the month of April 4500 Principal for the month of May 6730 Principal for the month of June 1730 Principal for the month of July 7730 Total Principal for 1 month ( ) 9690 P 9690, T 1 years, R 6% 1 \ Interest P R T Rs

9 Sol. 7 (a) The vertices of D ABC are A(3, 5), B(7,8) and C(1, -10). Coordinates of the mid-point D of BC are æx 1 x y1 y ö + +, ç è ø æ (- 10) ö ç, çè ø æ8 - ö ç, çè ø ( 4, - 1) Slope of AD y - y1 x - x Now, the equation of median is given by: y - y1 m(x - x1) y - 5-6(x - 3) y - 5-6x x + y (b) Since, the shopkeeper sells the article for 1500 and charges sales-tax at the rate of 1%. Tax charged by the shopkeeper 1% of VAT Tax charges - Tax paid Tax paid Tax paid 144 If the shopkeeper buys the article x. Tax on it 1% on x 144 x Thus, the price (inclusive of tax) paid by the shopkeeper

10 (i) In D AEC, tan 30 0 AE EC 1 y 3 x x y...(1) 3 In D DBA, cot 60 0 BD AB 1 x x m 3 Therefore, the horizontal distance between AB and CD m. (ii) Substituting the value of x in (1), 0 3 y 3 0 m Height of the lamp post CD (60-0) m 40 m Sol. 8 éx 3x ù é ù 5 é (a) êy 4y ú1 ê ú 1ê ë û ë û ë éx + 3x ù 5é êy + 4y ú 1 ê ë û ë x + 3x 5 x 1 and y + 4y 1 y (b) Sphere : R 15 cm Cone : r.5 cm, h 8 cm Let the number of cones recasted be n. n Volume of one cone Volume of solid sphere n p r h p R 3 3 n (.5) ( 8) 4 3 ( 15) n n 70 Thus, 70 cones were recasted. x + (p - 3) x + p 0 Here, A 1, B (p - 3), C p Since, the roots are real and equal, D 0

11 B - 4ac 0 (p - 3) - 4(1) (p) 0 p + 9-6p - 4p 0 p - 10p (p - 1)(p - 9) 0 p 1 or p 9 Sol. 9 (a) Area of the quadrant OACB 1 p r cm Area of the triangle OAD 1 base height cm Shaded Area Area of quadrant OACB - area of triangle OAD cm 6.15 cm (b) Number of white balls in the bag 30 Let the number of black balls in the box be x. Total number of balls x + 30 x P(drawing a black ball) x P(drawing a white ball) x + 30 It is given that: P(drawing a black ball) x P(drawing a white ball) 5 x 30 x x+ 30 x 1 x + 30 x+ 30 x 1 Therefore, number of black balls in the box is 1.

12 Class interval Frequency (f) Class mark (x) d x - A h (A 55) A Total 50-7 fd Here, A 55, h 10 å Mean A+ fd å f h Sol. 10 (a) (i) Steps of constructions: (a) Draw a line segment BC 6 cm. (b) At B, draw a ray BX making an angle of 10 o with BC. From point B cut an arc of radius 3.5 cm to meet ray BX at C. (d) Join AC. ABC is the required triangle. (ii) (a) Bisect BC and draw a circle with BC as diameter. (b) Draw perpendicular bisectors of AB. Let the two bisectors meet the ray of angle bisector of Ð ABC at point P. P is equidistant from AB and BC. (iii) On measuring Ð BCP 30 o

13 (b) Marks No. of Students Cumulative Frequency n (i) Through marks 60, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a line perpendicular to x-axis meeting at B. Median 43 (ii) Through marks 75, draw a line segment parallel to y-axis which meets the curve at D. From D, draw a line perpendicular to y-axis which meets y-axis at 110. Number of students getting more than 75% students (iii) Through marks 40, draw a line segment parallel to y-axis which meets the curve at C. From C, draw a line perpendicular to y-axis which meets y-axis at 5. Number of students who did not pass 5.

14 Sol. 11 (a) Let the coordinates of A and B be (x, 0) and (0, y) respectively. Given P divides AB is the ratio :3, Using section formula, we have : x y x y x 0 y x - 5 y 10 Thus, the coordinates of A and B are (-5, 0) and (0, 10) respectively. (b) 4 x x 8 Using componendo and dividendo, 4 x + 1+ x x + 1- x 17-8 ( + ) ( x - 1) x æ x 1 ö + æ5 ö ç ç ç èx - 1 ø çè 3 ø x x x + 1+ x x + 1- x Using componendo and dividendo x 8 x 4 x ± Original cost of each book x \ Number of books brought for x In nd case: The cost of each book (x - 8) Number of books bought for x - 8 From the given information, we have: ( )

15 x - 8 x 960x - 960x x x 8 ( - ) x ( x- 8 ) x - 8x (x - 48) (x + 40) 0 x 48 or -40 But x can't be negative. \ x 48 Thus, the original cost of each book is 48.

Mathematics Class X Board Paper 2011

Mathematics Class X Board Paper Solution Section - A (4 Marks) Soln.. (a). Here, p(x) = x + x kx + For (x-) to be the factor of p(x) = x + x kx + P () = Thus, () + () k() + = 8 + 8 - k + = k = Thus p(x)