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1 CBSE-XII-017 EXAMINATION MATHEMATICS Paper & Solution Time: 3 Hrs. Max. Marks: 90 General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into four sections - A, B, C and D. (iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of marks each, Section C contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each. (iv) Use of calculators is not permitted. Question numbers 1 to 4 carry 1 mark each. SECTION A 1. What is the common difference of an A.P. in which a 1 - a 7 = 84? Given a 1 a 7 =84...(1) In an A.P a 1, a, a 3, a 4 a n = a 1 + (n 1)d where d = common difference a 1 = a d..() a 7 = a 1 + 6d.(3) substituting () & (3) in (1) a 1 + 0d a 1 6d = 84 14d = 84 d = 6 common difference = 6. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60, then find the length of OP. 1 /

2 Given that BPA 60 OB = OA = a PA = PB [radii] [length of tangents Equal] OP = OP PBO and PAO are congruent. 60 BPO OPA 30 In a a PBO sin30 OP OP = a 3. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of the sun? Angle of equation of sun = Height of tower TG = 30m GST Length of shadow GS = 10 3 m TGS is a right angled triangle 30 tan 10 3 tan /

3 4. The probability of selecting a rotten apple randomly from a heap of 900 apples is What is the number of rotten apples in the heap? Probability of selecting rotten apple = Number of rotten apples Total number of apples No. of rotten apples No. of rotten apples = SECTION B Question numbers 5 to 10 carry marks each. 5. Find the value of p, for which one root of the quadratic equation px - 14x + 8 = 0 is 6 times the other. Given Quadratic Equation px 14x + 8 = 0 Also, one root is 6 times the other Lets say one root = x Second root = 6x From the equation : Sum of the roots = 14 p Product of roots = 8 p 14 x 6x p x p 8 6x p 8 6 p p p p p /

4 6. Which term of the progression Given progression 0, ,18,17, This is an Arithmetic progression because ,19,18,17, Common difference (d) = d 3 4 is the first negative term? Any n th term a n = n ( n 1) 4 4 Any term an < 0 when 83 < 3n This is valid for n = 8 and 8 th term will be the first negative term. 7. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord. Need to prove that BAP ABP AB is the chord We know that OA = OB (radii) OBP OAP 90 Join OP and OP = OP 4 /

5 By RHS congruency OBP OAP By CPCT BP AP In ABP BP AP Angles opposite to equal sides are equal BAP ABP Hence proved. 8. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA ABCD is the Quadrilateral Circle touches the sides at P, Q, R,S points For the circle AS & AP are triangles AS AP...(1) In the similar way BP = BQ..() CQ = CR (3) RD = DS..(4) Now AB + CD = AP + PB + CR + RD BC + AD = BQ + QC + DS + AS Using (1), (), (3), (4) in above equation BC + AD = BP + CR + RD + AP 5 /

6 AB + CD = BC + AD Hence proved 9. A line intersects the y-axis and x-axis at the points P and Q respectively. If (, -5) is the mid-point of PQ, then find the coordinates of P and Q. x y Equation of a line: 1 a b Where a =x-intercept b = y-intercept Given that line intersects y-axis at P P lines on y-axis and P = (0, b) Line intersects x-axis at Q Q lines on x-axis and Q = (a, 0) Midpoint of PQ = (, -5) a b, (, 5) a b, 5 a 4 & b 10 P (0, 10) Q (4,0) 10. If the distances of P(x, y) from A(5, 1) and B(-1, 5) are equal, then prove that 3x = y. Given that PA = PB P(x, y), A(5, 1), B(-1, 5) PA ( x 5) ( y 1) PB ( x 1) ( y 5) ( x 5) ( y 1) ( x 1) ( y 5) Squaring on both sides 6 /

7 x x y y x x y y x y x 10y 8y 1x 3x4y SECTION C Question numbers 11 to 0 carry 3 marks each. 11. If ad bc, then prove that the equation (a + b ) x + (ac + bd) x + (c + d ) = 0 has no real roots. Given ad bc for the equation ( a b ) x ( ac bd) x ( c d ) 0 For this equation not to have real roots its discriminant < 0 D = 4(ac + bd) 4(a + b )(c + d ) D = 4a c + 4b d + 8acbd 4a c 4b d 4b c 4a d D a d b c acbd 4( ) D 4( ad bc) Given ad D 0 bc Quadratic equation has no real roots 1. The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference of the A.P. First term (a) = 5 Last term (l) = 45 Sum of all the terms = 400 n 400 ( a l ) 800 n 50 n 16 No. of terms = 16 l = 45 (16 th term) 7 /

8 a ( n 1) d d 45 d d Common difference On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower. Given CF = 4m DF = 16m TCF TDF 90 Lets say TDF TCF In a right angled triangle TCF TF TF tan CF 4 TF 4 tan...(1) In TDF TF tan (90 ) 16 TF 16cot...() Multiply (1) & () 8 /

9 ( ) 64 8 TF TF mt Height of lower = 8mt 14. A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag. Bag contains 15 white balls Lets say there are x black balls Probability of drawing a black ball P(B) = 15 x x Probability of drawing a white ball P(W) = x Given that P(B) = 3P(A) x x 15 x x 45 No. of black balls = In what ratio does the point Also find the value of y. 4, 11 y divide the line segment joining the points P(, -) and Q(3, 7)? Lets say ratio is m + n Then 9 /

10 4 3m n 7m n, y, 11 m n m n 4 3m n 7m n, y 11 m n m n 4( m n) 11(3m n) 4m 4n 33m n n 9m m 9 ratio 9 : n m9, n 79 y y Three semicircles each of diameter 3 cm, a circle of diameter 4 5 cm and a semicircle of radius 4 5 cm are drawn in the given figure. Find the area of the shaded region. Given that AB = BC = CD = 3 cm Circle c has diameter = 4.5 cm Semicircle s 1 has diameter = 9cm Area of shaded region = Area of s 1 Area of (s +s 4 ) Area of c + Area of s /

11 Area of shaded region cm 17. In the given figure, two concentric circles with centre O have radii 1 cm and 4 cm. If AOB = 60, find the area of the shaded region. Use 7 Given OC = OD = 1 cm OA = OB = 4 cm Area of ACDB region = Area of sector OAB Area sector OCD (4) (1) /

12 Area of ACDB region = cm Area of shaded region= area of c 1 Area of c Area of ACDB region (4) ,465 cm 18. Water in a canal, 5 4 m wide and 1 8 m deep, is flowing with a speed of 5 km/hour. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation? Given canal width = 5.4 mt Depth = 1.8 mt Water flow speed = 5 km/hr Distance covered by water in 4 onion km Volume of water flows through pipe = m Area irrigate with 10mt of water standing m The slant height of a frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum. 1 /

13 Given: r 6cm R 18cm l 4cm Curved surface area of frustum of a cone 1 R r l 1 (6 18)4 48cm 0. The dimensions of a solid iron cuboid are 4 4 m 6 m 1 0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe. Volume of cuboid = m length = l Inner radius = 30 cm outer radius = 35 cm Volume of cuboid = volume of cylindrical pipe 13 /

14 l l cm l km (35 30 ) Question numbers 1 to 31 carry 4 marks each. 1. Solve for x : , x 1,, 4 x 1 5x 1 x 4 5 SECTION D ; x 1,, 4 x 1 5x 1 x 4 5 Take L.C.M. on the left hand side of equation 5x1 3( x1) 5 ( x 1)(5x 1) x 4 ( x 4)(8x 4) 5(5x 1)( x 1) x x x x x x 17x 6x x 17x 11x x( x 1) 11( x 1) 0 ( x1)(17 x11) 0 x 11,1 17. Two taps running together can fill a tank in 3 1 hours. If one tap takes 3 hours more than the other to 13 fill the tank, then how much time will each tap take to fill the tank? Two taps when run together fill the tank in Say taps are A, B and A fills the tank by itself in x hrs B fills tank in (x+3) hrs 1 3 hrs /

15 Portion of tank filled by A (in 1hr) = 1 x Portion of tank filled by B (in 1hr) = 1 x 3 Portion of tank filled by A & B (both in 1hr) = x x 3 40 ( x 3 x)40 13( x)( x 3) x 41x10 0 x 5 x x x A fills tank in 5hrs B fills tank in 8hrs 3. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 7), then find the ratio of their 9 th terms. Given two A.P s with n terms each A.P I = first term = a 1 A.P II = first term = a Common difference = d Sum of first n terms for A.P I = S 1 n S a ( n 1) d n Similarly S a ( n 1) d S S 1 7n 1 4n 7 ( n 1) a1 d1 7n 1 ( n 1) a 4n 7 d 15 /

16 Ratio of their 9 th a1 8d1 terms a 8d Comparing ( n 1) a1 d1 a1 8d1 & ( n 1) a a 8d d Upon comparing n 1 8 ( n 17) substituting n value ( n 1) a d a d n a ( 1) 8d n a 4n 7 4(17) 7 95 d ratio 60 : (17) Prove that the lengths of two tangents drawn from an external point to a circle are equal. PA & PB are the length of the tangents drawn from an external point P to circle C with radius r OA OB r OA PA OB PB Join O & P In the triangles OAP & OBP OA = OB OP = OP (radii) (common side) OAP OBP 90 (Right angle) By RHS congruency 16 /

17 OAP OBP By CPCT PA = PB 5. In the given figure, XY and X Y are two parallel tangents to a circle with centre O and another tangent AB with point of contact C, is intersecting XY at A and X Y at B. Prove that AOB = 90. Prove that AOB 90 In AOC and AOP OA = OA OP = OC (hypotenuse) (radii) ACO APO (right angle) AOC AOP By RHS congruency By CPCT AOC AOP...(1) Similarly In BOC & BOQ OC = OQ OB = OB BCO BQO 90 By RHS congruency BOC BOQ 17 /

18 By CPCT BOC BOQ...() PQ is a straight line AOP AOC BOC BOQ 180 From equations (1) and () AOC BOC AOB AOB Construct a triangle ABC with side BC = 7 cm, B = 45, A = 105. Then construct another triangle whose sides are 3 4 times the corresponding sides of the ABC. In the ABC, A B C 180 C 30 To construct the similar triangle first we need to construct For ABC 1). Draw BC 7cm with help of a ruler ABC ) Take a protractor measure angle 45 from point B and draw a ray BX 3) From point c, mark 30 with help of protractor & draw a ray CY 4) Now both BX and BY intersect at a point and this point is A Now we have ABC To construct similar triangle with corresponding sides 3 4 of the sides of ABC Step 1: Draw any raw making an acute angle with BC Step : Along the ray BZ mark 4 points B 1, B, B 3, B 4 such that BB 1 = B 1 B = B B 3 = B 3 B /

19 Step 3: Now join B 4 to C and draw a line parallel to B 4 C from B 3 intersecting the line BC line BC at C Step 4: Draw a line through C parallel to CA which intersects BA at A A BC is the required tri Justification: C' A' CA By construction A' BC ' ~ ABC [using AA similarity] A' B BC ' A' C ' [corresponding sides ratio will be proportional AB BC AC B4C B3C ' [By construction] BB C ~ BB C ' [By AA similarity] BC ' BC 4 3 BB BB 3 [By BPT] 4 BB3 3 But we know BB 4 A' B BC ' A'. C ' 3 AB BC AC /

20 7. An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45 and 60 respectively. Find the width of the river. [Use 3 = 1 73] Given aeroplane is at height pf 300m AB 300m and XY PQ Angles of depression of the two points P & Q are 30 and 43 XAP 30 & YAQ 45 XAP APB 30 [Alternative Interior angles] YAQ AQB 45 In PAB AB tan30 PB PB mt In BAQ AB tan45 BQ BQ 300m Width of the river = PB + BQ 300(1 3)mt 8. If the points A(k + 1, k), B(3k, k + 3) and C(5k - 1, 5k) are collinear, then find the value of k. Given A( k 1, k), B(3 k,k 3), C(5k 1,5 k) are collinear. If three points are collinear then the area of the triangle will be zero. For any 3 points (x 1, y 1 ), (x, y ), (x 3, y 3 ) Area will be 1 A x1( y y3) x( y3 y1) x3( y1 y) 1 O ( k 1)( k 3 5 k ) 3 k (5 k k ) (5 k 1)( k k 3) O ( k 1)(3 3 k) 3 k(3 k) 15k /

21 3k 3 9k 3 15k 0 6k 15k 0 5 k 0, 9. Two different dice are thrown together. Find the probability that the numbers obtained have (i) even sum, and (ii) even product. Two dice are through together total possible outcomes (i) Sum of outcomes = even This can be possible when Both outcomes are even Both outcomes are odd For both outcomes to be Even number of cases = 3 3 = 9 Similarly Both outcomes odd = 9 cases Total favourable cases = =18 18 Probability that 36 1 Sum of the outcomes is even (ii) Product of outcomes is even This is possible when Both outcomes are even first outcome even & the other odd first outcome odd & the other even Number of cases where both outcomes are even = 9 Number of cases for first outcome odd = 9 and the other Even No. of cases for first outcome odd & the other even = 9 Total favourable cases = = 7 Probability = In the given figure, ABCD is a rectangle of dimensions 1 cm 14 cm. A semicircle is drawn with BC as diameter. Find the area and the perimeter of the shaded region in the figure. 1 /

22 Area of shaded region = Area of rectangle Area of semicircle (7) cm Perimeter of shaded region = AB + AD + CD + length of arc BC cm 31. In a rain-water harvesting system, the rain-water from a roof of m 0 m drains into a cylindrical tank having diameter of base m and height 3 5 m. If the tank is full, find the rainfall in cm. Write your views on water conservation. Water from the roof drains into cylindrical tank Volume of when from roof flows into the tank of the rainfall is x cm and given the tank is full we can write volume of water collected on roof = volume of the tank 0x 100 x 5 cm rainfall is of 5cm /

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1 / 24 CBSE-XII-017 EXAMINATION CBSE-X-01 EXAMINATION MATHEMATICS Paper & Solution Time: 3 Hrs. Max. Marks: 90 General Instuctions : 1. All questions are compulsory.. The question paper consists of 34 questions

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