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1 CBSEXII07 EXAMINATION CBSEX009 EXAMINATION MATHEMATICS Series: HRL Paper & Solution Code: 0/ Time: Hrs. Max. Marks: 80 General Instuctions : (i) All questions are compulsory. (ii) The question paper consists of 0 questions divided into four sections A, B, C and D. Section A comprises of ten questions of mark each, Section B comprises of five questions of marks each, Section C comprises of ten questions of marks each and Section D comprises of five questions of 6 marks each (iii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. (iv) There is no overall choke. However, an internal choice has been provided in one question of marks each, three questions of marks each and two questions of 6 marks each. You have to attempt only one of the alternatives in all such questions. (v) In question on construction, the drawings should be neat and exactly as per the given measurement. (vi) Use of calculators is not permitted. SECTION A Questions number to 0 carry mark each.. Find the ( HCF LCM ] for the numbers 00 and 90. Given two numbers 00 and 90 Product of 00 and 90 HCF LCM If is a zero of the polynomial p(x) = ax  (a  )x , then find the value of a. If x = is the zero of the polynomial p x ax a x Then p() 0 a () ( a ) 0 a 0 a ( ) ( ). In LMN, L = 50 and N = 60. If LMN ~ PQR, then find Q. /
2 CBSEX009 EXAMINATION Given LMN ~ PQR In similar triangles, corresponding angles are equal. L P M Q x N R In LMN L M N 80 M M 70 Q If sec sec sin sin k, then find the value of k. sec ( sin )( sin ) k sec sin κ sec.cos κ cos k cos k 5. If the diameter of a semicircular protractor is 4 cm, then find its perimeter. Given diameter of semicircular protractor (AB) = 4cm d Perimeter of a semicircle d 4 Perimeter of protractor cm 6. Find the number of solutions of the following pair of linear equations : /
3 CBSEX009 EXAMINATION x + y 8 = 0 x + 4y = 6 xy 8 0 x 4y6 0 For any pair linear equations a x b y c 0 a x b y c 0 a b c, then a b c If There exists infinite solutions 8 Here a, b, c a b 4 c 6 a b c a b c Lines are coincident and will have infinite solutions. 7. Find the discriminant of the quadratic equation x 0x 0. x 0x 0 Discriminant for ax bx c 0 will be b For the given quadratic equation (0) 4( )( ) ac 8. If 4, a, are three consecutive terms of an A.P., then find the value of a. 5 Given 4, a, are in A.P. 5 4 a a 5 4 a 5 4 a 5 7 a 5 /
4 CBSEX009 EXAMINATION 9. In Figure, ABC is circumscribing a circle. Find the length of BC. Given BR = cm, AR = 4cm & AC = cm BP BR AR AQ CP CQ {Lengths of tangents to circle from external point will be equal} AQ 4 cm and BP cm As AC = cm QC + AQ = cm QC 7cm PC 7cm We know BC BP PC BC 7 BC 0cm 0. Two coins are tossed simultaneously. Find the probability of getting exactly one head. Two coins are tossed simultaneously Total possible outcomes { HH, HT, TH, TT} No. of total outcomes = /
5 CBSEX009 EXAMINATION Favourable outcomes for getting exactly One head = { Ht, TH } Probability = 4 Questions number to 5 carry marks each. SECTION B. Find all the zeroes of the polynomial x + x x 6, if two of its zeroes are and. x x x 6 0 Given two zeros are, Sum of all zeros =  Let the third zero be x x x All zeros will be,,. Which term of the A.P., 5, 7, 9,... will be 0 more than its st term? Given an A.P., 5, 7, 9, Lets say n th term is 0 more than st term T 0 T n a ( n ) d 0 ( a 0 d) ( n ) 0 0 n 0 n st term is 0 more than st term.. In Figure, ABD is a right triangle, rightangled at A and AC BD. Prove that AB = BC. BD. 5 /
6 CBSEX009 EXAMINATION In ABC AB AD BD In ABC AC BC AB In ACD AC CD AD...()...()...() Subtracting () from () AB AD BC CD Adding () and (4) AB BD BC CD...(4) AB BC CD BC CD AB BC CD BC. CD BC CD AB BC BC CD AB BC. BD ( ) 5 4. If cot, then evaluate ( sin )( sin ). 8 ( cos )( cos ) Find the value of tan 60, geometrically. 5 cot 8 ( sin )( sin ) ( sin ) ( cos)( cos ) ( cos ) cos sin cot 5 64 OR OR 6 /
7 CBSEX009 EXAMINATION Consider a right triangle with A 60, AC cm and AB = cm BC tan60 AC By Pythagoras theorem BC 4 BC tan60 5. If the points A (4, ) and B (x, 5) are on the circle with the centre O (, ), find the value of x. OA = OB (radii) OA ( 4) (0) OB x x ( ) ( 5) ( ) 4 ( x) 4 4 ( x) 4 x Questions number 6 to 5 carry marks each. 6. Prove that is an irrational number. Lets assume is a rational number p {p, q are integers and q 0} q SECTION C 7 /
8 CBSEX009 EXAMINATION p q p q q Pq is rational number but we know q Irrational rational is not a rational number. is an irrational. 7. Solve for x and y ax by a b b a ax by ab OR The sum of two numbers is 8. Determine the numbers if the sum of their. reciprocals is 8. 5 ax by a b. b a...() ax by ab...() Multiply () with b and subtract () from () a x y a b a by a b b a b a y a b a ya Substituting y = a in () a b x ( a) a b b a a x a b x b x b and y a Lets assume numbers are x, y Given x y 8 x 8 y...() OR 8 /
9 CBSEX009 EXAMINATION 8 x y 5 x y xy 5 xy 5 xy 5 From () xy y(8 y) 5 y y y, 5 x 5, The numbers are and 5 8. The sum of first six terms of an arithmetic progression is 4. The ratio of its 0 th term to its 0 th term is :. Calculate the first and the thirteenth term of the A.P. Lets say first term of given A.P. = a Common difference = d Sum of first six terms = 4 6 a 5 d 4 a5d4...() Also given T0 : T0 : a 9d a 9d a 7d a 9d ad ad...() Substituting () in () a 5a 4 a and d T a d 4 T 6 9. Evaluate : 5 cosec 58 cot58tan tantan7tan45tan5tan77 5 cosec 58 cot58tan tantan7tan45tan /
10 CBSEX009 EXAMINATION cot58 tan(90 58 ) tan tan77 cot(90 77 ) cot tan5 cot(90 5 ) cot7 tan45 Substituting in the given expression 5 cosec 58 cot 58 5 cosec 58 cot Draw a right triangle in which sides (other than hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are 4 times the corresponding sides of the first triangle. Given ABC which is a right angled triangle B 90 Steps:. Draw line segment BC = 8cm, draw a ray BX making 90 with BC. Draw an arc with radius 6cm from B so that it cuts BX at A. Now join AC to form ABC 4. Draw a ray by making an acute angle with BC, opposite to vertex A 5. Locate 4 points B, B, B, B 4, on by such that BB = B B = B B = B B 4 6. Join B 4 C and now draw a line from B parallel to B 4 C so that it cuts BC at C 0 /
11 CBSEX009 EXAMINATION 7. From C draw a line parallel to AC and cuts AB at A 8. A' BC ' is the required triangle Justification: ABC and A' BC ' ABC A' BC ' (Common) ACB A' C ' B (Corresponding angles) By AA criterion ABC ~ A' BC ' AB BC AC A' B BC ' A' C ' BB C and BB C ' In 4 B C B C ' [By construction] 4 BB C 4 ~ BB C ' BB4 BC B4C BB BC ' B C ' BB4 4 We know that BB A' B BC ' A' C ' AB BC AC 4. In Figure, AD BC and BD = CD. Prove that CA = AB + BC OR In Figure 4, M is midpoint of side CD of a parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced at E. Prove that EL = BL. /
12 CBSEX009 EXAMINATION BD CD; BD CD BC CD BC 4 BD BC 4 In right ACD, AC = AD + CD () In right ABD, (Pythagoras Theorem) AB AD BD...() (Pythagoras Theorem) From () and (), we get AC = AB BD + CD BC BC AC AB 4 4 BC 9BC AC AB 6 6 9BC BC AC AB 6 8BC AC AB 6 BC AC AB AB BC AC /
13 CBSEX009 EXAMINATION OR In DME and CMD EDM MCB [Alternate angles] DM = CM [mis midpoint of CD] DME BMC [VOA] By ASA congruency DME CMB By CPCT BM = ME DE = BC Now in ALE and BLC ALE BLC LAE LCB By AA similarly ALE ~ BLC AE AL LE BC BL LC EL AE BL BC EL AD DE BL BC EL BC BC BL BC EL BL [VOA] [Alternate angles]. Find the ratio in which the point (, y) divides the line segment joining the points A (, ) and B (, 7). Also find the value of y. Lets say ratio = m : n /
14 CBSEX009 EXAMINATION m n n 7m (, y), m n m n m n m n m n m n m: n 4 : 74 y 5 0 y 5 y 6 p(,6). Find the area of the quadrilateral ABCD whose vertices are A ( 4, ), B (, 6), C (, ) and D (, ). Joining AC Area of Quadrilateral ABCD ar( ABC ) ar( ADC) Area of triangle ABC = 4( 5 ( )) ( )( ( )) ( ( 5)) 4( 5 ) ( )( ) ( 5) [ 4( ) (0) ()] [ 0 9] []square units Area of triangle ADC = 4( ( )) ( ( )) ( ) 4 /
15 CBSEX009 EXAMINATION 4( ) ( ) ( ) 4(5) (0) ( 5) sq. units Area of quadrilateral (ABCD) = 8 sq.units 5 4. The area of an equilateral triangle is 49 cm. Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of triangle not included in the circles. [Take =.7] OR Figure 5 shows a decorative block which is made of two solids a cube and a hemisphere. The base of the block is a cube with edge 5 cm and the hemisphere, fixed on the top, has a diameter of 4. cm. Find the total surface area of the block. [Take ] 7 Let be the side of equilateral triangle. a 49 ; 4 a 49*4; a= 7 * = 4 cm; Radius of circle = 4/ = 7 cm 5 /
16 CBSEX009 EXAMINATION Area of the first circle occupied by triangle = area of sector r *7 * * cm 80 Area of all the sectors = 77/ * = 77 cm Area of triangle not Included In the circle = area of triangle area of all the sectors cm OR The total surface area of the cube = 6 (edge) = cm = 50 cm Note that the part of the cube where the hemisphere is attached is not Included in the surface area. So, the surface area of the block = TSA of cube base area of hemisphere + CSA of hemisphere 50 r r (50 r ) cm cm cm cm 6.86 cm 5. Two dice are thrown simultaneously. What is the probability that (i) 5 will not come up on either of them? (ii) 5 will come up on at least one? (iii) 5 will come up at both dice? Total outcomes = (i). Total outcomes when 5 comes up on either dice are (5, ) (5, ) (5, ) (5, 4) (5, 5) (5, 6) (6, 5) (4, 5) (, 5) (, 5) (, 5) P (5 will come up on either time) 6 P (5 will not come up) = (ii) P (5 will come at least once) = 6 (iii) P (5 will come up on both dice = 6 ) Questions number 6 to 0 carry 6 marks each. SECTION D 6 /
17 CBSEX009 EXAMINATION 6. Solve the following equation for x : 9x 9 (a + b) x + (a + 5ab + b ) = 0 OR If (5) is a root of the quadratic equation x + px 5 = 0 and the quadratic equation p (x + x) + k = 0 has equal roots, then find the values of p and k. 9x 9( a b) x (a 5ab b ) 0 Discriminant D 8( a b) 6 a 5ab b D 9 9a 9b 8ab 8a 8b 0ab D 9 a b ab D 9( a b) 9( a b) 9( a b) x 9 9( a b) ( a b) x 8 a b a b a b a b x, 6 6 a b a b x ; 5 is root of x + 9x 5 =0 ( 5) p( 5) p 0 p 7 p(x + x) + k = 0 has equal roots 7x + 7x + k = 0 [as we know p =7] Discriminant = 0 D49 8k 8k 49 k 7 4 OR 7. Prove that the lengths of the tangents drawn from an external point to a circle are equal. Using the above theorem prove that: If quadrilateral ABCD is circumscribing a circle, then AB + CD = AD + BC. PT and TQ are two tangent drawn from an external pant T to the circle C (O, r) 7 /
18 CBSEX009 EXAMINATION To prove:. PT = TQ. OTP = OTQ Construction Join OT Proof: We know that, a tangent to circle is perpendicular to the radius through the point of contact OPT = OQT = 90 In OPT and OQT OT = OT (Common) OP = OQ (Radius of the circle) OPT = OQT (90 ) OPT OQT (RHS congruence criterion) PT = TQ and OTP = OTQ (CPCT) PT = TQ The lengths of the tangents drawn from an external point to a circle are equal. OTP = OTQ Centre lies on the bisector of the angle between the two tangents. Let AB touches the circle at P.BC touches the circle at Q.DC touches the circle at R.AD. touches the circle at S THEN, PB = QB ( Length of the tangents drawn from the external point are always equal) QC =RC' AP : AS DS = DP NOW, AB + CD = AP + PB+DR+RC = AS+QB+DS+CQ = AS + DS + OB + CQ 8 /
19 CBSEX009 EXAMINATION = AD + BC HENCE PROVED 8. An aeroplane when flying at a height of 5 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 0 and 60 respectively. Find the distance between the two planes at that instant. Let the distance between the two planes = h m Given that: AD = 5 m and ABC 60 ABD 0 In ABD AD tan0 AB 5 AB AB 5...() ABC AC tan60 AB AD DC AB 5 h AB 5 h AB...() Equating equation () and (), we have 9 /
20 CBSEX009 EXAMINATION 5 h 5 h 5 5 h 650 Hence, distance between the two planes is 650 m. 9. A juice seller serves his customers using a glass as shown in Figure 6. The inner diameter of the cylindrical glass is 5 cm, but the bottom of the glass has a hemispherical portion raised which reduces the capacity of the glass. If the height of the glass is 0 cm, find the apparent capacity of the glass and its actual capacity. (Use =.4) OR A cylindrical vessel with internal diameter 0 cm and height 0.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of (i) water displaced out of the cylindrical vessel. (ii) water left in the cylindrical vessel. [Take ] 7 Apparent capacity of the glass = Volume of cylinder Actual capacity of the glass = Volume of cylinder Volume of hemisphere Volume of the cylindrical glass =.4 (.5) cm rh 0 /
21 CBSEX009 EXAMINATION Volume of hemisphere r (.5).7 cm Apparent capacity of the glass = Volume of cylinder = 96.5 cm Actual capacity of the glass = Total volume of cylinder volume of hemisphere = = 6.54cm Hence, apparent capacity = 96.5cm Actual capacity of the glass = 6.54cm 9. OR Given, internal diameter of the cylinder =0 cm Internal radius of the cylinder = 5cm and height of the cylinder = 0.5 cm Similarly, diameter of the cone = 7 cm radius of the cone =.5 cm and Height of the cone = 6cm i) Volume of water displaced out of cyindrical vessel = volume of cone rh cm 7 ii) Volume of water left In the cylindrical vessel = volume of cylinder  volume of cone RHVolume of cone = = =748cm 0. During the medical checkup of 5 students of a class their weights were recorded as follows: Weight (in Number of students kg) /
22 CBSEX009 EXAMINATION Draw a less than type and a more than type ogive from the given data. Hence obtain the median weight from the graph. Weight cumulative (more than type) More than 8 5 More than 40 More than 4 0 More than 44 6 More than 46 More than 48 7 More than 50 More than 5 0 Weight (in kg) Upper class limits Number of students (cumulative frequency) Less than 8 0 Less than 40 Less than 4 5 Less than 44 9 Less than 46 4 Less than 48 8 Less than 50 More than 5 5 Taking upper class limits on xaxis and their respective cumulative frequency on yaxis its give can be drawn as follows: Here, n = 5 So, /
23 CBSEX009 EXAMINATION n = 7.5 Mark the point A whose ordinate is 7.5 and its xcoordinate is Therefore, median of this data is It can be observed that the difference between two consecutive upper class limits is. The class marks with their respective frequencies are obtained as below: Weight (In kg) Frequency (f) Cumulative, frequency Less than = = = = = = = 5 Total (n) 5 The Cumulative frequency just greater than n 5 ie.., 7.5 is 8. Belonging to class interval Medan class = Lower class lima (l) of median class = 46 Frequency (f) of median class = 4 Cumulative frequency (cf) of class preceding median class = 4 Class size (h) = n cf Median l h f =46.5 Therefore, median of this data is 46.5 Hence, the value of median is verified. /
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