CONTENTS. v - viii Model Question Paper 2015 (Issued by KSEEB) in Oct Solutions On Tips Notes 25-48

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4 ONTENTS q q q Design & lueprint of the Question Paper (Issued by KSEE) in Oct. 05 v - viii Model Question Paper 05 (Issued by KSEE) in Oct Solutions - On Tips Notes 5-48 Sample Question Papers (Solved) Sample Question Paper 49-5 (KSEE June 05 Modified* Paper with Model nswers) Model nswers June Sample Question Paper 6-65 (KSEE pril 05 Modified* Paper with Model nswers) Model nswers pril Sample Question Paper Sample Question Paper Sample Question Paper Sample Question Papers (Self ssessment) Sample Question Paper Sample Question Paper Sample Question Paper Sample Question Paper Sample Question Paper Solutions Sample Question Paper 08-7 Sample Question Paper Sample Question Paper q Solutions for Sample Question Paper 6 to 0 an be downloaded from * The Marking Scheme provided in the solutions is as per the Marking Scheme followed by KSEE in June / pril 05 Model nswers.

5 PREFE Since 966 Karnataka Secondary Education Examination oard has been conducting SSL and other examinations. lbert Einstein said Education is not the learning of facts, but the training of mind to think. So to train the mind and strengthen the roots of students, the KSEE implements ontinuous & omprehensive Evaluation (E). The term ontinuous in the E refers to the periodicity and regularity in assessment and the term omprehensive refers to overall assessment of the learner, in both curricular & co curricular scheme of things. The KSEE initiated the change and to support it, Oswaal Publications started an exclusive series of books which comprises of Sample Question Papers along with other titles. The thought was to give students enough questions to solve rather than feeding them with answers to memorize. These Sample Question Papers strictly follow the SE guidelines, syllabus and marking scheme. This book has a total of 0 sample question papers, out of which 5 question papers are solved, and the answers follow the word limit specified by the KSEE. The answers have been written in steps and each step specifies marks as per the KSEE marking scheme. This will enable students to judiciously choose what to write, how much to write and how to manage the task in the allotted time which will reduce their stress during examination. However, as mentioned, Oswaal Publications believes that students should practise with enough number of questions, hence this book has 5 more sample papers which are unsolved and meant for self-assessment. fter solving the unsolved sample question papers, students can download solutions from our website and can evaluate their performance. In this way students can take proactive measures for examinations by learning from their mistakes. Each Sample Question Paper covers all typologies specified by KSEE. Wherever required, well labelled and high quality figures/diagrams are given for easy and quick learning. One of the major attractions is On Tips Notes which will help students in revision of the whole syllabus very fast and effectively. t last we would like to thank our authors, editors, reviewers and specially students who regularly send us suggestions which helps in continuous improvement of this book and makes this book stand in the category of One of the est. Wish you all Happy Learning. Publisher

10 MODEL QUESTION PPER 05-6 (Solved) (Issued by KSEE) in October-05 SUJET-MTHEMTIS LSS - X Time : Hours 45 Min. Max. Marks : 80 I. Four alternatives are given for each of the following questions / incomplete statements. Only one of them is correct or most appropriate. hoose the correct alternative and write the complete answer along with its letter in the space provided against each question. 8 = 8. In a G.P. if T = and T =, then common ratio is : 9 () () () (D) 4. The value of n po + n cn is : () n () n () (D) 0. Probability of a certain event : () 0 () () less than 0 (D) more than 4. X =,,, 4, 5 its standard deviation is, then the value of variance is : () () () (D) 5. If p (x) = x 4x x + 0 the factor for this polynomial is : () x + () x () x (D) x + 6. The slope of an equation x + y + = 0 is : () / () / () / (D) / 7. The distance between origin and a point (0,4) is : () () 4 () 8 (D) 6 8. If cos θ = and θ is an cute angle then the value of 'θ' is : () 0 () 0 () 45 (D) 60 II. nswer the following : 6 = In Venn diagram the value n( )?

11 0 OSWL Karnataka (SSL), Sample Question Papers, Mathematics, lass 0 0. In prime factorization of 09 write the highest prime factor.. If p(x) = x find the value of p( )?. In the adjoining figure the perimeter of PQR is 0 cm then calculate the measure of ( + )? P X Q. In the adjoining figure = 90 and D, express (D D), in terms of D? 90 D Write the formula to find Volume of a sphere. III. nswer the following : 6 = 5. In a Group of 58 people, 8 people drink rea only, 8 people drink coffee only and 0 people drink both coffee and tea then find the number of people who drink neither of these? 6. Prove that is an irrational number. 7. Write the value of (a) n P n n P n, (b) n n r n r 8. Find the value of n if n : n = : 9. In a Random experiment there is a chance of win or loss. ut the probability of winning is four times the loss then find the probability of win? 0. Simplify If x 4 x + = + x, then find the value of x.. When Polynomials (x + ax + x 5) and (x + x 4x a) are divisible by (x ) leaves the same reminder find the value of a. Find the quotient and remainder when (x 6 x 5 x + ) is divided by (x ).. Solve by using formula method x ax + (a + ) (a )= 0 4. onstruct a pair of tangents to a circle of radius cm. Such that angle between the tangents is In, DE and D EF. Prove that D = F. F D E

12 Solved Paper If ˆ = 5 then prove that 4 sin. cos 4. sin 6 =. 7. The Vertices of a triangle are (8, 4), (9, 5) and (0, 4). Prove that 'triangle is an isosceles triangle.' 8. The diameter of the base of ylinder is m and height is.8 m is method to recast a cone of diameter m, find the height of a cone. Diameter of garden roller is.4 m and length m takes 5 revolutions to complete the garden, find the area of the garden? 9. Dust bin in the form of frustum having radii 5 cm and 8 cm respectively and its depth is 6 cm. Find its volume? 0. Draw a scale drawing for the following information. Scale : 0 m = cm Towards D In metre 00 To E To To 50 From III. nswer the following : 6 = 8. Find the sum of all natural numbers between 00 and 00 which are divisible by 6. Find G.P. for which sum of the first terms is 4 and fifth term is 4 times the third term.. Find the standard deviation for the following Distribution by step deviation method. Marks (x) Number of students (f) In, =, D is the altitude for the base of a triangle. D = x units, D = x units, D = x + units, = (x + ) units. Find the measure of the sides of a triangle. Shweta takes 6 days less than the nkitha to complete the work. The same work is completed together in 4 days. How many days required to complete the work by nkitha alone? 4. "If two circles touch each other externally then the point of contact and the centres are collinear" Prove this. 5. In D, : D = :4. is a point on D, and is an equiangular triangle. Prove that D =. In, D : D = : and D. Prove hat ( ) =. 6. From the top of a building 6m high the angular elevation of the top of a hill is 60 and the angular depression of the foot of the hill is 0, find the height of the hill. Two windmills of height 50 m and 40 m are on either side of the field. person observes the top of the windmills from a point in between the towers the angle of elevation was found to be 45 in both the cases. Find the distance between the windmills. IV. nswer the following : 4 4 = 6 7. Five positive integers are in.p. The sum of middle terms is 4 and product of First and Fifth term is 48. Find the terms of.p. Find four terms of G.P. whose product of first three terms is 8 and sum of last three term is 4.

13 OSWL Karnataka (SSL), Sample Question Papers, Mathematics, lass 0 8. Solve graphically x 5x + 6 = 0 9. onstruct a transverage common tangent to two congruent circles of radii.5 cm, whose centres are 9 cm apart measure its length and verify by calculation. 40. "The areas of two-similar triangle are proportional to the squares on the corresponding sides" prove this.

14 Solved Paper-05 SOLUTIONS. () T = (Given) ar = ar = a = r T 9 = ar 9 =...(i) ar 8 =...(ii) Put value of a from eq. (i) in (ii) r r 8 = r 7 = r 7 = r 7 = 8 r 7 = ( ) 7 r =. (). () 4. () 5. () x + 6. (D) 7. () 4 8. (D) n ( ) = n ( ) n ( ) = 6 = P (x) = x. In given figure P P( ) = ( ) = ( ) = + = (Since tangents drawn from external points are equal) Q = QX...(ii) (Since tangents draw from external points are equal) P + PQ + Q = 0 cm P + PX + XQ + Q = 0 cm P + P + Q + Q = 0 cm (Given) [From eq. (i) and eq. (ii)] + = 0 cm. D. D = D 4 4. Volume of Sphere = π r 5. Total number of people = 58 No. of people drink tea only = 8 No. of people drink coffee only = 8 No. of people drink both coffee and tea = 0 T Number of people who drink none of these = Total number of people (No. of people drink tea only + No) of people drink coffee only + No. of people drink both coffee and tea) = 58 ( ) = = 6. Let us assume on the contrary that is a rational number. There exist co-prime positive integers a and b such that = a b a = b + Squaring on both sides, we get X ( ) = a + b Q P = PX...(i) = a b a + + b ½

15 4 OSWL Karnataka (SSL), Sample Question Papers, Mathematics, lass 0 a b a = b ( n ) ( n ) 6 n( n ) ( n ) = a b = b a b 6 4( n ) nn ( ) = = = b a b b a b a ab is a rational number. (Since, a and b are integers, So a b ab ½ is rational) 6 (8n 4) = n n 48n 4 = n n n 70n + 4 = 0 n (66 + 4) n + 4 = 0 n 66n 4n + 4 = 0 n (n 6) 4 (n 6) = 0 (n 6) (n 4) = 0 This contradicts that fact that assumption is wrong. is irrational, so our Hence, is irrational. ½ 7. (a) n P n n P n, n n = n n n n+ n n = 0 n n = ½ n = 6, n = 4 \ n = 6 9. Let the probability of winning = P (E) Let the probability of losing = P ( E ) Given, P (E) = 4P ( E ) We know that P (E) + P ( E ) = 4P ( E ) + P ( E ) = 5P ( E ) = = n n \ P ( E ) = 5 = 0 (b) n n r n r Probability of winning = 5 = n + n n r n n r r n r = 4 5 n n = n rr rn r = 0 = 8. n : n = : n n n n = + = + = n n = n n n ( n )( n ) n n 6 n ( n )( n ) n n =. If = 5 x 4 x + x 4 x + = + x = 4 + x

16 Solved Paper x 8 = ( x + ) 6x 8 = x x 6x x 8 4 = 4 x x = 4 x On Squaring both sides, we get (x ) = ( 4 x ) 9x x = 48x 9x 0x + 44 = 0 (x 40x + 48) = 0 x 40x + 48 = 0 x 6x 4x + 48 = 0 x(x ) 4 (x ) = 0 (x ) (x 4) = 0 \ x =, 4. Let when polynomial (x + ax + x+ 5) divided y (x ) leaves remainder = R x = 0 x = P(x) = x + ax + x + 5 P() = () + a() + ()+ 5 R = + a R = 0 + a ½ Let when polynomial x + x 4x a divided by x leaves remainder = R x = 0 x = q (x) = x + x 4x a q () = () + () 4() a = + 4 a q () = a R = a Given : R = R ½ 0 + a = a a + a = 0 a = a = \ a = 6 x ) x 6 x 5 x + (x 5 +x 6 x 5 + x + x Quotient = x 5 Remainder = 0. x ax + (a + ) (a ) = 0 4. x ax + a = 0 =, = a, = a x = x = x = x = a ± x = a ± 4 x = ± 4 ( a) ± ( a) 4 ( a ) a± 4 a 4a + 4 ( a ± ) x = a ± x = a +, O x = a Steps of construction : 40 (a) Draw a circle of radius cm, with centre 'O'. (b) Draw two radii O and O such that O = 40. ½ (c) t and, draw perpendicular P and P. \ P and P are the required tangents such that P = 40. ½ P ½ ½

17 6 OSWL Karnataka (SSL), Sample Question Papers, Mathematics, lass 0 5. = + 8 = 8 F = (0 9) + (4 5) D E = ( 9) + ( ) = 8 + = 8 In D, DE D = E D EF F D = E On comparing eq. () and eq. (), D = F D...() [by PT]...() [by PT] = (8 0) + ( 4 4) = (8) + ( 8) = = 8 = Hence D is isosceles. 8. ylinder : d =m, r = d Radius of the ylinder = r = m Height of the ylinder = h =.8 m Volume of the ylinder D = F. Hence Proved. 6. Given : = 5 To prove : 4 sin. cos 4. sin 6 = Proof : L.H.S. = 4 sin. cos 4. sin 6 = 4 sin 5. cos 4 5. sin 6 5 = 4 sin 0. cos 60. Sin 90 = 4 For one : V = pr h = p() (.8) =.8p m ½ d = m Radius of the one r = m =.5m Let the height of the one = h Volume of the one = π rh = 4 4 = π (.5) h m = = R.H.S. Hence Proved. 7. (8, 4) Volume of the one = Volume of the ylinder π.5.5 h =.8 p.8 h =.5.5 \ h =.4 m Height of the one =.4 m (9, 5) To Prove : Triangle is an isosceles. Proof : = (0, 4) (9 8) + (5 + 4) = () + (9).4 Radius of garden roller is = m = 0.7m Height of the garden roller is = m Number of revolutions = 5 rea of the garden = No, of revolutions urved surface area of roller

18 Solved Paper r = 8 cm R = 5 cm h = 6 cm V =? = 5 prh = = 44 m. R Q = 0 0 = 6 cm QE = 60 0 = cm R = 40 0 = 7 cm R = 60 0 = cm D = 00 0 = 0 cm ER = 60 0 = cm 0. Scale 0m = cm h r Volume = π [ + + ] h r R Rr = 6 [(8) + (5) + 5 8] 7 = [ ] 7 = 409 = 6994 cm D l. 04, 0, 6,..., 94 a = 04, d = 0 04 = 6, l = 94 l = a + (n )d 94 = 04 + (n )6 94 = n 6 94 = 6n n = n = 96 n = 96 6 \ n = 6 S n = n ( a + l ) 6 S 6 = ( ) 60 S 6 = 8 (498) S 6 = E 60 R 0 Q \ S 6 = 984 Given in G.P. S = 4 P P = 40 0 = cm P = 50 =.5 cm 0 and T 5 = 4T S = 4 ar ( ) = 4 r ar ( + )( r ) = 4 ( r ) a(r + ) = 4...() T 5 = 4T ar 5 = 4ar

19 8 OSWL Karnataka (SSL), Sample Question Papers, Mathematics, lass 0 ar 4 = 4ar ar ar 4 = 4 r = 4 r = () \ r = Volume of r put in eq. () a (r + ) = 4 a ( + ) = 4 a () = 4 \ G.P.: a, ar, ar,... a = ,,,... are in G.P.. x f fx d = x x d fd n = 0 Sfx = 580 Sfd =80.M. x = S.D. = =. 6 8, Σ fx 580 = = 9, n 0 Σfd n,80 0 = 59 =.609 D 4. (x) + () + (x) () = (x) + () (x) 4x + + 4x = 4x + 4x + x x 8x = 0 x (x 8) = 0 () + x \ x = 0 ½ and x = 8 = x + = 8 + ½ = 6 + \ = 7 unit \ = 7 unit. = [Given] Let a work done by nkitha = x days work done by Shweta = x 6 days work done by together = 4 days x + = x x 6 4 x 6+ x = xx ( 6) 4 x 6 = x 6x 4 = 4 x 6 4 (x 6) = x 6x ½ 8x 4 = x 6x x 4x + 4 = 0 x ( + )x + 4 = 0 x x x + 4 = 0 x(x ) (x ) = 0 (x ) (x ) = 0 \ x =, \ x = is not possible So, nkitha will take days to complete the work. X ½ x x + In D D, D = 90 x = D + D (x + ) = (x ) + x P Y

20 Solved Paper-05 9 Given : and are the centres of touching circles. P is the point of contact. To prove :, P and are collinear. onstruction : Draw the tangent XPY... Proof : In the figure, PX = 90...() and PX = 90...() y adding equations () and (), PX + PX = = 80 P = 80 \,P and are collinear. 5. Given : : D = : 4 = = To prove : D = D (E + D) = E D E D E. D = E D (4x) E. D = x D 6x (x ) = x [E.D = x 4x = x ] D 6x 4x = x D 0x = x D = x + 0x D = x \ D = [from eq ()] Hence proved x x x E x 4x D x onstruction : Draw E D Proof : : D = : 4 Let = x, D = 4x = = = x...() [Given, equiangular D] E = E = x = [E is a perpendicular bisector of ] In D ED, E = D ED...() [y Pythagoras theorem] In D E, E = E...() [y Pythagoras theorem] On comparing eq. (I) and eq (II) D ED = E Given : D D and D : D = : To prove : ( ) = D In DD, = D + D...(i) In DD, = D + D...(ii) Subtracting eq. (ii) from (i), we get = D D = = 6 = 8 6

21 Oswaal Karnataka (SSL) Sample Question Papers For lass 0 Mathematics For March 06 5% OFF Publisher : Oswaal ooks ISN : uthor : Panel Of Experts Type the URL : 8 Get this eook

KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE S. S. L. C. EXAMINATION, MARCH/APRIL, » D} V fl MODEL ANSWERS

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