CCE RR KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE G È.G È.G È.. Æ fioê, d È 2017

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1 CCE RR O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl KRNTK SECONDRY EDUCTION EXMINTION BORD, MLLESWRM, BNGLORE G È.G È.G È.. Æ fioê, d È 07 S. S. L. C. EXMINTION, JUNE, 07» D} V fl MODEL NSWERS MO : ] MOÊfi} MSÊ : 8-E Date : ] CODE NO. : 8-E Œ æ fl : V {} Subject : MTHEMTICS ( Ê Æ p O» fl / New Syllabus ) ( Æ» ~%} % / Regular Repeater ) (BMW«ŒÈ Œ M} / English Version ) [ V Œ V fl : 80 [ Max. : 80 ns. Key I.. B { 6, 7, 8 }. C D x y 5. B 8 6. C an acute angle 7. D cm 8. units RR-XXII-800 [ Turn over

2 8-E CCE RR II P 0 0. Probability of a certain event is. Mid-point of the class-interval Method : Method : cos 48 sin 4 cos 48 sin 4 sin 4 sin 4 cos 48 cos y x comparing with y mx + c slope m y-intercept c 0 4. Total surface area of a solid hemi-sphere π r sq.units III. 5. Solution : n ( ) 7, n ( B ) 6, n ( U B ) 5 n ( I B )? n ( U B ) n ( ) + n ( B ) n ( I B ) n ( I B ) n ( I B ) 6 5 n ( I B ) 6. a) rithmetic mean.m. a + b b) Harmonic mean H.M. ab a + b RR-XXII-800

3 CCE RR 8-E 7. Solution : Here a, r / / S? a S r / S 8. Let us assume, + 5 is a rational number + 5 q p where p, q z, q 0 + q p 5 q + p q 5 q + p 5 is a rational number Q q is rational but 5 is not a rational number this gives us contradiction our assumption + 5 is a rational number is wrong + 5 is an irrational number 9. triangle is formed by joining non-collinear points. Total number of triangles that can be drawn out of 8 non-collinear 8 points C Here n 8, r n! n C r ( n r )! r! 8! 8 C (8 )!! ! 5! 56 RR-XXII-800 [ Turn over

4 8-E 4 CCE RR lternate method : Number of triangles n C n( n )( n ) 6 If n / 8 C 6/ Solution : x + 8! 9! 0! + 8! 9 8! 8! 0 9/. Solution : + 9 x 0 9/ x 00 x 0 9 8! x 0 9 8! There are 7 marbles, out of these 4 marbles can be drawn in 7 C 4 5 ways n ( S ) 5 Two marbles out of 4 red marbles can be drawn in 4 C 6 ways The remaining marbles must be black and they can be drawn in C ways n ( ) 4 C C 6 8 P ( ) n ( ) n ( S ) 8 5 RR-XXII-800

5 CCE RR 5 8-E. Direct method : x x table 5 5 Standard deviation 6 6 σ Σx N Σx N Σ x 5 Σ x 55 σ.4 N 5 ctual mean method : x d x x d Σ x Mean x N Σ x 5 Σ d 0 standard deviation σ Σ d N 0 5 σ.4 RR-XXII-800 [ Turn over

6 8-E 6 CCE RR ssumed mean method : ssumed mean 6 ( any score can be taken ) x d x d N 5 Σ d 5 Σ d 5 Standard deviation σ Σd Σd N N σ.4 Step deviation method : ssumed mean 7, Common factor of the scores C x x d d C N 5 Σ d 0 Σ d 0 Standard deviation σ Σd N Σd N C σ.4 RR-XXII-800

7 CCE RR 7 8-E. The equation is in the form of ax + bx + c 0 where a, b, c 4 x b ± b a 4ac ( ) ± ( ) 4 4 ± ± 5 / ( ± 5 ) / ( + 5 ) and ( 5 ) are the roots of the given quadratic equation OR This is in the form of ax + bx + c 0 where a, b, c Δ b 4ac ( ) 4 ( ) Δ > 0 roots are real and distinct RR-XXII-800 [ Turn over

8 8-E 8 CCE RR 4. radius r 5 cm angle between the radii 80 circle angle between the radii tangents at and B RR-XXII-800

9 CCE RR 9 8-E 5. In Δ BC and Δ DC B C D C C B C D given common angle Δ CB ~ Δ DC equiangular triangles C CB criteria DC C C BC DC OR In BC, B C 90 and BD C B D C () corollary similarly BC CD C () corollary dividing () by () B BC D CD C C B BC D CD 6. sin 0 cos 60, tan 45 sin 0. cos 60 tan 45 () RR-XXII-800 [ Turn over

10 8-E 0 CCE RR 7. Solution : ( x, y ) ( 5, 4 ) ( x, y ) ( 7, ) d ( x x ) + ( y y ) radius of the circle [ 7 ( 5)] + ( 4) ( ) + ( ) ( ) + ( ) r 8. ratio between the radii of two cylinders r : r : ratio between their curved surface areas S : S 5 : 6 S S π r π r h h 5 h 6 h h h 5 / 6/ 4 5 ratio between their heights 5 : 4 RR-XXII-800

11 CCE RR 8-E 9. Sphere - radius r 0 cm Cone - height h 0 cm - radius r 5 cm volume Number of small cones formed volume of of the sphere each small cone 4 π r / / π r h / / / 5/ 0 6 Number of small cones formed 6 0. Scale : 5 m cm 50 m cm 75 m cm 00 m 4 cm 5 m 5 cm 00 m 8 cm. Calculation Plan drawing RR-XXII-800 [ Turn over

12 8-E CCE RR IV.. Rationalising factor of 6 is ( 6 + ) ( 6 + ) ( 6 ) ( / + / ) +. x + x 8 x + x + 4x 5x + 6 x + x ( ) ( ) x 5x + 6 x + x ( ) ( ) 8x + 6 8x 8 (+) (+) 4 Quotient q (x) x + x 8 remainder r (x) 4 RR-XXII-800

13 CCE RR 8-E Verification : g ( x ) q ( x ) + r ( x ) ( x + ) ( x + x 8 ) + 4 x + x 8x + x + x x + 4 x 5x + 6 p ( x ) p ( x ) [ g ( x ) q ( x ) ] + r ( x ) OR Synthetic division : The quotient is 4x 4x + 9 remainder r ( x ) 4. Let the three conscutive + ve integers be x, ( x + ) and ( x + ) from the statement, x + ( x + ) ( x + ) 9 x + x + x + x + 9 x + x + 9 x + x x + x x x + 5x 90 0 x ( x 6 ) + 5 ( x 6 ) 0 5 ( x 6 ) ( x + 5 ) 0 5 x 6, or x The three consecutive + ve integers are 6, 7, 8 OR RR-XXII-800 [ Turn over

14 8-E 4 CCE RR Let the numbers be x, y and x > y sum of their squares is 80 i.e. x + y 80 () Square of the smaller number is equal to 8 times the bigger number y 8x () Substituting () in () we get x + 8x 80 x + 8x x + 8x 0x 80 0 x ( x + 8 ) 0 ( x + 8 ) ( x 0 ) ( x + 8 ) 0 x 0 or x 8 If x 0 then y 8x y 8 0 y The numbers are 0 and Data : and B are the centres of touching circles. P is the point of contact RR-XXII-800

15 B N + BN + CCE RR 5 8-E To prove :, P and B are collinear Construction : Tangent XY is drawn at P Proof : In the figure P X 90 () radius drawn at the point of contact B P X 90 () is perpendicular to the tangent P X + B P X by adding () and () P B 80 P B is a straight angle PB is a straight line, P and B are collinear 5. In Δ BC, B BC C N BC BN NC BC B In BN, N B 90 N B BN B B B B 4 N 4B B 4 4N B OR RR-XXII-800 [ Turn over

16 8-E 6 CCE RR In BD, D B 90 B D + BD D B BD () In DC, D C 90 C D + DC D C DC () from () and () B BD C DC B + DC C + BD 6. LHS tan sin sin sin Q tan cos sin cos sin sin cos cos sin ( cos cos ) but cos sin sin. cos sin cos sin. sin tan. sin. LHS RHS RR-XXII-800

17 CCE RR 7 8-E lternate method : LHS tan sin ( sec ) sin Q tan sec ( cos ) Q cos sec cos sin cos cos cos + cos 4 cos cos + cos 4 cos ( cos ) cos Q cos + cos 4 ( cos ). ( sin ) cos Q cos sin sin cos. sin tan. sin. LHS RHS. OR RR-XXII-800 [ Turn over

18 8-E 8 CCE RR Let B represents height of the building B 50 m BC be the distance between the building and the object ngle of depression is 0 Since M BC, So M C C B 0 In BC, B C 90, C B 0 tan 0 B BC 50 BC BC distance between the building and the object 50 m V. 7. In an P T + T 5 0 a + d + a + 4d 0 a + 6d 0 a + d 5 () and T 4 + T 8 46 a + d + a + 7d 46 a + 0d 46 a + 5d () RR-XXII-800

19 CCE RR 9 8-E Subtracting () from () a + 5d a + d 5 ( ) ( ) d 8 d 4 If d 4 then a + d 5 a a + 5 a 5 If a and d 4 then the P is, 7,, 5,... OR In a GP T 4 8 ar 8 () and T ar 7 8 () dividing () by () we get ar 7 / 8 6 ar 8/ / r 4 6 r If r then ar 8 a () 8 8a 8 a If a and r then a r n S n r S S RR-XXII-800 [ Turn over

20 8-E 0 CCE RR 8. x x 0 y x x x 0 4 y table Drawing parabola 4 identifying roots Scale : x-axis cm unit y-axix cm unit. x, RR-XXII-800

21 CCE RR 8-E lternate method : x x 0 y y x y + x + x x 0 y y x + x 0 y Table Drawing parabola Identifying roots 4 RR-XXII-800 [ Turn over

22 8-E CCE RR RR-XXII-800

23 CCE RR 8-E 9. d 8 cm R 4 cm r cm R r 4 cm Length of the tangent KL MN 7 8 cm Drawing B and marking mid-point Drawing circles C, C, C 4 Joining BP, BQ, MN, KL Measuring and writing the length of tangents RR-XXII-800 [ Turn over

24 8-E 4 CCE RR 40. Data : In Δ BC and Δ DEF B C E D F, B C D E F To prove : B DE BC EF C DF Construction : Points G and H are marked on B and C such that G DE and H DF. G and H joined. Proof : In Δ GH and Δ DEF G DE G H E D F construction data H DF construction Δ GH Δ DEF SS postulate 4 GH EF CPCT G H D E F but D E F B C data G H B C alternate angles GH BC B G BC C GH H cor. BPT but G DE, GH EF, H DF B DE BC C EF DF RR-XXII-800

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