O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE

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1 CCE PF CCE PR O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE G È.G È.G È.. Æ fioê,» ^È% / HØ È 0 S. S. L. C. EXAMINATION, MARCH/APRIL, 0» D} V fl MODEL ANSWERS MO : ] MOÊfi} MSÊ : 8-E Date : ] CODE NO. : 8-E Œ æ fl : V {} Subject : MATHEMATICS ( Ê Æ p O» fl / New Syllabus ) ( S % Æ» ~%} S % / Private Fresh Private Repeater ) (BMW«ŒÈ Œ M} / English Version) [ V fl : 00 [ Max. : 00 Ans. Key I.. C. A 5. C D 5. B 5 units. D 7. B 5 8. A. [ Turn over

2 8-E CCE PFPR II. 9. A l U A A l {,,, 4, 5 } {, 4, 5 } A l {, } 0. ( a, b ) L.C.M. L.C.M. a b ( a, b ) H.C.F. 8 L.C.M.. f ( x ) x x f ( ) ( ) ( ) 8. d 0 cm d 4 cm R cm r cm Distance between centres d R r d 5 7 cm. In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. 4. Total surface area of cylinder πr ( r h ) sq.units. III. 5. n 8 Number of diagonals n C n 8 C 8 8 P Number of diagonals 0 OR

3 CCE PFPR 8-E n 8 Number of diagonals n C n n ( n ) 8 ( 8 ) Number of diagonals 0.. If possible, let us assume 5 is a rational number. 5 q p where p, q z, q 0, q p 5 q q p 5. 5 is a rational number Q q q p is a rational number But 5 is not a rational number. Our supposition 5 is a rational number is wrong. 5 is an irrational number. 7. Total number of watches n ( S ) 500 Number of watches defective n ( A ) 50 P ( A ) n ( A ) n ( S ) 50 P ( A ) Probability of watches to be defective 0 or [ Turn over

4 8-E 4 CCE PFPR 8. L.C.M. of and is / / OR ) ( ) ( ) ( 5.

5 CCE PFPR 5 8-E 0. x x x 8 x 4x 5x x x ( ) ( ) x 5x x x ( ) ( ) 8x 8x 8 () () 4 q ( x ) x x 8 R ( x ) 4. OR Alternate method : R ( x ) 4 q ( x ) x x 8 Note : Finding remainder by using remainder theorem method give mark only OR [ Turn over

6 8-E CCE PFPR x 4 x x x x x x x 4 x x ( ) ( ) () x x x x ( ) ( ) () x r ( x ) x { r ( x ) } x Hence we should add ( x ) to P ( x ) so that the resulting polynomial is exactly divisible by g ( x ). In Δ ABC, DE AB CD CE CA CB ( Corollary to B.P.T. ) 5 CE 8 CE 5 8 CE CE 7 5 cm. tan θ tan θ We know that tan 0 θ 0 sin θ sin ( 0 ) sin θ sin 90.

7 CCE PFPR 7 8-E. ( x, y ) (, ) and ( x, y ) ( 4, 7 ) Mid-point x x y y, 4 7, 0, (, 5 ). 4. r 7 cm l 0 cm C.S.A. of cone πrl sq.cm. OR Volume of cylinder πr h c.c. 5. x 4x 0 a, b 4, c x b ± b 4ac a x ( 4 ) ± ( 4 ) 4 ( ) ( ) ( ) x 4 ± 8 x 4 ± 8 x 4 ± ( ± ) x ±. [ Turn over

8 8-E 8 CCE PFPR. 7. Scale : 0 m cm 40 m cm 80 m 4 cm 0 m cm 0 m 8 cm 0 m cm Construction of circle Radius OP Arcs Tangent at P Calculation Field drawing

9 CCE PFPR 9 8-E 8. n ( M ), n ( D ) 5, n ( M I D ) 7 n ( M U D )? n ( M ) n ( D ) n ( M U D ) n ( M I D ) OR n (M U D ) n ( M ) n ( D ) n (M I D ) n (M U D ) Number of people in the group 0 9. Volume of hemisphere Volume of cylinder π r π r h h h h cm a 4, d 7 4 S? 0 n S n [ a ( n ) d ] S 0 0 [ 4 ( 0 ) ] 0 [ 8 57 ] 0 5 S 50 0 Sum of the first 0 terms of the series 50. A U ( B I C ) {,,, 4 } U { 5 } {,,, 4, 5 } [ Turn over

10 8-E 0 CCE PFPR., 4,... H.P., 4,... AP. a, d 4 T? 0 T a ( n ) d n T ( 0 ) 0 8 T 0 AP 0 T 0 HP 0 OR a, d, T? 0 T n a ( n ) d T 0 ( 0 ) up to 9 a,. 0 r a S r S? S S S / S.

11 CCE PFPR 8-E ( 5 ) S { ( HH ) ( TT ) ( HT ) ( TH ) } n ( S ) 4 Exactly one tail ( A ) ( HT ) ( TH ) n ( A ) n ( A ) P ( A ) n ( S ) P ( A ) 4 or. Total number of students Football 0 o 0 Tennis 0 0 o Hockey 8 0 o 80 Football [ Turn over

12 8-E CCE PFPR 7. Let m 5, n 7 m n 5 7 mn k x x ( m n ) mn 0 x x ( ) 5 0 x x 5 0 mv mv k v v k m k ± m v ± 00 ± 00 v ± In Δ ABC, B 90, BD AC BD AD CD 8 4 CD 4CD 4 CD 4 cm r 7 cm T.S.A. of sphere 4 πr sq.cm.

13 CCE PFPR 8-E IV. 4. Data : A and B are the centres of touching circles. P is point of contact. To prove : A, P and B are collinear Construction : Draw the common tangent XPY Proof : APX 90 (i) ( AP XY ) BPX 90 (ii) ( BP XY ) Add (i) and (ii) APX BPX 80 APB is a straight line APB A, P, B are collinear. C.I. f x fx x x D D f D N 0 fx 80 f D 50 [ Turn over

14 8-E 4 CCE PFPR Mean fx 80 x 8 N 0 f D 50 σ 5 5 N 0 For any other correct method full marks is to be given. 4. Given digits :,,, 4, 5, a) 4-digit number can be formed in P 4 ways P P 4 0. b) Less than 5000 : Th H Ten U 4 P ways 5 P ways 4 P ways P ways Unit s place can be filled in P ways 4 Ten s place can be filled in P ways 5 Hundred s place can be filled in P ways 4 Thousand s place can be filled in P ways Total number of ways P 4 P 5 P 4 P The number less than n P n P OR. n ( n ) ( n ) ( n ) n ( n ) ( n ) ( n ) n n n n n 45 n 45 n 5.

15 CCE PFPR 5 8-E 44. LHS sin ( 90 o sin θ θ ) cos θ cos ( 90 o θ ) cos θ sin θ cos θ sin θ cos θ ( sin θ ) cos θ sin θ ( θ sin θ ) cos θ cos θ. sin θ cos θ cos θ. sin θ cos θ cos cos θ θ sin sec θ LHS RHS OR LHS cos ( A B ) cos ( 0 0 ) cos 90 0 RHS cos A. cos B sin A. sin B cos 0. cos 0 sin 0. sin LHS RHS. [ Turn over

16 8-E CCE PFPR 45. Let the number of pupils x Total cost Rs. 000 Each share 000 Rs. x If 0 of them are failed to join the function then number of pupils joined x 0 Total cost Rs. 000 Each share 000 Rs. x 0 Each would have to pay Rs. 5 more x 0 x 000x 000 ( x 0 ) x ( x 0 ) 5 000x 000x x 50x 5 x 50x by 5 x 0x x 50x 40x x ( x 50 ) 40 ( x 50 ) 0 ( x 50 ) ( x 40 ) 0 x 50, 40 x 50 Number of pupils of the class 50. OR x 5x 0 a, b 5, c b ( 5 ) m n 5 a c m n a

17 CCE PFPR 7 8-E i) ( m n ) ( m n ) ( m n ) [ ( m n ) 4 mn ] ( ) ii) ( m n ) 4 mn ( 5 ) 4 ( ) 5 7. AM CN AB BM BN BC ( AB BC AC ) AM CN AC BM BN BC AB AC AM CN BC AB AC 4 4 4AC BC AB 4 4 (AM CN ) 4AC BC AB ( BC AB AC ) 4 (AM CN ) 4AC AC 4 (AM CN ) 5 AC. In Δ AOB OR O 90 o AB OA OB AB AC BD AC BD 4 4 AC BD 4 4AB AC BD. [ Turn over

18 8-E 8 CCE PFPR V. 47. Data : In Δ ABC and Δ DEF BAC ABC EDF DEF To prove : AB DE BC CA EF FD Construction : Mark points G and H on AB and AC such that AG DE and AH DF. Join G and H. Proof : In Δ AGH and Δ DEF AG DE Q Construction GAH EDF Q Data AH DF Construction Δ AGH Δ DEF Q SAS AGH DEF But, ABC AGH DEF ABC GH BC. In Δ ABC AB BC CA AG GH HA Hence AB BC CA. DE EF FD 4

19 CCE PFPR 9 8-E 48. d 8 cm, R 4 cm r cm R r 4 cm Drawing AB and marking mid-point M Drawing circles C, C, C Joining BK, BL, PQ, RS Measuring and writing the length of tangents PQ RS ( PQ RS 7 8 cm) a, a d, a d... A.P. a a d a 4d 9 a d 9 ( a d ) 9 a d a d a d a 5d 5 a 9d 5 ( a d ) (i) [ Turn over

20 8-E 0 CCE PFPR a d (ii) From eq. (i) and (ii) a d a d 7 ( ) ( ) ( ) d 4 d 4 From eq. (i) a d a ( 4 ) a 8 a 5 Now a 5, d 4 T n a ( n ) d T 0 5 ( 0 ) 4 T 0 5 T a, ar, a ar a ( r ar... G.P. OR ar 7 r r ) 7 r 7 a (i) ar 4 5 ar ar 5 ar ( r r ) 5 ar 7 5 a r r r 5 7 8

21 CCE PFPR 8-E r r 7 a 4 a 7 a 7 7 T a, T, T 4,, 4, 8... G.P. Alternate method : S 7 r a 7 r r 7 r a... (i) r S S a 7 5 r a ( r ) ( r ) ( r ) 7 a ( r ) a r 9 7 r 9 8 r 5 7 r 7 a r r 7 a 7 a 7 7 T a, T, T 4 [ Turn over

22 8-E CCE PFPR,, 4... G.P x x 0 x x y x ; y x y x x 0 y y x x 0 y 0 4 Table ( ) Parabola Straight line Drawing perpendiculars and identifying roots 4

23 CCE PFPR 8-E Alternate Method : x x 0 y x x x 0 y Table Drawing parabola Identifying roots 4 [ Turn over

24 8-E 4 CCE PFPR

O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE

CCE RF CCE RR O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl 50 00 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALE 50 00 G È.G È.G È.. Æ fioê,» ^È% / HØ È 0 S. S. L. C. EXAMINATION,