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81-E_Ans

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KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE – 560 003

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S. S. L. C. EXAMINATION, MARCH, 2007

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MODEL ANSWERS å»Ê¢∑ :
22. 03. 2007 Date : 22. 03. 2007

‚¢xË∆ ‚¢zW : 81-E CODE NO. : 81-E ê·ÂŒÈ : πÂä∆Â
Subject : MATHEMATICS
[ …¬ÂÆÂ⁄ÕÂå˘ •¢∑π›ÂÈ : 100
[ Max. Marks : 100 ( English Version )

Qn. Letter of Nos. the answer

Value Points

Marks Allotted

Part – A 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. D B C D C A A C B D C A 2 1 1 1 1 1 1 1 1 1 1 1 1 [ Turn over

{ 0, 1, 4 }
4 7 55 2 a ( 1 – r ) 8

 3     2        1     4 
1 AB 22

81-E
Qn. Letter of Nos. the answer

2
Value Points Marks Allotted

13. 14. 15. 16. 17. 18. 19. 20.

B C D A C B D D

3P 3 × 2P 2 1 0·02 2·56 3b P –2 15 (  x 2 – 9 )

1 1 1 1 1 1 1

∑  (  x
xyz

 2 – x 

)

1

21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35.

A B A C B D D B A A C C C D A

p  2 + q 2 + r 2 33 ab
3

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

40
b –c

a

3 5 5 ± 3 x 2 + 2x – 5 = 0 x 2 – 6x + 4 = 0 ∆ > 0 5 2 9 + 5 3 17 ≡ 5 ( mod 12 )

3
Qn. Letter of Nos. the answer Value Points

81-E
Marks Allotted

36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60.

B D B A B A B A D C A D B D B C C A D A C C C C D

55 5 cm PQ OB OA 7·5 m 9, 12, 18 and 3, 4, 6 1 : 2 Proportional AX AC 5m Acute angles 3·5 cm 3 cm 6 cm 5 cm 20˚ 2πrh 396 sq.cm volume 1540 c.c. 3 πr 2 N +R =A +2 square 5 10. = AY AB = XY CB

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 [ Turn over

81-E
Qn. Nos.

4
Value Points Marks Allotted

Part – B 61. To find A I ( B I C ) : — B I C = { 5, 8

} }
I { 5, 8 } …… (i)

1 2

A I ( B I C ) = { 3, 4, 5, 9 A I (B I C)= {5} To find ( A I B ) I C :— A I B = { 4, 5

1 2

}

1 2

( A I B ) I C = { 4, 5 } I { 5, 7, 8, 9 } (A I B) I C ={5} …… (ii)
1 2

2

62.

 1     2   A =      0     3   1     2   1     2      A 2 =      0     3    0     3 
A 2 = 

 1 + 0  0 + 0

2 + 6 

 0 + 9 

1 2

 1     8   A 2 =      0     9   1     2   2     4   2A = 2       =      0     3   0     6 
∴

1 2

1 2

 1     8   2     4   A 2 – 2A =       –      0     9   0     6   – 1     4   A 2 – 2A =        0     3   .
1 2

2

5
Qn. Nos. Value Points

81-E
Marks Allotted

63.

There are 7 players. Ashaya is one among them. Among 5 players Ashaya is there. ∴ Form a team of 4 players out of 6 players. This can be done in
6 6

C  4 ways

1

C  4 = C  2

6

2 ! 6 ×  5 30 = = = 15. 2 ×  1 2
7

6

C  2 =

6  P

 2

1 C  5 –
6

2

( or any other method i.e. ∴ 64. Classinterval 1 – 5 6 – 10 11 – 15 16 – 20 Frequency f 2 3 4 1 N = 10 Average upto 'd' To find fd 2 ∴ 65. Variance ∑ fd 2 N = ∑ fx 100 = = 10 N 10 15 teams can be formed.

C  5 ) = 15

Mid-point x 3 8 13 18

fx 6 24– 2 52+ 3

d – 7 4 9

d 2 49 12 36 64

fd 2 98

18+ 8 64 ∑ fx = 100

∑ fd 2 = 210
1 2 1 2 1 2

210 = 21. 10

1 2

2

2m 2 + 2m + m  3 + 1 and 2m + 1 + m  2 Ascending order, m 3 + 2m  2 + 2m + 1 and m  2 + 2m + 1 m m 2 + 2m + 1 m 2 + 1m m + 1 m 3 + 2m  2 + 2m + 1 m 3 + 2m  2 + 1m m + 1 m + 1 0 ∴ HCF = ( m + 1 )
1 2 1 2

m
1 2

1
1 2

2

[ Turn over

81-E
Qn. Nos.

6
Value Points Marks Allotted

66.

a +b +c =0 L.H.S. : = = a 2 b  2 c 2 + + bc ca ab a 3 + b  3 + c 3 abc ...   3abc abc a +b +c =0 a 3 + b  3 + c 3 = 3abc. =
1 2 1 2

L.C.M. = abc. 1

= 3 = R.H.S. 67. 3 ×
3

2

6
1 2

L.C.M. of order of 2 and 3 is 6. 3 = 32 = 36 =
3
  1   3

6
2  6

33
=

=

6

27
=

1 2

6
3

= 6

1  3

=6 =

6

62
×

6

36
=

1 2

∴ 68. B = 4B = a 2 =

3 ×

6

6

27

6

36

6

972

1 2

2

3 a 2 4 3 a  2 4B 3 4B 3
1 2 1 2

a

=±

If B = 16 3 , then a =± a =± 4 ×  16 3 3 64
1 2 1 2

a = ± 8.

2

7
Qn. Nos. Value Points

81-E
Marks Allotted

69.

x 2 – 7x + 12 = 0 a = 1, b = – 7, c = 12 x = – b ±  b  2 – 4ac 2a ( formula )
1 2

x =

– ( – 7 ) ±   ( – 7 ) 2 – 4 ( 1 ) ( 12 )  2 ( 1 ) + 7 ±  49 – 48 2 7 ± 1 2 x= 7 + 1 7 – 1 or x = 2 2

1 2

x = x =

1 2

∴ ∴ 70.

x = 4 or x = 3.

1 2 1 2

2

Let altitude of the triangle be x cms, then base is ( x + 4 ) cms. Area of triangle = 1 × base × height 2 1 (x+4) x 2

48 96

=

1 2

= x  2 + 4x
1 2

x 2 + 4x – 96 = 0 x 2 + 12x – 8x – 96 = 0 x ( x + 12 ) – 8 ( x + 12 ) = 0 ( x – 8 ) ( x + 12 ) = 0 x – 8 = 0 or x + 12 = 0 x = 8 or x = – 12. ∴ Height of the triangle is 8 cm.

1 2

2 [ Turn over

81-E
Qn. Nos.

8
Value Points Marks Allotted

71.
B

O 5 cm

A

C1

C2
1 2 1 2 1 2 1 2

Circle C 1 Pointing OA from a distance of 5 cm Circle C 2 Tangent AB. 72.

2

D 150

Y

C

Rough sketch

1

X

100

B

1 × 150 × 100 = 7500 sq. m. 2 1 Area of ∆DYC = × 150 × 100 = 7500 sq. m. 2 1 Area of trapezium XBCY = × 50 ( 100 + 150 ) 2 1 = × 50 ( 250 ) 2 = 6250 sq. m. Area of ∆AXB = ∴ Area of ABCD = 7500 7500 6250 21,250 sq. m.

A

1 2

1 2

2

9
Qn. Nos. Value Points

81-E
Marks Allotted

73.

There are 25 terms

∴

Middle term is 13

∴

T  13 = 20

1 2

T  n = a + ( n – 1 ) d

( formula )

1 2

20 = a + ( 13 – 1 ) d

∴

a + 12d = 20

............. ( i )

1 2

Sum S n =

n [ 2a + ( n – 1 ) d ] 2

( formula )

1 2

S n =

25 [ 2a + ( 25 – 1 ) d ] 2

1 2

S n =

25 [ 2a + 24d ] 2

1 2

S n =

25 2

× 2 ( a + 12d )

1 2

S n =

25 2

× 2 × 20

from eqn. ( i )

S n = 500.

1 2

4

[ Turn over

81-E
Qn. Nos.

10
Value Points Marks Allotted

74.

A
90˚

B

D

C

Data : In ∆ ABC, ∠ BAC = 90˚ To prove : BC 2 = AB 2 + AC 2 Construction : Draw AD ⊥ BC Proof : In ∆ ABC and ∆ DBA ∠ BAC = ∠ BDA = 90˚ ∠ ABC = ∠ ABD ∆ ABC  ∆ DBA ∴ AB BC = DB BA Proportion of the sides related to similar triangles, ∴ BC . BD = AB 2 .............. ( i ) .. . Data & construction .. . Common triangles are equiangular

1 2 1 2 1 2

1 2

1 2

∆ ABC and ∆ DAC ∠ BAC = ∠ ADC = 90˚ ∠ ACB = ∠ ACD ∴ ∴ ∆ ABC  ∆ DAC BC . DC = AC 2 .............. ( ii )
1 2 1 2

By adding (i) and (ii) BC . DB + BC . DC = AB 2 + AC 2 BC ( BD + DC ) = AB 2 + AC 2 BC . BC = AB 2 + AC 2 ∴ BC 2 = AB 2 + AC 2 .. . BD + DC = BC
1 2

Hence in a right-angled triangle, square on the hypotenuse is equal to the sum of the squares on the other two sides.

4

11
Qn. Nos. Value Points

81-E
Marks Allotted

75.

C 1

circle

1 2

C 2

,,

1 2

C 3

( R – r ) 3·5 – 2 = 1·5 cm

1 2

C 4

1 2

Tangent to C 3

1 2

To draw AX and AP

1 2

BC ⊥ BY and BD ⊥ BQ

1 2

∴

Tangents are XY and PQ.

1 2

4

[ Turn over

81-E
Qn. Nos.

12
Value Points Marks Allotted

76.

x 2 + x – 2 = 0 x 2 = – x + 2 y = x  2 x y y =2–x x y 0 2 1 1 2 0 1 0 0 1 1 – 1 1 2 4 – 2 4 1
1 2

( Roots of x 2 + x – 2 = 0 are – 2 and 1 )

1 2

4


				
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