ha~u& S,@ &Bernd AodQ, Ace&,do, 7,30r(%?nd2 - 560 oos KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAFJI,

BANGALORE - 560 003

azF.azF.aP9.b. &Beg,, @a,e - 2013 S. S. L. C. EXAMINATION, APRIL, 2013

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MODEL ANSWERS

am06 : 08. 04. 2013 I &oBed &oa3, : 8 1-E Date : 08. 04. 2013 I CODE NO. : 8 1-E

ado53 : ?t&d Subject : MATHEMATICS

[ dddndg wodr(Q : 100

[ Max. Marks : 100

( English Version )

[ Turn over

Marks Allotted

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1

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1

Qn. Nos.

I . 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Letter of the answer

A

C

A

D

B

D

C

C

A

C

Value Points

A n ( B U C ) = ( A n B ) U ( A f l C )

m

[: :I 13

5x2y 2

p 2 + q 2 + r 2

0

a 2 + b 2 - a b

5

2 s

Qn. Nos.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

n.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

Letter of the answer

B

A

C

B

B

D

C

D

A

B

Value Points

* 4.5

x 2 - 4 x + 1 = O

5

{ o , 1, 2, 3 1

A ABC 1

AY -

CY

2.G

8 cm.

Is = x 100 X

880 sq.cm

Marks Allotted

1

1

1

1

1

1

1

1

1

1

A ' n B'

m x p

1

I/x+y

a x 2 + b x + c = 0

Imaginary

t = d d 2 - ( R + r I 2

Right angle ( 90" )

V = n r 2 h

Regular pentagon.

1

1

1

1

1

1

1

1

1

1

[ Turn over

9n. Nos.

31.

32.

Value Points

2 , 2 f i , 4 .......

Here a = 2, T n = 64

2* r = -- = 2

T n = a . r n - I

64 = 2 . ( & I n - ' :. ( a ) n - l = 32 .

1 2 2 ( n - 1 I

= 32

n - 1 - 2 2 = 2 5

n - 1 . - - " 2 - 5

:. n - 1 = 5 x 2

:. n - 1 = 10

:. n = 1 0 + 1

:. -1

1 + 2 + 4 + ...... up to 9 terms.

2 4 Here a = l , r = - = - = 2 1 2

n = 9

:. S , = a

z 9 - 1 = l [ 2 - 1 I - - 256 - 1

1

= 255.

Marks Allotted

1 - 2

- 2 I

1 - 2

1 - 2

1 - 2

I - 2

1 - 2

1 - 2

2

2

Qn. Nos.

33.

34.

35.

Value Points

Let the three terms in harmonic progression are x, y, z.

2ab ff = --

a + b

Here a = x = 22

b = z

H = y = 2 0

2 x 2 z x z :. H = 22 + z

- 4 z 2 4 32 - z 3

4 :. 20 = - z 3

:. 42 = 20 x 3

= 60

6 0 :. z = - = 15 4

:. x = 2 z = 2 x 15 = 30

:. The three terms are 30. 20. 15.

The matrix obtained by interchanging rows into columns or

columns into rows is called transposing of a matrix.

Example A = !: :I . . . I ; [ ' 3 ]

2 4

Note : Any suitable example may be considered.

a) If one event occurs in rn different ways and another event

occurs independently in n different ways then the two events

together can be done in ( rn x n ) different ways. This is called

fundamental counting principle.

b) P r means the number of permutations of n things taking r

things at a time.

Allotted

I 5

1 - 2

1 - 2

1 - 2

1

1

1

1

Marks

2

2

2

Qn. Nos.

36.

37.

5

Value Points

There are 5 red flowers. The number of ways in which 2 flowers can be removed from 5 red flowers is C . Then remaining 2 flowers from 3 white flowers in C ways.

:. Total number of ways

= 5 ~ 2 X 3c2

- 5 x p 2 3 x 7 x -

- 7 x 1 y x 1

= 1 0 x 3

= 30

H = ( a - 7 )

L = ( a 3 - 10a2 + 11a+ 7 0 )

A = ( a 2 - 1 2 a + 3 5 )

B = ?

A x B = H x L

H x L ... B = ------ - L - H x - A A

L a 3 - 1 0 a 2 + 1 1 a + 7 0 - - A - a 2 - 12a+ 35

a + 2

a 2 - 1 2 a + 3 5 ) a - 1 0 a 2 + 1 1 a + 7 0

- 12a2 + 35a i -1 (+I (-1

ZrZ:% - (+I (-1

0 0 0

:. B = ( a - 7 ) ( a + 2 )

= a 2 - 5a - 14.

8 1-E

[ Turn

Marks Allotted

1 - 2

1 - 2

- 2 1

1 - 2

1 - 2

1

1 - 2

over

2

2

!4n. Nos.

38.

39.

40,

Value Points

6+v5 45 - <2

Rationalising factor of denominator is <5 + <2

&+a 6+* G - & X ~ + . T z

- - ( 6 + ~ 5 ) ~ ( ~ ) ~ - ( f i ) ~

- - ( 1 r s ) ~ + 2 z G x * + ( * ) ~ 5 - 2

- - 5 + 2 m + 2 3

- - 7 + 2 m 3

Let the cost price be Rs. x

Selling price = Rs. 18.75

Loss = Cost price - Selling price

X -- x x = x - 18.75 100

x2 - - 100

- x - 18.75

x 2 = 1 0 0 ~ - 1875

:. x - 1 00x + 1875 = 0

( x - 7 5 ) ( ~ - 2 5 ) = 0

:. x = 75 or x = 25

The cost price is Rs. 75 or Rs. 25.

2 x - 8 x + 1 = O

Compare the equation with standard form

a = l , b = -8 , c = l

Marks Allotted

1 Z

1 Z

- 1 2

- 1 2

1 - 2

1 Z

1 - 2

1 - 2

2

2

[ Turn over

Qn. Nos.

4 1 .

42.

Value Points

X = - b + . i G c

2 a

- - - ( - 8 ) ? 4 ( - ~ ) ~ - 4 x 1 x 1 2 x 1

- - 8 + 464-4 2

- - 8?= 2

- - 8 ? 4-5 2

- - 8 c 2- 2

- - 7 ( 4 + m )

7 = 4 ? E .

An equation that can be expressed in the form ax2 + c = 0, where

a and c are real numbers and a # 0 is a pure quadratic equation.

Or the quadratic equation having only second degree variable is

called a pure quadratic equation.

One example : x2 - 9 = 0

Note : Any suitable example may be considered.

k x 2 + 6 x + 1 = O

This is in the form ax2 + bx + c = 0

Here a = k, b = 6 , c = 1

~ = b ~ - 4 a c

Since the roots are equal,

b 2 - 4 a c = 0

~ 6 ) ~ - 4 x k x 1 = 0

36 - 4k = 0

4k = 36 36 ... k = - = g 4

:. /k=91

Allotted

- 1 2

I Z

I Z

1 - 2

1

1

1 - 2

1 - 2

1 - 2

1 Z

Marks

2

2

2

9n- Nos.

43.

44.

Value Points

P

AP and BP are tangents

For 3.5 cm circle

OP drawn

Midpoint C construction

AP and BP construction.

A

B D

C

ABCD is a rhombus.

AC and BD are diagonals. They bisect perpendicularly at 0.

1 1 :. A 0 = 3 AC and BO = B D

Allotted

1 - 2

1 - 2

1 - 2

1 Z

I - 2

Marks

2

[ Turn over

9". Nos.

45.

Value Points

A In right angled A AOB, AOB = 90"

According to Pythagoras theorem,

A B ~ = A O ~ + B O ~

= (;.c)' + ( ~ B D ) '

1 1 = - 4 A C ~ + ; T ~ ~ 2

1 :. A B ~ = 4 - ( A C ' + B D ' )

:. I A C 2 + B D 2 = 4 A B 2 1

T

Q

O P I F T

:. L OPT = 90"

Let L OPQ = xo

L QPT = 90° - xo

Similarly L PQT = 90' - xo

Allotted

i: I

z 1

1 - 2

1 - 2

I - 2

Marks

2

Qn. Nos.

46.

Value Points

In A PQT

LPTQ+LQPT+LPQT= 180'

L PTQ + 90' - x0 + 90' - xo = 180"

L PTQ + 180' - 2x0 = 180'

:. L PTQ = 2x0

:. L PTQ = 2 L OPQ.

Scale 20 m = 1 cm

i) - - - 8 cm 20

ii) 120 - - 6 cm

100 - 5 cm ) 20 -

60 iv) - - 20 - 3 cm

V) = 2 cm

80 vi) - 20 - - 4 cm.

D

C

E

A

Marks Allotted

1 - 2

1 - 2

1 Z

1 1,

2

2

[ Turn over

Qn. Nos.

47.

48.

49.

Value Points

A@ C

P

bD B C

Here F = 5

v = 5 E = 8

F + V = 5 + 5 = 1 0

E + 2 = 8 + 2 = 1 0

From (i) and (ii)

I F + v = E + 2 I

Number of candidates appeared for examination = n ( U ) = 100

Number of candidates passsed in Maths = n ( M ) = 82%

Number of candidates passed in Science = n ( S ) = 72%

Number of candidates passed in both = n ( M S ) = 55%

Marks Allotted

1 Z

l 2 1

1 - 2

1 - 2

1 - 2

- 2 1

2

2

!&- Nos.

50.

Marks Allotted

- 2 1

1 - 2

1 - 2

1 - 2

1

1

- I 2

1 - 2

- 2 1

1 - 2

3

3

Value Points

Total number of candidates passed = n ( M U S )

n ( M U S ) = n ( M ) + n ( S ) - n ( ~ ' n ~ )

= 82% + 72% - 55%

= 99%

:. Number of failed candidates

= n ( U ) - n ( M U S )

= 100-99

= 1%

N = 20 C fx = 240 1 fd2 = 550

- X = N

TO find X 240 - - To find - 20

= 12 C f o 2

Standard deviation = o = pg = -

= m 5

= 5.24

- x - x = D

- 10

- 5 0

5

10

f x

4

2 1

120

5 1

44

C.I.

0 - 4

5 - 9

10- 14

15 - 19

20 - 24

D 2

100

25

0

25

100 .

f 2

3 10

3

2

jD2

200

75

0

75

200

x

2

7

12

17

22

[ Turn over

CP". Nos.

51.

Marks Allotted

1

1 12

1 - 2 3

Value Points

A B

1 X

- 4

:. H.C.F. = x 2 + 3x + 2 To find HCF

A x B . L=-- - - . . H

B A = A x g or -xB. H

B x 3 - x 2 - lox- 8 - H = x 2 + 3 x + 2

x - 4

x 2 + 3x + 2 x - x 2 - l o x - 8 C + 3 x 2 + 2x (-1 (-1 (-1

Division 1 :z::g:t (+I (+I (+I

0 0 0

. - - x - 4 " H -

B :. L = A X - H

- - ( ~ ~ - 2 ~ ~ - 13x- 10) ( x - 4 )

OR

!a Nos.

52.

Value Points

A L = - x B H

A - - x3 - 2x2 - 13x- 10 H - x2 + 3x + 2

x - 5

x 2 + 3 x + 2

A . - .. * = ( x - 5 )

:. L = ( x - 5 ) ( x 3 - x 2 - l o x - 8 )

a ( b 2 c 2 - 1 ) b ( c 2 a 2 - 1 ) c ( a 2 b 2 - 1 ) L.H.S. = bc+ 1 + +

c a + 1 a b + 1

- - a ( b c - l ) ( M ) + b ( c a - l ) ( m ) &?+-I- a%+-l- +

c ( a b - l l ( W ) Jab++

= a ( b c - l ) + b ( c a - l ) + c ( a b - 1 )

= abc - a + abc - b + abc - c

= 3 a b c - a - b - c

= 3abc- abc . a + b + c = a b c ]

= 2abc.

- 1 2

- 2 1

1 - 2

1 - 2

- 1

1 - 2

Marks Allotted

3

3

[ Turn over

!3". Nos.

53 .

54 .

Value Points

fig.

Data : Two circles with centres A and B touch each other at P externally.

To Prove : A, B and P are collinear.

Proof: L APQ = 90' (i) ( Radius is perpendicular

L BPQ = 90" (ii) to tangent )

Adding (i) and (ii)

L A P Q + L BPQ= 180'

:. APB is a straight line

A, B and P are collinear. . . Volume o f sphere = volume of cone

4 . - 1 .. .r3 = - .r2 h 3

4 22- x - x r 3 = L x ~ x 1 2 x 1 2 x 6 5 7- 3 + 4 r 3 = 1 2 x 1 2 ~ 6

l / n 3 x 1 2 x 6 r 3 =

qc :. r 3 = 2 1 6

:. r =

p G T

Total surface are of a sphere = 4 n r 2

2 2 = 4 x 7 ~ 6 ~ 6

I T.S.A. of a sphere = 452.57 sq.cm /

Allotted

1

I - 2

1 - 2

1 - 2

1 -

2

1 - 2

1 1 P + Z

- 1 2

1 - 2

1 - 2

1 Z

Marks

3

3

gn. Nos.

55.

Value Points

In an Arithmetic Progression a = 2

a + ( a + d ) + ( a + 2 d ) + [ a + 3 d ) + ( a + 4 d ) =

1 - [ ( a + 5 d ) + ( a + 6 d ) + ( a + 7 d ) + ( a + 8 d ) + ( a + 9 d ) ] 4

1 :. 5a + 10d = 4 1 5a + 35d I

Substitute a = 2

1 5 x 2 + 10d=z[5x2+35dl

1 10 + 10d=4 [ 10 + 35dI

4 ( 10 + 10d ) = 10 + 35d

40+40d = 10+35d

:. 40d - 35d = 10 - 40

5d = - 30

- 30 ... d = - - - = - 6 5

:. pTT

20th term in A.P. = a + 19d

= 2 + 19(-6)

= 2+(-114)

= - 112.

:.

Note : Any other correct method may be considered and for

correct answer full marks may be given.

Marks Allotted

1

- 2 I

1 - 2

1 - 2

1 - 2

1 - 2

I - 2 4

[ Turn over

Qn. Nos. 56.

Value Points

c 2

c1

Here d = l O c m , R = 4 c m , r = 2 c m and R - r = 4 - 2 = 2 c m .

For C , , C , , C circles

Distance between two centres of circles - AB

To find mid-point C

Construction of tangents BD and BE

Construction of tangents FH and GI

Length of tangents

FH = GI = 9.7 cm or 9.8 cm.

Marks Allotted

1 I T

1 - 2

1 - 2

1 - 2

1 - 2

1 - 2 4

9n. Nos.

57.

Value Points

A

B A%, c Eb F

Data : In triangles A ABC and A DEF

L BAC = L EDF

LABC=LDEF

and L ACB = L DFE

AB BC AC To prove : - - - DE - EF =

Construction : Mark X on AB. Yon AC, such that

AX = DE and AY = DF. Join XY.

Proof:

Allotted

- 2 I

1 - 2

1 - 2

1 - 2

1 - 2

1 - 2

1 - 2

1 - 2

Statement

In A AXY and A DEF, L A = L D ,

AX = DE, AY = DF

:. A AXY= A DEF

:. X Y = EF and L AXY = L DEF

LAXY=LDEF=LABC

i.e., L AXY = L ABC

:. XY I I BC . . . (i)

AB AC BC . " AX-AY-XY

AB AC BC i.e., DE - DF - EF

Marks

4

Reasons

Data

Construction

S.A.S. postulate

Congruent triangle property

Data

L A X Y and L ABC are

. corresponding angles

B.P.T. and corollary

From construction and (i)

9'. Nos.

58.

Value Points

y = x 2

Marks Allotted

1

1

1 z 1 5 1

1 -

X

Y

4

y=3-2x

Scale : x-axis : 1 cm = 1 unit

y-axis : 1 cm = 1 unit

Roots 1 and-3

To draw parabola

To draw straight line

To draw perpendicular

To write the roots

0

0 1

2

4

1 - 1

1

- 2

4

3

9

- 3

9