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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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
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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
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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
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Marks Allotted
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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
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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
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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
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1 - 2
1 Z
1 1,
2
2
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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
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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
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3
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!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
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I - 2
1 - 2
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1 -
2
1 - 2
1 1 P + Z
- 1 2
1 - 2
1 - 2
1 Z
Marks
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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
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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)
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