CCE PR UNREVISED
PR (D) - 710 [ Turn over
O⁄´¤%lO⁄ ÆË√v⁄ ÃO⁄–y Æ⁄¬fiO¤– »⁄flMs⁄ÿ, »⁄fl≈Ê«fiÀ⁄ ¡⁄M, ∑ÊMV⁄◊⁄‡¡⁄fl — 560 003
KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE – 560 003
G—È.G—È.G≈È.“. Æ⁄¬fiOÊ⁄–, »⁄·¤^È% / HØ√≈È — 2019 S. S. L. C. EXAMINATION, MARCH/APRIL, 2019
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MODEL ANSWERS
¶´¤MO⁄ : 25. 03. 2019 ] —⁄MOÊfi}⁄ —⁄MSÊ¿ : 81-U
Date : 25. 03. 2019 ] CODE NO. : 81-U
…Œ⁄æ⁄fl : V⁄{}⁄ Subject : MATHEMATICS
( ‘⁄◊Ê⁄ Æ⁄p⁄¿O⁄√»⁄fl / Old Syllabus ) ( Æ⁄‚¥´⁄¡¤»⁄~%}⁄ S¤—⁄W @∫⁄¥¿£% / Private Repeater )
(BMW«ŒÈ ∫¤Œ¤M}⁄¡⁄ / Urdu Version )
[ V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 100
[ Max. Marks : 100
Qn. Nos.
Ans. Key
Value Points Marks
allotted
I. 1.
Ans. :
(B) { 4, 20 } 1
D
81-U 2 CCE PR
PR (D) - 710
Qn. Nos.
Ans. Key
Value Points Marks
allotted
2.
Ans. :
(A) raS−
=∞ 1
1
3.
Ans. :
(B) A × B = L × H 1
4.
Ans. :
(C) 3 1
5.
Ans. :
(D) 02 =++ cbxax 1
CCE PR 3 81-U
PR (D) - 710 [ Turn over
Qn. Nos.
Ans. Key
Value Points Marks
allotted
6.
Ans. :
(D) 40° 1
7.
Ans. :
(A) 5 1
8.
Ans. :
(C) 25 cm 1
81-U 4 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
II. 6 × 1 = 6
9.
Ans. :
A,M. = 2
ca + ½
= 2
2016 +
= 236 ½
= 18 1
10.
Ans. :
!)(!rn
nPr
n−
=
½
!)35(!5
35
−=P
= !2
!2345 ×××
= 60 ½ 1
11.
Ans. :
)( AP = 1 – P ( A ) ½
= 1 – 0·8
= 0·2 ½ 1
CCE PR 5 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
12.
Ans. :
C.V. = 100×σ
X ½
= 603 × 100
= 5 ½ 1
13.
Ans. :
4x + 1
x – 2 4 2x – 7x + 9
4 2x – 8x (–) (+)
x + 9 1
x – 2 (–) (+)
+ 11 1
+ 11
f ( x ) = 4 2x – 7x + 9
f ( 2 ) = 4 ( 2 ) 2 – 7 ( 2 ) + 9 ½
= 4 ( 4 ) – 14 + 9
= 16 – 14 + 9 = 11 ½ 1
81-U 6 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
x – 2 = 0 ⇒ x = 2
2 4 – 7 9
8 2 1
4 1 11 1
11.
14.
Ans. :
Δ = – 4 ac 1
III.
15.
Ans. :
n ( U ) = 60, n ( N ) = 40, n ( M ) = 35, n ( N I M ) = 26.
n ( M ) + n ( N ) = n ( M U N ) + n ( M I N ) ½
35 + 40 = n ( M U N ) + 26 ½
n ( M U N ) = 75 – 26 = 49 ½
M U N =
( M U N ) l =
∴ n ( M U N ) l = n ( U ) – n ( M U N )
= 60 – 49
= 11 ½ 2
CCE PR 7 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
16.
Ans. :
HP = 31,
51 , 1, – 1, ...
In AP 5, 3, 1, – 1, ... ½
a = 5, d = 3 – 5 = – 2, n = 10
nT = a + ( n – 1 ) d ½
10T = 5 + ( 10 – 1 ) ( – 2 )
= 5 + 9 ( – 2 )
= 5 – 18 ½
= – 13.
HP, 10T = 131
− ½ 2
17. Ans. :
3 + 5
⇒ 3 + 5 = qp
p, q ∈ z q ≠ 0 ½
⇒ qp
=5 – 3
⇒ q
qp 35
−=
½
½
∴
½ 2
81-U 8 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
18.
Ans. :
1 + 1 2
19.
Ans. :
½
½
!!)(
!rrn
nC r
n−
=
!2!)28(
!82 −
=Cn ½
= 282
56!2!6!678
==×××
= 28 – 8 = 20 ½ 2
n = 8 ∴ ½
CCE PR 9 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
= 2
)3( −nn ½
= 2
)38(8 − ½
= 2
58 ×
= 20 ½ 2
20.
Ans. :
: S = { ( HT ), ( HH ), ( TT ), ( TH ) }
n ( S ) = 4 ½
A =
= { ( HH ) }
∴ n ( A ) = 1
P ( A ) = 41
)()(=
SnAn
½
B = Event of getting exactly one tail
= { ( HT ), ( TH ) } ½
∴ n ( B ) = 2
P ( B ) = 42
)()(=
SnBn
½ 2
21.
Ans. :
81-U 10 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
½
∴ = = ½
3 = = ½
∴ × 3 = × = ½ 2
22.
Ans. :
½
23
23
23
3
23
3
−
−×
+=
+ ½
= 22 )2()3(
)23(3
−
− ½
= 631
6323
69−=
−=
−
− ½
2
23.
Ans. :
CCE PR 11 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
x – 1 = 0 ⇒ x = 1
1 1 1 – 3 5
1 2 – 1 1
1 2 – 1 4
Q ( x ) = 2x + 2x – 1 ½
R ( x ) = 4 ½ 2
p ( x ) = 2x – x – ( 2k + 2 )
p ( x ) – 4
∴ p ( – 4 ) = 0 ½
p ( x ) = 2x – x – ( 2k + 2 )
0 = ( – 4 )2 – ( – 4 ) – ( 2k + 2 ) ½
0 = 16 + 4 – 2k – 2 ½
0 = 18 – 2k
⇒ 2k = 18 or k =
218 = 9 ½
2
81-U 12 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
24.
Ans. :
— ½
— ½
— 1 2
25.
CCE PR 13 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
Ans. :
XY || BC Δ ABC
YCAY
XBAX
=
½
41
223
=−−
pp
½
4 ( p – 3 ) = 2p – 2
4p – 12 = 2p – 2 ½
4p – 2p = 12 – 2
2p = 10 ; p = 2
10 = 5 ½ 2
Δ AOB Δ COD
AOB = COD ( )
CDO = OBA ( )
∴ AA
Δ AOB ~ Δ COD 1
81-U 14 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
2
2
CDAB
= ½
14
14)2(
2
2
2
2===
CDCD
CDDC
⇒ 4 × Δ COD = 84
Δ COD = 484 = 21 cm2. ½
2
26.
Ans. :
tan A = 43
222 ACBABC =+
222 43 AC=+
⇒ 2AC = 25 ⇒ AC = 5 ½
sin A = 5
3=
cos A = 54
= ½ 2
1
CCE PR 15 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
27.
Ans. :
θ = 45°, m = tan θ c = 2 ½
m = tan 45° = 1 ½
y = mx + c ½
y = ( 1 ) x + 2 ⇒ y = x + 2 or x – y + 2 = 0 ½ 2
28.
Ans. :
),( 11 yx ),( 22 yx
A ( 6, 5 ) B ( 4, 4 )
AB = 212
212 )()( yyxx −+− ½
= 22 )54()64( −+− ½
= 22 )1()2( −+− ½
= 514 =+ ½ 2
29.
Ans. :
( CSA ) = 4070
l = 37 cm
r = ?
CSA = πrl ½
4070 = 722 × r × 37 ½
Rearranging, r = 22
71103722
74070 ×=
××
½
r = 35 cm ½ 2
110
1
81-U 16 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted 30.
Ans. :
80 m = 2080 = 4 cm
120 m =
20120 = 6 cm
160 m =
20160 = 8 cm ½
220 m =
20220 = 11 cm
60 m =
2060 = 3 cm
100 m =
20100 = 5 cm
2
1½
2
CCE PR 17 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
31.
Ans. :
— ½
— 1
— ½ 2
32.
Ans. :
42 += nTn
200 = 42 +n ½
⇒ 42 +n = 200 ½
2n = 200 – 4 = 196 ½
n = 196 ⇒ n = 14 ½ 2
33.
Ans. :
)35()64( yxyx −++ ½
= yyxx 3654 −++ ½
= ( 4 + 5 ) x + ( 6 – 3 ) y ½
= yx 39 + ½ 2
81-U 18 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
34.
Ans. :
3 183 × 360°
= 3 × 20° = 60°
6 120°
5 100°
4 80°
Total = 18
Calculation — ½
Pi-chart — 1½ 2
CCE PR 19 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
35.
Ans. :
p ( x ) = 2x + 14x + 48
2x + 14x + 48 = 0 48 ½
( x + 6 ) ( x + 8 ) = 0 6 8 ½
x + 6 = 0 ⇒ x = – 6 ½
x + 8 = 0 ⇒ x = – 8
½ 2
36.
Ans. :
XYZ PQX
XYZ = PQX = 90°
X ½
AA
XYZ ~ PQX
XPXZ
XQXY
= ½
81-U 20 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
⇒ 4
2416=
XQ ⇒
38
24416
=×
=XQ ½
= 322 cm ½
2
37.
Ans. :
AC
leΔ ADC,
222 DCADAC += ½
= 122 + 122 = 144 + 144
= 2 × 144 ½
AC = 1442×
∴ AC = 12 2 cm ½ 2
38.
Ans. :
m = 3, n = 5 ½
2x – ( m + n ) x + mn = 0 ½
2x – ( 3 + 5 ) x + ( 3 ) ( 5 ) = 0 ½
2x – 8x + 15 = 0 ½ 2
½
CCE PR 21 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
39.
Ans. :
),()6,5(),( 2211 yxyx = = ( – 3, 8 )
P ( x, y ) = ⎥⎥⎦
⎤
⎢⎢⎣
⎡ ++
2,
22121 yyxx
½
= ⎥⎦
⎤⎢⎣
⎡ +−2
86,
235
½
= ⎥⎦
⎤⎢⎣
⎡2
14,22 ½
= [ 1, 7 ] ½ 2
40.
Ans. :
tan 2A = cot ( A – 18° )
cot ( 90° – 2A ) = cot ( A – 18° ) ½
⇒ 90° – 2A = A – 18° ½
⇒ 3A = 90° + 18° = 108° ½
A = 3
108o = 36° ½ 2
IV.
41.
Ans. :
81-U 22 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
:
BQ BP ½
AP, AQ, AB are joined.
: a) BQBP =
b) PAB = QAB ½
c) PBA = QBA
:
In Δ APB and Δ AQB APB = AQB = 90°
AB = AB 1½
3
∴ AP = AQ
∴ Δ APB ≅ Δ AQB
∴ a) BP = BQ
b) PAB = QAB CPCT
c) PBA = QBA
½
CCE PR 23 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
42.
Ans. :
C = 132 cm, h = 25 cm, r = ? V = ?
C = 2πr ½
132 = 2 × 722 × r ½
r = 44
71322227132 ×
=××
= 21 cm ½
V = hr 2π ½
= 25)21(722 2 ××
= 252121722
××× ½
= 34650 cm3 ½ 3
h = 20 cm, r = 5 cm ½
hrV 2
31π=
= 205722
31 2 ×××
r = ?
3
34 rπ= ½
81-U 24 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
V ( ) = V ( ) ½
32722
34205
722
31 r××=××× ½
322
3 5554
205=×=
×=r ½
r = 5 cm. ½ 3
43.
Ans. :
x f f x d = x – x 2d 2df
35
40
45
50
55
2
4
8
4
2
70
160
360
200
110
– 10
– 5
0
5
10
100
25
0
25
100
200
100
0
100
200
N = ∑ f = 20 ∑ f x = 900 ∑ 2df = 600
1½
CCE PR 25 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
,
Nxf
x∑
=
= 20900 ½
= 45
= σ = Ndf 2∑
½
= 20600
= 30 ½
= 5·5 3
C = 5
Let A = 45
x f Step deviation
d = C
AX −
f d 2d 2df
35
40
45
50
55
2
4
8
4
2
– 2
– 1
0
+ 1
+ 2
– 4
– 4
0
4
4
4
1
0
1
4
8
4
0
4
8
N = 20 ∑ f d = 0 ∑ 2df = 24
1½
81-U 26 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
σ =
22
⎟⎟⎠
⎞⎜⎜⎝
⎛ ∑−
∑N
dfNdf
× C ½
=
2
200
2024
⎟⎟⎠
⎞⎜⎜⎝
⎛− × 5 ½
= 521 ×⋅
= 1·1 × 5 ½
= 5·5 3
x f f x 2x 2xf
35
40
45
50
55
2
4
8
4
2
70
160
360
200
110
1225
1600
2025
2500
3025
2450
6400
16200
10000
6050
∑ f x = 900 ∑ 2xf = 41,100
σ = 22
⎟⎟⎠
⎞⎜⎜⎝
⎛ ∑−
∑N
xfN
xf ½
= 2
20900
2041100
⎟⎟⎠
⎞⎜⎜⎝
⎛− ½
= 2)45(2055 −
= 20252055 −
= 30
= 5·5 ½ 3
1½
CCE PR 27 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
Let assumed mean, A = 45
x f d = x – A f d 2d 2df
35
40
45
50
55
2
4
8
4
2
– 10
– 5
0
5
10
– 20
– 20
0
20
20
100
25
0
25
100
200
100
0
100
200
N = 20 ∑ f d = 0 ∑ 2df = 600
σ =
22
⎟⎟⎠
⎞⎜⎜⎝
⎛ ∑−
∑N
dfNdf
½
=
2
200
20600
⎟⎟⎠
⎞⎜⎜⎝
⎛− ½
= 030 −
= 30
= 5·5 ½ 3
44.
Ans. :
1½
81-U 28 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
Δ ABD,
tan 60° = BDAB
BD503 = ½
⇒ BD = 3
50 m
Δ BDC,
tan 30° = BDCD ½
3503
1 CD= ½
⇒ CD = 3
13
50× ½
= 3216
350
= ½ 3
LHS = A
Asin1sin1
−+
= A
A
A
A
sin1
sin1
sin1
sin1
+
+×
−
+ ½
½ CD = ?
CCE PR 29 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
=
A
A
2
2
sin1
)sin1(
−
+ ½
= A
A2cos
sin1 + ½
= A
Acos
sin1 +
½
= AA
A cossin
cos1
+ ½
= sec A + tan A. ½ 3
45.
Ans. :
2x – 2x + 3 = 3x + 1
2x – 2x + 3 – 3x – 1 = 0 ½
2x – 5x + 2 = 0 ½
a 2x + bx + c = 0, a = 1, b = – 5, c = 2
x =
aacbb
242 −±−
½
81-U 30 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
=
)1(2)2()1(4)5()5( 2 −−±−− ½
= 2
8255 −± ½
= 2
175 ± ½
x = 2
175 + or x =
2175 −
3
2x – 6x + 2 = 0
a 2x + bx + c = 0, a = 1, b = – 6, c = 2
m + n = 1
)6( −−=
−ab
= 6 ½
mn = 12
=ac = 2 ½
a) 2611
=+
=+mn
nmnm
= 3 1
b) ( m + n ) ( mn ) = ( 6 ) ( 2 ) = 12 1 3
46.
Ans. :
CCE PR 31 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
ACBCAB == = a ΔABC
∴ 22aBCPCBP === units
ABP, ½
222 BPAPAB +=
222
2 ⎟⎟⎠
⎞⎜⎜⎝
⎛+=
aAPa
22
24
APaa =−
4
34
4 2222 aaa
AP =−
= ½
AP =
23
43 2 aa
= units ½
Δ ABC = 21 × × ½
= APBC ××21
= 2
321 a
a ××
= 4
32a ½
3
½
81-U 32 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
ABC, 222 BCACAB += ½
( )
CDE, 222 CECDDE += ½
DCB, 222 CBDCDB +=
ACE, 222 CEACAE += ½
LHS = 22 DEAB +
= 2222 CECDBCAC +++ ½
= )()( 2222 CDBCCEAC +++ ( )
= 22 DBAE + ½
= RHS 3
½
CCE PR 33 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
V.
47.
Ans. :
— 2
— 1
— 1 4
Y
81-U 34 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
48.
Ans. :
, 4T = a + 3d
13 = a + 3d ... (i) ½
, 8T = a + 7d
29 = a + 7d ... (ii) ½
(ii) — (i) ⇒
29 = a + 7d
13 = a + 3d ½ (–) (–) (–)
16 = 4d
4d = 16 ; d = 4
16 = 4 ½
a + 7d = 29
a + 7 ( 4 ) = 29
a + 28 = 29
a = 29 – 28 = 1 a = 1, d = 4 ½
2nSn = { 2a + ( n – 1 ) d } ½
2
1010 =S { 2 ( 1 ) + ( 10 – 1 ) ( 4 ) } ½
= 5 { 2 + 9 ( 4 ) }
= 5 [ 38 ] = 190 19010 =S ½ 4
CCE PR 35 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
ra , a , ar ... (i) ½
,
ra + a + ar = 14 ... (ii) ½
, ⎟⎟⎠
⎞⎜⎜⎝
⎛ra a ( ar ) = 64 ½
⇒ 3a = 64, a =
⇒ a = 4 ½
Eq. (ii) ⇒
r4 + 4 + 4r = 14
r
rr 2444 ++ = 14 ½
⇒ 4 + 4r + 24r = 14r
⇒ 24r – 10r + 4 = 0 ( ) ½
22r – 5r + 2 = 0
22r – 4r – r + 2 = 0
2r ( r – 2 ) – 1 ( r – 2 ) = 0
( 2r – 1 ) ( r – 2 ) = 0
r = 21
r = 2 ½
81-U 36 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
r = 21
, r = 2
214
=ra
= 8 24
=ra
= 2
a = 4 a = 4
ar = 4 )(21
= 2 ar = 4 ( 2 )
ar = 8 ½ 4
49.
Ans. :
: Δ ABC Δ DEF,
DEF = ABC ½
ACB = DFE
: EFBC
DFAC
DEAB
== ½
: H G AC AB
H G DFAHDEAG == , ½
½
CCE PR 37 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
Δ AGH Δ DEF ,
DEAG = GAH
= EDF
DFAH =
∴ Δ AGH ≅ Δ DEF SAS AGH = DEF CPCT
ABC = DEF
2
⇒ AGH = ABC Axiom-1
∴ GH || BC
∴ Δ ABC, AHAC
GHBC
AGAB
== Corollary of Thale’s theorem 4
Hence, DFAC
EFBC
DEAB
== Δ AGH ≅ Δ DEF
50.
Ans. :
2x – x – 2 = 0
2x = x + 2
y = 2x y = x + 2
y = 2x
x 0 1 – 1 2 – 2 3 – 3
y 0 1 1 4 4 9 9
y = x + 2
x 0 2 3
y 2 4 5
81-U 38 CCE PR
PR (D) - 710
Qn. Nos. Value Points Marks
allotted
Tables — 2
Drawing parabola — 1
Drawing line — ½
Identifying roots — ½ 4
CCE PR 39 81-U
PR (D) - 710 [ Turn over
Qn. Nos. Value Points Marks
allotted
2x – x – 2 = 0
y = 2x – x – 2
x 0 1 – 1 2 – 2 3 – 3 4
y – 2 – 2 0 0 4 4 10 10
XY table — 2
Parabola — 1
Identifying roots — 1 4