81-U PR D - 710 Modelkseeb.kar.nic.in/docs/MODEL-ANSWERS-2019... · CCE PR 3 81-U PR (D) - 710 [...

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

»⁄·¤•⁄¬ D}⁄ °¡⁄V⁄◊⁄fl

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


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



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


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



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


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



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


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



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


allotted

24.

Ans. :

— ½

— ½

— 1 2

25.

CCE PR 13 81-U



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


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



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


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



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


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



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


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



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


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



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


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



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


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



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


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



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


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



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


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



allotted

V.

47.

Ans. :

— 2

— 1

— 1 4

Y

81-U 34 CCE PR

PR (D) - 710


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



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


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



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


allotted

Tables — 2

Drawing parabola — 1

Drawing line — ½

Identifying roots — ½ 4

CCE PR 39 81-U



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

81-U PR D - 710 Modelkseeb.kar.nic.in/docs/MODEL-ANSWERS-2019... · CCE PR 3 81-U PR (D) - 710 [...

Documents

Transcript of 81-U PR D - 710 Modelkseeb.kar.nic.in/docs/MODEL-ANSWERS-2019... · CCE PR 3 81-U PR (D) - 710 [...