UNREVISEDkseeb.kar.nic.in/docs/JUNE-2019-MODEL-ANSWERS/MATHEMATICS... · 2020. 7. 3. · cce pr 13 81-k pr (d) - 7040 [ turn over xp xz xq xy 4 16 24 xq xq = 24 16 4 xq = 3 8 ½ =

PR (D) - 7040 [ Turn over

— 560 003

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

/ 2020 S.S.L.C. EXAMINATION, MARCH / APRIL, 2020

MODEL ANSWERS

: 07. 04. 2020 ] : 81-K

Date : 07. 04. 2020 ] CODE NO. : 81-K

Subject : MATHEMATICS

/ Old Syllabus

/ Private Repeater

/ Kannada Version

[ : 100

[ Max. Marks : 100

I. 1.

A B

(A) ( A B )l = A

l B

l

(B) ( A B )l = A

l B

l

(C) ( A B )l = A

l B

l

(D) ( A B )l = A B

(A) ( A B )l = A

l B

l 1

D CCE PR UNREVISED

81-K 2 CCE PR

PR (D) - 7040

2. a b ( G )

(A) G = 2

ba (B) G = ab

(C) G = 2

ba (D) G = ab

(B) G = ab 1

3. 8 12 24

(A) 4 (B) 24

(C) 8 (D) 12

(A) 4 1

4. P ( x ) = 4

2 x P ( 2 )

(A) 8 (B) 4

(C) 0 (D) 2

(C) 0 1

5. 0

2 cbxax

(A) acb 42 (B) acb 42

(C) acb 42 (D) acb 42

(B) acb 42 1

CCE PR 3 81-K


6.

AB P

OPA

(A) 60º (B) 0º

(C) 180º (D) 90º

(D) 90º 1

7.

sin 30º

(A) 1 (B) 3

(C) 2

1 (D) 2

(C) 2

1

1

8.

(A) 6, 8, 9 (B) 3, 4, 6

(C) 7, 8, 9 (D) 6, 8, 10

(D) 6, 8, 10 1

81-K 4 CCE PR

PR (D) - 7040

II. 6 1 = 6

9 14

9. 52 nTn 10

nT = 2n – 5

10T = 2 ( 10 ) – 5

= 20 – 5

10T = 15

1

10. 00 CPnn

00CP nn = 1 + 1 = 2

1

11.

0

12.

= 100 C.V. = X

100

1

13. p ( x ) = 865423 xxx

3 1

14. 0562 xx

= 1

)6(

a

b = 6 ½ + ½

1

CCE PR 5 81-K


III.

15. 50

42 24

= n ( A ) = 50

= n ( B ) = 42 ½

= n ( A B ) = 24

= n ( A B ) = ? ½

n ( A B ) = n ( A ) + n ( B ) – n ( A B )

= 50 + 42 – 24

= 92 – 24 ½

= 68

= 68. ½ 2

16. 6

1,

4

1,

2

1 , .... 20

2

1 ,

4

1 ,

6

1 , .....

2, 4, 6, .... ½

a = 2, d = 2 n = 20 n

T = ?

nT = a + ( n – 1 ) d

20T = 2 + ( 20 – 1 ) 2 ½

= 2 + ( 19 ) 2 ½

= 2 + 38

= 40

20 =40

1. ½

2

81-K 6 CCE PR

PR (D) - 7040

17. 32

2 + 3

2 + q

p3 , p, q z, q 0 ½

q

p3 – 2

q

qp 23

3 ½

q

qp 2

3 ½

32 ½ 2

18.

10 ½

n = 10

= 2

)3( nn ½

= 2

)310(10 ½

= 2

710 ½

= 35. 2

CCE PR 7 81-K


10 ½

n = 10

= nCn 2

= 102

10 C ½

= 10!2!)210(

!10

½

= 10!2!8

!8910

= 1012

910

= 45 – 10 ½

= 35 2

19. ( n + 3 ) ! = 20 ( n + 1 ) ! n

( n + 3 ) ! = 20 ( n + 1 ) !

( n + 3 ) ( n + 2 ) ( n + 1 ) ! = 20 ( n + 1 ) ! ½

( n + 3 ) ( n + 2 ) = 20

2n + 5n + 6 – 20 = 0 ½

2n + 5n – 14 = 0

2n + 7n – 2n – 14 = 0

n ( n + 7 ) – 2 ( n + 7 ) = 0

( n + 7 ) ( n – 2 ) = 0 ½

n + 7 = 0 n – 2 = 0

n = – 7 n = 2

– 7 ½

n = 2 2

81-K 8 CCE PR

PR (D) - 7040

20. 1 6

a)

b)

S = { 1, 2, 3, 4, 5, 6 } ½ n ( S ) = 6

a) A =

A = { 2, 4, 6 } n ( A ) = 3

P ( A ) = 6

3

)(

)(

Sn

An

2

1 1½

b) B =

B = { 1, 4 }

n ( B ) = 2

P ( B ) = 6

2

)(

)(

Sn

Bn

3

1

2

21.

4 3 = 12 ½

= = ½

= = ½

= ½ 2

22. 183

1

2

13 .

3 183

1

2

1

= 293

1

2

13 ½

= 233

1

2

3

2

CCE PR 9 81-K


= 22

2

2

3 ½

= 22

23 ½

= 2

2223

= 2

25 ½

23.

)3(5723 23 xxxx .

p ( x ) = 2x – 2x – 15

x – 3 = 0 x = 3

3 3 – 2 7 – 5

9 21 84 1

3 7 28 79

q ( x ) = 32x + 7x + 28 ½

r ( x ) = 79 ½ 2

2x – 2x – 15 = 2x – 5x + 3x – 15

= x ( x – 5 ) + 3 ( x – 5 ) 1

= ( x – 5 ) ( x + 3 )

2x – 2x – 15 = 0

( x – 5 ) ( x + 3 ) = 0 1

x – 5 = 0 x + 3 = 0

x = 5 x = – 3

x = 5 x = – 3. 2

81-K 10 CCE PR

PR (D) - 7040

24. 4 8

PA PB

— ½

— ½

2 — ½

— ½ 2

CCE PR 11 81-K


25. 50

30º

½

AB = = 50

BC = ½

ABC A = 30°

tan 30° = AB

BC ½

503

1 BC

BC = 3

50

BC = 3

350 ½

= 3

350

2

81-K 12 CCE PR

PR (D) - 7040

26. XYZ P XY PQ XZ

XP = 4 XY = 16 XZ = 24 XQ

ABCD AB || CD AB = 3 CD AOB

COD

1

XYZ PQX

XYZ = PQX

= 90°

XYZ ~ PQX

CCE PR 13 81-K


XP

XZ

XQ

XY

4

2416

XQ

XQ = 24

416

XQ = 3

8 ½

= 3

22 = 2.66 ½

2

AOB COD

AOB = COD

OAB = OCD

AOB ~ COD

2

2

CD

AB

= 1

99)3(

2

2

2

2

CD

CD

CD

CD ½

AOB : COD = 9 : 1 ½ 2

81-K 14 CCE PR

PR (D) - 7040

27. ( 2, 3 ) ( 4, 7 )

),(11

yx = ( 2, 3 ) ),(22

yx = ( 4, 7 )

=

2,

2

2121yyxx

½

=

2

73,

2

42 ½

=

2

10,

2

6 ½

= ( 3, 5 ) ½ 2

28. P ( 2, – 3 ) Q ( 10, y ) 10 y

),(11

yx = ( 2, – 3 ) ),(22

yx = ( 10, y )

d = 212

212

)()( yyxx ½

10 = 22 ])3([)210( y ½

10 = 22 )3()8( y

100 = ( 8 ) 2 + ( y + 3 ) 2

100 = 64 + ( y + 3 ) 2

100 – 64 = ( y + 3 ) 2 ½

36 = ( y + 3 ) 2

( y + 3 ) 2 = 36

CCE PR 15 81-K


y + 3 = 36

y = – 3 6

y = – 3 + 6 y = – 3 – 6 ½

y = 3 y = – 9

y = 3 – 9 2

29. 20

7

r = 7 cm h = 20 cm = ?

TSA = 2r ( r + h ) ½

= 2 )207(77

22 ½

= 2 2777

22 ½

= 44 27

= 1188 cm2 . ½ 2 30.

2

20 = 1

D

160

100 C 80

E 100 60

40 B 60

A

20 m = 1 cm

160 m = 20

160 = 8 cm

100 m = 20

100 = 5 cm ½

81-K 16 CCE PR

PR (D) - 7040

60 m = 20

60 = 3 cm

40 m = 20

40 = 2 cm

80 m = 20

80 = 4 cm

1½ 2

31. 2

i) A B

ii) A \ B

i) A B

1

CCE PR 17 81-K


ii) A \ B

1 2

32. 13 nT

n 26

nT n

13 nTn

½

26 = 3n – 1

26 + 1 = 3n ½

27 = 3n

n = ½

n = 3 ½ 2

33.

23

23

.

23 23 ½

23

23

23

23

23

23

½

=

222

23

23

½

=

23

2.3.22322

2

81-K 18 CCE PR

PR (D) - 7040

= 1

6223 ½

= 5 + 2 6

34.

2

14 6 2 18

14

40

14 360° = 126°

6

40

6 360° = 54°

2

40

2 360° = 18°

½

18

40

18 360° = 162°

40

1½ 2

35. ( x – a ) 223 xxax a

CCE PR 19 81-K


p ( x ) = 223 xxax

p ( x ) = 2)(23 aaaa ½

0 = 233 aaa ½

0 = a + 2 ½

a = – 2 ½ 2

36. 2x + 3 = 2x 0

2 cbxax

a, b c

2x + 3 = 2x

2x – 2x + 3 = 0 ½

a = 1 b = – 2 c = 3 1½ 2

37. ABC AMP

MP

BC

PA

CA 2

ABC AMP 1

ABC = AMP = 90°

A

ABC ~ AMP 1 2

81-K 20 CCE PR

PR (D) - 7040

AM

AB

MP

BC

AP

AC

MP

BC

PA

AC

38.

5 12

½

ABC B = 90°

222 BCABAC

= 122 + 5 2 1½

= 144 + 25

2AC = 169

AC = 169

AC = 13 2

39. tan 2A = cot ( A – 18º ) 2A A

tan 2A = cot ( A – 18° ) ... (i)

tan 2A = cot ( 90° – 2A ) ... (ii) ½

(i) (ii) 2

CCE PR 21 81-K


cot ( A – 18° ) = cot ( 90° – 2A )

A – 18° = 90° – 2A 1½

A + 2A = 90° + 18

3A = 108°

A = 3

108

A = 36°

40. ( 3, 4 ) ( 7, 8 )

),(11

yx = ( 3, 4 ) ),(22

yx = ( 7, 8 ) ½

= 12

12

xx

yy

= 37

48

1½

= 4

4

= 1 2

IV.

41. 3

( C-I ) ( f )

1 — 5 4

6 — 10 3

11 — 15 2

16 — 20 1

n = 10

81-K 22 CCE PR

PR (D) - 7040

C.I. f x f x d =

x – x

2d 2df

1 — 5 4 3 12 – 5 25 100

6 — 10 3 8 24 0 0 0

11 — 15 2 13 26 5 25 50

16 — 20 1 18 18 10 100 100

n = 10 fx = 80 2df = 250

10

80)(

n

fxx = 8 1½

= 2510

2502

n

df = 5 1½

3

C.I. f x d 2d f d

2df

1 — 5 4 3 – 1 1 – 4 4

6 — 10 3 8 0 0 0 0

11 — 15 2 13 1 1 2 2

16 — 20 1 18 2 4 2 4

n = 10 f d = 0 2df = 10

= A = 8 1½

= 510

0

10

10222

i

n

df

n

df 1½

= 51501

= 5 3

CCE PR 23 81-K


C.I. f x d = A – x 2d f d

2df

1 — 5 4 3 – 5 25 – 20 100

6 — 10 3 8 0 0 0 0

11 — 15 2 13 5 25 10 50

16 — 20 1 18 10 100 10 100

n = 10 fd = 0 2df = 250

= A = 8 1½

= 02510

0

10

250222

n

df

n

df

1½

= 25

= 5 3

C.I. f x 2x f x

2xf

1 — 5 4 3 9 12 36

6 — 10 3 8 64 24 192

11 — 15 2 13 169 26 338

16 — 20 1 18 324 18 324

n = 10 fx = 80 2xf = 890

=

22

n

xf

n

xf

1½

=

2

10

80

10

890

1½

= 6489

= 25

= 5 3

81-K 24 CCE PR

PR (D) - 7040

42.

½

A B P ½

A, P B ½

XPY ½

APX = 90° ... (i)

BPX = 90° ... (ii)

BPXAPX = 90° + 90° (i) (ii)

APB = 180° APB

APB

A, P B

3

1

CCE PR 25 81-K


½

½

43. 3

x ( x + 1 ) = 6x + 24.

m n 01272 xx

a) ( m + n ) mn

b) nm

11

x ( x + 1 ) = 6x + 24 2x + x = 6x + 24

2x + x – 6x – 24 = 0 1

2x – 5x – 24 = 0

a = 1 b = – 5 c = – 24

x = a

acbb

2

42

x = )1(2

)24()1(4)5()5( 2 ½

x =

2

115

5

1215

2

96255

x = 2

115 x =

2

115

x = 2

16 x =

2

6

x = 8 x = – 3. ½ 3

2x – 7x + 12 = 0

a = 1 b = – 7 c = 12

m + n = 1

7

1

)7(

a

b = 7 1

mn = 1

12

a

c = 12 1

a) ( m + n ) mn = 7 12 = 84

b) 12

711

mn

mn

nm. 1

3

81-K 26 CCE PR

PR (D) - 7040

44. BD =

3

1 BC ABC D BC

22 79 ABAD 3

ABC ABC = 90º, BD AC. AB = c BC = a

BD = p CA = b 222111

pca

½

ABC AB = BC = AC

AP BC

BP = PC = 2

1 BC ½

BD = 3

1 BC

APD P = 90° ½

2AD = 22 DPAP 3

CCE PR 27 81-K


= 22 )( BDBPAP

= BDBPBDBPAP ..2222

= BCBCBCAB3

1

2

12)

3

1( 22

= 2223

1

9

1BCBCAB 1½

= 9

39 222 BCBCAB

2AD = 9

39 222 ABABAB AB = BC = AC

9 2AD = 7 2AB .

ABC B = 90°

2AB = AC . AD

ADACAB .

11

2 ... (i) ½

2BC = AC . CD

CDACBC .

11

2 ... (ii) ½

2BD = AD . CD

2P = AD . CD ... (iii) ½

(i) (ii)

CDACADACBCAB .

1

.

111

22

3

81-K 28 CCE PR

PR (D) - 7040

CDADACac

1111122

1½

=

CDAD

ADCD

AC .

1

= CDAD

AC

AC .

1

CDADac .

11122

222

111

pac .

ABC = 2

1 AC BD

= 2

1 b p

= 2

bp ... (i) ½

ABC = 2

1 BC AB

= 2

1 a c

= 2

ac ... (ii) ½

(i) (ii)

22

acbp ½

bp = ac

b = p

ac

222 ACBCAB

CCE PR 29 81-K


2c + 2a = 2b 1

2c + 2a = 2

22

p

ca

22ca ½

222

22

22

2

22

2

pca

ca

ca

a

ca

c

222

111

pca .

3

45.

A

A

cos1

cos1

= cosec A + cot A 3

A

A

A

A

cos

sin1

sin1

cos

= 2 sec A

A

A

A

A

A

A

cos1

cos1

cos1

cos1

cos1

cos1

½

= )cos1()cos1(

)cos1( 2

AA

A

½

=

A

A

2

2

cos1

)cos1(

½

=

A

A

2

2

sin

)cos1( ½

= A

A

sin

cos1

= A

A

A sin

cos

sin

1 ½

= cosec A + cot A. ½ 3

81-K 30 CCE PR

PR (D) - 7040

)sin1(cos

)sin1(cos

cos

sin1

sin1

cos 22

AA

AA

A

A

A

A

½

= )sin1(cos

sin.1.2sin1cos 222

AA

AAA

½

= )sin1(cos

sin211

AA

A

[ 1cossin 22 AA ] ½

= )sin1(cos

sin22

AA

A

½

= )sin1(cos

)sin1(2

AA

A

½

= 2 Acos

1 ½

= 2 sec A. 3 46.

6 10

14

3

1·5

7 2

CCE PR 31 81-K


= 10 cm, = 2

10 = 5 cm = R

= 6 m, = 2

6 = 3 cm = r ½

r = 2

14 = 7 cm h = ?

=

hrrR 2333

1)(

3

2 ½

h =

2

33

3

1

)(3

2

r

rR

½

= 2

33 )(2

r

rR

= 2

33

7

)35(2 1

= 49

)27125(2

= 49

982

= 2 2

h = 4 cm. ½

= 4 cm. 3

81-K 32 CCE PR

PR (D) - 7040

= 2)51(

7

22

3

1

3

1 22 hr

= 251517

22

3

1

= 4·71 cm3 .

= 53)51(

7

22 22 hr 1½

= 5351517

22

= 24·75 cm3 .

= 33 )51(

7

22

3

2

3

2r

= 5151517

22

3

2

= 7

5049

= 7·07 cm3

= + +

= 4·71 + 24·75 + 7·07 1½

= 36·53 cm3 . 3

V.

47. 4 8 24

6 10 44

4

4 24 8 384

10

84TT = 24

a + 3d + a + 7d = 24

2a + 10d = 24 ... (i)

CCE PR 33 81-K


106TT = 44

a + 5d + a + 9d = 44

2a + 14d = 44 ... (ii)

(i) (ii)

2a + 10d = 24 ... (i)

2a + 14d = 44 ... (ii)

(–) (–) (–)

– 4d = – 20 1

d = – 20

– 4

d = 5

d = 5 (i)

2a + 10d = 24

2a + 10 (5) = 24

2a + 50 = 24

2a = 24 – 50

2a = – 26

a = 2

26

a = – 13

a = – 13

a + d = – 13 + 5 = – 8 1

a + d = – 13 + 2 (5) = – 3 4

81-K 34 CCE PR

PR (D) - 7040

4T = 24

3ar = 24 ... (i) ½

8T = 384

7ar = 384 ... (ii) ½

(i) (ii)

384

24

7

3

ar

ar 1

16

11

4

r

4r = 16

r =

r = 2

3ar = 24

a (2) 3 = 24 1

a (8) = 24

a = 8

24

a = 3

1

)1(

r

raS

n

n

102331

)11024(3

12

)12(3 10

10

S 1

10S = 3069

4

CCE PR 35 81-K


48. 9 4 2

PQ RS

— 2

— 1

— 1 4

81-K 36 CCE PR

PR (D) - 7040

49.

ABC DE || BC ½

EC

AE

DB

AD ½

D, C B, E ½

EL AB DN AC ½

ELDB

ELAD

2121

A =

2

1 bh

DBAD

DNEC

DNAE

2121

A =

2

1 bh

ECAE

EC

AE

DB

AD BDE = CDE

1.

CCE PR 37 81-K


50. 2x – x – 6 = 0.

2x – x – 6 = 0

y = 2x y = 6 + x

y = 2x

x 0 1 2 3 – 1 – 2 – 3

y 0 1 4 9 1 4 9

y = 6 + x

x 0 1 2

y 6 7 8

x- 1 = 1

y- 1 = 1

81-K 38 CCE PR

PR (D) - 7040

2x – x – 6 = 0

y = 2x – x – 6

x 0 1 2 3 4 – 1 2 – 3

y – 6 – 6 – 4 0 6 – 4 0 6

x- 1 = 1

y- 1 = 1

2

CCE PR 39 81-K


4

Documents

UNREVISEDkseeb.kar.nic.in/docs/JUNE-2019-MODEL-ANSWERS/MATHEMATICS... · 2020. 7. 3. · cce pr 13 81-k pr (d) - 7040 [ turn over xp xz xq xy 4 16 24 xq xq = 24 16 4 xq = 3 8 ½ =