5 mark questions

X-STD

MATHEMATICS

PREPARED BY:R.RAJENDRAN. M.A., M. Sc., M. Ed.,K.C.SANKARALINGA NADAR HR. SEC. SCHOOL,CHENNAI-21

Prove That A – (B C) = (A – B) (A – C)

AA A

A A

B B B

B B

C C C

C

C

(1)BC (2) A-(BC) (3) A – B

(4) A – C (5) (A – B) (A – C)

From the diagrams (2) and (5)

A – (B C) = (A – B) (A – C)

Prove that A – (B C) = (A – B) (A – C) using Venn diagram

A B

C

A A

A A

B B

B B

CC

C C

(1) BC (2) A – ( BC) (3) A – B

(4) A – C (5) (A – B)(A – C)

From the diagram (2)and(5) A – (BC) = (A – B)(A – C)

Prove that A (B C) = (A B) (A C) using Venn diagram.

A B

C

A A

A A

B B

BB

C C

C C

B C A (B C) A B

A C (A B)(A C)

From the figures 2 and 5

A(B C)=(AB)(AC)

Prove that A (B C) = (A B) (A C) using Venn diagram.

A B

C

A A

A A

B B

BB

C C

C C

B C A (B C) A B

A C (AB) (AC)

From the figures 2 and 5

A(BC)=(AB)(AC)

Prove that (A B)’ = A’ B’ using Venn diagram.

A BB

AA

AA B

B B

U U U

UU

1. AB 2. (AB)’ 3. A’

4. B’ 5. A’ B’

From the diagrams 2 and 5 (A B)’ = A’ B’

Prove that (A B)’ = A’ B’ using Venn diagram.

A BB

AA

AA B

B B

U U U

UU

1. A B 2. (A B)’ 3. A’

4. B’ 5. A’ B’

From the diagrams 2 and 5 (AB)’ = A’B’

Given, A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8}, show that (i) A (B C) = (A B) C. (ii) Verify (i) using Venn diagram.

Solution(i) B C = {3, 4, 5, 6} {5, 6, 7, 8}

= {3, 4, 5, 6, 7, 8}` A (B C) = {1, 2, 3, 4, 5} { 3, 4, 5, 6, 7, 8}

= {1, 2, 3, 4, 5, 6, 7, 8}………… (1) A B = {1, 2, 3, 4, 5} {3, 4, 5, 6}

= {1,2,3,4,5,6}`(A B) C = {1,2,3,4,5,6} {5,6,7,8}

= {1, 2, 3, 4, 5, 6, 7, 8}……………. (2)

From (1) and (2), we have A (B C) = (A B) C.

Using Venn diagram, we have

A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8}

A B

C

5 6

7 8

34

A B

C

5 6

7 8

341

2

(1) B C (2) A (B C)

Let A = {a,b,c,d}, B = {a,c,e} and C = {a,e}.

(i) Show that A (B C) = (A B) C. (ii) Verify (i)

using Venn diagram.

Solution

Given A = {a,b,c,d}, B = {a,c,e} and C = {a,e}.

B C = {a,c,e} {a,e} = {a,e}

A (B C) = {a,b,c,d} {a,e} = {a}…… (1)

A B = {a,b,c,d.} {a,c,e} = {a,c}.

(A B) C = {a,c} {a,e} = {a}…….. (2)

From (1) and (2) A (B C) = (A B) C.

A = {a,b,c,d}, B = {a,c,e} and C = {a,e}.

A B

C

a e

(1) BC

A B

C

a

(2) A(BC)

A = {a,b,c,d}, B = {a,c,e} and C = {a,e}.

A B

C

a

c

(1) AB

A B

C

a

(2) (AB)C

from (2) and (4) , it is verified that A (B C) =(A B) C

Let A = {0,1,2,3,4}, B = {1, - 2, 3,4,5,6} and C = {2,4,6,7}.

(i) Show that A (B C) = (A B) (A C). (ii) Verify using Venn

diagram.

Solution

(i) B C = {1, - 2, 3, 4, 5, 6} {2, 4, 6, 7 } = {4, 6};

A (B C) = {0,1, 2, 3, 4} {4, 6} = {0,1,2,3,4,6}…….(1)

A B = {0,1,2,3,4} {1, - 2, 3,4,5,6} = {- 2, 0,1, 2, 3, 4, 5, 6},

A C = {0,1,2,3,4} {2,4,6,7} = {0,1, 2, 3, 4, 6, 7}.

(A B) (A C) = {- 2, 0,1, 2, 3, 4, 5, 6} {0,1, 2, 3, 4, 6, 7}

= {0,1, 2, 3, 4, 6}. …….(2)

From (1) and (2) ,we get A (B C) = (A B) (A C).

A = {0, 1, 2, 3, 4}, B = {1, –2, 3, 4, 5, 6},C = {2, 4, 6, 7}.

A B

C

4 6

(1) BC

A B

C

4

(2) A(BC)

6

3

01

2

A B

C

4

3

(3) AB

A B

C

0

(4) A C

from (2) and (4) , it is verified that A (B C) =(A B) C

A = {0, 1, 2, 3, 4}, B = {1, –2, 3, 4, 5, 6},C = {2, 4, 6, 7}.

1

2

4

3

01

26

7

-2

6

5

A B

C

4

(5) (AB)(AC)

6

3

01

2

from (2) and (5) , it is verified that

A (B C) = (A B) (A C)

Given that U = {a,b,c,d,e, f,g,h}, A = {a, b, f, g}, and B = {a, b, c}, verify De Morgan’s laws of complementation.

U = {a, b, c, d, e, f, g, h} A = {a, b, f, g} B = {a, b, c}.De Morgan’s laws(AB)’ = A’ B’(AB)’ = A’ B’VERIFICATION A B = {a, b, c, f, g}(A B)’ = {d, e, h}……..(1)A’ = {c, d, e, h}B’ = {d, e, f, g, h}A’B’ = {d, e, h}……….(2)From (1) and (2) (A B)’ = A’B’

A B = {a, b}(A B)’ = {c, d, e, f, g, h}....(1)A’ = {c, d, e, h}B’ = {d, e, f, g, h}A’ B’ = {c, d, e, f, g, h}….(2)From (1) and (2)

(A B)’ = A’ B’

Verify De Morgan’s laws for set difference using the sets given below: A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} and C = {3,9,10,12,13}..A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7}

C = {3,9,10,12,13}..

De Morgan’s laws

(1) A\(BC) = (A\B)(A\C) (2) A\(BC) = A\B) (A\C)

VERIFICATION

B C = {1, 2, 3, 5, 7, 9, 10, 12, 13}

A\(BC) = {11, 15}……..(1)

A\B = {3, 9, 11, 13, 15}

A\C = {1, 5, 7, 11, 15}

(A\B)(A\C) = {11, 15}……….(2)

From (1) and (2) A\(BC) = (A\B)(A\C)

Verify De Morgan’s laws for set difference using the sets given below: A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} and C = {3,9,10,12,13}..

A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7}

C = {3,9,10,12,13}..

BC = { }

A\(B C) = {1, 3, 5, 7, 9,11,13,15}……..(1)

A\B = {3, 9, 11, 13, 15}

A\C = {1, 5, 7, 11, 15}

(A\B) (A\C) = {1, 3, 5, 7, 9,11,13,15}……….(2)

From (1) and (2) A\(B C) = (A\B) (A\C)

Let A = {10,15, 20, 25, 30, 35, 40, 45, 50}, B = {1, 5,10,15, 20, 30}and C = {7, 8,15,20,35,45, 48}. Verify A\(B C) = (A\B) (A\C)

A = {10,15, 20, 25, 30, 35, 40, 45, 50},

B = {1, 5, 10, 15, 20, 30}

C = {7, 8, 15, 20, 35, 45, 48}

BC = {15, 20}

A\(B C) = {10, 25, 30, 35, 40, 45, 50}……..(1)

A\B = {25, 30, 35, 40, 45, 50}

A\C = {10, 25, 30, 40, 50}

(A\B) (A\C) = {10, 25, 30, 35, 40, 45, 50}……….(2)

From (1) and (2) A\(B C) = (A\B) (A\C)

13. Let A = { 6, 9, 15, 18, 21 }; B = { 1, 2, 4, 5, 6 } and f : A "

B be defined by f(x) = . Represent f by (i) an arrow

diagram (ii) a set of ordered pairs (iii) a table (iv) a graph .3

3x

3

3)(

xxf

13

3

3

36)6(

f

23

6

3

39)9(

f

43

12

3

315)15(

f

53

15

3

318)18(

f

63

18

3

321)21(

f

6

9

15

18

21

1

2

4

5

6

A B

ARROW DIAGRAM

SET OF ORDERED PAIR

f = {(6, 1), (9, 2), (15, 4), (18, 5), (21, 6)}

X 6 9 15 18 21Y 1 2 4 5 6

Table

Graph

| | | | | | | | |

3 6 9 12 15 18 21 24 27

– 7

– 6

– 5

– 4

– 3

– 2

– 1 (6, 1)

(9, 2)

(15, 4)

(18, 5)

(21, 6)

Let A= { 0, 1, 2, 3 } and B = { 1, 3, 5, 7, 9 } be two sets. Let f : A B be a function given by f (x) = 2x + 1. Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow diagram and (iv) a graph.

Given A= {0, 1, 2, 3},

B = {1, 3, 5, 7, 9}

f(x) = 2x + 1

f(0) = 2(0) + 1 = 0 + 1 = 1

f(1) = 2(1) + 1 = 2 + 1 = 3

f(2) = 2(2) + 1 = 4 + 1 = 5

f(3) = 2(3) + 1 = 6 + 1 = 7

(i)Set of ordered pairs

{(0, 1), (1, 3), (2, 5), (3, 7)}

(ii) Table

x

f(x)

0

1

1

3

2

5

3

7

4

6

8

10

3

4

5

6

7

ABARROW

DIAGRAM

Graph

| | | | | | | |

0 2 4 6 8 10 12 14 16

– 7

– 6

– 5

– 4

– 3

– 2

– 1

(4, 3)

(6, 4)

(8, 5)

(10, 6)

A function f: [1, 6)R is defined as follows

f(x) =

6 x 4 103

4 x 2 12

2 x 1 1

2x

x

x Find the value of (i) f(5), (ii) f(3), (iii) f(1), (iv) f(2) – f(4) (v) 2f(5) – 3f(1)

Since 5 lies between 4 and 6

f(x) = 3x2 – 10

f(5) = 3(5)2 – 10

= 75 – 10

= 65

Since 3 lies between 2 and 4

f(x) = 2x – 1

= 2(3) – 1 = 6 – 1

= 5

Since 1 lies in the interval 1 x < 2

f(x) = 1 + x

f(1) = 1 + 1 = 2

Since 2 lies in the interval 2 x < 4

f(x) = 2x – 1

= 2(2) – 1 = 4 – 1 = 3

Since 4 lies in the interval 4 x < 6

f(x) = 3x2 – 10

f(4) = 3(4)2 – 10

= 48 – 10 = 38

f(2) – f(4) = 3 – 38 = – 34

2f(5) – 3 f(1) = 2 (65) – 3(2)

= 130 – 6

= 124

A function f: [-3, 7)R is defined as follows

6 x 4 ; 32

4 x 2 ; 23

2 x 3- 1;4

)(

2

x

x

x

xf

Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv)

(i) Since 5 lies between 4 and 6

f(x) = 3x – 2

f(5) = 3(5) – 2

= 15 – 2 = 13

Since 6 lies in the interval 4 < x 6

f(x) = 2x – 3

f(6) = 2(6) – 3 = 12 – 3

= 9

f(5) + f(6) = 13 + 9 = 22

)1()6(2

)1()3(

ff

ff

A function f: [-3, 7)R is defined as follows

f(x) =

6 x 4 ; 32

4 x 2 ; 23

2 x 3- 1;4 2

x

x

x Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4)(iv)

(ii) Since 1 lies in the interval -3 x < 2

f(x) = 4x2 – 1

f(1) = 4(1)2 – 1

= 4 – 1 = 3

Since –3 lies in the interval-3 x < 2

f(x) = 4x2 – 1

= 4(-3)2 – 1

= 36 – 1 = 35

f(1) – f(-3) = 3 – 35 = – 32

)1()6(2

)1()3(

ff

ff

A function f: [-3, 7)R is defined as follows

f(x) =

6 x 4 ; 32

4 x 2 ; 23

2 x 3- 1;4 2

x

x

x Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv)

(iii) Since -2 lies in the interval -3 x < 2

f(x) = 4x2 – 1

f(1) = 4(-2)2 – 1

= 16 – 1 = 15

Since 4 lies in the interval 2 x 4 f(x) = 3x – 2

= 3(4) – 2

= 12 – 2 = 10

f(-2) – f(4) = 15 – 10 = 5

)1()6(2

)1()3(

ff

ff

A function f: [-3, 7)R is defined as follows

f(x) =

6 x 4 ; 32

4 x 2 ; 23

2 x 3- 1;4 2

x

x

x Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv)

(iv) Since 3 lies in the interval 2 x 4 f(x) = 3x – 2

f(3) = 3(3) – 2

= 9 – 2 = 7

Since -1 lies in the interval -3 x < 2

f(x) = 4x2 – 1

= 4(-1)2 – 1

= 4 – 1 = 3

f(6) = 9, f(1) = 3

)1()6(2

)1()3(

ff

ff

3)9(2

37

)1()6(2

)1()3(

ff

ff

318

10

15

10

3

2

A function f: [-7, 6)R is defined as follows

6 x 2 ; 1

2 x 5- ; 5

5- x 7- ;12

)(

2

x

x

xx

xf

Find the value of

(i) 2f(-4) + 3f(2)

(ii) f(-7) – f(– 3),

(iii) (i) Since -4 lies in the interval -5 x 2 f(x) = x + 5

f(-4) = – 4 + 5

= 1

Since 2 lies in the interval -5 x 2 f(x) = x + 5

f(2) = 2 + 5

= 7

)1(3)6(

)4(2)3(4

ff

ff

2f(-4) + 3f(2)

= 2(1) + 3(7)

= 2 + 21

= 23

A function f: [-7, 6)R is defined as follows

f(x) =

6 x 2 ; 1

2 x 5- ; 5

5- x 7- ;122

x

x

xx Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)

(ii) Since -7 lies in the interval -7 x < -5

f(x) = x2 + 2x + 1

f(-7) = (–7)2 + 2(–7) + 1

= 49 – 14 + 1 = 36

Since -3 lies in the interval -5 x 2 f(x) = x + 5

f(2) = -3 + 5

= 2

)1(3)6(

)4(2)3(4

ff

ff

f(-7) – f(–3)

= 36 – 2

= 34

A function f: [-7, 6)R is defined as follows

f(x) =

6 x 2 ; 1

2 x 5- ; 5

5- x 7- ;122

x

x

xx Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)

(iii) Since -3 lies in the interval -5 x 2 f(x) = x + 5 f(-3) = -3 + 5 = 2

Since -6 lies in the interval -7 x < -5

f(x) = x2 + 2x + 1

f(-6) = (–6)2 + 2(–6) + 1

= 36 – 12 + 1 = 25

)1(3)6(

)4(2)3(4

ff

ff

4 lies in the interval 2 < x < 6

f(x) = x – 1

f(4) = 4 – 1

= 3

Since 1 lies in the interval -5 x 2 f(x) = x + 5

f(1) = 1 + 5

= 6

A function f: [-7, 6)R is defined as follows

f(x) =

6 x 2 ; 1

2 x 5- ; 5

5- x 7- ;122

x

x

xx Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)

(iii) f(-3) = 2

f(-6) = 25

f(4) = 3

f(1) = 6

)1(3)6(

)4(2)3(4

ff

ff

)6(325

)3(2)2(4

)1(3)6(

)4(2)3(4

ff

ff

1825

68

27

14

If verify that (AB)T = BTAT .

11

12 ,

37

25BA

LHS = RHS (AB)T = BTAT .

11

12 ,

37

25BA

11

12

37

25AB

13)1(7)1(327

12)1(5)1(225

37314

25210

411

38

LHSAB T ........43

118)(

11

12 ,

32

75 TT BA

32

75

11

12TT AB

317)1(215)1(

3)1(722)1(52

3725

314210

RHSAB TT ........43

118

If verify that (AB)T = BTAT . 631 ,

5

4

2

BA

LHSAB T ........

302412

15126

542

)(

631 ,

5

4

2

BA

631

5

4

2

BA

)6(53515

)6(43414

)6(23212

BA

30155

24124

1262

BA

If verify that (AB)T = BTAT . 631 ,

5

4

2

BA

LHS = RHS (AB)T = BTAT .

LHSAB T ........

302412

15126

542

)(

6

3

1

,542 TT BA

542

6

3

1

TT AB

5)6(46)2(6

5343)2(3

5141)2(1TT AB

RHSAB TT ....

302412

15126

542

If then show that A2 – 4A + 5I2 = O

32

11A

32

11A

32

11

32

112 AAA

33)1(22312

3)1()1(12)1(11

9262

3121

78

41

128

44

32

1144A

50

05

10

0155 2I

50

05

128

44

78

4154 2

2 IAA

5127088

044541

00

00

A2 – 4A + 5I2 = O

7. Find X and Y if 2X + 3Y =

)........(104

3232

YX

)2(..........51

222Y3X

)3......(04

322642)1(

YX

)4.......(51

2236Y9X3(2)

51

223

04

3225X)4((3)

153

66

08

64

15038

6664

1511

1225 X

1511

122

5

1 X

51

222Y3X and

04

32

)........(104

3232

YX

)2(..........51

222Y3X

)3......(04

323963)1(

YX

)4.......(51

222Y4X62(2)

51

222

04

3235Y)4((3)

102

44

012

96

100212

4946

1014

1325 Y

1014

132

5

1 Y

11. An electronic company records each type of entertainment device sold at three of their branch stores so that they can monitor their purchases of supplies. The sales in two weeks are shown in the following spreadsheets.

TV DVD VIDEOGAMES CD players

WEEK I

Store I 30 15 12 10

Store II 40 20 15 15

Store III 25 18 10 12

WEEK II

Store I 25 12 8 6

Store II 32 10 10 12

Store III 22 15 8 10

Find the sum of the items sold out in two weeks using matrix addition.

1081522

12101032

681225

,

12101825

15152040

10121530

Let BA

1081522

12101032

681225

12101825

15152040

10121530

BA

101281015182225

1215101510203240

61081212152530

22183347

27253072

16202755

If verify (AB)C = A(BC) .

12 and

2

1

0

,321

121

CBA

LHS = RHS (AB)C =A(BC)

2

1

0

321

121AB

231201

211201AB

620

220

8

4

128

4)(

CAB

1828

1424LHS.....

816

48

12

2

1

0

BC

1222

1121

1020

24

12

00

24

12

00

321

121)(BCA

231201432201

211201412201

6201240

220440RHS.....

816

48

If then prove that

(A + B)2 A2 + 2AB + B2.

23

61 ,

32

41BA

23

61 ,

32

41BA

23

61

32

41BA

11

20

2332

6411

11

20

11

20)( 2BA

11211101

12201200

1210

2020LHS...

31

22

32

41

32

412A

33)4()2()2(31)2(

3)4()4(1)2()4(11

9862

12481

178

169

23

61

23

612B

)2()2(633)2()1(3

)2(66)1(36)1()1(

41863

126181

229

1819

If then prove that

(A + B)2 A2 + 2AB + B2.

23

61 ,

32

41BA

23

61 ,

32

41BA

23

61

32

41AB

)2(36)2(33)1()2(

)2()4(613)4()1(1

61292

86121

1811

1413

229

1819

1811

14132

178

1692 22 BABA

229

1819

3622

2826

178

169

2236179228

18281619269

36391722

34282628

RHS....35

62

LHS RHS

(A + B)2 A2 + 2AB + B2.

If find (A + B)C and

AC + BC. Is (A + B)C = AC + BC ?

64

32 and

90

78 ,

67

33CBA

64

32 and

90

78 ,

67

33CBA

90

78

67

33

BA

157

1011

9607

7383

64

32

157

1011)(

CBA

615)3(741527

610)3(11410211

90216014

60334022

6974

2762

64

32

67

33AC

66)3(74627

63)3(34323

36212414

189126

1538

918

64

32

90

78BC

69)3(04920

67)3(84728

540360

42242816

5436

1844

If find (A + B)C and

AC + BC. Is (A + B)C = AC + BC ?

64

32 and

90

78 ,

67

33CBA

5436

1844

1538

918BCAC

54153638

1894418

6974

2762

(A+B)C = AC + BC

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find the 27th term.

Given t10 = 41

t18 = 73

tn = a + (n – 1)d

t10 = 41

a + (10 – 1)d = 41

a + 9d = 41…………(1)

t18 = a + (18 – 1)d = 73

a + 17d = 73 …….(2)

(2) a + 17d = 73

(1) a + 9d = 41

(2) – (1) 8d = 32

d = 32/8

d = 4

Sub d = 4 in (1)

a + 9(4) = 41

a + 36 = 41

a = 41 – 36

a = 5

t27 = a + (27 – 1)d

= 5 + (26)4

= 5 + 104 = 109(-) (-) (-)

The sum of three consecutive terms in an A.P. is 6 and their product is –120. Find the three numbers.

Let a – d, a, a + d be three consecutive terms in an A.P.

Sum = 6

a – d + a + a + d = 6

3a = 6

a = 6/3 = 2

Product = – 120

(a – d) a (a + d) = – 120

(2 – d) 2 (2 + d) = – 120

(22 – d2) 2 = – 120

4 – d2 = – 120/2 = – 60

– d2 = – 60 – 4

– d2 = – 64

d2 = 64

d = 8The given numbers are

2 – 8, 2, 2 + 8

– 6, 2, 10 (or)

10, 2, –6

A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the tenth year. Assuming that the production increases uniformly by a fixed number every year, find the number of TVs produced in the first year and in the 15th year.

Given t7 = 1000

t10 = 1450 tn = a + (n – 1)d

t7 = 1000a + (7 – 1)d = 1000 a + 6d = 1000…………(1)

t10 = a + (10 – 1)d = 1450 a + 9d = 1450 …….(2)(2) a + 9d = 1450(1) a + 6d = 1000

(2) – (1) 3d = 450

d = 450/3

d = 150

Sub d = 150 in (1)

a + 6(150) = 1000

a + 900 = 1000

a = 1000 – 900

a = 100

t15 = a + (15 – 1)d

= 100 + (15)150

= 100 + 2250 = 2350

Find the sum of the first 40 terms of the series 12 – 22 + 32 – 42 + ……. .

12 – 22 + 32 – 42 + ……. . = 1 - 4 + 9 - 16 + 25 ….. to 40 terms

= (1 – 4) + (9 – 16) + (25 – 36) + …… to 20 terms.

= (-3) + (-7) + (-11) +………20 terms

It is an AP with a = –3, d = –4, n = 20

dn )1(2a2

nSn

)4)(120(2(-3)2

20S20

)4(19)62

20

7662

20

822

20

10

= 10(-82)

= – 820

Find the sum of all 3 digit natural numbers, which are divisible by 9.

First number = 100 + (9 – 1) = 108

Last number = 999

The AP is

108 + 117 +……+ 999

a = 108, l = 999, d = 9 1d

a-ln

19

108-999n

110919

981

n = 110

)al(2

nsn

)1107(2

110

)999108(2

110s110

= 55 1107

= 60885

55

1 1

9 1 0 0

9

1 0

9

1

1 1 1

9 9 9 9

9

9

9

9

9

0

The three digit numbers are from 100 to 999

Find the sum of all natural numbers between 300 and 500 which are divisible by 11.

First number = 300 + (11 – 3) = 308

Last number = 500 – 5 = 495

The AP is

308 + 319 +……+ 495

a = 308, l = 495, d = 111

d

a-ln

111

308495n

117111

187

n = 18

)al(2

nsn

)803(2

18

)495308(2

18s18

= 9 803

= 7227

9

2 7

11 3 0 0

2 2

8 0

7 7

3

4 5

11 5 0 0

4 4

6 0

5 5

5

A sum of Rs. 1000 is deposited every year at 8% simple interest. Calculate the interest at the end of each year. Do these interest amounts form an A.P.? If so, find the total interest at the end of 30 years.

Every year deposit = Rs. 1000

Rate of interest = 8% = 0.08

First year interest = 1000 0.08 = 80

Second year interest = 2000 0.08 = 160

Third year interest = 3000 0.08 = 240

The interest 80, 160, 240,…. Forms an AP.

Total interest = 80 + 160 + 240 +……30terms

= 15 2480

= Rs.37200

})1(2{2n dnan

S

}80)130(802{2

3030 S

}8029160{2

30

}2320160{2

30

24802

30

Find the sum of all numbers between 100 and 200 which are not divisible by 5.

The required sum

= (100 +101 + 102+ ……+ 200) – (100 + 105 + 110 + … + 200)

100 + 101 + 102 + ………… + 200

= (1 + 2 + 3 + ………….+ 200) – (1 + 2 + 3+ ………+ 99)

2

1)n(nn

= 100 201 – 99 50

= 20100 – 4950

= 15150

100 + 105 + …… + 200

Here a = 100, d = 5, l = 200

2

1)99(99

2

1)200(200

2

10099

2

201200

100 50

1n

d

al

15

100200n

15

100

21120

Find the sum of all numbers between 100 and 200 which are not divisible by 5.

)(2

nsn al

)300(2

21

)100200(2

21s21

= 21 150

= 3150

150

The required sum

= 15150 – 3150

= 12000

Calculate the standard deviation of the following data. 10, 20, 15, 8, 3, 4

35.66,6655

25106610

.9

9981

8566118

7

78309

10

20

15

8

3

4

60

x d = x –x d2 n

xx 10

6

60

10 – 10 = 0

20 – 10 = 10

15 – 10 = 5

8 – 10 = -2

3 – 10 = -7

4 – 10 = -6

0

0

100

25

4

49

36

214

n

dDS

2

.

6

214

6666.35

97.5

257

Calculate the standard deviation of the following data. 38, 40, 34 ,31, 28, 26, 34

22.00,0044

166008

.6

6516

840092

9

98361

38

40

34

31

28

26

34

231

x d = x –x d2 n

xx 33

7

231

38 – 33 = 5

40 – 33 = 7

34 – 33 = 1

31 – 33 = -2

28 – 33 = -5

26 – 33 = -7

34 – 33 = 1

0

25

49

1

4

25

49

1

154

n

dDS

2

.

7

154

22

69.4

39

Calculate the standard deviation of the following data.

3

7

8

10

13

15

18

10

23

8

50

x f d = x – 3 d2 fd

fd2 3 – 3 = 0

8 – 3 = 5

13 – 3 = 10

18 – 3 = 15

23 – 3 = 20

0

25

100

225

400

X 3 8 13 18 23

f 7 10 15 10 8

0

50

150

150

160

510

0

250

1500

2250

3200

7200

Standard deviation

22

f

fd

f

fd

2

50

510

50

7200

22.10144

04.104144

96.39

39.9666

36396123369

27

.3

3.6

11. A group of 45 house owners contributed money towards green environment of their street. The amount of money collected is shown in the table below. Calculate the variance and standard deviation.

0 – 20

20 – 40

40 – 60

60 – 80

80 – 100

Time f mid x d = (x – 30)/20 d2 fd fd2

Amount 0-20 20-40 40-60 60-80 80-100

No of house owners 2 7 12 19 5

2

7

12

19

5

45

10

30

50

70

90

-2

-1

0

1

2

4

1

0

1

4

-4

-7

0

19

10

28

8

7

0

19

20

58

Standard deviation 20

22

f

fd

f

fd

2045

28

45

582

0.904490.9

8194418925

19

5

2062.02888.1 2

203844.02888.1 209044.0

2095.0 19

5Variance = (S.D)2

= (19)2= 361

Calculate the coefficient of variation of the following data: 20, 18, 32, 24, 26.

24.00,0044

168008

.8

8704

960096

9

98721

20

18

32

24

26

120

x d = x –x d2 n

xx 24

5

120

20 – 24 = -4

18 – 24 = -6

32 – 24 = 8

24 – 24 = 0

26 – 24 = 2

0

16

36

64

0

4

120

n

dDS

2

.

5

120

24

%100x

CV

879

%10024

89.4

89.4

%24

489

%375.20

The time (in seconds) taken by a group of people to walk across a pedestrian crossing is given in the table below. Calculate the variance and standard deviation of the data.

5 – 10

10 – 15

15 – 20

20 – 25

25 – 30

Time f mid x d = x –17.5 d2 fd fd2

Time (sec) 5-10 10-15 15-20 20-25 25-30

No of people 4 8 15 12 11

4

8

15

12

11

50

7.5

12.5

17.5

22.5

27.5

-10

-5

0

5

10

100

25

0

25

100

-40

-40

0

60

110

90

400

200

0

300

1100

2000

Variance

22

f

fd

f

fd

2

50

90

50

2000

24.340)8.1(40 2

76.36

76.36. DS36.7666

367600120

6 7236374

.06

06.6

Show that the points A(2 , 3), B(4 , 0) and C(6, -3) are collinear.

Area of the ABC =

½ {(x1y2 + x2 y3 + x3y1) – (x1y3 + x3y2+ x2 y1)}sq. units

2 4 6

2

3 0 -3

3

= ½{(2)(0) + (4)(–3) + (6)(3)} – {(2)(–3) + (6)(0) + (4)(3)}

= ½ {(0 – 12 + 18) – (–6 + 0 + 12)}

= ½ {(18 – 12) – (12 – 6)}

= ½ (6 – 6)

= 0

The given points are collinear.

find the value of k for which the given points are collinear. (2, - 5),(3, - 4) and (9, k)

If the points are collinear, Area of the ABC = 0

{(x1y2 + x2 y3 + x3y1) – (x1y3 + x3y2+ x2 y1)} = 0

2 3 9 2

-5 -4 k -5

{(2)(-4) + (3)(k) + (9)(-5)} – {(2)(k) + (9)(-4) + (3)(-5)} = 0

{(–8 + 3k – 45) – (2k – 36 – 15)} = 0

{(3k – 53) – (2k – 51)} = 0

3k – 2k – 53 + 51 = 0

k – 2 = 0

k = 2

Find the area of the quadrilateral whose vertices are(6, 9), (7, 4), (4,2) and (3,7)

Area of the quadrilateral ABCD

= ½ {(x1y2 + x2 y3 + x3y4 + x4y1)

– (x2y1 + x3 y2 + x4y3 + x1y4)}

4 7 6 3 4

2 4 9 7 2

= ½ {(4)(4) + (7)(9) + (6)(7) + (3)(2)} –

{(2)(7) + (4)(6) + (9)(3) + (7)(4)}

= ½ {(16 + 63 + 42 + 6) – (14+ 24 + 27 + 28)}

A(4,2)

D(3,7)

B(7,4)

C(6,9)

= ½ {127 – 93}

= ½ (34)

= 17 sq. units

Find the area of the quadrilateral whose vertices are(-3, 4), (-5,- 6), (4,- 1) and (1, 2)

Area of the quadrilateral ABCD =

= ½ {(x1y2 + x2 y3 + x3y4 + x4y1)

– (x2y1 + x3 y2 + x4y3 + x1y4)}

-5 4 1 -3

-5

-6 -1 2 4

-6

= ½ {(-5)(-1) + (4)(2) + (1)(4) + (-3)(-6)} –

{(-5)(4) + (-3)(2) + (1)(-1) + (4)(-6)}

= ½ {(5 + 8 + 4 + 18) – (-20 – 6 – 1 – 24)}

A(-5,-6)

D(-3,4)

B(4,-1)

C(1,2)

= ½ {35 – (–51)}

= ½ (35 + 51)

= ½ (86)

= 43 sq. units

Find the area of the quadrilateral whose vertices are(-4, 5), (0, 7), (5,- 5) and (-4,- 2)

Area of the quadrilateral ABCD =

= ½ {(x1y2 + x2 y3 + x3y4 + x4y1)

– (x2y1 + x3 y2 + x4y3 + x1y4)}

-4

-2

5

-5

0

7

-4

5

-4

-2

= ½ {(-4)(-5) + (5)(7) + (0)(5) + (-4)(-2)} –

{(-4)(5) + (-4)(7) + (0)(-5) + (5)(-2)}

= ½ {(20 + 35 + 0 + 8) – (-20 – 28 – 0 – 10)}

= ½ {63 – (–58)}

= ½ (63 + 58)

= ½ (121)

= 60.5 sq. units

A(-4,-2)

D(-4,5)

B(5,-5)

C(0,7)

A coin is tossed three times. Find the probability of getting (i) head and tail alternatively (ii) at least two heads (iii) exactly two heads (iv) no heads

Tossing three coins once is as same as a coin is tossed three times

Sample space S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

n(S) = 8

(i) Let the event of getting head and tail alternatively be A

A = {HTH, THT} n(A) = 2

The probability of getting head and tail alternatively is

4

1

8

2

n(S)

n(A) P(A)

(ii) Let the event of getting at least two heads be B

B = {HHH, HHT, HTH, THH} n(B) = 4

The probability of getting at least two heads is 2

1

8

4

n(S)

n(B) P(B)

(iii) Let the event of getting exactly two heads be C

C = {HHT, HTH, THH} n(C) = 3

The probability of getting exactly two heads is

8

3

n(S)

n(C) P(C)

(iv) Let the event of getting no head be D

D = {TTT} n(D) = 3

The probability of getting no head is

8

1

n(S)

n(D) P(D)

A card is drawn from a pack of 52 cards. Find the probability that it is either red card or king card.

Total number of cards = 52

n(S) = 52

Let A be the event drawing a red card

n(A) = 26

The probability of drawing a red card is

2

1

52

26

n(S)

n(A) P(A)

Let B be the event drawing a king card

n(B) = 4

The probability of drawing a king card is

13

1

52

4

n(S)

n(B) P(B)

n(A B) = 2

26

1

52

2

n(S)

B)n(A B)P(A

P(AB) = P(A) + P(B) – P(A B)

52

2

52

4

52

26

52

2-426

13

7

52

28

Two dice are rolled once. Find the probability of getting an even number on the second die or the total of face numbers 10.

When two dice are rolled the sample space

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

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} n(S) = 36

Let the event of getting even in the second die be A

A = {(1, 2), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6) (3, 2), (3, 4), (3, 6), (4, 2), (4, 4), (4, 6) (5, 2), (5, 4), (5, 6), (6, 2), (6, 4), (6, 6)}

n(A) = 18

2

1

36

18

n(S)

n(A) P(A)

Let the event of getting the total of the face numbers is 10 be B

B = {(4, 6), (5, 5), (6, 4)} n(B) = 3

12

1

36

3

n(S)

n(B) P(B)

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

18

1

36

2

n(S)

B)n(A B)P(A

P(AB) = P(A) + P(B) – P(A B)

36

2

36

3

36

18

36

2-318

36

19

16. Two persons X and Y appeared in an interview for two vacancies in an office. The chance for X’s selection is 1/5 and the chance for Y’s selection is 1/7. Find the chance that (i) both of them are selected (ii) only one of them is selected, (iii) none of them is selected.

The probability of X’S selection = P(X) = 1/5

The probability of Y’S selection = P(Y) = 1/7

5

4

5

11 )XP(

7

6

7

11 )YP(

(i) The probability of both of them to be selected

35

1

7

1

5

1P(Y)P(X) Y)P(X

(ii) The probability of only one of them to be selected

P(Y))XP( )YP(P(X) Y)XP()YP(X

7

2

35

10

35

4

35

6

7

1

5

4

7

6

5

1

35

24

7

6

5

4)YP()XP( )YXP(

(iii) The probability of none of them to be selected