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#### Question O1:

Mention the numerals used in base-2 system.

The numerals used in base-2 system are 0 and 1.

#### Question W1:

Fill in the blanks with suitable answers:

(1) The total number of numerals used in base 10 system is ………

(2) Total numerals used in base-2 system is ……..

(3) The biggest numeral used is base 10 system is ……

(1) The total number of numerals used in base 10 system is 10.

Reason: The numerals used in base 10 system are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.

(2)Total numerals used in base-2 system is 2.

Reason: The numerals used in base 2 system are 0 and 1.

(3)The biggest numeral used in base 10 system is 9.

(4)The biggest numeral used in base 2 system is 1.

#### Question O2:

Which is the greatest numeral used in base-2 system.

The greatest numeral used in base-2 system is 1.

#### Question W2:

Observe the following figures and write the same in base 2 system by grouping.

Example:

(2) The number in base 2 systems is 1010(2)

(3) The number in base 2 systems is 1100(2)

(4) The number in base 2 systems is 1111(2)

#### Question W3:

Convert the base 10 numbers into base 2 by division method.

(1) 2

(2) 6

(3) 18

(4) 22

(5) 30

(6) 44

(7) 51

(8) 63

(9) 77

(10) 100

(1) 2

 2 2 Remainder 1 0

2(10) = 10(2)

(2) 6

 2 6 Remainder 2 3 0 1 1

6(10) = 110(2)

(3) 18

 2 18 Remainder 2 9 0 2 4 1 2 2 0 1 0

18(10) = 10010(2)

(4) 22

 2 22 Remainder 2 11 0 2 5 1 2 2 1 1 0

22(10) = 10110(2)

(5) 30

 2 30 Remainder 2 15 0 2 7 1 2 3 1 1 1

30(10) = 11110(2)

6 (44)

 2 44 Remainder 2 22 0 2 11 0 2 5 1 2 2 1 1 0

44(10) = 101100(2)

(7) 51

 2 51 Remainder 2 25 1 2 12 1 2 6 0 2 3 0 1 1

51(10) = 110011 (2)

(8)63

 2 63 Remainder 2 31 1 2 15 1 2 7 1 2 3 1 1 1

63(10) = 111111 (2)

(9) 77

 2 77 Remainder 2 38 1 2 19 0 2 9 1 2 4 1 2 2 0 1 0

77(10) = 1001101 (2)

(10) 100

 2 100 Remainder 2 50 0 2 25 0 2 12 1 2 6 0 2 3 0 1 1

100(10) = 1100100 (2)

#### Question W5:

(a) Which of the following is greater? 10(2) or 11(2)

(b) Write in ascending order: 1011(2), 110(2), 1000(2)

(c) Find the value of: (1 × 24) + (0 × 23) + (1 × 22) + (0 × 21) + (0 × 20)

(d) Expand and write: (1) 10101(2) (2) 11110(2)

(a) 10(2)

 21 20 1 0

10(2) = 1 × 21 + 0 × 20 = 1 × 2 + 0 × 1 = 2 + 0 = 2

11(2)

 21 20 1 1

11(2) = 1 × 21 + 1 × 20 = 1 × 2 + 1 ×1 = 2 + 1 = 3

11(2) > 10(2)

(b) 110(2)

 22 21 20 1 1 0

110(2) =1 × 22 + 1 × 21 + 0 × 20 = 1× 4 + 1 × 2 + 0 ×1 = 4 + 2 + 1 = 7

1000(2)

 23 22 21 20 1 0 0 0

1000(2) = 1 × 23+ 0 × 22 + 0 × 21 + 0 × 20 = 1 × 8 + 0 × 4 + 0 × 2 + 0 × 1 = 8 + 0 + 0 + 0 = 8

1011(2)

 23 22 21 20 1 0 1 1

1011(2) =1 × 23+ 0 × 22 +1 × 21 + 1 × 20

= 1 × 8 + 0 × 4 + 1 × 2 + 1 ×1

= 8 + 0 + 2 + 1

= 11

Hence, the ascending order of the given numbers is 110(2), 1000(2), 1011(2).

(c) (1 × 24) + (0 × 23) + (1× 22) + (0 × 21) + (0 × 20)

= (1 × 16) + (0 × 8) + (1 × 4) + (0 × 2) + (0 × 1)

= 16 + 0 + 4 + 0 + 0

= 16 + 4 = 20

(d) (1) 10101(2)

 24 23 22 21 20 1 0 1 0 1

10101(2) = 1 × 24 + 0 × 23+ 1 × 22 + 0 × 21 + 1 × 20

(2) 11110(2)

 24 23 22 21 20 1 1 1 1 0

11110(2) = 1 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20

#### Question W4:

Complete the cross word puzzle.

(1) Left to Right

Example:

(a) 5 in base 2 (3 sq)

(b) 7 in base 2 (3)

(d) 12 in base 2 (4)

(e) 8 in base 2 (4)

(f) 4 in base 2 (3)

(i) 13 in base 2 (4)

(2) Down

(a) 15 in base 2 (4 sq)

(c) 10 in base 2 (4)

(e) 14 in base 2 (4)

(f) 2 in base 2 (2)

(g) 3 in vase 2 (2)

(h) 11 in base 2 (4)

(i) 6 in base 2 (3)

 a1 0 g1 b e d e h i f

(1) 1(2) = 1 × 20 = 1 × 1 = 1

1(2) = 1

(2) 11(2)

 21 20 1 1

11(2) = 1 × 21 + 1 × 20 = 1 × 2 + 1 × 1 = 2 + 1 = 3

11(2) = 3

(3) 101(2)

 22 21 20 1 0 1

Therefore,

101(2) =1×22 +0×21 + 1×20

= 1 × 4 + 0 × 2 + 1 ×1

= 4 + 0 + 1

= 5

101(2) = 5

(4) 100(2)

 22 21 20 1 0 0

100(2) =1×22 +0×21 + 0×20 = 1 × 4 + 0 × 2 + 0 ×1 = 4 + 0 + 0 = 4

100(2) = 4

(5) 111(2)

 22 21 20 1 1 1

111(2) = 1 × 22 + 1 × 21 + 1 × 20 = 1 × 4 + 1 × 2 + 1 ×1 = 4 + 2 + 1 = 7

111(2) = 7

(6) 1001(2)

 23 22 21 20 1 0 0 1

1001(2) =1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 1 × 8 + 0 × 4 + 0 × 2 + 1 × 1 = 8 + 0 + 0 + 1 = 9

1001(2) = 9

(7) 1110(2)

 23 22 21 20 1 1 1 0

1110(2) =1 × 23 + 1 × 22 + 1 × 21 + 0 × 20

= 1 × 8 + 1 × 4 + 1 × 2 + 0 × 1

= 8 + 4 + 2 + 0

= 14

1110(2) = 14

(8) 1010(2)

 23 22 21 20 1 0 1 0

1010(2) = 1 × 23+ 0 × 22 + 1 × 21 + 0 × 20

= 1 × 8 + 0 × 4 + 1 × 2 + 0 ×1

= 8 + 0 + 2 + 0

= 10

1010(2) = 10

(9) 1111(2)

 23 22 21 20 1 1 1 1

1111(2) = 1× 23+ 1 × 22 +1 × 21 + 1 × 20

= 1 × 8 + 1 × 4 + 1 × 2 + 1 ×1

= 8 + 4 + 2 + 1

= 15

1111(2) = 15

(10) 11001(2)

 24 23 22 21 20 1 1 0 0 1

11001(2) = 1× 24 + 1 × 23+ 0 × 22 + 0 × 21 + 1 × 20

= 1 × 16 + 1 × 8 + 0 × 4 + 0 × 2 + 1 × 1

= 16 + 8 + 0 + 0 + 1

= 25

11001(2) = 25

(1)

(b) 7 = (1 × 22) + (1 × 21) + (1 × 20) = 111(2)

(d) 12 = (1 × 23) + (1 × 22) + (0 × 21) + (0 × 20) = 1100(2)

(e) 8 = (1 × 23) + (0 × 22) + (0 × 21) + (0 × 20) = 1000(2)

(f) 4 = (1 × 22) + (0 × 21) + (0 × 20) = 100(2)

(i) 13 = (1 × 23) + (1 × 22) + (0 × 21) + (1 × 20) = 1101(2)

(2)

(a) 15 = (1 × 23) + (1 × 22) + (1 × 21) + (1 × 20) = 1111(2)

(c) 10 = (1 × 23) + (0 × 22) + (1 × 21) + (0 × 20) = 1010(2)

(e) 14 = (1 × 23) + (1 × 22) + (1 × 21) + (0 × 20) = 1110(2)

(f) 2 = (1 × 21) + (0 × 20) = 10(2)

(g) 3 = (1 × 21) + (1 × 20) = 11(2)

(h) 11 = (1 × 23) + (0 × 22) + (1 × 21) + (1 × 20) = 1011(2)

(i) 6 = 110(2)

Therefore, the solution of the puzzle is

 a 1 0 g 1 b 1 1 c 1 1 1 d 1 1 0 0 1 1 1 e 1 0 0 0 0 1 h 1 1 i 1 1 0 1 f 1 0 0 1 1 0 0 1

#### Question O1:

Mention the numerals used in base-5 system.

The numerals used in base-5 system are 0, 1, 2, 3, and 4.

#### Question W1:

Fill in the blanks with suitable answers.

(1) The biggest numerals used in the base five system is ……..

(2) 17 has ……. group of 5 and …….. units.

(3) 2 × 52 + 1 × 51 + 0 × 50 This expression is in ……. system.

(4) Expansion of 432(5) is ……..

(5) 28 in base 5 system is …….

(1) The biggest numeral used in base five systems is 4.

(2) 17 has 3 groups of 5 and 2 units.

(3) 2 × 52 + 1 × 51 + 0 × 50. This expression is in base-5 system.

(4) Expansion of 432(5) is 4 × 52 + 3 × 51 + 2 × 50.

Explanation:

 52 51 50 4 3 2

432(5) = 4 × 52 + 3 × 51 + 2 × 50

(5) 28 in base 5 system is 103(5).

Explanation:

#### Question O2:

Convert 5 into base-5 system.

 5 = 10(5) [5 = 1 × 51 + 0 × 50 ]

#### Question W2:

Convert the following into base 5 system.

(1) 12

(2) 15

(3) 29

(4) 49

(5) 60

(6) 84

(7) 101

(8) 150

(9) 285

(10) 400

#### Question W3:

Expand the following in base 5 system.

(1) 32(5)

(2) 124(5)

(3) 432(5)

(4) 3021(5)

(1) 32(5)

 51 50 3 2

32(5) = 3 × 51 + 2 × 50

(2) 124(5)

 52 51 50 1 2 4

124(5) = 1 × 52 + 2 × 51 + 4 × 50

(3) 432(5)

 52 51 50 4 3 2

432(5) = 4 × 52 + 3 × 51 + 2 × 50

(4) 3021(5)

 54 52 51 50 3 0 2 1

3021(5) = 3 × 54 + 0 × 52 + 2 × 51 + 1 × 50

#### Question W4:

Convert the following base 5 numbers into base 10.

(1) 20(5)

(2) 44(5)

(3) 304(5)

(4) 403(5)

(5) 1042(5)

(6) 2010(5)

(7) 2000(5)

(8) 2341(5)

(1) 20(5)

 51 50 2 0

20(5) = 2 × 51 + 0 × 50 = 2 × 5 + 0 × 1 = 10 + 0 = 10

(2) 44(5)

 51 50 4 4

44(5) = 4 × 51 + 4 × 50 = 4 × 5 + 4× 1= 20 + 4 = 24

(3) 304(5)

 52 51 50 3 0 4

304(5) = 3 × 52 + 0 × 51 + 4 × 50 = 3 × 25 + 0 × 5 + 4 × 1 = 75 + 0+ 4 = 79

(4) 403(5)

 52 51 50 4 0 3

403(5) = 4 × 52 + 0 × 51 + 3 × 50 = 4 × 25 + 0 × 5 + 3 × 1 = 100 + 0 + 3 = 103

(5) 1042(5)

 53 52 51 50 1 0 4 2

1042(5) = 1 ×53 + 0 × 52 + 4 × 51 + 2 × 50

= 1 × 125 + 0 × 25 + 4 × 5 + 2 × 1

= 125 + 0 + 20 + 2

= 147

(6) 2010(5)

 53 52 51 50 2 0 1 0

2010(5) = 2 × 53 + 0 × 52 + 1 × 51 + 0 × 50

= 2 × 125 + 0 × 25 + 1 × 5 + 0 × 1

= 250 + 0+ 5 + 0

= 255

(7) 2000(5)

 53 52 51 50 2 0 0 0

2000(5) = 2 × 53 + 0 × 52 + 0 × 51 + 0 × 50

= 2 × 125 + 0 × 25 + 0 × 5 + 0 × 1

= 250 + 0+ 0 + 0

= 250

(8) 2341(5)

 53 52 51 50 2 3 4 1

2341(5) = 2 × 53 + 3 × 52 + 4 × 51 + 1 × 50

= 2 × 125 + 3 × 25 + 4 × 5 + 1 × 1

= 250 + 75 + 20 + 1

= 346

#### Question W5:

Convert the following numbers into base 5 and verify:

(1) 52

(2) 70

(3) 95

(4) 120

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Verification: 202(5)

 52 51 50 2 0 2

202(5) = 2 × 52 + 0 × 51 + 2 × 50 = 52

Verification: 240(5)

 52 51 50 2 4 0

240(5) = 2 × 52 + 4 × 51 + 0 × 50 = 70

Verification: 340(5)

 52 51 50 3 4 0

340(5) = 3×52 +4×51 + 0×50 = 95

Verification: 440(5)

 52 51 50 4 4 0

440(5) = 4 × 52 + 4 × 51 + 0 × 50 = 120

#### Question W6:

Fill the number puzzle.

Left to Right

(a) 5 as base 5 (2sq)

(b) 31 in base 5 (3)

(e) 38 in base 5 (3)

(f) 15 in base 5 (2)

(g) 106 in base 5 (3)

(i) 7 in base 5 (3)

(k) 9 in base 5 (2)

Down

(a) 26 in base 5 (3 sq)

(c) 25 in base 5 (3)

(d) 69 in base 5 (3)

(f) 16 in base 5 (2)

(h) 44 in base 5 (3)

(i) 8 in base 5 (2)

(j) 24 in base 5 (2)

 a1 0 b c d e f g h i j k

Left to Right

(b) 31 = 1 × 52 + 1 × 51+ 1 ×50 = 111(5)

(e) 38 =1 × 52 + 2 × 51+ 3 ×50 = 123(5)

(f) 15 = 3 × 51+ 1 ×50 = 30(5)

(g) 106 = 4 × 52 + 1 × 51 + 1 × 50 = 411(5)

(i) 7 = 1 × 51+ 2 × 50 = 12(5)

(k) 9 = 1 × 51+ 4 × 50 = 14(5)

Down

(a) 26 = 1 × 52 + 0 × 51 + 1 × 50 = 101(5)

(c) 25 = 1 × 52 + 0 × 51 + 0 × 50 = 100(5)

(d) 69 = 2 × 52 + 3 × 51 + 4 × 50 = 234(5)

(f) 16 = 3 × 51 + 1 × 50 = 31(5)

(h) 44 = 1 × 52 + 3 × 51+ 4 × 50 = 134(5)

(i) 8 = 1 × 51 + 3 ×50 = 13(5)

(j) 24 = 4 × 51 + 4 × 50 = 44(5)

Therefore, the solution of the puzzle is

 a 1 0 b 1 1 c 1 0 d 2 0 e 1 2 3 f 3 0 g 4 h 1 1 i 1 2 3 j 4 3 k 1 4 4

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