Fill in the blanks

Question 1

The binary system consists of two digits 0 and 1.

Question 2

A decimal number system uses the digits from 0 to 9.

Question 3

The base in a decimal number system is 10.

Question 4

A binary number system is written with 2 as the base.

Question 5

In a decimal to binary conversion, the first remainder is known as Least Significant Bit (LSB) and the last remainder is Most Significant Bit (MSB).

Question 6

2⁰ = 1

Question 7

Dr. John Von Neuman developed the concept of Binary Number system.

Question 8

The number represented with base eight (8) is known as Octal Number.

Question 9

The binary equivalent of hexa-decimal digit E is 1110.

Question 10

A number system that uses alphanumeric digits, is known as Hexadecimal Number System.

Complete the table

Tick the appropriate answer

Question 1

Binary multiplication is possible with:

1. 0 and 1
2. 0 and 0
3. 1 and 1
4. All of the above ✓

Question 2

The base of an octal number is represented by:

1. 2
2. 8 ✓
3. 7
4. None

Question 3

To convert an Octal number to its binary equivalent, each Octal digit is expressed into:

1. 3 bits form ✓
2. 4 bits form
3. 8 bits form
4. All of the above

Question 4

Sixteen raised to the power zero (16⁰) is equivalent to:

1. 0
2. 1 ✓
3. 0 and 1
4. None

Question 5

An octal number system uses the digits from:

1. 0 to 8
2. 1 to 8
3. 0 to 7 ✓
4. All of the above

Question 6

The base of a Hexadecimal number is represented by:

1. H16
2. 16 ✓
3. 15
4. None

Question 7

In a Hexadecimal number system 'B' represents the digit:

1. 11 ✓
2. 12
3. 14
4. 13

Question 8

To express a Hexadecimal number to its Binary equivalent, each Hexadecimal digit is expressed into:

1. 2 bits form
2. 3 bits form
3. 4 bits form ✓
4. None

Question 9

The binary equivalent of a Hexadecimal digit 12 (C) is represented by:

1. 1010
2. 1011
3. 1101
4. 1100 ✓

Question 10

The Hexadecimal equivalent digit of 1011( 4 bits form) is:

1. E
2. F
3. B ✓
4. C

Convert the following to their binary equivalents

Question 1

(784)₁₀

Answer

Therefore, (784)₁₀ = (1100010000)₂

Question 2

(999)₁₀

Answer

Therefore, (999)₁₀ = (1111100111)₂

Question 3

(1401)₁₀

Answer

Therefore, (1401)₁₀ = (10101111001)₂

Question 4

(1423)₁₀

Answer

Therefore, (1423)₁₀ = (10110001111)₂

Question 5

(478.75)₁₀

Answer

Converting integral part

Converting fractional part

Therefore, (478.75)₁₀ = (111011110.11)₂

Question 6

(165.35)₁₀

Answer

Converting integral part

Converting fractional part

Therefore, (165.35)₁₀ = (10100101.0101)₂

Question 7

(277.27)₁₀

Answer

Converting integral part

Converting fractional part

Therefore, (277.27)₁₀ = (100010101.0100)₂

Question 8

(322.2)₁₀

Answer

Converting integral part

Converting fractional part

Therefore, (322.2)₁₀ = (101000010.0011)₂

Perform the following

Question 1

(1010110)₂ to ( )₁₀

Answer

Equivalent decimal number = 2 + 4 + 16 + 64 = 86

Therefore, (1010110)₂ = (86)₁₀

Question 2

(1000011)₂ to ( )₁₀

Answer

Equivalent decimal number = 1 + 2 + 64 = 67

Therefore, (1000011)₂ = (67)₁₀

Question 3

(1100111)₂ to ( )₁₀

Answer

Equivalent decimal number = 1 + 2 + 4 + 32 + 64 = 103

Therefore, (1100111)₂ = (103)₁₀

Question 4

(10101011)₂ to ( )₁₀

Answer

Equivalent decimal number = 1 + 2 + 8 + 32 + 128 = 171

Therefore, (10101011)₂ = (171)₁₀

Question 5

(1010.001)₂ to ( )₁₀

Answer

Integral part

Fractional part

Equivalent decimal number = 2 + 8 + 0.125 = 10.125

Therefore, (1010.001)₂ = (10.125)₁₀

Question 6

(10100.11)₂ to ( )₁₀

Answer

Integral part

Fractional part

Equivalent decimal number = 4 + 16 + 0.5 + 0.25 = 20.75

Therefore, (10100.11)₂ = (20.75)₁₀

Question 7

(1010.111)₂ to ( )₁₀

Answer

Integral part

Fractional part

Equivalent decimal number = 2 + 8 + 0.5 + 0.25 + 0.125 = 10.875

Therefore, (1010.111)₂ = (10.875)₁₀

Question 8

(1010.0111)₂ to ( )₁₀

Answer

Integral part

Fractional part

Equivalent decimal number = 2 + 8 + 0.25 + 0.125 + 0.0625 = 10.4375

Therefore, (1010.0111)₂ = (10.4375)₁₀

Add the following binary numbers

Question 1

(1101111)₂ + (1011011)₂

Answer

$\begin{matrix} & & \overset{1}{1} & \overset{1}{1} & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & \overset{1}{1} & 1 \\ + & & 1 & 0 & 1 & 1 & 0 & 1 & 1 \\ \hline & \bold{1} & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{0} & \bold{1} & \bold{0} \end{matrix}$

Therefore, (1101111)₂ + (1011011)₂ = (11001010)₂

Question 2

(1111011)₂ + (1011101)₂

Answer

$\begin{matrix} & & \overset{1}{1} & \overset{1}{1} & \overset{1}{1} & \overset{1}{1} & \overset{1}{0} & \overset{1}{1} & 1 \\ + & & 1 & 0 & 1 & 1 & 1 & 0 & 1 \\ \hline & \bold{1} & \bold{1} & \bold{0} & \bold{1} & \bold{1} & \bold{0} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (1111011)₂ + (1011101)₂ = (11011000)₂

Question 3

(10111)₂ + (110111)₂

Answer

$\begin{matrix} & & \overset{1}{0} & 1 & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & 1 \\ + & & 1 & 1 & 0 & 1 & 1 & 1 \\ \hline & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{1} & \bold{1} & \bold{0} \end{matrix}$

Therefore, (10111)₂ + (110111)₂ = (1001110)₂

Question 4

(1101101)₂ + (1110010)₂

Answer

$\begin{matrix} & & \overset{1}{1} & 1 & 0 & 1 & 1 & 0 & 1 \\ + & & 1 & 1 & 1 & 0 & 0 & 1 & 0 \\ \hline & \bold{1} & \bold{1} & \bold{0} & \bold{1} & \bold{1} & \bold{1} & \bold{1} & \bold{1} \end{matrix}$

Therefore, (1101101)₂ + (1110010)₂ = (11011111)₂

Question 5

(10111101)₂ + (1101101)₂

Answer

$\begin{matrix} & & \overset{1}{1} & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & \overset{1}{1} & 1 & \overset{1}{0} & 1 \\ + & & & 1 & 1 & 0 & 1 & 1 & 0 & 1 \\ \hline & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{0} & \bold{1} & \bold{0} & \bold{1} & \bold{0} \end{matrix}$

Therefore, (10111101)₂ + (1101101)₂ = (100101010)₂

Subtract the following

Question 1

(110101)₂ - (011010)₂

Answer

$\begin{matrix} & \overset{0}{\cancel{1}} & \overset{0}{\cancel{1}} & 0 & \overset{0}{\cancel{1}} & 0 & 1 \\ - & 0 & 1 & 1 & 0 & 1 & 0 \\ \hline & \bold{0} & \bold{1} & \bold{1} & \bold{0} & \bold{1} & \bold{1} \end{matrix}$

Therefore, (110101)₂ - (011010)₂ = (11011)₂

Question 2

(110110)₂ - (01010)₂

Answer

$\begin{matrix} & 1 & \overset{0}{\cancel{1}} & 0 & 1 & 1 & 0 \\ - & & 0 & 1 & 0 & 1 & 0 \\ \hline & \bold{1} & \bold{0} & \bold{1} & \bold{1} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (110110)₂ - (01010)₂ = (101100)₂

Question 3

(101011)₂ - (11101)₂

Answer

$\begin{matrix} & \overset{0}{\cancel{1}} & \overset{1}{\cancel{0}} & \overset{0}{\cancel{1}} & 0 & 1 & 1 \\ - & & 1 & 1 & 1 & 0 & 1 \\ \hline & \bold{0} & \bold{0} & \bold{1} & \bold{1} & \bold{1} & \bold{0} \end{matrix}$

Therefore, (101011)₂ - (11101)₂ = (1110)₂

Question 4

(11111)₂ - (01101)₂

Answer

$\begin{matrix} & 1 & 1 & 1 & 1 & 1 \\ - & 0 & 1 & 1 & 0 & 1 \\ \hline & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{0} \end{matrix}$

Therefore, (11111)₂ - (01101)₂ = (10010)₂

Question 5

(10000)₂ - (01111)₂

Answer

$\begin{matrix} & \overset{0}{\cancel{1}} & \overset{1}{\cancel{0}} & \overset{1}{\cancel{0}} & \overset{1}{\cancel{0}} & 0 \\ - & 0 & 1 & 1 & 1 & 1 \\ \hline & \bold{0} & \bold{0} & \bold{0} & \bold{0} & \bold{1} \end{matrix}$

Therefore, (10000)₂ - (01111)₂ = (1)₂

Perform the following arithmetic

Question 1

Add octal numbers (1546)₈ and (641)₈

Answer

$\begin{matrix} & \overset{1}{1} & \overset{1}{5} & 4 & 6 \\ + & & 6 & 4 & 1 \\ \hline & \bold{2} & \bold{4} & \bold{0} & \bold{7} \end{matrix}$

Therefore, (1546)₈ + (641)₈ = (2407)₈

Explanation

1. 6 + 1 = 7.
2. 4 + 4 = 8, as sum is equal to 8 so we will do 8 - 8 = 0. Hence, sum = 0, carry over = 1.
3. 1 + 5 + 6 = 12, as sum is greater than 8 so we will do 12 - 8 = 4. Hence, sum = 4, carry over = 1.
4. 1 + 1 = 2.

Question 2

Subtract octal number (2546)₈ from (5142)₈

Answer

$\begin{matrix} & \overset{4}{\cancel{5}} & \overset{0}{\cancel{1}} & \overset{3}{\cancel{4}} & 2 \\ - & 2 & 5 & 4 & 6 \\ \hline & \bold{2} & \bold{3} & \bold{7} & \bold{4} \end{matrix}$

Therefore, (5142)₈ - (2546)₈ = (2374)₈

Explanation

1. 2 - 6, here 6 is greater than 2 so we borrow 1 from next significant digit 4 and it becomes 3. Now 2 becomes 2 + 8 = 10 and 10 - 6 = 4.
2. 3 - 4, again 4 is greater than 3 so we borrow 1 from next significant digit 1 and it becomes 0. Current digit 3 becomes 8 + 3 = 11. Lastly, 11 - 4 = 7.
3. 0 - 5, here 5 is greater than 0. Borrowing 1 from next significant digit 5 makes it 4. Current digit 0 becomes 0 + 8 = 8. Lastly, 8 - 5 = 3.
4. 4 - 2 = 2.

Question 3

Add hexa-decimal numbers (1F5C)₁₆ and (AC2B)₁₆

Answer

$\begin{matrix} & \overset{1}{1} & F & \overset{1}{5} & C \\ + & A & C & 2 & B \\ \hline & \bold{C} & \bold{B} & \bold{8} & \bold{7} \end{matrix}$

Therefore, (1F5C)₁₆ + (AC2B)₁₆ = (CB87)₁₆

Explanation

1. C + B = 12 + 11 = 23. As 23 is greater than 16 so 23 - 16 = 7. Hence, sum = 7 and carry over = 1.
2. 1 + 5 + 2 = 8.
3. F + C = 15 + 12 = 27. As 27 is greater than 16 so 27 - 16 = 11 or B. Hence, sum = B and carry over = 1.
4. 1 + 1 + A = 1 + 1 + 10 = 12 or C.

Question 4

Subtract hexa-decimal number (A68D)₁₆ from (D53F)₁₆

Answer

$\begin{matrix} & \overset{C}{\cancel{D}} & \overset{4}{\cancel{5}} & 3 & F \\ - & A & 6 & 8 & D \\ \hline & \bold{2} & \bold{E} & \bold{B} & \bold{2} \end{matrix}$

Therefore, (D53F)₁₆ - (A68D)₁₆ = (2EB2)₁₆

Explanation

1. F - D = 15 - 13 = 2.
2. 3 - 8, here 8 is greater than 3 so we borrow 1 from next significant digit 5 and it becomes 4. Now 3 becomes 3 + 16 = 19 and 19 - 8 = 11 or B.
3. 4 - 6, here again 6 is greater than 4 so we borrow 1 from next significant digit D and it becomes C. Now 4 becomes 4 + 16 = 20 and 20 - 6 = 14 or E.
4. C - A = 12 - 10 = 2.

Question 5

Subtract hexa-decimal number (ABCD)₁₆ from (EA9C)₁₆

Answer

$\begin{matrix} & \overset{D}{\cancel{E}} & \overset{9}{\cancel{A}} & \overset{8}{\cancel{9}} & C \\ - & A & B & C & D \\ \hline & \bold{3} & \bold{E} & \bold{C} & \bold{F} \end{matrix}$

Therefore, (EA9C)₁₆ - (ABCD)₁₆ = (3ECF)₁₆

Explanation

1. C - D = 12 - 13, as 13 is greater than 12 so we borrow 1 from next significant digit 9 and it becomes 8. Now 12 becomes 12 + 16 = 28 and 28 - 13 = 15 or F.
2. 8 - C = 8 - 12, as 12 is greater than 8 so we borrow 1 from next significant digit A and it becomes 9. Now 8 becomes 8 + 16 = 24 and 24 - 12 = 12 or C.
3. 9 - B = 9 - 11, as 11 is greater than 9 so we borrow 1 from next significant digit E and it becomes D. Now 9 becomes 9 + 16 = 25 and 25 - 11 = 14 or E.
4. D - A = 13 - 10 = 3.

Perform the following multiplication

Question 1

(10110)₂ x (110)₂

Answer

$\begin{matrix} & & & 1 & 0 & 1 & 1 & 0 \\ \times & & & & & 1 & 1 & 0 \\ \hline & & & 0 & 0 & 0 & 0 & 0 \\ & & 1 & 0 & 1 & 1 & 0 & * \\ & 1 & 0 & 1 & 1 & 0 & * & * \\ \hline \bold{1} & \bold{0} & \bold{0} & \bold{0} & \bold{0} & \bold{1} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (10110)₂ x (110)₂ = (10000100)₂

Question 2

(10101)₂ x (100)₂

Answer

$\begin{matrix} & & 1 & 0 & 1 & 0 & 1 \\ \times & & & & 1 & 0 & 0 \\ \hline & & 0 & 0 & 0 & 0 & 0 \\ & 0 & 0 & 0 & 0 & 0 & * \\ 1 & 0 & 1 & 0 & 1 & * & * \\ \hline \bold{1} & \bold{0} & \bold{1} & \bold{0} & \bold{1} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (10101)₂ x (100)₂ = (1010100)₂

Question 3

(11000)₂ x (101)₂

Answer

$\begin{matrix} & & 1 & 1 & 0 & 0 & 0 \\ \times & & & & 1 & 0 & 1 \\ \hline & & 1 & 1 & 0 & 0 & 0 \\ & 0 & 0 & 0 & 0 & 0 & * \\ 1 & 1 & 0 & 0 & 0 & * & * \\ \hline \bold{1} & \bold{1} & \bold{1} & \bold{1} & \bold{0} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (11000)₂ x (101)₂ = (1111000)₂

Question 4

(10.001)₂ x (0.11)₂

Answer

$\begin{matrix} & 1 & 0 & . & 0 & 0 & 1 \\ \times & & & 0 & . & 1 & 1 \\ \hline & & 1 & 0 & 0 & 0 & 1 \\ & 1 & 0 & 0 & 0 & 1 & * \\ \hline & \bold{1\medspace.} & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{1} \end{matrix}$

Therefore, (10.001)₂ x (0.11)₂ = (1.10011)₂

Question 5

(101.10)₂ x (1.10)₂

Answer

$\begin{matrix} & & 1 & 0 & 1 & . & 1 \\ \times & & & & 1 & . & 1 \\ \hline & & & 1 & 0 & 1 & 1 \\ & & 1 & 0 & 1 & 1 & * \\ \hline & \bold{1} & \bold{0} & \bold{0} & \bold{0\medspace.} & \bold{0} & \bold{1} \end{matrix}$

Therefore, (101.10)₂ x (1.10)₂ = (1000.0100)₂

Question 6

(10.101)₂ x (10.1)₂

Answer

$\begin{matrix} & & & 1 & 0 & . & 1 & 0 & 1 \\ \times & & & & & 1 & 0 & . & 1 \\ \hline & & & & 1 & 0 & 1 & 0 & 1 \\ & & & 0 & 0 & 0 & 0 & 0 & * \\ & & 1 & 0 & 1 & 0 & 1 & * & * \\ \hline & & \bold{1} & \bold{1} & \bold{0\medspace.} & \bold{1} & \bold{0} & \bold{0} & \bold{1} \end{matrix}$

Therefore, (10.101)₂ x (10.1)₂ = (110.1001)₂

Question 7

(100.01)₂ x (11.1)₂

Answer

$\begin{matrix} & & & 1 & 0 & 0 & . & 0 & 1 \\ \times & & & & & 1 & 1 & . & 1 \\ \hline & & & & 1 & 0 & 0 & 0 & 1 \\ & & & 1 & 0 & 0 & 0 & 1 & * \\ & & 1 & 0 & 0 & 0 & 1 & * & * \\ \hline & & \bold{1} & \bold{1} & \bold{1} & \bold{0\medspace.} & \bold{1} & \bold{1} & \bold{1} \end{matrix}$

Therefore, (100.01)₂ x (11.1)₂ = (1110.111)₂

Question 8

(101.001)₂ x (0.11)₂

Answer

$\begin{matrix} & & 1 & 0 & 1 & . & 0 & 0 & 1 \\ \times & & & & & 0 & . & 1 & 1 \\ \hline & & & 1 & 0 & 1 & 0 & 0 & 1 \\ & & 1 & 0 & 1 & 0 & 0 & 1 & * \\ \hline & & \bold{1} & \bold{1\medspace.} & \bold{1} & \bold{1} & \bold{0} & \bold{1} & \bold{1} \end{matrix}$

Therefore, (101.001)₂ x (0.11)₂ = (11.11011)₂

Convert to Octal equivalents

Question 1

(4507)₁₀

Answer

Therefore, (4507)₁₀ = (10633)₈

Question 2

(7757)₁₀

Answer

Therefore, (7757)₁₀ = (17115)₈

Question 3

(7606)₁₀

Answer

Therefore, (7606)₁₀ = (16666)₈

Question 4

(10110111)₂

Answer

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{111}$

Therefore, (10110111)₂ = (267)₈

Question 5

(110101101)₂

Answer

Grouping in bits of 3:

$\underlinesegment{110} \quad \underlinesegment{101} \quad \underlinesegment{101}$

Therefore, (110101101)₂ = (655)₈

Question 6

(100001111)₂

Answer

Grouping in bits of 3:

$\underlinesegment{100} \quad \underlinesegment{001} \quad \underlinesegment{111}$

Therefore, (100001111)₂ = (417)₈

Question 7

(11010.0101)₂

Answer

Grouping in bits of 3:

$\underlinesegment{011} \quad \underlinesegment{010} \quad \bold{.} \quad \underlinesegment{010} \quad \underlinesegment{100}$

Therefore, (11010.0101)₂ = (32.24)₈

Question 8

(1000.011)₂

Answer

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{000} \quad \bold{.} \quad \underlinesegment{011}$

Therefore, (1000.011)₂ = (10.3)₈

Question 9

(451.125)₁₀

Answer

Converting integral part

Converting fractional part

Therefore, (451.125)₁₀ = (703.1)₈

Question 10

(245.53)₁₀

Answer

Converting integral part

Converting fractional part

Therefore, (245.53)₁₀ = (365.4172)₈

Question 11

(430.26)₁₀

Answer

Converting integral part

Converting fractional part

Therefore, (430.26)₁₀ = (656.2050)₈

Question 12

(242.24)₁₀

Answer

Converting integral part

Converting fractional part

Therefore, (242.24)₁₀ = (362.1727)₈

Convert the following to Decimal numbers

Question 1

(5100)₈

Answer

Equivalent decimal number = 64 + 2560 = 2624

Therefore, (5100)₈ = (2624)₁₀

Question 2

(7070)₈

Answer

Equivalent decimal number = 56 + 3584 = 3640

Therefore, (7070)₈ = (3640)₁₀

Question 3

(4720)₈

Answer

Equivalent decimal number = 16 + 448 + 2048 = 2512

Therefore, (4720)₈ = (2512)₁₀

Question 4

(A452)₁₆

Answer

Equivalent decimal number = 2 + 80 + 1024 + 40960 = 42066

Therefore, (A452)₁₆ = (42066)₁₀

Question 5

(ABC)₁₆

Answer

Equivalent decimal number = 12 + 176 + 2560 = 2748

Therefore, (ABC)₁₆ = (2748)₁₀

Question 6

(CD7)₁₆

Answer

Equivalent decimal number = 7 + 208 + 3072 = 3287

Therefore, (CD7)₁₆ = (3287)₁₀

Question 7

(27.64)₈

Answer

Integral part

Fractional part

Equivalent decimal number = 7 + 16 + 0.75 + 0.0625 = 23.8125

Therefore, (27.64)₈ = (23.8125)₁₀

Question 8

(57.55)₈

Answer

Integral part

Fractional part

Equivalent decimal number = 7 + 40 + 0.625 + 0.078 = 47.703

Therefore, (57.55)₈ = (47.703)₁₀

Question 9

(301.26)₈

Answer

Integral part

Fractional part

Equivalent decimal number = 1 + 192 + 0.25 + 0.0936 = 193.3436

Therefore, (301.26)₈ = (193.3436)₁₀

Question 10

(707.71)₈

Answer

Integral part

Fractional part

Equivalent decimal number = 7 + 448 + 0.875 + 0.0156 = 455.8906

Therefore, (707.71)₈ = (455.8906)₁₀

Perform the following

Question 1

(3402)₈ to ( )₂

Answer

Therefore, (3402)₈ = ($\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{100}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{010}}$)₂

Question 2

(1507)₈ to ( )₂

Answer

Therefore, (1507)₈ = ($\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{111}}$)₂

Question 3

(2003)₈ to ( )₂

Answer

Therefore, (2003)₈ = ($\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{011}}$)₂

Question 4

(ABC)₁₆ to ( )₂

Answer

Therefore, (ABC)₁₆ = ($\bold{\underlinesegment{1010}}\medspace\bold{\underlinesegment{1011}}\medspace\bold{\underlinesegment{1100}}$)₂

Question 5

(9AD)₁₆ to ( )₂

Answer

Therefore, (9AD)₁₆ = ($\bold{\underlinesegment{1001}}\medspace\bold{\underlinesegment{1010}}\medspace\bold{\underlinesegment{1101}}$)₂

Question 6

(DE)₁₆ to ( )₂

Answer

Therefore, (DE)₁₆ = ($\bold{\underlinesegment{1101}}\medspace\bold{\underlinesegment{1110}}$)₂

Question 7

(345.33)₈ to ( )₂

Answer

Therefore, (345.33)₈ = ($\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{100}}\medspace\bold{\underlinesegment{101}}\medspace\bold{.}\medspace\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{011}}$)₂

Question 8

(501.75)₈ to ( )₂

Answer

Therefore, (501.75)₈ = ($\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{001}}\medspace\bold{.}\medspace\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{101}}$)₂

Question 9

(265.55)₈ to ( )₂

Answer

Therefore, (265.55)₈ = ($\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{110}}\medspace\bold{\underlinesegment{101}}\medspace\bold{.}\medspace\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{101}}$)₂

Question 10

(334.33)₈ to ( )₂

Answer

Therefore, (334.33)₈ = ($\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{100}}\medspace\bold{.}\medspace\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{011}}$)₂

Convert the following to their Hexa-decimal equivalents

Question 1

(110011101111)₂

Answer

Grouping in bits of 4:

$\underlinesegment{1100} \quad \underlinesegment{1110} \quad \underlinesegment{1111}$

Therefore, (110011101111)₂ = (CEF)₁₆

Question 2

(100101101110)₂

Answer

Grouping in bits of 4:

$\underlinesegment{1001} \quad \underlinesegment{0110} \quad \underlinesegment{1110}$

Therefore, (100101101110)₂ = (96E)₁₆

Question 3

(11010111100)₂

Answer

Grouping in bits of 4:

$\underlinesegment{0110} \quad \underlinesegment{1011} \quad \underlinesegment{1100}$

Therefore, (11010111100)₂ = (6BC)₁₆

Question 4

(9894)₁₀

Answer

Therefore, (9894)₁₀ = (26A6)₁₆

Question 5

(89392)₁₀

Answer

Therefore, (89392)₁₀ = (15D30)₁₆

Question 6

(4966)₁₀

Answer

Therefore, (4966)₁₀ = (1366)₁₆

Question 7

(11001110.00100111)₂

Answer

Grouping in bits of 4:

$\underlinesegment{1100} \quad \underlinesegment{1110} \medspace . \medspace \underlinesegment{0010} \quad \underlinesegment{0111}$

Therefore, (11001110.00100111)₂ = (CE.27)₁₆

Question 8

(10010110.00111110)₂

Answer

Grouping in bits of 4:

$\underlinesegment{1001} \quad \underlinesegment{0110} \medspace . \medspace \underlinesegment{0011} \quad \underlinesegment{1110}$

Therefore, (10010110.00111110)₂ = (96.3E)₁₆

Question 9

(1554.115)₈

Answer

(1554.115)₈ = (001101101100.001001101)₂

Grouping in bits of 4:

$\underlinesegment{0011}\medspace\underlinesegment{0110}\medspace\underlinesegment{1100}\medspace . \medspace\underlinesegment{0010}\medspace\underlinesegment{0110}\medspace\underlinesegment{1000}$

Therefore, (1554.115)₈ = (36C.268)₁₆

Question 10

(7013.2011)₈

Answer

(7013.2011)₈ = (111000001011.010000001001)₂

Grouping in bits of 4:

$\underlinesegment{1110}\medspace\underlinesegment{0000}\medspace\underlinesegment{1011}\medspace . \medspace\underlinesegment{0100}\medspace\underlinesegment{0000}\medspace\underlinesegment{1001}$

Therefore, (7013.2011)₈ = (E0B.409)₁₆

Convert the following to their Binary equivalents followed by Octal equivalents

Question 1

(ABC)₁₆

Answer

(ABC)₁₆ = (101010111100)₂

Grouping in bits of 3:

$\underlinesegment{101}\medspace\underlinesegment{010}\medspace\underlinesegment{111}\medspace\underlinesegment{100}$

(ABC)₁₆ = (5274)₈

Question 2

(2CDE)₁₆

Answer

(2CDE)₁₆ = (10110011011110)₂

Grouping in bits of 3:

$\underlinesegment{010}\medspace\underlinesegment{110}\medspace\underlinesegment{011}\medspace\underlinesegment{011}\medspace\underlinesegment{110}$

(2CDE)₁₆ = (26336)₈

Question 3

(B45)₁₆

Answer

(B45)₁₆ = (101101000101)₂

Grouping in bits of 3:

$\underlinesegment{101}\medspace\underlinesegment{101}\medspace\underlinesegment{000}\medspace\underlinesegment{101}$

(B45)₁₆ = (5055)₈

Question 4

(4DC3)₁₆

Answer

(4DC3)₁₆ = (0100110111000011)₂

Grouping in bits of 3:

$\underlinesegment{100}\medspace\underlinesegment{110}\medspace\underlinesegment{111}\medspace\underlinesegment{000}\medspace\underlinesegment{011}$

(4DC3)₁₆ = (46703)₈

Question 5

(786A)₁₆

Answer

(786A)₁₆ = (0111100001101010)₂

Grouping in bits of 3:

$\underlinesegment{111}\medspace\underlinesegment{100}\medspace\underlinesegment{001}\medspace\underlinesegment{101}\medspace\underlinesegment{010}$

(786A)₁₆ = (74152)₈

Question 6

(2345)₁₆

Answer

(2345)₁₆ = (0010001101000101)₂

Grouping in bits of 3:

$\underlinesegment{010}\medspace\underlinesegment{001}\medspace\underlinesegment{101}\medspace\underlinesegment{000}\medspace\underlinesegment{101}$

(2345)₁₆ = (21505)₈

Answer the following questions

Question 1

Give a brief concept of counting system of primitive people.

Answer

Humans started counting since pre-historic times. Primitive people used fingers, sticks, toes and bones for counting. The primitive techniques of using fingers and sticks for counting are in use even today.

Question 2

What do you understand by Decimal Odometer?

Answer

Odometer is the device present in the automobile that is used to measure the speed of the vehicle and distance covered by it. Decimal Odometer is based on the principle of decimal number system. It uses the digits from 0 to 9 and contains a set of 6 wheels.

Question 3

What are the different types of number systems that a computer deals with?

Answer

The different types of number systems are:

1. Binary Number System
2. Octal Number System
3. Decimal Number System
4. Hexadecimal Number System

Question 4

What do you understand by Binary Number system?

Answer

A number system that uses only two types of digits (i.e. 0 and 1) to represent a number is known as Binary Number System. Its base is 2.

Question 5

Explain Decimal Number system with an example.

Answer

In Decimal Number system, a number is represented by using 10 digits — 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Its base is 10. Numbers written without base are by default decimal numbers. For example, 2457 would mean (2457)₁₀, where 10 represents the base of decimal number.

Question 6

How will you convert:

(a) A decimal number to a binary number

Answer

1. Divide a decimal number by 2 and obtain quotient and remainder. The remainder will be the least significant bit (LSB) of the binary number.
2. Divide quotient by 2 and again obtain the next quotient and remainder.
3. Repeat the above step unless quotient becomes 0.
4. The last remainder will be the most significant bit (MSB) of the binary number.
5. Arrange all the remainders from MSB to LSB. It will be the binary equivalent of the given decimal number.

(b) A binary number to a decimal number

Answer

1. Multiply each digit of binary number with 2⁰, 2¹, 2², .... respectively from right side towards left side of the number.
2. Add all the products to get the equivalent decimal number.

(c) An octal number to a binary number

Answer

To convert an octal number into its binary equivalent, write each octal digit into binary in a combination of 3 bits. The complete set of bits will be the binary equivalent of the given octal number.

(d) A binary number to a Hexa-decimal number

Answer

To convert a binary number into its equivalent hexadecimal form, group the binary digits in 4 bits form from right hand side. If required, add zeros to the left to make the group in 4 bits. Write the hexadecimal equivalent of each binary group. The result will be the hexadecimal equivalent of the binary number.

(e) A binary number to an octal number

Answer

To convert a binary number into its equivalent octal number, group the binary numbers in 3 bits form from right hand side towards left hand side. If required, add zeros to the left to make the group in 3 bits. Convert each binary group of 3 bits into its octal equivalent.

(f) A hexa-decimal number to an octal number

Answer

To convert a hexa-decimal number to an octal number, first convert the hexa-decimal number to its binary equivalent by writing the binary of the corresponding hexa-decimal digit in 4 bit form. After that group the binary numbers in 3 bits form from right hand side towards left hand side. Convert each binary group of 3 bits into its octal equivalent.

Question 7

What are the rules to perform:

(a) Binary Addition

Answer

1. 0 + 0 = 0
2. 0 + 1 = 1
3. 1 + 0 = 1
4. 1 + 1 = 0 and carry over 1
5. 1 + 1 + 1 = 1 and carry over 1

(b) Binary Subtraction

Answer

1. 0 - 0 = 0
2. 1 - 0 = 1
3. 1 - 1 = 0
4. 0 - 1 = 1 with a borrow from the next higher significant digits.

Question 8

Write down the rules for:

(a) Binary Multiplication

Answer

1. 0 x 0 = 0
2. 0 x 1 = 0
3. 1 x 0 = 0
4. 1 x 1 = 1

(b) Binary Division

Answer

1. 0 ÷ 0 = 0
2. 0 ÷ 1 = 0
3. 1 ÷ 0 = ∞
4. 1 ÷ 1 = 1

Question 9

What do you understand by

(a) An Octal number

Answer

An octal number uses 8 types of digits — 0, 1, 2, 3, 4, 5, 6, 7. It is represented with base 8.

(b) A Hexa-decimal number

Answer

A Hexa-decimal number uses 16 types of digits (0 to 15). To represent digits from 10 to 15 it uses letters from A to F respectively. It is represented with base 16.

Question 10

Give two differences between:

(a) Binary number and Decimal number

Answer

(b) Octal number and Binary number

Answer

(c) Hexa-decimal and Octal number

Answer

Question 11

Name the different types of operations that can be performed in Binary arithmetics.

Answer

1. Binary Addition
2. Binary Subtraction
3. Binary Multiplication
4. Binary Division

Question 12

Complete the following table:

Answer