Numbers

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).

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

 0 + 0 = 0 1 + 0 = 1 1 + 1 = 0 and carry over 1 0 + 1 = 1
 0 - 0 = 0 1 - 0 = 1 1 - 1 = 0 0 - 1 = 1 with a borrowfrom next higher digit
 0 x 0 = 0 1 x 1 = 1 1 x 0 = 0 0 x 1 = 0
 0 ÷ 0 = 0 0 ÷ 1 = 0 1 ÷ 1 = 1 1 ÷ 0 = Infinite

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

2QuotientRemainder
27840 (LSB)
23920
21960
2980
2491
2240
2120
260
231
211 (MSB)
0

Therefore, (784)10 = (1100010000)2

Question 2

(999)10

2QuotientRemainder
29991 (LSB)
24991
22491
21240
2620
2311
2151
271
231
211 (MSB)
0

Therefore, (999)10 = (1111100111)2

Question 3

(1401)10

2QuotientRemainder
214011 (LSB)
27000
23500
21751
2871
2431
2211
2100
251
220
211 (MSB)
0

Therefore, (1401)10 = (10101111001)2

Question 4

(1423)10

2QuotientRemainder
214231 (LSB)
27111
23551
21771
2880
2440
2220
2111
251
220
211 (MSB)
0

Therefore, (1423)10 = (10110001111)2

Question 5

(478.75)10

Converting integral part
2QuotientRemainder
24780
22391
21191
2591
2291
2140
271
231
211
0
Converting fractional part
Multiply=ResultantCarry
0.75 x 2=0.51
0.5 x 2=01

Therefore, (478.75)10 = (111011110.11)2

Question 6

(165.35)10

Converting integral part
2QuotientRemainder
21651
2820
2411
2200
2100
251
220
211
0
Converting fractional part
Multiply=ResultantCarry
0.35 x 2=0.70
0.7 x 2=0.41
0.4 x 2=0.80
0.8 x 2=0.61

Therefore, (165.35)10 = (10100101.0101)2

Question 7

(277.27)10

Converting integral part
2QuotientRemainder
22771
21380
2691
2340
2171
280
240
220
211
0
Converting fractional part
Multiply=ResultantCarry
0.27 x 2=0.540
0.54 x 2=0.081
0.08 x 2=0.160
0.16 x 2=0.320

Therefore, (277.27)10 = (100010101.0100)2

Question 8

(322.2)10

Converting integral part
2QuotientRemainder
23220
21611
2800
2400
2200
2100
251
220
211
0
Converting fractional part
Multiply=ResultantCarry
0.2 x 2=0.40
0.4 x 2=0.80
0.8 x 2=0.61
0.6 x 2=0.21

Therefore, (322.2)10 = (101000010.0011)2

Perform the following

Question 1

(1010110)2 to ( )10

Binary
No
PowerValueResult
0 (LSB)2010x1=0
12121x2=2
12241x4=4
02380x8=0
124161x16=16
025320x32=0
1 (MSB)26641x64=64

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

Therefore, (1010110)2 = (86)10

Question 2

(1000011)2 to ( )10

Binary
No
PowerValueResult
1 (LSB)2011x1=1
12121x2=2
02240x4=0
02380x8=0
024160x16=0
025320x32=0
1 (MSB)26641x64=64

Equivalent decimal number = 1 + 2 + 64 = 67

Therefore, (1000011)2 = (67)10

Question 3

(1100111)2 to ( )10

Binary
No
PowerValueResult
1 (LSB)2011x1=1
12121x2=2
12241x4=4
02380x8=0
024160x16=0
125321x32=32
1 (MSB)26641x64=64

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

Therefore, (1100111)2 = (103)10

Question 4

(10101011)2 to ( )10

Binary
No
PowerValueResult
1 (LSB)2011x1=1
12121x2=2
02240x4=0
12381x8=8
024160x16=0
125321x32=32
026640x64=0
1 (MSB)271281x128=128

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

Therefore, (10101011)2 = (171)10

Question 5

(1010.001)2 to ( )10

Binary
No
PowerValueResult
02010x1=0
12121x2=2
02240x4=0
12381x8=8
Fractional part
Binary
No
PowerValueResult
02-10.50x0.5=0
02-20.250x0.25=0
12-30.1251x0.125=0.125

Equivalent decimal number = 2 + 8 + 0.125 = 10.125

Therefore, (1010.001)2 = (10.125)10

Question 6

(10100.11)2 to ( )10

Binary
No
PowerValueResult
02010x1=0
02120x2=0
12241x4=4
02380x8=0
124161x16=16
Fractional part
Binary
No
PowerValueResult
12-10.51x0.5=0.5
12-20.251x0.25=0.25

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

Binary
No
PowerValueResult
02010x1=0
12121x2=2
02240x4=0
12381x8=8
Fractional part
Binary
No
PowerValueResult
12-10.51x0.5=0.5
12-20.251x0.25=0.25
12-30.1251x0.125=0.125

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

Binary
No
PowerValueResult
02010x1=0
12121x2=2
02240x4=0
12381x8=8
Fractional part
Binary
No
PowerValueResult
02-10.50x0.5=0
12-20.251x0.25=0.25
12-30.1251x0.125=0.125
12-40.06251x0.0625=0.0625

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

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

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

Question 2

(1111011)2 + (1011101)2

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

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

Question 3

(10111)2 + (110111)2

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

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

Question 4

(1101101)2 + (1110010)2

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

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

Question 5

(10111101)2 + (1101101)2

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

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

Subtract the following

Question 1

(110101)2 - (011010)2

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

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

Question 2

(110110)2 - (01010)2

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

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

Question 3

(101011)2 - (11101)2

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

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

Question 4

(11111)2 - (01101)2

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

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

Question 5

(10000)2 - (01111)2

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

Therefore, (10000)2 - (01111)2 = (1)2

Perform the following arithmetic

Question 1

Add octal numbers (1546)8 and (641)8

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

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

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

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

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

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

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

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

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

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

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

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

Question 2

(10101)2 x (100)2

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

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

Question 3

(11000)2 x (101)2

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

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

Question 4

(10.001)2 x (0.11)2

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

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

Question 5

(101.10)2 x (1.10)2

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

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

Question 6

(10.101)2 x (10.1)2

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

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

Question 7

(100.01)2 x (11.1)2

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

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

Question 8

(101.001)2 x (0.11)2

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

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

Convert to Octal equivalents

Question 1

(4507)10

8QuotientRemainder
845073 (LSB)
85633
8706
880
811 (MSB)
0

Therefore, (4507)10 = (10633)8

Question 2

(7757)10

8QuotientRemainder
877575 (LSB)
89691
81211
8157
811 (MSB)
0

Therefore, (7757)10 = (17115)8

Question 3

(7606)10

8QuotientRemainder
876066 (LSB)
89506
81186
8146
811 (MSB)
0

Therefore, (7606)10 = (16666)8

Question 4

(10110111)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
1117
1106
0102

Therefore, (10110111)2 = (267)8

Question 5

(110101101)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
1015
1015
1106

Therefore, (110101101)2 = (655)8

Question 6

(100001111)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
1117
0011
1004

Therefore, (100001111)2 = (417)8

Question 7

(11010.0101)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
0102
0113
..
0102
1004

Therefore, (11010.0101)2 = (32.24)8

Question 8

(1000.011)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
0000
0011
..
0113

Therefore, (1000.011)2 = (10.3)8

Question 9

(451.125)10

Converting integral part
8QuotientRemainder
84513
8560
877
0
Converting fractional part
Multiply=ResultantCarry
0.125 x 8=01

Therefore, (451.125)10 = (703.1)8

Question 10

(245.53)10

Converting integral part
8QuotientRemainder
82455
8306
833
0
Converting fractional part
Multiply=ResultantCarry
0.53 x 8=0.244
0.24 x 8=0.921
0.92 x 8=0.367
0.36 x 8=0.882

Therefore, (245.53)10 = (365.4172)8

Question 11

(430.26)10

Converting integral part
8QuotientRemainder
84306
8535
866
0
Converting fractional part
Multiply=ResultantCarry
0.26 x 8=0.082
0.08 x 8=0.640
0.64 x 8=0.125
0.12 x 8=0.960

Therefore, (430.26)10 = (656.2050)8

Question 12

(242.24)10

Converting integral part
8QuotientRemainder
82422
8306
833
0
Converting fractional part
Multiply=ResultantCarry
0.24 x 8=0.921
0.92 x 8=0.367
0.36 x 8=0.882
0.88 x 8=0.047

Therefore, (242.24)10 = (362.1727)8

Convert the following to Decimal numbers

Question 1

(5100)8

Octal
No
PowerValueResult
0 (LSB)8010x1=0
08180x8=0
182641x64=64
5 (MSB)835125x512=2560

Equivalent decimal number = 64 + 2560 = 2624

Therefore, (5100)8 = (2624)10

Question 2

(7070)8

Octal
No
PowerValueResult
0 (LSB)8010x1=0
78187x8=56
082640x64=0
7 (MSB)835127x512=3584

Equivalent decimal number = 56 + 3584 = 3640

Therefore, (7070)8 = (3640)10

Question 3

(4720)8

Octal
No
PowerValueResult
0 (LSB)8010x1=0
28182x8=16
782647x64=448
4 (MSB)835124x512=2048

Equivalent decimal number = 16 + 448 + 2048 = 2512

Therefore, (4720)8 = (2512)10

Question 4

(A452)16

Number
PowerValueResult
216012x1=2
5161165x16=80
41622564x256=1024
A (10)163409610x4096=40960

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

Therefore, (A452)16 = (42066)10

Question 5

(ABC)16

Number
PowerValueResult
C (12)160112x1=12
B (11)1611611x16=176
A (10)16225610x256=2560

Equivalent decimal number = 12 + 176 + 2560 = 2748

Therefore, (ABC)16 = (2748)10

Question 6

(CD7)16

Number
PowerValueResult
716017x1=7
D (13)1611613x16=208
C (12)16225612x256=3072

Equivalent decimal number = 7 + 208 + 3072 = 3287

Therefore, (CD7)16 = (3287)10

Question 7

(27.64)8

Octal
No
PowerValueResult
78017x1=7
28182x8=16
Fractional part
Octal
No
PowerValueResult
68-10.1256x0.125=0.75
48-20.01564x0.0156=0.0625

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

Therefore, (27.64)8 = (23.8125)10

Question 8

(57.55)8

Octal
No
PowerValueResult
78017x1=7
58185x8=40
Fractional part
Octal
No
PowerValueResult
58-10.1255x0.125=0.625
58-20.01565x0.0156=0.078

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

Therefore, (57.55)8 = (47.703)10

Question 9

(301.26)8

Octal
No
PowerValueResult
18011x1=1
08180x8=0
382643x64=192
Fractional part
Octal
No
PowerValueResult
28-10.1252x0.125=0.25
68-20.01566x0.0156=0.0936

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

Therefore, (301.26)8 = (193.3436)10

Question 10

(707.71)8

Octal
No
PowerValueResult
78017x1=7
08180x8=0
782647x64=448
Fractional part
Octal
No
PowerValueResult
78-10.1257x0.125=0.875
18-20.01561x0.0156=0.0156

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

Octal
Number
Binary
Equivalent
2010
0000
4100
3011

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

Octal
Number
Binary
Equivalent
7111
0000
5101
1001

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

Octal
Number
Binary
Equivalent
3011
0000
0000
2010

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

Number
Binary
Equivalent
C (12)1100
B (11)1011
A (10)1010

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

Question 5

(9AD)16 to ( )2

Number
Binary
Equivalent
D (13)1101
A (10)1010
91001

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

Question 6

(DE)16 to ( )2

Number
Binary
Equivalent
E (14)1110
D (13)1101

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

Question 7

(345.33)8 to ( )2

Octal
Number
Binary
Equivalent
5101
4100
3011
.
3011
3011

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

Octal
Number
Binary
Equivalent
1001
0000
5101
.
7111
5101

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

Octal
Number
Binary
Equivalent
5101
6110
2010
.
5101
5101

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

Octal
Number
Binary
Equivalent
4100
3011
3011
.
3011
3011

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

Grouping in bits of 4:

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

Binary
Number
Equivalent
1111F (15)
1110E (14)
1100C (12)

Therefore, (110011101111)2 = (CEF)16

Question 2

(100101101110)2

Grouping in bits of 4:

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

Binary
Number
Equivalent
1110E (14)
01106
10019

Therefore, (100101101110)2 = (96E)16

Question 3

(11010111100)2

Grouping in bits of 4:

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

Binary
Number
Equivalent
1100C (12)
1011B (11)
01106

Therefore, (11010111100)2 = (6BC)16

Question 4

(9894)10

16QuotientRemainder
1698946
16618A (10)
16386
1622
0

Therefore, (9894)10 = (26A6)16

Question 5

(89392)10

16QuotientRemainder
16893920
1655873
16349D (13)
16215
1611
0

Therefore, (89392)10 = (15D30)16

Question 6

(4966)10

16QuotientRemainder
1649666
163106
16193
1611
0

Therefore, (4966)10 = (1366)16

Question 7

(11001110.00100111)2

Grouping in bits of 4:

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

Binary
Number
Equivalent
1110E (14)
1100C (12)
.
00102
01117

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

Question 8

(10010110.00111110)2

Grouping in bits of 4:

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

Binary
Number
Equivalent
01106
10019
.
00113
1110E (14)

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

Question 9

(1554.115)8

Octal
Number
Binary
Equivalent
4100
5101
5101
1001
.
1001
1001
5101

(1554.115)8 = (001101101100.001001101)2

Grouping in bits of 4:

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

Binary
Number
Equivalent
1100C (12)
01106
00113
.
00102
01106
10008

Therefore, (1554.115)8 = (36C.268)16

Question 10

(7013.2011)8

Octal
Number
Binary
Equivalent
3011
1001
0000
7111
.
2010
0000
1001
1001

(7013.2011)8 = (111000001011.010000001001)2

Grouping in bits of 4:

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

Binary
Number
Equivalent
1011B (11)
00000
1110E (14)
.
01004
00000
10019

Therefore, (7013.2011)8 = (E0B.409)16

Convert the following to their Binary equivalents followed by Octal equivalents

Question 1

(ABC)16

Number
Binary
Equivalent
C (12)1100
B (11)1011
A (10)1010

(ABC)16 = (101010111100)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
1004
1117
0102
1015

(ABC)16 = (5274)8

Question 2

(2CDE)16

Number
Binary
Equivalent
E (14)1110
D (13)1101
C (12)1100
20010

(2CDE)16 = (10110011011110)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
1106
0113
0113
1106
0102

(2CDE)16 = (26336)8

Question 3

(B45)16

Number
Binary
Equivalent
50101
40100
B (11)1011

(B45)16 = (101101000101)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
1015
0000
1015
1015

(B45)16 = (5055)8

Question 4

(4DC3)16

Number
Binary
Equivalent
30011
C (12)1100
D (13)1101
40100

(4DC3)16 = (0100110111000011)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
0113
0000
1117
1106
1004

(4DC3)16 = (46703)8

Question 5

(786A)16

Number
Binary
Equivalent
A (10)1010
60110
81000
70111

(786A)16 = (0111100001101010)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
0102
1015
0011
1004
1117

(786A)16 = (74152)8

Question 6

(2345)16

Number
Binary
Equivalent
50101
40100
30011
20010

(2345)16 = (0010001101000101)2

Grouping in bits of 3:

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

Binary
Number
Equivalent
Octal
1015
0000
1015
0011
0102

(2345)16 = (21505)8

Answer the following questions

Question 1

Give a brief concept of counting system of primitive people.

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?

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?

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?

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.

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

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

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

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

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

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

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:

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

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

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

(b) Binary Division

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

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

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

Binary numberDecimal number
It uses 2 digits — 0 and 1.It uses 10 digits — 0 to 9.
It uses base 2.It uses base 10.

(b) Octal number and Binary number

Octal numberBinary number
It uses 8 digits — 0 to 7.It uses 2 digits — 0 and 1.
It uses base 8.It uses base 2.

(c) Hexa-decimal and Octal number

Hexa-decimalOctal number
It uses 16 digits — 0 to 15.It uses 8 digits — 0 to 7.
It uses base 16.It uses base 8.

Question 11

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

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

Question 12

Complete the following table:

Octal
Digit
Binary
Equivalent
5
7
1
6
3
Digit
Binary
Equivalent
8
11
4
15
9

Octal
Digit
Binary
Equivalent
5101
7111
1001
6110
3011