Octal Number System

The number with base eight is called octal number. We can generate these numbers with the combination of 0, 1, 2 3, 4, 5, 6, 7. This note provides an information about Octal Number System and its conversions.

Summary

The number with base eight is called octal number. We can generate these numbers with the combination of 0, 1, 2 3, 4, 5, 6, 7. This note provides an information about Octal Number System and its conversions.

Things to Remember

  • The number with base eight is called octal number. It is represented by Q or O.
  • When decimal number is repetitively divided by eight and remainders are arranged in the form of octal numbers,then decimal number are converted into octal.
  • Each octal is multiplied by its weighted position. The sum of all products is known as decimal form of octal.
  • The three digit format of binary digits is used for octal to binary conversions or vice versa.

MCQs

No MCQs found.

Subjective Questions

Q1:

The frequency distribution table for 231 people of a village classified according to their ages is as follows:

Age Group (year) No.of people
0 - 5 18
5 - 10 15
10 - 15 14
15 - 20 31
20 - 25 33
25 - 30 60
30 - 35 23
35 - 40 20
40 - 45 12
45 - 50 5

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Representing the above table in a cumulative frequency:<br><br></p> <table width="201"><tbody><tr><td>Age group (in years)</td> <td>Cumulative Frequency</td> </tr><tr><td>0 - 5</td> <td>18</td> </tr><tr><td>5 - 10</td> <td>33</td> </tr><tr><td>10 - 15</td> <td>47</td> </tr><tr><td>15 - 20</td> <td>111</td> </tr><tr><td>20 - 25</td> <td>171</td> </tr><tr><td>25 - 30</td> <td>194</td> </tr><tr><td>30 - 35</td> <td>214</td> </tr><tr><td>35 - 40</td> <td>226</td> </tr><tr><td>45 - 50</td> <td>231</td> </tr></tbody></table>

Q2:

figure
figure

The total materials expenditure of Rs. 9800 for a student to study in higher education in Kathmandu is given in the pie chart below. Answer the following questions on the basis of the pie chart.
(i) how much is spent for books?
(ii). How much is the expenditure for house rent and food more than that of education?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>(i) . Expenditure on books = 7% of 9800 <br> = \(\frac{7}{100}\) \(\times\) 9800 <br> = Rs. 686<br><br>(ii) Percentage of expenditure on house rent and food = 100% - (7% + 22% + 7% + 16%)<br> = 100% - 52%<br> = 48%. <br>Hence , <br>expendituren on house rent and food = 48% of Rs. 9800<br> = Rs. 9800 \(\times\) \(\frac{48}{100}\) <br> = Rs. 4.704<br>Expenditure on education = 22% of Rs. 9800<br> = Rs. 2156<br><br>\(\therefore\) (rs.4704 - Rs. 2156) = Rs. 2548 is more expenditure on house rent and food than education Ans.</p>

Q3:

figure
figure

The different materials in a bundle of 100 meters of cloths weighing 25 kg  were found as shown in th pie chart given below :
Answers the following questions on the basis of the pie chart :
(i). What is the weight of cotton in the bundle of cloth ?
(ii). Which of two raw materials have equal weight ? What is the weight of those  items ?
(iii). How many times is the weight of Nylon heavier than cotton ?What is the weight of Nylom ?
(iv). What is the percentage of polyster ?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here, total weight = 360<br>i.e. 360 = 25kg<br>or 1\(^{o}\) = \(\frac{25}{360}\) kg<br><br>(i). Here , the angle subtended at centre for cotton<br>= 360 -(198 + 36 + 36 +18 ) = 72<br><br>But , 1 weight = \(\frac{25}{360}\)k.g<br><br>\(\therefore\) 72 weight = \(\frac{25}{360}\) \(\times\) 72 =5 k.g<br><br>Hence , weight of cotton = 5 k.g Ans.</p>

Q4:

The blood groups of 900 patients who came to Bir Hospital , Ktahmandu being injured in accidents throughout as follows.
Draw a pie chart on the basis of the given data. (There are 4 types of human blood i.e , A , B , AB and O.)

Types Of blood A B AB O
Percentage 40% 10% 5% 45%

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <figure class="inline-right" style="width: 250px;"><img src="/uploads/712.png" alt="figure" width="250" height="156"><figcaption>figure</figcaption></figure><p>Here , total percentage = 100%<br>Now , let 100% = total angle formed at the centre. <br>i.e. 100% = 360<br>or , 1% = \(\frac{360}{100}\) = 3.6<br><br>Now , to find out the central angle formed at centre of each for the blood group are as follows :<br>(i). 40% blood of group A = 40 \(\times\) 3.6 = 144<br>(ii). 10% blood of group B = 10 \(\times\) 3.6 = 36<br>(iii). 5% blood of group AB = 5 \(\times\) 3.6 = 18<br>(iv). 45% blood of group O = 45 \(\times\) 3.6 = 162</p>

Q5:

The percentage distribution of different elements in the body of a man of 60 kg is given below. Express the weight in percentage ond represent them in a pie.

Element Muscles Bones Blood Fats Others
Percentage 45% 18% 12% 20% 5%

 


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <figure class="inline-right" style="width: 250px;"><img src="/uploads/810.png" alt="figure" width="250" height="156"><figcaption>figure</figcaption></figure><p>Here, total percentage = 100%<br>Here , let 100% = 360<br>\(\therefore\) 1% \(\frac{360}{100}\) = 3.6<br>Now , to find angle for each part of the body formed at the centre of pie chart are as follows :<br>(i). 45% of muscles = 45 \(\times\) 3.6 = 162<br>(ii). 18% of bones = 18 \(\times\) 3.6 = 64.8<br>(iii). 12% of blood = 12 \(\times\) 3.6 = 43.2<br>(iv). 20% of fats = 20 \(\times\) 3.6 = 72<br>(v). 5% of others = 5 \(\times\) 3.6 = 18<br><br>Here , <br>360\(^o\) = 60 k.g<br>\(\therefore\) 1 = \(\frac{60}{360}\) = \(\frac{1}{6}\) k.g. <br>\(\therefore\) 162 muscles = \(\frac{1}{6}\) \(\times\) 162 = 27 Kg. <br><br>64. 8 bones = \(\frac{1}{6}\) \(\times\) 64.8 = 10.8 k.g<br>43.2 blood = \(\frac{1}{6}\) \(\times\) 43.2 =7.2k.g.<br>72 fats = \(\frac{1}{6}\) \(\times\) 72 = 12 k.g.<br>18 others = \(\frac{1}{6}\) \(\times\) 18 = 3 k.g .</p>

Q6:

The daily schedule of class 9 students is as follows : Convert these data into percentage and then represent them in a pie chart :

Work School Sports Homework Food Misc Sleeping
Time (hour) 7 2 3 1 \(\frac{1}{2}\) 2 \(\frac{1}{2}\) 8

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <figure class="inline-right" style="width: 250px;"><img src="/uploads/99.png" alt="figure" width="250" height="156"><figcaption>figure</figcaption></figure><p>Here , total time = 24 hrs<br>wE Covert given data into percentage as follows :<br>School's book = \(\frac{7}{24}\) \(\times\) 100% = \(\frac{175}{6}\)%<br>Sleeping work = \(\frac{8}{24}\) \(\times\) 100% = \(\frac{100}{3}\)%<br>Sport's work = \(\frac{2}{24}\) \(\times\) 100% =\(\frac{25}{3}\)%<br>Home work = \(\frac{3}{24}\) \(\times\) 100% = \(\frac{25}{2}\)%<br>Food = 1 \(\frac{1}{2}\) \(\times\) 100% = \(\frac{25}{4}\)%<br><br>Misc. work = 2 \(\frac{1}{2}\) \(\times\) 100% = \(\frac{125}{12}\)%<br>Total angle formed at the centre of each pie chart = 360<br>Here , let 100% = 360<br>1% = \(\frac{360}{100}\) = 3.6<br>Now , we find the angles formed at the centre for each of work as follows :<br>School \(\frac{175}{6}\) =\(\frac{175}{6}\) \(\times\) 3.6 = 105<br>Sleeping \(\frac{100}{3}\) =\(\frac{100}{3}\) \(\times\) 3.6 = 120<br>Sports \(\frac{25}{3}\) =\(\frac{25}{3}\) \(\times\) 3.6 = 30<br>Home work \(\frac{25}{2}\) =\(\frac{25}{2}\) \(\times\) 3.6 = 45<br>Fooding \(\frac{25}{4}\) = \(\frac{25}{4}\) \(\times\) 3.6 = 22.5<br>Misc. \(\frac{125}{12}\) = \(\frac{125}{12}\) \(\times\) 3.6 = 37.5</p>

Q7:

A survey of 360 household of a municipality was conducted regarding the use of fuel for cooking. The data were found as follows :

Fuel Type wood electricity gas Kerosene
Percentage 12.5% 22.5% 25% 40%

Represent this data in a pie chart . 


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <figure class="inline-right" style="width: 250px;"><img src="/uploads/1111.png" alt="figure" width="250" height="156"><figcaption>figure</figcaption></figure><p></p> <table width="275"><tbody><tr><td>Fuel</td> <td>Percent</td> </tr><tr><td>Wood</td> <td>12.5%</td> </tr><tr><td>Electricity</td> <td>22.5%</td> </tr><tr><td>Gas</td> <td>25%</td> </tr><tr><td>Kerosene</td> <td>40%</td> </tr><tr><td>Total</td> <td>100%</td> </tr></tbody></table><p>Total percentage of consumed fuel = 100%<br>Angle formed at the centre of pie chart = 360<br>Here , 100% = 360<br>Now , we convert fuel percet into degree as follows :<br>Wood (12.5%) = 12.5 \(\times\) 3.6 = 45<br>Electricity (22.5%) = 22.5 \(\times\) 3.6 = 81<br>Gas (25%) = 25 \(\times\) 3.6 = 90<br>Kerosene (40%) = 40 \(\times\) = 144<br><br></p>

Q8:

Draw the histogram for the following data given in the following frequency table. 

Number Frequency
0 - 20
10 -20
20 - 30
30 - 40
40 - 50		 2
3
4
3
1

 


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <figure class="inline-left" style="width: 350px;"><img src="/uploads/262_1_i.jpg" alt="figure" width="350" height="222"><figcaption>figure</figcaption></figure>

Q9:

Draw the histogram for the following data given in the following frequency table

Number Frequency
20 - 30
30 - 40
40 - 50
50 - 60
60 - 70		 1
4
3
2
3

 


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <figure class="inline-left" style="width: 400px;"><img src="/uploads/262_ii.png" alt="figure" width="400" height="300"><figcaption>figure</figcaption></figure>

Q10:

Draw the histogram for the following data given in the following frequency table. 

Number Frequency
5 - 10
10 - 15
15 - 20
20 - 25
25 - 30
30 - 35		 3
2
4
1
5
2

 


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <figure class="inline-left" style="width: 400px;"><img src="/uploads/263_i.png" alt="figure" width="400" height="300"><figcaption>figure</figcaption></figure>

Q11:

Draw the histogram for the following data given in the following frequency table 

Number Frequency
5 - 15
15 - 25
25 - 35
35 - 45
45 - 55
55 -65		 2
4
3
3
2
4

 


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <figure class="inline-left" style="width: 400px;"><img src="/uploads/263_ii.png" alt="figure" width="400" height="300"><figcaption>figure</figcaption></figure>

Q12:

Construct the histogram for each of the following data with equal size of class intervals and make a less than type cumulative frequency table for each of them. Then draw the ogive :

Class Frequency
10 -15
15 -20
20 - 25
25 - 30
30 - 35		 2
4
3
2
3

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>As the scale on X - axis start from 10 , a break is drawn near the ogive and the graph is drawn using scale beginning from 10 but not the origin itself. In order to draw the histogram , take 1cm = 5 units on X- axis representing the class intervals and 1 cm = 1 unit on Y-axis representing the frequencies of given data , and draw rectangles as shown given below :<br><br>Now , convert the above table in cumulative frequency table.</p> <table width="238"><tbody><tr><td>Class</td> <td>Frequency</td> <td>Cumulative<br> Frequency</td> </tr><tr><td>10 - 15<br>15 - 20<br>20 - 25<br>25 - 30<br>30 - 35</td> <td>2<br>4<br>3<br>2<br>3</td> <td>2<br>6<br>9<br>11<br>14</td> </tr></tbody></table><p>Now , to draw the CF curve , convert the above table in less than type CF table by including imagined class interval 5 - 10 woth frequency as O show below :</p> <table width="248"><tbody><tr><td>Class</td> <td>Cumulative<br> Frequency</td> </tr><tr><td>Less than `10<br>Less than 15<br>Less than 20<br>Less than 25<br>Less than 30<br>Less than 35</td> <td>0<br>2<br>6<br>9<br>11<br>14</td> </tr></tbody></table><p>In order to draw the cumulative frequency curve of the data from above 'less than' frequency table let us take point 1cm = 5 units on X- axis and 1 cm = 2 unit on y- axis and plot the points and joined smothly. </p>

Q13:

Construct the histogram for the data with equal size of class intervals and make a less than type cumulative frequency table for each of them. Then draw the ogive :

Class Frequency
12 -15
15 - 18
18 - 21
21 - 24
24 - 27
27 - 30		 2
4
3
6
4
1

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>To draw the histogram from the above table , let us mark the points by taking the scale 2cm = 3 units and 1 cm = 1 unit in Y- axis and draw the rectangle as shown in the graph given below :<br><br><br>Now , converting the above table into cumulative frequency table , we have</p> <table width="284"><tbody><tr><td>Class</td> <td>Frequency</td> <td>Cumulative Frequency</td> </tr><tr><td>12 - 15<br>15 - 18<br>18 - 21<br>21 - 24<br>24 - 27<br>27 - 30</td> <td>2<br>4<br>3<br>6<br>4<br>1</td> <td>2<br>6<br>9<br>15<br>19<br>20</td> </tr></tbody></table><p>Again for drawing the ogive , we convert the above table into 'less than' type cumulative frequency table by including imagined class interval 9 - 12 with frequency 0 as follows :</p> <table width="240"><tbody><tr><td>Class</td> <td>Cumulative Frequency</td> </tr><tr><td>Less than 12<br>Less than 15<br>Less than 18<br>Less than 21<br>Less than 24<br>Less than 27<br>Less than 30</td> <td>0<br>2<br>6<br>9<br>15<br>19<br>20</td> </tr></tbody></table><p>In order to draw the ogive of the data from above 'less than' frequency table , let us plot the points by taking the scale 1cm = 3 units on X - axis representing less than class interval and 1 cm = 2 units on Y- axis representing the cumulative frequencies and join freely as shown in the given below ,</p>

Q14:

The pulse rate per minute of 34 students of class IX was measured and found to be as follows:
61 56 72 65 67 59 68 82 60 66 52 59 73 68
66 61 51 59 84 68 79 75 68 58 71 62 57 53
74 70 66 71 54
 
Represent the above data by frequency distribution table and cumulative frequency table by taking the class intervals 55 - 60, 60 - 65,.....

Class interval

Tally Marks

 Frequency

50 – 55

////

4

55 – 60

//// /

6

60 – 65

////

5

65 – 70

////          ////

9

70 – 75

////         /

6

75 – 80

/ /

2

80 – 85

//

2

 

 

Total = 34

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Cumulative frequency Table:</p> <table width="221"><tbody><tr><td>Class - Interval</td> <td>Cumulative Frequency</td> </tr><tr><td>50 - 55</td> <td>4</td> </tr><tr><td>55 - 60</td> <td>10</td> </tr><tr><td>60 - 65</td> <td>15</td> </tr><tr><td>65 - 70</td> <td>24</td> </tr><tr><td>70 - 75</td> <td>30</td> </tr><tr><td>75 - 80</td> <td>32</td> </tr><tr><td>80 - 85</td> <td>34</td> </tr></tbody></table>

Q15:

The marks obtained in a unit test of 25 full marks by 23 students of class IX are as follows :
17  15  12  5 18  24 20  10 14 
20  23  17  22  21  14  9  13
21  24  20  25  16
Represent the above data in a frequency distribution table by classifying them into the class 0 - 5 , 5 - 10. . .  ,


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <table width="336"><tbody><tr><td width="213"> <p>Numbers</p> </td> <td width="213"> <p>Tally Marks</p> </td> <td width="213"> <p>Frequency</p> </td> </tr><tr><td width="213"> <p>0 &ndash; 5</p> </td> <td width="213"> <p>/</p> </td> <td width="213"> <p>1</p> </td> </tr><tr><td width="213"> <p>5 &ndash; 10</p> </td> <td width="213"> <p>//</p> </td> <td width="213"> <p>2</p> </td> </tr><tr><td width="213"> <p>10 &ndash; 15</p> </td> <td width="213"> <p>////</p> </td> <td width="213"> <p>5</p> </td> </tr><tr><td width="213"> <p>15 &ndash; 20</p> </td> <td width="213"> <p>////</p> </td> <td width="213"> <p>5</p> </td> </tr><tr><td width="213"> <p>20 &ndash; 25</p> </td> <td width="213"> <p>//// ////</p> </td> <td width="213"> <p>9</p> </td> </tr><tr><td width="213"> <p>25 - 30</p> </td> <td width="213"> <p>/</p> </td> <td width="213"> <p>1</p> </td> </tr><tr><td width="213"> <p>Total</p> </td> <td width="213"> <p></p> </td> <td width="213"> <p>23</p> </td> </tr></tbody></table><p></p>

Q16:

figure
figure

The votes secured by three candidates A , B and C from the total polling of 12000 for one seat of the parliament member in the general parliamentary election in the year 2056 are presented in the following pie chart :
Study the pie chart and answer the following questions :
(i). How many votes did B get ?
(ii). Who secured the highest vote ?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Votes secured by A = 90<br>(i). Central angle for votes secured by B = 360 - 120 - 90 = 150<br>We have given<br>360 =12000 votes<br>1 = \(\frac{12000}{360}\) votes<br>150 = \(\frac{12000}{360}\) \(\times\) 150 votes<br>= 5000 votes.<br><br>(ii) . Candidate who secured the highest vote<br>By question , votes secured by A = 90<br>90 = \(\frac{12000}{360}\) \(\times\) 90 votes = 3000 votes.<br>Similarly , votes secured by C = 120<br>120 = \(\frac{12000}{360}\) \(\times\) 120 votes = 4000 votes.<br><br>Hence , B gets highest vote.</p>

Q17:

Find the arithmetic mean of given discrete data. 

Wages(rs.) 120 150 175 200 225 300
No. of workers 12 17 11 13 4 3

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <table width="331"><tbody><tr><td>Wages(in rs.)</td> <td>Frequency(f)</td> <td>fx</td> </tr><tr><td>120<br>150<br>175<br>200<br>225<br>300</td> <td>12<br>17<br>11<br>13<br>4<br>3</td> <td>1440<br>2250<br>1925<br>2600<br>900<br>900</td> </tr><tr><td></td> <td>N = \(\sum\)f = 60</td> <td>\(\sum\)fx = 10315</td> </tr></tbody></table><p>By formula ,<br>Mean = \(\overline {x}\) = \(\frac{&sum;fx}{N}\) = \(\frac{10315}{60}\) = 179.12 Ans.</p>

Q18:

Find the arithmetic mean. 

Marks Obtained 32 40 45 50 55 65 82
No. of students  9 12 17 13 11 5 4

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <table width="299"><tbody><tr><td>Marks obtained(x)</td> <td>Frequency(f)</td> <td>fx</td> </tr><tr><td>32<br>40<br>45<br>50<br>55<br>65<br>82</td> <td>9<br>12<br>17<br>13<br>11<br>5<br>4</td> <td>288<br>480<br>765<br>650<br>605<br>325<br>328</td> </tr><tr><td></td> <td></td> <td></td> </tr></tbody></table><p>By formula ,<br>Mean = \(\overline {x}\) = \(\frac{&sum;fx}{N}\) = \(\frac{3441}{71}\) = 48.46 Ans.</p>

Q19:

The mean of the given data is 17 , what is the value of x ?

Marks Obtained 5 10 15 20 `25 30
No. of students  2 5 10 x 4 2

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <table width="424"><tbody><tr><td>Marks Obtained(m)</td> <td>Frequency(f)</td> <td>fm</td> </tr><tr><td>5<br>10<br>15<br>20<br>25<br>30</td> <td>2<br>5<br>10<br>x<br>4<br>2</td> <td>10<br>50<br>150<br>20x<br>100<br>60</td> </tr><tr><td></td> <td>N = \(\sum\)f = 23 + x</td> <td>\(\sum\)fm = 370 + 20x</td> </tr></tbody></table><p>By formula<br>Mean = \(\overline {m}\) = \(\frac{\sum fx}{N}\) = \(\frac{370 + 20x}{23 + x}\)<br>According to question ,<br>Mean = \(\overline{m}\) = 17<br>17 = \(\frac{370 + 20x}{23 + x}\)<br>or , 391 + 17x = 370 + 20x<br>or , 391 - 370 = 20x - 17x<br>or , 21 = 3x<br>x = \(\frac{21}{3}\) = 7 Ans.</p>

Q20:

If the mean of the given data below is 5 . 6, find the value of p:

x 2 4 6 8 10
f 7 4 p 5 4

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <table width="245"><tbody><tr><td>x</td> <td>f</td> <td>fx</td> </tr><tr><td>2<br>4<br>6<br>8<br>10</td> <td>7<br>4<br>p<br>5<br>4</td> <td>14<br>16<br>6p<br>40<br>40</td> </tr><tr><td></td> <td>N = \(\sum\)f = 20 + p</td> <td>\(\sum\)fx = 110 + 6p</td> </tr></tbody></table><p><br>By formula ,<br>Mean \(\overline {x}\) = \(\frac{\sum fx}{N}\)<br>or , 5.6 = \(\frac{110 + 6P}{20 + P}\)<br>OR , 112 + 5.6p = 110 + 6p<br>or , 112 -110 = 6p - 5.6p<br>or, 2 = 0.4 \(\times\) p<br>\(\therefore\) p = 5 Ans.</p>

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Octal Number System

Octal Number System

The number with base eight is called octal number. It is represented by Q or O. We can generate these numbers with the combination of 0, 1, 2 3, 4, 5, 6, 7. We can represent these numbers wit suffix eight. Eg (5432)8


Weighted value

85 84 83 82 81 80
32768 4096 512 64 8 1

Conversions from Decimal to Octal

Decimal number is repetitively divided by eight and remainders are arranged in the form of octal numbers.

Example

Convert (240)10into octal.

8 240 0
8 30 6
3

(240)10= (360)8

Convert (356)10into octal.

8 356 4
8 44 4
4

(365)10=(444)8


Conversions from Octal to Decimal.

Each octal is multiplied by its weighted position. The sum of all products is known as decimal form of octal.

Example

Convert the octal numbers into decimal

  1. (340)8= 3 x 82+ 4 x 81+ 0 x 80=192 + 32 + 0 = (224)10
  2. (175)8= 1 x82+7 x 81+ 5 x80= (125)10

Octal to Binary and Binary to Octal Conversions

The three digit format of binary digits is used for octal to binary conversions or vice versa.

Octal to Binary

3-bits binary numbers are written for each octal digit.

Example

Convert (56)8 (octal) into binary.

Algorithm:

  • Convert each octal digit into binary and make three digits grouping.
    5= 101
    6= 110

Now our equivalent binary number is (56)8=(101110)2

Convert (64102)8 into Binary number.

6= 110
4= 100
1= 001
0= 000
2= 010

Now our equivalent binary number is (64102)8= (110100001000010)2

Binary to Octal

The binary numbers are broken into 3-bits section from last bit and convert into octal equivalent of each binary section. Table shows the decimal, octal and equivalent binary bits.

Decimal Octal Binary
0 0 000
1 1 001
2 2 010
3 3 011
4 4 100
5 5 101
6 6 110
7 7 111

Example

Convert (10011)2 into octal number.

Algorithm

  • make 3-bits grouping from last bit.
    010 011
  • convert each group into decimal numbers.
    010 = 2
    011 = 3
    = (23)8

Convert (1011010011)2 into octal number.

001 011 010 011

001= 1
011= 3
010= 2
011= 3

= (1323)8

Octal to Hexadecimal

Algorithm

  • Convert octal into binary.
  • make groups of 4-bits from last bit.
  • convert each group into decimal numbers.

Example

1. (34765)8into Hexadecimal number.

  • Convert into binary
    3= 011, 4= 100, 7=111, 6= 110, 5=101
    (011100111110101)2
  • Make a four digits group from the last bit of the binary number.
    0011 1001 1111 0101 (Add 0 before the group to make four digit)
    3 9 15 5 (Convert into equivalent decimal numbers)
    =(39F5)16Where F= 15

Lesson

Computer Number System

Subject

Computer Science

Grade

Grade 10

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