Question

7. Conversion.
1 1. Convert (1010110100), into its octal equivalent.
2. Convert (6 FE 4), into its binary equivalent.
3. Convert (493), into its octal equivalent.​

Answer 1

Answer:

$\sf 1) (1010110100)_2=(1264)_8$

$\sf 2) (6FE4)_{16}=(110111111100100)_2$

$\sf 3) (493)_{10}=(755)_8$

Concept:

• For converting binary to octal , we first make three pairs of the binary number . After making pairs , we convert the pairs of three digits into octal numbers.

• For converting hexadecimal to binary, we first write the binary equivalent of the hexadecimal.

• For converting decimal to octal , we have to first write the decimal into binary . And then , we make pairs of three digits and write the octal equivalent.

Explanation:

1).

$\begin{array}{cccc}\underline{001}&\underline{010}&\underline{110}&\underline{100}\\ \downarrow&\downarrow&\downarrow&\downarrow\\1&2&6&4\end{array}$

Therefore, the answer is :

$(1010110100)_2\:=\:(1264)_8$

2).

$\begin{array}{cccc}6&F&E&4\\\downarrow&\downarrow&\downarrow&\downarrow\\0110&1111&1110&0100\end{array}$

Therefore, the answer is:

$(6FE4)_{16}\:=\:(110111111100100)_2$

3).

First converting Decimal to Binary:

$\setlength{\unitlength}{1cm}\thicklines \begin{picture}(10,10) \multiput(0,0)(0,0.8){3}{ \line(1,0){3}} \put(1, - 0.8){ \line(0,1){3}} \multiput(0.2,0.2)(0,0.8){3}{ \huge{8}} \put(1.5,1.8){ \huge{493}}\put(1.5,1){ \huge{60} \quad \: \vector(1,0){1} 3}\put(1.5,0.2){ \huge{7} \quad \: \: \vector(1,0){1} 4}\put(1.5, - 0.6){ \huge{0} \qquad \vector(1,0){1} 7} \put(6, - 0.7){\vector(0,1){3}} \end{picture}$

Now converting Binary to Octal:

$\begin{array}{ccc}\underline{111}&\underline{101}&\underline{101}\\\downarrow&\downarrow&\downarrow\\7&5&5\end{array}$

Therefore, the answer is:

$(493)_{10}\:=\:(743)_8$

Other Information:

You can refer to the below table for:

Binary code of all Decimal digits (0-9):

$\begin{array}{|c|c|} \cline{1-2} \sf Decimal \:Number& \sf Binary\: Equivalent\\ \cline{1-2} 0 & 0\\ \cline{1-2} 1 & 1\\ \cline{1-2} 2 & 10\\ \cline{1-2} 3 & 11\\ \cline{1-2} 4 & 100\\ \cline{1-2} 5 & 101\\ \cline{1-2} 6 & 110\\ \cline{1-2} 7 & 111\\ \cline{1-2} 8 & 1000\\ \cline{1-2} 9 & 1001\\ \cline{1-2} \end{array}$

Binary Code of all Octal Digits (0-7):

$\begin{array}{|c|c|} \cline{1-2} \sf Octal\: Number& \sf Binary \: Equivalent \\ \cline{1-2} 0 & 000\\ \cline{1-2} 1 & 001\\ \cline{1-2} 2 & 010\\ \cline{1-2} 3 & 011\\ \cline{1-2} 4 & 100\\ \cline{1-2} 5 & 101\\ \cline{1-2} 6 & 110\\ \cline{1-2} 7 & 111\\ \cline{1-2} \end{array}$

Binary Code of all Hexadecimal Digits (0-9&A-F)

$\begin{array}{|c|c|} \cline{1-2} \sf Hexadecimal \: Number& \sf Binary \: Equivalent \\ \cline{1-2} 0 & 0000\\ \cline{1-2} 1 & 0001\\ \cline{1-2} 2 & 0010\\ \cline{1-2} 3 & 0011\\ \cline{1-2} 4 & 0100\\ \cline{1-2} 5 & 0101\\ \cline{1-2} 6 & 0110\\ \cline{1-2} 7 & 0111\\ \cline{1-2} 8 & 1000\\ \cline{1-2} 9 & 1001\\ \cline{1-2} A(10) & 1010\\ \cline{1-2} B(11) & 1011\\ \cline{1-2} C(12) & 1100\\ \cline{1-2} D(13) & 1101\\ \cline{1-2} E(14) & 1110\\ \cline{1-2} F(15) & 1111 \\ \cline{1-2} \end{array}$

Answer 2

Answer:

2 . 6 E F4

answer = 0110111111100100

Explanation:

6= 0110 , F =1111 ,E=1110 , 4= 0100

now , write it opposite