Question

Try the following conversions.

(i) $(514)_{8}=\left(\right. ? {) }_{10}$

(ii) $(220)_{8}=(\text { ? })_{2}$

(iii) $(76 \mathrm{~F})_{16}=(\text { ? })_{10}$

(iv) $(4 \mathrm{D} 9)_{16}=(?)_{10}$

(v) $(11001010)_{2}=(?)_{10}$

(vi) $(1010111)_{2}=(\text { ? })_{10}$

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Answer

Conversion (i) $ (514)_8 = (?)_{10} $

To convert from octal to decimal:

  • $ 514_8 = 5 \times 8^2 + 1 \times 8^1 + 4 \times 8^0 $

  • $ 5 \times 64 + 1 \times 8 + 4 \times 1 $

  • $ 320 + 8 + 4 = 332 $

So, $(514)_8 = (332)_{10}$

Conversion (ii) $ (220)_8 = (?)_{2} $

To convert from octal to binary:

  • Replace each octal digit with its 3-bit binary equivalent:

    • 2 → 010

    • 2 → 010

    • 0 → 000

  • Combine the binary values: 0 100 100 000

So, $(220)_8 = (100100000)_2$

Conversion (iii) $ (76F)_{16} = (?)_{10} $

To convert from hexadecimal to decimal:

  • $ 76F_{16} = 7 \times 16^2 + 6 \times 16^1 + 15 \times 16^0 $

  • $ 7 \times 256 + 6 \times 16 + 15 $

  • $ 1792 + 96 + 15 = 1903 $

So, $(76F)_{16} = (1903)_{10}$

Conversion (iv) $ (4D9)_{16} = (?)_{10} $

To convert from hexadecimal to decimal:

  • $ 4D9_{16} = 4 \times 16^2 + 13 \times 16^1 + 9 \times 16^0 $

  • $ 4 \times 256 + 13 \times 16 + 9 $

  • $ 1024 + 208 + 9 = 1241 $

So, $(4D9)_{16} = (1241)_{10}$

Conversion (v) $ (11001010)_{2} = (?)_{10} $

To convert from binary to decimal:

  • Compute the positional value:

    • $ 1 \times 2^7 + 1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 $

    • $ 1 \times 128 + 1 \times 64 + 0 \times 32 + 0 \times 16 + 1 \times 8 + 0 \times 4 + 1 \times 2 + 0 \times 1 $

    • $ 128 + 64 + 8 + 2 = 202 $

So, $(11001010)_2 = (202)_{10}$

Conversion (vi) $ (1010111)_{2} = (?)_{10} $

To convert from binary to decimal:

  • Compute the positional value:

    • $ 1 \times 2^6 + 0 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 $

    • $ 1 \times 64 + 0 \times 32 + 1 \times 16 + 0 \times 8 + 1 \times 4 + 1 \times 2 + 1 \times 1 $

    • $ 64 + 16 + 4 + 2 + 1 = 87 $

So, $(1010111)_2 = (87)_{10}$

Summary:

  • $(514)_8 = (332)_{10}$

  • $(4D9)_{16} = (1241)_{10}$

  • $(220)_8 = (100100000)_2$

  • $(11001010)_2 = (202)_{10}$

  • $(76F)_{16} = (1903)_{10}$

  • $(1010111)_2 = (87)_{10}$


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