Computer Science, asked by sumanvipin, 5 hours ago

Find out the values of X. Y and Z in the following
(82.875)10 = (X)2 = (Y)8 = (Z)16

Answers

Answered by amitnrw
2

Given :   (82.875)₁₀  = X₂ =  Y₈  = Z₁₆

To Find : Values of X , Y , Z

Solution:

(82.875)₁₀

         82

 2       41      0

 2       20     1

 2       10      0

  2        5     0

   2        2     1

   2       1       0

   2        0      1

(82)₁₀  =  (1010010)₂

0.875  * 2  =     1.75          1

0.75    * 2   =     1.5             1

0.5      * 2    =     1                1

(0.875)₁₀ = (0.111)₂

Hence  (82.875)₁₀ = ( 1010010.111)₂

          82

   8     10        2

   8       1        2

   8       0        1

(82)₁₀  =  (122)₈

0.875  * 8  =    7          1

 (82.875)₁₀ = ( 122.7)₈

            82

   16      5        2

    16      0       5

0.875  * 16  =    14    = E

 (82.875)₁₀ = ( 52.E)₁₆

X = 1010010.111

Y  =  122.7

Z =  52.E

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amansharma264: Perfect answer sir
Answered by pulakmath007
12

SOLUTION

TO DETERMINE

The value of X, Y, Z when

 \sf{(82.875)_{10} =  (X)_2 = ( Y)_8 = (Z)_{16}}

EVALUATION

Here the given decimal number is

 \sf{(82.875)_{10} }

Now

 \sf{(82.875)_{10} =  (82)_{10}  +  (0.875)_{10}}

CONVERT TO BINARY

\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:82 \:\:} \:  \: \:  \:  \:  }}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:41 \:\:} \boxed{0}} \\\underline{\sf{2}}&\underline{\sf{\:\:20\:\:} \boxed{1}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:10 \:\:} \boxed{0}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:5 \:  \: \:} \boxed{0}}\\\underline{\sf{2}}& \underline{\sf{\:\:2\:\:} \boxed{1}} \\ \underline{\sf{}}&{\sf{\:\:1 \:\:} \boxed{0}} \end{array}\end{gathered}\end{gathered}\end{gathered}

We write down the Remainders in the boxes

Therefore

 \sf{(82)_{10} =  (1010010)_{2} }

Again

 \sf{0.875 \times 2 =  \boxed{1}.750}

 \sf{0.750 \times 2 =  \boxed{1}.500}

 \sf{0.500 \times 2 =  \boxed{1}.000}

Therefore

 \sf{(0.875)_{10} = (0.111)_2}

Combining we get

 \sf{(82.875)_{10} =  (1010010.111)_{2} }

Thus we get

 \boxed{ \sf{ \:  \: X =  1010010.111 \:  \: } }

CONVERT TO OCTAL

 \sf{(82.875)_{10} =  (1010010.111)_{2} }

PROCESS :

In this method we group the binary digits into three ( 3 bits form ) beginning from the right to left, writing exta zeros on the left side, if needed to make the number of bits into a multiple of 3 . Then we can directly write the equivalent decimal number for each group of binary to get the equivalent octal number.

Computation :

 \sf{ (1010010.111)_{2} }

 \sf{   = \big(  \: \boxed{ 001} \:  \boxed{010} \:  \boxed{010}. \boxed{111} \:  \big)_{2} }

 \sf{   = (  122.7 )_{8} }

This gives

 \sf{(82.875)_{10} =  (122.7)_{8} }

Thus we get

  \boxed{\sf{ \:  Y  = 122.7\:  \: }}

CONVERT TO HEXADECIMAL

 \sf{(82.875)_{10} =  (1010010.111)_{2} }

PROCESS :

In this method we group the binary digits into four ( 4 bits form ) beginning from the right to left, writing exta zeros on the left side, if needed to make the number of bits into a multiple of 4 . Then we can directly write the equivalent decimal number for each group of binary to get the equivalent hexadecimal number.

Computation :

 \sf{ (1010010.111)_{2} }

 \sf{   = \big(  \: \boxed{ 0101} \:  \boxed{0010}. \boxed{1110} \:  \big)_{2} }

 \sf{   = \big(\: 52.E  \: \big)_{16} }

Since in hexadecimal E = 14

This gives

 \sf{(82.875)_{10} =  (52.E   )_{16}}

Thus we get

 \boxed{ \:  \:  \sf{Z\:  = 52.E  } \:  \: }

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amansharma264: Excellent answer sir
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