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2021

Mathematics

POST UTME

The decimal number (127.25)10(127.25)_{10} , when converted to Binary number, takes the form

A.

111111.01

B.

111111.01

C.

1111.11

D.

11111.11

Correct Answer: 111111.01

Explanation

127.25 = 127 + 0.25

We divide the whole number part by 2 and multiply the decimal part by 2

thus,

2127263  R  1231  R  1215  R  127  R  123  R  121  R  10  R  1Next, we multiply the decimal part by 2:0.25×2=0.500.50×2=1.000.012And finally, we combine:11111112+0.012=1111111.012\begin{array}{r|l} 2 & 127 \\ 2 & 63 \; \text{R} \; 1 \\ 2 & 31 \; \text{R} \; 1 \\ 2 & 15 \; \text{R} \; 1 \\ 2 & 7 \; \text{R} \; 1 \\ 2 & 3 \; \text{R} \; 1 \\ 2 & 1 \; \text{R} \; 1 \\ & 0 \; \text{R} \; 1 \\ \end{array} \text{Next, we multiply the decimal part by 2:} \begin{array}{r} 0.25 \times 2 = 0.50 \\ 0.50 \times 2 = 1.00 \\ \end{array} \Rightarrow 0.01_2 \text{And finally, we combine:} \Rightarrow 1111111_2 + 0.01_2 = 1111111.01_2


127.25 = 127 + 0.25

We divide the whole number part by 2 and multiply the decimal part by 2

thus,

2127263  R  1231  R  1215  R  127  R  123  R  121  R  10  R  1Next, we multiply the decimal part by 2:0.25×2=0.500.50×2=1.000.012And finally, we combine:11111112+0.012=1111111.012\begin{array}{r|l} 2 & 127 \\ 2 & 63 \; \text{R} \; 1 \\ 2 & 31 \; \text{R} \; 1 \\ 2 & 15 \; \text{R} \; 1 \\ 2 & 7 \; \text{R} \; 1 \\ 2 & 3 \; \text{R} \; 1 \\ 2 & 1 \; \text{R} \; 1 \\ & 0 \; \text{R} \; 1 \\ \end{array} \text{Next, we multiply the decimal part by 2:} \begin{array}{r} 0.25 \times 2 = 0.50 \\ 0.50 \times 2 = 1.00 \\ \end{array} \Rightarrow 0.01_2 \text{And finally, we combine:} \Rightarrow 1111111_2 + 0.01_2 = 1111111.01_2


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