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Statement I: The conversion of binary nu...

Statement I: The conversion of binary number 1101 to decimal number can be written as `2^(4) + 2^(3) + 0 + 2^(1) = 26`.
Statement II : In binary system, base is 2 and the digits used are 0 and 1.

A

Statement I is true, statement II is true, statement II is a correct explanation for statement I.

B

Satement I is true, statement II is true, statement II is not a correct explanation for statement I.

C

Statement I is true, statement II is false.

D

Statement I is false, statement II is true.

Text Solution

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The correct Answer is:
D
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Knowledge Check

  • Statement I: The conversion of binary fraction 0.101 to decimal fraction can be written as 2^(-1) + 0 + 2^(-3) = 0.625. Statement II: In binary system, base is 2 and the digits used are 0 and 1.

    A
    Statement I is true, statement II is true, statement II is a correct explanation for statement I.
    B
    Satement I is true, statement II is true, statement II is not a correct explanation for statement I.
    C
    Statement I is true, statement II is false.
    D
    Statement I is false, statement II is true.
  • Stetement I: Decimal value of the binary number 111 is 7. Therefore, (0.111)_(2) = (7/(2^(3)))_(10) . Statement II: Decimal fraction 0.111 can be written as (111)/(10^(3)) .

    A
    Statement I is true, statement II is true, statement II is a correct explanation for statement I.
    B
    Satement I is true, statement II is true, statement II is not a correct explanation for statement I.
    C
    Statement I is true, statement II is false.
    D
    Statement I is false, statement II is true.
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