UGC-NET | UGC-NET CS 2017 Dec 2 | Question 7

The Octal equivalent of the binary number 1011101011 is :
(A) 7353
(B) 1353
(C) 5651
(D) 5657


Answer: (B)

Explanation: First we will make pair of 3 bits from LSB :
ie- 1 011 101 011
now convert these bits into decimal and that will be
1 – 1
011 – 3
101 – 5
011 – 3
so the answer will be 1353.
So, option (B) is correct.

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