Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 3 (Q.No. 3)
3.
Decimal number 46 in excess 3 code =
Answer: Option
Explanation:
Decimal 46 in excess 3 code = 46 + 33 = 79 in decimal = 0111 1001 in 4 bit binary.
Discussion:
7 comments Page 1 of 1.
Krishna said:
3 years ago
Thanks @Rohit.
Rohit said:
4 years ago
0100 0110
+ 0011 0011 individual 3 binary code add then;
---------------------------
0111 1001
+ 0011 0011 individual 3 binary code add then;
---------------------------
0111 1001
(1)
Jagdish sonagara said:
5 years ago
Right answer is 1001111.
Ch@ndu said:
5 years ago
Excess 3 is obtained by 3 to the individual digit.
Sunil said:
5 years ago
Why we can not write it 46+3=49?
Please tell me.
Please tell me.
Ashok said:
7 years ago
You are right @Shreekar.
But we add 3 for individual digit.
46 = 4+3 and 6+3 so its 79.
But we add 3 for individual digit.
46 = 4+3 and 6+3 so its 79.
(1)
Shreekar said:
7 years ago
The excess 3 code should be 3 more than decimal code not 33.
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