Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 1 (Q.No. 19)
19.
Decimal 43 in hexadecimal and BCD number system is respectively.
Answer: Option
Explanation:
(43)10 = (2B)16
(43)10 = (01000011)2 .
Discussion:
17 comments Page 2 of 2.
AMIR AHMAD said:
10 years ago
00101011 = 2B(H)?
For hexadecimal.
0010 = 2.
1011 = B.
For hexadecimal.
0010 = 2.
1011 = B.
B.Mallesh goud said:
10 years ago
How 00101011 = 2B ? Please explain.
Sateesh said:
1 decade ago
One simple logic among answers is 43 is odd number. So always LSB is 1 and Hexadecimal also odd number should come as left nibble like (B).
Arora rox said:
1 decade ago
For hexadecimal to binary conversion of (43) which is in decimal form, convert 4 and the 3 into 4 bits separately, and then combine the answer.
4 = 0100.
3 = 0011.
43 = 01000011.
4 = 0100.
3 = 0011.
43 = 01000011.
Surekha said:
1 decade ago
Hex to decimal:
Divide 43 by 2 as shown:
43/2 = 21 rem 1 again divide 21 with 2.
21/2 = 10 rem 1.
10/2 = 5 rem 0.
5/2 = 2 rem 1.
2/2 = 1 rem 0.
00101011 = 2B.
Divide 43 by 2 as shown:
43/2 = 21 rem 1 again divide 21 with 2.
21/2 = 10 rem 1.
10/2 = 5 rem 0.
5/2 = 2 rem 1.
2/2 = 1 rem 0.
00101011 = 2B.
Ashmita Prakash said:
1 decade ago
Can you please elaborate the explanation?
Rathi said:
1 decade ago
In a binary system all possible combinations of bits are valid.
In a BCD system there are 4 bits per digit. And the only valid combinations are 0(0000 Binary) through to 9(1001 Binary). If you want more digits then you have to add another 4-bit set.
So the number 43 would be represented by 101011 in binary and 0100 0011 in BCD.
In a BCD system there are 4 bits per digit. And the only valid combinations are 0(0000 Binary) through to 9(1001 Binary). If you want more digits then you have to add another 4-bit set.
So the number 43 would be represented by 101011 in binary and 0100 0011 in BCD.
(1)
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