Digital Electronics - Number Systems and Codes - Discussion
Discussion Forum : Number Systems and Codes - Filling the Blanks (Q.No. 2)
2.
Hex 4B5258 is ________ in ASCII code.
Discussion:
4 comments Page 1 of 1.
KIRAN V said:
5 years ago
First, make a group of 2 - digits each.
4B5258 (4B) (52) (58)
Now convert Hexa-decimal to Decimal of each group.
4B = 4 * 16^1 + B * 16^0 = 64 + 11 = 75
52 = 5 * 16^1 + 2 * 16^0 = 80 + 2 = 82
58 = 5 * 16^1 + 8 * 16^0 = 80 + 8 = 88
Now Write the ASCII Equivalent of Each.
75 = K 82 = R 88 = X.
Therefore, Ans. = KRX.
4B5258 (4B) (52) (58)
Now convert Hexa-decimal to Decimal of each group.
4B = 4 * 16^1 + B * 16^0 = 64 + 11 = 75
52 = 5 * 16^1 + 2 * 16^0 = 80 + 2 = 82
58 = 5 * 16^1 + 8 * 16^0 = 80 + 8 = 88
Now Write the ASCII Equivalent of Each.
75 = K 82 = R 88 = X.
Therefore, Ans. = KRX.
Navneet said:
5 years ago
They are given in hexadecimal form.
So in Hexa form,
58=X.
52=R.
4B=K.
So in Hexa form,
58=X.
52=R.
4B=K.
Bhagya K S said:
6 years ago
Can anyone explain the logic behind this question?
Chandrika pradeep said:
8 years ago
Can anyone explain?
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