Electronics and Communication Engineering - Digital Electronics - Discussion

Discussion Forum : Digital Electronics - Section 2 (Q.No. 18)
18.
AECF116 + 15ACD16 = __________ .
C47BB16
C47BE16
A234F16
A111116
Answer: Option
Explanation:

Convert to decimal, add and change the result to hexadecimal.

Discussion:
2 comments Page 1 of 1.

M.chandrashekar said:   8 years ago
AECF1=716017
15ACD=88781
88781+716017=804798,
804798=C47BE.

SWATHI said:   10 years ago
Just write question correctly ACEF1+15ACD = C47BE.

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