A decimal number has 25 digits. The number of bits needed for its equivalent binary representation is, approximately
A. | 50 |
B. | 55 |
C. | 85 |
D. | 80 |
Option: C Explanation : The largest decimal number with 25 digits is 9,999,999,999,999,999,999,999,999. The smallest decimal number in the form 2n-1 which is greater than or equal to that is 19,342,813,113,834,066,795,298,815. That corresponds to 284-1. So, the minimum number of binary bits required to represent the decimal number 25 nines in a row is 84. This is 84 ones in a row. If you want to support negative as well as positive numbers, you will need 85. Click on Discuss to view users comments. ARUMUGAM said: (4:18pm on Wednesday 15th May 2013)
Generalised Expalanation:Let k bits are required to represent d digit decimal number.(ie) Let 2 power k = 10 power dApply base 2 logarithm on both sideslog(2 power k) base 2 = log (10 power d) base 2k log2 base 2 = d log 10 base 2As log10 base 2=3.32192 and log2 base2 =1,k=3.32192dHere d=25,k=3.32192*25k=83
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A. | 10000 |
B. | 01011 |
C. | 01100 |
D. | 01101 |
Option: B Explanation : Click on Discuss to view users comments. |
The decimal number 80 can be represented in BCD code as
A. | 1000 0001 |
B. | 0101 0000 |
C. | 0010 0001 |
D. | 1000 0000 |
Option: D Explanation : Click on Discuss to view users comments. |
How many fractional digits 2-48 will have?
A. | 24 |
B. | 47 |
C. | 49 |
D. | 48 |
Option: D Explanation : Click on Discuss to view users comments. |
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