OperatingSystems
October 3, 2023OperatingSystems
October 3, 2023Nielit Scientific Assistance CS 15102017
Question 1

A decimal has 25 digits. the number of bits needed for its equivalent binary representation is approximately
50


74


40


None of the above

Question 1 Explanation:
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10^{ 3} 1 which is 999.
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2 ^{25} 1
So we can write 10 ^{25} –1 = 2^{ n} – 1 → 10^{ 25} = 2^{ n}
After taking log _{2} on both sides
log _{2} 2 ^{n} =log _{ 2} 10 ^{25}
n log _{2} 2=25 log _{2} 10
n = 25 log _{2} 10
n = 25 x 3.3 [ log _{2} 2=1 & log _{2} 10 =3.322]
n = 82.5
Note: Original question paper given option D is 60. But actual answer is 82.5.
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2 ^{25} 1
So we can write 10 ^{25} –1 = 2^{ n} – 1 → 10^{ 25} = 2^{ n}
After taking log _{2} on both sides
log _{2} 2 ^{n} =log _{ 2} 10 ^{25}
n log _{2} 2=25 log _{2} 10
n = 25 log _{2} 10
n = 25 x 3.3 [ log _{2} 2=1 & log _{2} 10 =3.322]
n = 82.5
Note: Original question paper given option D is 60. But actual answer is 82.5.
Correct Answer: D
Question 1 Explanation:
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10^{ 3} 1 which is 999.
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2 ^{25} 1
So we can write 10 ^{25} –1 = 2^{ n} – 1 → 10^{ 25} = 2^{ n}
After taking log _{2} on both sides
log _{2} 2 ^{n} =log _{ 2} 10 ^{25}
n log _{2} 2=25 log _{2} 10
n = 25 log _{2} 10
n = 25 x 3.3 [ log _{2} 2=1 & log _{2} 10 =3.322]
n = 82.5
Note: Original question paper given option D is 60. But actual answer is 82.5.
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2 ^{25} 1
So we can write 10 ^{25} –1 = 2^{ n} – 1 → 10^{ 25} = 2^{ n}
After taking log _{2} on both sides
log _{2} 2 ^{n} =log _{ 2} 10 ^{25}
n log _{2} 2=25 log _{2} 10
n = 25 log _{2} 10
n = 25 x 3.3 [ log _{2} 2=1 & log _{2} 10 =3.322]
n = 82.5
Note: Original question paper given option D is 60. But actual answer is 82.5.
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