Data Representation — Question 9

(a) 10011011101

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0100} \quad \underlinesegment{1101} \quad \underlinesegment{1101}$

Therefore, (10011011101)₂ = (4DD)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{011} \quad \underlinesegment{011} \quad \underlinesegment{101}$

Therefore, (10011011101)₂ = (2335)₈

(b) 1111011101011011

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1111} \quad \underlinesegment{0111} \quad \underlinesegment{0101} \quad \underlinesegment{1011}$

Therefore, (1111011101011011)₂ = (F75B)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{101} \quad \underlinesegment{011} \quad \underlinesegment{011}$

Therefore, (1111011101011011)₂ = (173533)₈

(c) 11010111010111

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0011} \quad \underlinesegment{0101} \quad \underlinesegment{1101} \quad \underlinesegment{0111}$

Therefore, (11010111010111)₂ = (35D7)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{011} \quad \underlinesegment{010} \quad \underlinesegment{111} \quad \underlinesegment{010} \quad \underlinesegment{111}$

Therefore, (11010111010111)₂ = (32727)₈

(d) 1010110110111

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1011} \quad \underlinesegment{0111}$

Therefore, (1010110110111)₂ = (15B7)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{110} \quad \underlinesegment{111}$

Therefore, (1010110110111)₂ = (12667)₈

(e) 10110111011011

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0010} \quad \underlinesegment{1101} \quad \underlinesegment{1101} \quad \underlinesegment{1011}$

Therefore, (10110111011011)₂ = (2DDB)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{011}$

Therefore, (10110111011011)₂ = (26733)₈

(f) 1111101110101111

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1111} \quad \underlinesegment{1011} \quad \underlinesegment{1010} \quad \underlinesegment{1111}$

Therefore, (1111101110101111)₂ = (FBAF)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{111} \quad \underlinesegment{101} \quad \underlinesegment{110} \quad \underlinesegment{101} \quad \underlinesegment{111}$

Therefore, (1111101110101111)₂ = (175657)₈