Arrange the following in ascending order of magnitude \(110_{two}, 31_{eight}, 42_{five}\)
A.
\(110_{two}, 31_{five}, 42_{eight}\)
B.
\(42_{five}, 110_{two}, 31_{eight}\)
C.
\(42_{five}, 31_{eight}, 110_{two}\)
D.
\(110_{two}, 42_{five}, 31_{eight}\)
Correct answer is D
Converting each number to base 10
\(110_{two} = 1 × 2^2 + 1 × 2^1 + 0 × 2^0\)
= 1 × 4 + 1 × 2 + 0 × 1
= 4 + 2 + 0
= \(6_{ten}\)
\(31_{eight} = 3 × 8^1 + 1 × 8^0\)
= 3 × 8 + 1 × 1
= 24 + 1
= \(25_{ten}\)
\(42_{five} = 4 × 5^1 + 2 × 5^0\)
= 4 × 5 + 2 × 1
= 20 + 2
= \(22_{ten}\)
Hence, \(31_{eight} > 42_{five} > 110_{two}\)
In ascending order, \(110_{two}, 42_{five}, 31_{eight}\)
