Arrange the following in ascending order of magnitude $110_{two}, 31_{eight}, 42_{five}$

$110_{two}, 31_{five}, 42_{eight}$

$42_{five}, 110_{two}, 31_{eight}$

$42_{five}, 31_{eight}, 110_{two}$

$110_{two}, 42_{five}, 31_{eight}$

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, $110_{two}, 42_{five}, 31_{eight}$

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