The LCM of the numbers $0.6, 9.6$ and $0.12$ is

0.10

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0.6

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0.12

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9.60

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Solution

The correct option is D $9.60$
$0.6 = \frac{60}{100}, 9.6 = \frac{96}{100}$ and $0.12 = \frac{12}{100}$ .
$60 = 2 \times 2 \times 3 \times 5$
$960 = 2 \times 2 \times 2 \times 2 \times 2 \times 3$
$12 = 2 \times 2 \times 3$
LCM of numerators for the three numbers i.e. $60, 960$ and $12$ is $960$ .
LCM of denominators of the three numbers $= 100$
Required LCM $= 9.60$ .
So, option $D$ is correct.

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Similar questions

0.12, 9.60

0.60

\frac{1}{\sqrt{0. ¯ ¯ ¯ 2} + \sqrt{0. ¯ ¯ ¯ 4}} + \frac{1}{\sqrt{0. ¯ ¯ ¯ 4} + \sqrt{0. ¯ ¯ ¯ 6}} + \frac{1}{\sqrt{0. ¯ ¯ ¯ 6} + \sqrt{0. ¯ ¯ ¯ 8}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 10} + \sqrt{0. ¯ ¯¯¯ ¯ 12}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 12} + \sqrt{0. ¯ ¯¯¯ ¯ 14}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 14} + \sqrt{0. ¯ ¯¯¯ ¯ 16}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 16} + \sqrt{0. ¯ ¯¯¯ ¯ 18}} + . . . . . .

. . . . . . . . . . . . . . . + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 38} + \sqrt{0. ¯ ¯¯¯ ¯ 40}} = \frac{3}{2} (\sqrt{a} + \sqrt{b}),

\frac{1}{\sqrt{0. ¯ ¯ ¯ 2} + \sqrt{0. ¯ ¯ ¯ 4}} + \frac{1}{\sqrt{0. ¯ ¯ ¯ 4} + \sqrt{0. ¯ ¯ ¯ 6}} + \frac{1}{\sqrt{0. ¯ ¯ ¯ 6} + \sqrt{0. ¯ ¯ ¯ 8}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 10} + \sqrt{0. ¯ ¯¯¯ ¯ 12}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 12} + \sqrt{0. ¯ ¯¯¯ ¯ 14}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 14} + \sqrt{0. ¯ ¯¯¯ ¯ 16}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 16} + \sqrt{0. ¯ ¯¯¯ ¯ 18}} + . . . . . .

. . . . . . . . . . . . . . . + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 38} + \sqrt{0. ¯ ¯¯¯ ¯ 40}} = \frac{3}{2} (\sqrt{a} + \sqrt{b}),

(\frac{0.12 \times (0.0104 - 0.002) + 0.36 \times 0.002) \times 100}{0.12 \times 0.12}

\frac{(0.0104 - 0.002) + 0.366 \times 0.002}{0.12 \times 0.12}

Simplify :

0.12×(0.0104-0.002)+0.36×0.002

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0.12×0.12

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