The correct option is D 111/12 51/6 31/4 =31/6 13/4 =62 39/12 (LCM of 6 and 4 is 12) =23/12=111/12 ...

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Subtraction of Fractions

51/6-31/4 =

Question

$5 \frac{1}{6} - 3 \frac{1}{4}$ =_____

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$1 \frac{5}{12}$

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$1 \frac{7}{12}$

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$1 \frac{11}{12}$

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Solution

The correct option is D
$1 \frac{11}{12}$

$5 \frac{1}{6} - 3 \frac{1}{4} = \frac{31}{6} - \frac{13}{4} = \frac{62 - 39}{12} (LCM of 6 and 4 is 12) = \frac{23}{12} = 1 \frac{11}{12}$

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Similar questions

\frac{1}{2} < \frac{9}{13} < \frac{3}{4} < \frac{12}{17}

\frac{3}{4} < \frac{9}{13} < \frac{1}{2} < \frac{12}{17}

\frac{1}{2} < \frac{3}{4} < \frac{9}{13} < \frac{12}{17}

\frac{1}{2} < \frac{9}{13} < \frac{12}{17} < \frac{3}{4}

\frac{2 \frac{5}{4} - 5 \frac{1}{6} + 3 \frac{1}{3}}{0.087 + 0.3717 + 0.0413}

Distance(km)	$1 - 11$	$11 - 21$	$21 - 31$	$31 - 41$	$41 - 51$
Number of persons	$4$	$6$	$10$	$12$	$18$

3 \times 5 \frac{1}{6} = 31

3 \times 5 \frac{1}{6} = \frac{31}{2}

1 + (\frac{1}{2} + \frac{1}{3}) \frac{1}{4} + (\frac{1}{4} + \frac{1}{5}) \frac{1}{4^{2}} + (\frac{1}{6} + \frac{1}{7}) \frac{1}{4^{3}} + . . . \infty = {log}_{e} \sqrt{b}

\frac{b}{4}

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