Question
Find the sum:
$3\frac{1}{3}+4\frac{1}{4}+6\frac{1}{6}$
$3\frac{1}{3}+4\frac{1}{4}+6\frac{1}{6}$
✓
Answer
We have,
L.C.M. of 3, 4 and 6 = (2 × 2 × 3) = 12
$\begin{array}{c|c}3&3,4,6\\\hline2&1,4,2\\\hline2&1,2,1\\\hline&1,1,1\end{array}$
Therefore,
$3\frac{1}{3}+4\frac{1}{4}+6\frac{1}{6}$
$=\frac{10}{3}+\frac{17}{4}+\frac{37}{6}$
$=\frac{(40+51+74)}{12}$
$\Big(\frac{12}{3}=4,4\times10=40\Big)$
$\Big(\frac{12}{4}=3,3\times17=51\Big)$
and $\Big(\frac{12}{6}=2,2\times37=74\Big)$
$=\frac{165}{12}$
$=\frac{55}{4}$
$=13\frac{3}{4}$
L.C.M. of 3, 4 and 6 = (2 × 2 × 3) = 12
$\begin{array}{c|c}3&3,4,6\\\hline2&1,4,2\\\hline2&1,2,1\\\hline&1,1,1\end{array}$
Therefore,
$3\frac{1}{3}+4\frac{1}{4}+6\frac{1}{6}$
$=\frac{10}{3}+\frac{17}{4}+\frac{37}{6}$
$=\frac{(40+51+74)}{12}$
$\Big(\frac{12}{3}=4,4\times10=40\Big)$
$\Big(\frac{12}{4}=3,3\times17=51\Big)$
and $\Big(\frac{12}{6}=2,2\times37=74\Big)$
$=\frac{165}{12}$
$=\frac{55}{4}$
$=13\frac{3}{4}$
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