Question

Find the sum:
$3\frac{1}{3}+4\frac{1}{4}+6\frac{1}{6}$

Answer 1

We have:



$$\begin{gathered} 3\frac{1}{3}+4\frac{1}{4}+6\frac{1}{6} \\ =\frac{10}{3}+\frac{17}{4}+\frac{37}{6}\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}3,4\mathrm{and}6=(2\times 2\times 3)=12 \\ =\frac{\left(40+51+74\right)}{12}\left\{\left[12÷3=4,4\times 10=40\right],\left[12÷4=3,3\times 17=51\right]\mathrm{and}\left[12÷6=2,2\times 37=74\right]\right\} \\ =\frac{{\cancel{165}}^{55}}{{\cancel{12}}_4}=\frac{55}{4}=13\frac{3}{4} \end{gathered}$$
234+556