Question

Find the sum:
$2\frac{1}{3}+1\frac{1}{4}+2\frac{5}{6}+3\frac{7}{12}$

Answer 1

We have:


$$\begin{gathered} 2\frac{1}{3}+1\frac{1}{4}+2\frac{5}{6}+3\frac{7}{12} \\ =\frac{7}{3}+\frac{5}{4}+\frac{17}{6}+\frac{43}{12}\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}3,4,6\mathrm{and}12=(2\times 2\times 3)=12 \\ =\frac{\left(28+15+34+43\right)}{12}\left\{\left[12÷3=4,4\times 7=28\right],\left[12÷4=3,3\times 5=15\right],\left[12÷6=2,2\times 17=34\right]\mathrm{and}\left[12÷12=1,1\times 43=43\right]\right\} \\ =\frac{{\cancel{120}}^{10}}{{\cancel{12}}_1}=10 \end{gathered}$$
234+556