Question
Find the sum:
$\frac{7}{12}+\frac{9}{16}$
$\frac{7}{12}+\frac{9}{16}$
✓
Answer
L.C.M. of 12 and 16 = (2 × 2 × 2 × 2 × 3) = 48
$\begin{array}{c|c}2&12,16\\\hline2&6,8\\\hline2&3,4\\\hline2&3,2\\\hline3&3,1\\\hline&1,1\end{array}$
Now, we know:
$\frac{7}{12}=\frac{7\times4}{12\times4}=\frac{28}{48}$
$\frac{9}{16}=\frac{9\times3}{16\times3}=\frac{27}{48}$
Therefore,
$\frac{7}{12}+\frac{9}{16}$
$=\frac{28}{48}+\frac{27}{48}$
$=\frac{(28+27)}{48}$
$=\frac{55}{48}$
$=1\frac{7}{48}$
$\begin{array}{c|c}2&12,16\\\hline2&6,8\\\hline2&3,4\\\hline2&3,2\\\hline3&3,1\\\hline&1,1\end{array}$
Now, we know:
$\frac{7}{12}=\frac{7\times4}{12\times4}=\frac{28}{48}$
$\frac{9}{16}=\frac{9\times3}{16\times3}=\frac{27}{48}$
Therefore,
$\frac{7}{12}+\frac{9}{16}$
$=\frac{28}{48}+\frac{27}{48}$
$=\frac{(28+27)}{48}$
$=\frac{55}{48}$
$=1\frac{7}{48}$
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