Question Number 16840 by Sai dadon. last updated on 27/Jun/17
$$\mathrm{6}/\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right) \\ $$
Commented by RasheedSoomro last updated on 27/Jun/17
$$\mathrm{6}/\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right) \\ $$$$=\mathrm{6}/\left\{\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right)\right\} \\ $$$$=\mathrm{6}/\left(\mathrm{2}×\mathrm{3}\right) \\ $$$$=\mathrm{6}/\mathrm{6} \\ $$$$=\mathrm{1} \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{question}\:\mathrm{were} \\ $$$$\mathrm{6}/\mathrm{2}×\left(\mathrm{2}+\mathrm{1}\right) \\ $$$$=\mathrm{3}×\mathrm{3} \\ $$$$=\mathrm{9} \\ $$
Commented by Sai dadon. last updated on 27/Jun/17
$${what}\:{if}\:{will}\:{be} \\ $$$$\mathrm{6}\boldsymbol{\div}\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right) \\ $$
Commented by RasheedSoomro last updated on 27/Jun/17
$$\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{single}\:\mathrm{term}. \\ $$$$\mathrm{hidden}\:\mathrm{multiplication}\:\mathrm{between} \\ $$$$\mathrm{2}\:\mathrm{and}\:\left(\mathrm{2}+\mathrm{1}\right)\:\mathrm{has}\:\mathrm{priority}\:\mathrm{over}\: \\ $$$$\mathrm{division}. \\ $$