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Question

Replace $\overline{)}$ by the correct number: (i) $\overline{)}-\frac{5}{8}=\frac{1}{4}$ (ii) $\overline{)}-\frac{1}{5}=\frac{1}{2}$ (iii) $\frac{1}{2}-\overline{)}=\frac{1}{6}$

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Solution

$\left(\mathrm{i}\right)\square -\frac{5}{8}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\square =\frac{1}{4}+\frac{5}{8}\phantom{\rule{0ex}{0ex}}\square =\frac{1×2}{4×2}+\frac{5×1}{8×1}\left(\mathrm{Because}\mathrm{LCM}\mathrm{of}4&8\mathrm{is}8\right)\phantom{\rule{0ex}{0ex}}\square =\frac{2}{8}+\frac{5}{8}=\frac{2+5}{8}=\frac{7}{8}\phantom{\rule{0ex}{0ex}}\square =\frac{7}{8}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ $\left(\mathrm{ii}\right)\square -\frac{1}{5}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\square =\frac{1}{2}+\frac{1}{5}\phantom{\rule{0ex}{0ex}}\square =\frac{1×5}{2×5}+\frac{1×2}{5×2}=\frac{5}{10}+\frac{2}{10}=\frac{5+2}{10}\phantom{\rule{0ex}{0ex}}\square =\frac{7}{10}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ $\left(\mathrm{iii}\right)\frac{1}{2}-\square =\frac{1}{6}\phantom{\rule{0ex}{0ex}}\square =\frac{1}{2}-\frac{1}{6}\phantom{\rule{0ex}{0ex}}\square =\frac{1×3}{2×3}-\frac{1×1}{6×1}\left(\mathrm{Because}\mathrm{LCM}\mathrm{of}2&6\mathrm{is}6\right)\phantom{\rule{0ex}{0ex}}\square =\frac{3}{6}-\frac{1}{6}=\frac{2÷2}{6÷2}\left(\mathrm{HCF}\mathrm{of}\mathrm{the}\mathrm{numerator}&\mathrm{denominator}\mathrm{is}2\right)\phantom{\rule{0ex}{0ex}}\square =\frac{1}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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