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An egg vendor calls on his first customer and sells half his eggs and half an egg. To the second customer, he sells half of what he has left with and half an egg, and to the third customer , he sells half of what he was then left with and half an egg. however, he did not break any egg. If in the end, the vendor was left with three eggs the what number of eggs did he have initially?

a)26 (b)31 (c)39 (d) none of these

Please find below the solution to the asked query:

$\mathrm{The}\mathrm{trick}\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{statement}\mathrm{that}\mathrm{he}\mathrm{does}\mathrm{not}\mathrm{break}\mathrm{any}\mathrm{egg}.\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{number}\mathrm{of}\mathrm{eggs}\mathrm{be}\mathrm{x}.\phantom{\rule{0ex}{0ex}}\mathrm{He}\mathrm{sells}\frac{\mathrm{x}}{2}+\frac{1}{2}\mathrm{eggs}\mathrm{to}\mathrm{first}\mathrm{customer}.\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{he}\mathrm{is}\mathrm{left}\mathrm{with}\phantom{\rule{0ex}{0ex}}\mathrm{x}-\left(\frac{\mathrm{x}}{2}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{x}-\frac{\mathrm{x}}{2}-\frac{1}{2}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{x}}{2}-\frac{1}{2}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{x}-1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly}\mathrm{after}\mathrm{dealing}\mathrm{with}\mathrm{second}\mathrm{customer},\mathrm{he}\mathrm{is}\mathrm{left}\mathrm{with}\frac{\mathrm{x}-1-2}{2\times 2}=\frac{\mathrm{x}-3}{4}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly}\mathrm{after}\mathrm{dealing}\mathrm{with}\mathrm{third}\mathrm{customer},\mathrm{he}\mathrm{is}\mathrm{left}\mathrm{with}\frac{\mathrm{x}-3-4}{2\times 4}=\frac{\mathrm{x}-7}{8}\phantom{\rule{0ex}{0ex}}\mathrm{According}\mathrm{to}\mathrm{question}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{x}-7}{8}=3\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{x}-7=24\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{x}=24+7=31\left(\mathbf{Answer}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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