Mathematics, 19.05.2020 01:58, landofliam30
From a number of apples a man sells half number existing apples plus 1 to first customer, sells 1/3rd of the remaining apples plus 1 to second customer and 1/5th of the remaining apples plus 1 to third customer. He finds that he has 3 apples remaining. How many apples he originally had
Answers: 2
![=\frac{1}{3}*(\frac{x}{2}-1)+1\\\\=[\frac{1}{3}*\frac{x}{2}]- \frac{1}{3}*1+1\\\\=\frac{x}{6}-\frac{1}{3}+1\\\\=\frac{x}{6}-\frac{1}{3}+\frac{3}{3}\\\\=\frac{x}{6}+\frac{2}{3}\\](/tpl/images/0654/0708/e6aa2.png)
![=\frac{1}{5}*[\frac{x}{3}-\frac{5}{3}]+1\\\\=\frac{1}{5}*\frac{x}{3}-\frac{1}{5}*\frac{5}{3}+1\\\\=\frac{x}{15}-\frac{1}{3}+1\\\\=\frac{x}{15}-\frac{1}{3}+\frac{3}{3}\\\\=\frac{x}{15}+\frac{2}{3}\\\\](/tpl/images/0654/0708/a113d.png)
![=\frac{x}{3}-\frac{5}{3}-[\frac{x}{15}+\frac{2}{3}]\\\\=\frac{x}{3}-\frac{5}{3}-\frac{x}{15}-\frac{2}{3}\\\\=\frac{x}{3}-\frac{x}{15}-\frac{5}{3}-\frac{2}{3}\\\\=\frac{5x}{15}-\frac{x}{15}-\frac{7}{3}\\\\=\frac{4x}{15}-\frac{7}{3}\\](/tpl/images/0654/0708/b0997.png)
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