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A fruit seller buys a large quantity of apples for $150\$$ . $200$ of the apples are rotten and he sells each of the remaining apples at $10$ cents more than what he paid and makes a profit of $50\$$. Find the number of apples that he brought originally.

Let $X$ be the number of apples

Let $Y$ be the original selling price of the apples

Remaining apples = $X-200$

First equation -

$ X - 200 = P + 0.10 $

I don't know how to find the second equation . Can I get help? Thanks in advance .

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  • $\begingroup$ How did you arrive at that first equation? $\endgroup$ – Henrik supports the community May 7 '16 at 8:09
  • $\begingroup$ Suppose he purchased the apples for $k$ dollars per apple. Then he is selling $X-200$ apples for $k+0.1$ each and making a profit of 50. Can you see what equations you need? $\endgroup$ – almagest May 7 '16 at 8:09
  • $\begingroup$ What is 150? Cost per Apple? $\endgroup$ – N.S.JOHN May 7 '16 at 8:12
  • $\begingroup$ Just realise the typing error and corrected it @N.S.JOHN $\endgroup$ – user307640 May 7 '16 at 8:16
  • $\begingroup$ Both the equations I can see are of the of the $\text{amount}*\text{price per apple}=\text{amount}$ form. $\endgroup$ – Henrik supports the community May 7 '16 at 8:23
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Supposing he bought at first $T$ apples for \$150. He sells the good ones, $T-200$, with 10¢ markup, and get a profit of \$50. So \begin{align} \underbrace{50}_{\text{profit}} = \underbrace{\underbrace{(T-200)}_{\text{sold apples}} \left(\underbrace{\frac{150}{T}}_{\text{original price}} + \underbrace{0.10}_{\text{10¢ markup}}\right)}_{\text{revenue}} - \underbrace{150}_{\text{cost}} \end{align} which yields $$\frac{T}{10}-\frac{30000}{T}-70=0,$$ with solutions $$T=\left\{-300,1000\right\}.$$

Obviously, $T\in \mathbb{N}^{*}$, so $$T = 1000 \text{ apples.}$$

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  • $\begingroup$ Thank you, it changes slightly the answer, but the reasoning behind was already valid. $\endgroup$ – Guilherme Thompson May 7 '16 at 8:29
  • $\begingroup$ If he bought 296 apples, and 200 were rotten, he would only have 96 to sell, with a 10 cent markup that's 9.6 dollars, that doesn't leave room for a 50 dollar profit. $\endgroup$ – Henrik supports the community May 7 '16 at 11:24
  • $\begingroup$ @Henrik, thank you. My bad. $\endgroup$ – Guilherme Thompson May 7 '16 at 15:58
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    $\begingroup$ If he brought $1000$ apples, $800$ are remaining with a $.10$ markup he would sell each for $.25=.15+.10$. Obviously he would have $.25(800)=200$ in his hand and would have made $50$ in profit. Why is this answer receiving down-votes? Did he just now fix it, because it looks correct.@Henrik @GuilhermeThompson $\endgroup$ – Ahmed S. Attaalla May 7 '16 at 16:35

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