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Say I have a item XYZ for sale on a auction site.

The auction site has fees (10%) associated with item XYZ upon selling it.

I would like to sell item XYZ for the (cost_of_item + selling fees) + 20%.

Is there a Formula that can calculate the selling price, of item XYZ?

Any help is appreciated. This is not homework. Its been a long time since college.

Thanks, Bryan

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    Im not sure if its even possible, but 20% of the (cost_of_item + selling_fees).2012-05-08

2 Answers 2

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Suppose that your cost is $c$. We will find the appropriate selling price $s$. Note that if we sell at price $s$, then the selling fee will be $\frac{10}{100}s$, which I would prefer to write as $0.1s$.

The selling price should be $20\%$ more than cost plus selling fee. So the selling price should be $(c+01.s)+\frac{20}{100}(c+0.1s)$. It is more pleasant to write this as $(c+0.1s)(1.2)$.

We therefore get the equation $s=(c+0.1s)(1.2),$ which we want to solve for $s$. By expanding the right-hand side, we get $s=1.2c +0.12s$, and therefore $0.88s=1.2c$. It follows that
$s=\frac{1.2}{0.88}c.$ The number $1.2/0.88$ is approximately $1.3636$. So we should add approximately $36.36\%$ to the cost price.

Here is a sample calculation. Your cost was $100$. The formula above says to sell for $136.36$. Let's see how that works out. You get charged a selling fee of $10\%$ of $136.36$, which is $13.64$. So your cost plus selling fee is $113.64$. Note that $20\%$ of this is $22.73$. Add $113.64$ and $22.73$. We get $136.37$, basically your selling price. (There will often be a difference of a cent, because prices and fees usually get rounded to the nearest cent.)

Remark: The selling fee may change, or your markup of $20\%$ may change. It should not be hard to modify the above formula to take account of such changes.

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    thats so good, exactly what I was looking for. Really appreciate it.2012-05-08
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Unless I misunderstood your problem, $price = (cost + 0.1 price) \cdot 1.2$ $price = 1.2 cost + 0.12price$ $price = \frac{1.2}{1-0.12}cost = \frac {15}{11}cost$

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    Really appreciate this, I will give it a try.2012-05-08