# profit and loss GRE question

I found this GMAT question on a forum, but couldn't find an answer nor do I understand it well.

An article costing C dollars is sold for \$100 at a loss of x percent of the selling price. It is then resold at a profit of x percent of the new selling price S. If the difference between S and C is$1.11..., then x is

A: Undetermined
B: 80/9
C: 10
D: 95/9
E: 100/9

From what I understand, the selling price is \$100. Loss of x% =$\frac{x}{100} * 100 = x$So the cost is the selling price + x, so$C = x + 100$? And we have$C - S = \$1.11$
$C = \$100 + \$1.11 = \$101.11$And x =$\$101.11 - \$100 = \$1.11$

But I think this is correct and I am misunderstanding the question. I would appreciate it if someone could explain what this question means and where I went wrong.

• You assumed $S=100$, which is not true. Dec 31, 2017 at 19:41

We are told that the article originally cost $C$ and when sold for $100$ there was a loss of $x\%$ of the $100$ selling price, so $C=100+x$. The profit on the second sale is $S-100$, which we are told is $\frac x{100}S$. We are also told $|C-S|=\frac {10}9$ The profit on the second sale is taken on the higher selling price, so $S \gt C$ and $S-C=\frac {10}9$ $$S-100=\frac {xS}{100}\\ S(1-\frac x{100})=100\\S=\frac {10000}{100-x}=C+\frac {10}9=100+x+\frac {10}9\\10000=(100-x)(100+x+\frac {10}9)=10000-x^2-\frac {10}9x+\frac {1000}9\\ x^2+\frac {10}9x-\frac {1000}9=0\\ x=10,-\frac {100}9$$