1965 AHSME Problems/Problem 32
Problem
An article costing dollars is sold for $100 at a loss of percent of the selling price. It is then resold at a profit of percent of the new selling price . If the difference between and is dollars, then is:
Solution
The magnitude of the loss after the first sale is , which equals % of the selling price, . Thus, , and so . The profit made after the second sale, , is % of the new selling price, and this quantity is represented by . Equating these two expressions, we see that , and so . Because we know that the difference between and is , we can solve the following equation: \begin{align*} \\ \frac{10,000}{100-x}-(x+100)&=\frac{10}{9} \\ 10,000-(100+x)(100-x)&=\frac{10}{9}(100-x) \\ 10,000-(10,000-x^2)&=\frac{1000}{9}-\frac{10}{9}x \\ x^2+\frac{10}{9}x-\frac{1000}{9}&=0 \\ 9x^2+10x-1000&=0 \\ (9x+100)(x-10)&=0 \\ \end{align*} Because , , and so we choose answer .
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 31 |
Followed by Problem 33 | |
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