Difference between revisions of "1954 AHSME Problems/Problem 11"
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== Solution == | == Solution == | ||
Let <math>x</math> be the marked price. Then <math>\frac{2}{3}x</math> would be the selling price of the dresses. The cost of the dresses was <math>\frac{3}{4}</math> of the selling price, or <math>\frac{3}{4}\cdot\frac{2}{3}x \implies \frac{1}{2}x</math>. So the ratio of the cost to the marked price is <math>\frac{\frac{1}{2}x}{x}</math>, or <math>\frac{1}{2} \implies \boxed{\textbf{(A)}\ \frac{1}{2}}</math>. | Let <math>x</math> be the marked price. Then <math>\frac{2}{3}x</math> would be the selling price of the dresses. The cost of the dresses was <math>\frac{3}{4}</math> of the selling price, or <math>\frac{3}{4}\cdot\frac{2}{3}x \implies \frac{1}{2}x</math>. So the ratio of the cost to the marked price is <math>\frac{\frac{1}{2}x}{x}</math>, or <math>\frac{1}{2} \implies \boxed{\textbf{(A)}\ \frac{1}{2}}</math>. | ||
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+ | ==See Also== | ||
{{AHSME 50p box|year=1954|num-b=10|num-a=12}} | {{AHSME 50p box|year=1954|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:41, 17 February 2020
Problem 11
A merchant placed on display some dresses, each with a marked price. He then posted a sign “ off on these dresses.” The cost of the dresses was of the price at which he actually sold them. Then the ratio of the cost to the marked price was:
Solution
Let be the marked price. Then would be the selling price of the dresses. The cost of the dresses was of the selling price, or . So the ratio of the cost to the marked price is , or .
See Also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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