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Difference between revisions of "1954 AHSME Problems/Problem 11"

(Created page with "== Problem 11== A merchant placed on display some dresses, each with a marked price. He then posted a sign “<math>\frac{1}{3}</math> off on these dresses.” The cost of t...")
 
 
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== Solution ==
 
== Solution ==
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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}\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 $\frac{1}{2} \implies \fbox\textbf{(A)} \frac{1}{2}\$.
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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 19:41, 17 February 2020

Problem 11

A merchant placed on display some dresses, each with a marked price. He then posted a sign “$\frac{1}{3}$ off on these dresses.” The cost of the dresses was $\frac{3}{4}$ of the price at which he actually sold them. Then the ratio of the cost to the marked price was:

$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$

Solution

Let $x$ be the marked price. Then $\frac{2}{3}x$ would be the selling price of the dresses. The cost of the dresses was $\frac{3}{4}$ of the selling price, or $\frac{3}{4}\cdot\frac{2}{3}x \implies \frac{1}{2}x$. So the ratio of the cost to the marked price is $\frac{\frac{1}{2}x}{x}$, or $\frac{1}{2} \implies \boxed{\textbf{(A)}\ \frac{1}{2}}$.

See Also

1954 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AHSME Problems and Solutions

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