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

(Solution)
 
(2 intermediate revisions by 2 users not shown)
Line 7: Line 7:
  
 
== 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}\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>.
 
 
 
 
  
 +
==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 “$\frac{1}{3}$ 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS