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Difference between revisions of "1954 AHSME Problems/Problem 7"

(Created page with "== Problem 7== A housewife saved <math>\textdollar{2.50}</math> in buying a dress on sale. If she spent <math>\textdollar{25}</math> for the dress, she saved about: <math>\...")
 
 
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== Solution ==
 
== Solution ==
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<math>25\cdot x=25-2.5\implies 100x=100-10\implies x=\frac{9}{10}</math>
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Since she saved <math>\textdollar{2.50}</math> and spent <math>\textdollar{25}</math>, the original price of the dress was <math>\textdollar{27.50}</math>. The percent saved can be modeled as <math>27.5x=2.5</math>, so the answer is <math>x=\frac{2.5}{27.5}</math> or <math>\frac{1}{11}</math>, approximately <math>9\% \implies \textbf{(B)}</math>
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<math>100(1-x)\implies 100(\frac{1}{10})=10\%</math>, so <math>\fbox{C}</math>
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{{AHSME 50p box|year=1954|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 16:49, 15 April 2017

Problem 7

A housewife saved $\textdollar{2.50}$ in buying a dress on sale. If she spent $\textdollar{25}$ for the dress, she saved about:

$\textbf{(A)}\ 8 \% \qquad \textbf{(B)}\ 9 \% \qquad \textbf{(C)}\ 10 \% \qquad \textbf{(D)}\ 11 \% \qquad \textbf{(E)}\ 12\%$

Solution

Since she saved $\textdollar{2.50}$ and spent $\textdollar{25}$, the original price of the dress was $\textdollar{27.50}$. The percent saved can be modeled as $27.5x=2.5$, so the answer is $x=\frac{2.5}{27.5}$ or $\frac{1}{11}$, approximately $9\% \implies \textbf{(B)}$


1954 AHSC (Problems • Answer Key • Resources)
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Problem 8
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