1954 AHSME Problems/Problem 7: Difference between revisions

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)
Preceded by Problem 6	Followed by Problem 8
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

@@ Line 6: / Line 6: @@
 == Solution ==
-<math>25\cdot x=25-2.5\implies 100x=100-10\implies x=\frac{9}{10}</math>
+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>
-<math>100(1-x)\implies 100(\frac{1}{10})=10\%</math>, so <math>\fbox{C}</math>
+{{AHSME 50p box|year=1954|num-b=6|num-a=8}}
+{{MAA Notice}}

Page

Toolbox

Search

1954 AHSME Problems/Problem 7: Difference between revisions

Latest revision as of 16:49, 15 April 2017

Problem 7

Solution