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Difference between revisions of "1965 AHSME Problems/Problem 32"

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== Solution ==
 
== Solution ==
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<math>\fbox{C}</math>
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The magnitude of the loss after the first sale is <math>C-100</math>, which equals <math>x</math>% of the selling price, <math>\$100</math>. Thus, <math>C-100=100*\frac{x}{100}</math>, and so <math>C=x+100</math>. The profit made after the second sale, <math>S'-100</math>, is <math>x</math>% of the new selling price, and this quantity is represented by <math>S'*\frac{x}{100}</math>. Equating these two expressions, we see that <math>S'-100=S'*\frac{x}{100}</math>, and so <math>S'=\frac{10,000}{100-x}</math>. Because we know that the difference between <math>S'</math> and <math>C</math> is <math>\frac{10}{9}</math>, we can solve the following equation:
 +
\begin{align*} \\
 +
\frac{10,000}{100-x}-(x+100)&=\frac{10}{9} \\
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10,000-(100+x)(100-x)&=\frac{10}{9}(100-x) \\
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10,000-(10,000-x^2)&=\frac{1000}{9}-\frac{10}{9}x \\
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x^2+\frac{10}{9}x-\frac{1000}{9}&=0 \\
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9x^2+10x-1000&=0 \\
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(9x+100)(x-10)&=0 \\
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\end{align*}
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Because <math>x \geq 0</math>, <math>x=10</math>, and so we choose answer <math>\boxed{\textbf{(C) }10}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 10:56, 19 July 2024

Problem

An article costing $C$ dollars is sold for $100 at a loss of $x$ percent of the selling price. It is then resold at a profit of $x$ percent of the new selling price $S'$. If the difference between $S'$ and $C$ is $1\frac {1}{9}$ dollars, then $x$ is:

$\textbf{(A)}\ \text{undetermined} \qquad \textbf{(B) }\ \frac {80}{9} \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ \frac{95}{9}\qquad \textbf{(E) }\ \frac{100}{9}$

Solution

The magnitude of the loss after the first sale is $C-100$, which equals $x$% of the selling price, $$100$. Thus, $C-100=100*\frac{x}{100}$, and so $C=x+100$. The profit made after the second sale, $S'-100$, is $x$% of the new selling price, and this quantity is represented by $S'*\frac{x}{100}$. Equating these two expressions, we see that $S'-100=S'*\frac{x}{100}$, and so $S'=\frac{10,000}{100-x}$. Because we know that the difference between $S'$ and $C$ is $\frac{10}{9}$, we can solve the following equation: \begin{align*} \\ \frac{10,000}{100-x}-(x+100)&=\frac{10}{9} \\ 10,000-(100+x)(100-x)&=\frac{10}{9}(100-x) \\ 10,000-(10,000-x^2)&=\frac{1000}{9}-\frac{10}{9}x \\ x^2+\frac{10}{9}x-\frac{1000}{9}&=0 \\ 9x^2+10x-1000&=0 \\ (9x+100)(x-10)&=0 \\ \end{align*} Because $x \geq 0$, $x=10$, and so we choose answer $\boxed{\textbf{(C) }10}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 31 		Followed by
Problem 33
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
All AHSME Problems and Solutions

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