1965 AHSME Problems/Problem 32: Difference between revisions
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== Solution == | == Solution == | ||
<math>\ | 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} \\ | |||
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 <math>x \geq 0</math>, <math>x=10</math>, and so we choose answer <math>\boxed{\textbf{(C) }10}</math>. | |||
== See Also == | == See Also == | ||
{{AHSME 40p box|year=1965|num-b=31|num-a=33}} | {{AHSME 40p box|year=1965|num-b=31|num-a=33}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | |||
Latest revision as of 10:56, 19 July 2024
Problem
An article costing 
dollars is sold for \$100 at a loss of 
percent of the selling price.
It is then resold at a profit of percent of the new selling price 
.
If the difference between and is 
dollars, then is:
Solution
The magnitude of the loss after the first sale is 
, which equals % of the selling price, 
. Thus, 
, and so 
. The profit made after the second sale, 
, is % of the new selling price, and this quantity is represented by 
. Equating these two expressions, we see that 
, and so 
. Because we know that the difference between and is 
, 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 
, 
, and so we choose answer 
.
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 | ||
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