Difference between revisions of "1952 AHSME Problems/Problem 34"
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== Solution == | == Solution == | ||
| − | <math>\fbox{\ | + | <math>\fbox{Correct Answer - (E)}</math> |
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| + | Converting the word problem in to an equation, we take x as the initial amount, we get <cmath>x(1+\dfrac{p}{100})(1-\dfrac{p}{100}) = 1</cmath> Using, <math>(a+b)(a-b) = a^{2} - b^{2}</math>. We now simplify it to, <cmath>x(1-\dfrac{p^{2}}{10000})=1</cmath> Simplifying to, <cmath>x(\dfrac{10000-p^{2}}{10000}) = 1</cmath> Dividing both sides by <math>\dfrac{10000-p^{2}}{10000}</math>, we get <math>\boxed{(E) | ||
| + | x = \dfrac{10000}{10000-p^{2}}}</math> | ||
== See also == | == See also == | ||
Revision as of 13:11, 16 December 2017
Problem
The price of an article was increased 
. Later the new price was decreased . If the last price was one dollar, the original price was:
Solution
Converting the word problem in to an equation, we take x as the initial amount, we get ![\[x(1+\dfrac{p}{100})(1-\dfrac{p}{100}) = 1\]](http://latex.artofproblemsolving.com/6/4/3/643b953a332c0866bc20b1a698d767fc33b467e1.png)
Using, 
. We now simplify it to, ![\[x(1-\dfrac{p^{2}}{10000})=1\]](http://latex.artofproblemsolving.com/d/d/f/ddff1910f8ab914e602f078a6122fff5390fd579.png)
Simplifying to, ![\[x(\dfrac{10000-p^{2}}{10000}) = 1\]](http://latex.artofproblemsolving.com/8/f/d/8fd34853bbd0afcbb9eb7c6c08be06fda85e67d8.png)
Dividing both sides by 
, we get 
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 33 |
Followed by Problem 35 | |
| 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. 
