Difference between revisions of "1952 AHSME Problems/Problem 34"
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== Solution == | == Solution == | ||
− | <math>\ | + | 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 the value of x. <math>\boxed{\textbf{(E)}\dfrac{10000}{10000-p^{2}}}</math> |
== See also == | == See also == |
Latest revision as of 13:14, 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 Using, . We now simplify it to, Simplifying to, Dividing both sides by , we get the value of x.
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 33 |
Followed by Problem 35 | |
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All AHSME Problems and Solutions |
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