Difference between revisions of "1991 AHSME Problems/Problem 2"

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==Problem==
 
<math>|3-\pi|=</math>
 
<math>|3-\pi|=</math>
  
(A) <math>\frac{1}{7}</math> (B) <math>0.14</math> (C) <math>3-\pi</math> (D) <math>3+\pi</math> (E) <math>\pi-3</math>
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<math> \textbf{(A)\ }\frac{1}{7}\qquad\textbf{(B)\ }0.14\qquad\textbf{(C)\ }3-\pi\qquad\textbf{(D)\ }3+\pi\qquad\textbf{(E)\ }\pi-3 </math>
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==Solution==
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Since <math>\pi>3</math>, the value of <math>\abs{3-\pi}</math> is negative. The absolute value of a negative quantity is the negative quantity multiplied by <math>-1</math>, or the negative of that quantity. Therefore <math>|3-\pi|=-(3-\pi)=\pi-3</math>, which is choice <math>\boxed{\textbf{E}}</math>
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:42, 22 July 2014

Problem

$|3-\pi|=$

$\textbf{(A)\ }\frac{1}{7}\qquad\textbf{(B)\ }0.14\qquad\textbf{(C)\ }3-\pi\qquad\textbf{(D)\ }3+\pi\qquad\textbf{(E)\ }\pi-3$

Solution

Since $\pi>3$, the value of $\abs{3-\pi}$ (Error compiling LaTeX. ! Undefined control sequence.) is negative. The absolute value of a negative quantity is the negative quantity multiplied by $-1$, or the negative of that quantity. Therefore $|3-\pi|=-(3-\pi)=\pi-3$, which is choice $\boxed{\textbf{E}}$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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