Difference between revisions of "1991 AHSME Problems/Problem 2"
(Added solution and changed format of answer choices.) |
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| + | ==Problem== | ||
<math>|3-\pi|=</math> | <math>|3-\pi|=</math> | ||
| − | (A) | + | <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> |
| + | |||
| + | ==Solution== | ||
| + | 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 13:42, 22 July 2014
Problem
Solution
Since 
, the value of $\abs{3-\pi}$ (Error compiling LaTeX. Unknown error_msg) is negative. The absolute value of a negative quantity is the negative quantity multiplied by 
, or the negative of that quantity. Therefore 
, which is choice 
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
