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

(Added solution and changed format of answer choices.) |
|||

Line 1: | Line 1: | ||

+ | ==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 14:42, 22 July 2014

## Problem

## Solution

Since , 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 , 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.