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

m
m (See also)
Line 10: Line 10:
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1991|num-b=16|num-a=18}}   
+
{{AHSME box|year=1991|num-b=1|num-a=3}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:13, 28 September 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 $\boxed{\textbf{E}}$


See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS