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

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==Problem==
 
==Problem==
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<math>|3-\pi|=</math>
 
<math>|3-\pi|=</math>
  
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==Solution==
 
==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>
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Since <math>\pi>3</math>, the value of <math>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>
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== See also ==
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{{AHSME box|year=1991|num-b=1|num-a=3}} 
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[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:34, 31 July 2016

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 $3-\pi$ 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

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