Difference between revisions of "1991 AHSME Problems/Problem 2"
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| + | ==Problem== | ||
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<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> |
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| + | ==Solution== | ||
| + | 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 == | ||
| + | {{AHSME box|year=1991|num-b=1|num-a=3}} | ||
| + | |||
| + | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:34, 31 July 2016
Problem
Solution
Since 
, the value of 
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 
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| 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. 
