1991 AHSME Problems/Problem 2

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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
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