2000 AMC 12 Problems/Problem 5: Difference between revisions

Revision as of 08:45, 8 November 2021

The following problem is from both the 2000 AMC 12 #5 and 2000 AMC 10 #9, so both problems redirect to this page.

If $|x - 2| = p$ , where $x < 2$ , then $x - p =$

$\textbf{(A) \ -2 } \qquad \textbf{(B) \ 2 } \qquad \textbf{(C) \ 2-2p } \qquad \textbf{(D) \ 2p-2 } \qquad \textbf{(E) \ |2p-2| }$

When $x < 2,$ $x-2$ is negative so $|x - 2| = 2-x = p$ and $x = 2-p$ .

Thus $x-p = (2-p)-p = 2-2p$ . $\boxed{\text{(C)2-2p}}$

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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
All AMC 10 Problems and Solutions

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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
All AMC 12 Problems and Solutions

@@ Line 4: / Line 4: @@
 If <math>|x - 2| = p</math>, where <math>x < 2</math>, then <math>x - p =</math>
-<math> \mathrm{(A) \ -2 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 2-2p } \qquad \mathrm{(D) \ 2p-2 } \qquad \mathrm{(E) \ |2p-2| }  </math>
+<math> \textbf{(A) \ -2 } \qquad \textbf{(B) \ 2 } \qquad \textbf{(C) \ 2-2p } \qquad \textbf{(D) \ 2p-2 } \qquad \textbf{(E) \ |2p-2| }  </math>
 == Solution ==