Difference between revisions of "2000 AMC 12 Problems/Problem 5"

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If <math>|x - 2| = p</math>, where <math>x < 2</math>, then <math>x - p =</math>
 
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>
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<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 ==
 
== Solution ==
 
When <math>x < 2,</math> <math>x-2</math> is negative so <math>|x - 2| = 2-x = p</math> and <math>x = 2-p</math>.
 
When <math>x < 2,</math> <math>x-2</math> is negative so <math>|x - 2| = 2-x = p</math> and <math>x = 2-p</math>.
  
Thus <math>x-p = (2-p)-p = 2-2p \Longrightarrow \mathrm{(C)} </math>.
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Thus <math>x-p = (2-p)-p = 2-2p</math>.
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<math>\boxed{\text{(C)2-2p}}</math>
  
 
== See also ==
 
== See also ==
 +
{{AMC10 box|year=2000|num-b=8|num-a=10}}
 
{{AMC12 box|year=2000|num-b=4|num-a=6}}
 
{{AMC12 box|year=2000|num-b=4|num-a=6}}
{{AMC10 box|year=2000|num-b=8|num-a=10}}
 
 
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:46, 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.

Problem

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

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

Solution

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

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

See also

2000 AMC 10 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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