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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>
  
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<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 ==
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Thus <math>x-p = (2-p)-p = 2-2p</math>.
 
Thus <math>x-p = (2-p)-p = 2-2p</math>.
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<math>\boxed{\text{C}}</math>
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<math>\boxed{\mathbf{(C)}\ \ensuremath{2-2p}}</math>
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== Solution 2 (guess and check/desperation) ==
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 +
If you did not find that slick Solution 1, all hope is not lost. We could still guess and check our way to the right answer.
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We first plug in <math>x=1</math>, and get that <math>p=1</math> too. Hence <math>x-p=0</math>, eliminating choices <math>A</math> and <math>B</math>.
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We then plug in <math>x=0</math>, and get <math>p=2</math>. Therefore, <math>x-p=-2</math>. The answer is negative, eliminating <math>E</math>. Furthermore, <math>2p-2=2(2)-2=4-2=2\neq-2</math>, so choice <math>D</math> is false. Hence, the answer must be <math>C</math>, which upon checking indeed still holds true.
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-Monkey_King
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==Video Solution by Daily Dose of Math==
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https://youtu.be/albUhCOwv3Y?si=4XcusOEp70EA6XKr
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 +
~Thesmartgreekmathdude
  
 
== See also ==
 
== See also ==

Latest revision as of 23:20, 5 September 2024

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 $x-p = (2-p)-p = 2-2p$. $\boxed{\mathbf{(C)}\ \ensuremath{2-2p}}$

Solution 2 (guess and check/desperation)

If you did not find that slick Solution 1, all hope is not lost. We could still guess and check our way to the right answer.

We first plug in $x=1$, and get that $p=1$ too. Hence $x-p=0$, eliminating choices $A$ and $B$.

We then plug in $x=0$, and get $p=2$. Therefore, $x-p=-2$. The answer is negative, eliminating $E$. Furthermore, $2p-2=2(2)-2=4-2=2\neq-2$, so choice $D$ is false. Hence, the answer must be $C$, which upon checking indeed still holds true. -Monkey_King

Video Solution by Daily Dose of Math

https://youtu.be/albUhCOwv3Y?si=4XcusOEp70EA6XKr

~Thesmartgreekmathdude

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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