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

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

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.

@@ Line 1: / Line 1: @@
+{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #5]] and [[2000 AMC 10 Problems|2000 AMC 10 #9]]}}
 == Problem ==
-If <math>\displaystyle |x - 2| = p,</math> where <math>\displaystyle x < 2,</math> then <math>\displaystyle 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>
+<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 ==
-When <math>\displaystyle x < 2,</math>, <math>x-2</math> is negative so <math>|x - 2| = 2-x</math> and <math>\displaystyle 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>.
-Therefore:
+Thus <math>x-p = (2-p)-p = 2-2p</math>.
+<math>\boxed{\mathbf{(C)}\ \ensuremath{2-2p}}</math>
+== Solution 2 (guess and check/desperation) ==
-<math>x=2-p</math>
+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.
-<math>\displaystyle x-p = (2-p)-p = 2-2p \Longrightarrow \mathrm{(C)} </math>
+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>.
+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.
+-Monkey_King
+==Video Solution by Daily Dose of Math==
+https://youtu.be/albUhCOwv3Y?si=4XcusOEp70EA6XKr
+~Thesmartgreekmathdude
 == See also ==
-* [[2000 AMC 12 Problems]]
+{{AMC10 box|year=2000|num-b=8|num-a=10}}
-*[[2000 AMC 12 Problems/Problem 4|Previous Problem]]
+{{AMC12 box|year=2000|num-b=4|num-a=6}}
-*[[2000 AMC 12 Problems/Problem 6|Next problem]]
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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