2000 AMC 12 Problems/Problem 5

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.

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2000 AMC 12 Problems/Problem 5

Contents

Problem

Solution

Solution 2 (guess and check/desperation)

Video Solution by Daily Dose of Math

See also