2000 AMC 12 Problems/Problem 5

Revision as of 00:20, 6 September 2024 by Monkey king (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
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 (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

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