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Difference between revisions of "2022 AMC 10A Problems/Problem 6"

m (→‎Solution 2)
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<math>\textbf{(A) } 3-2a \qquad \textbf{(B) } 1-a \qquad \textbf{(C) } 1 \qquad \textbf{(D) } a+1 \qquad \textbf{(E) } 3</math>
 
<math>\textbf{(A) } 3-2a \qquad \textbf{(B) } 1-a \qquad \textbf{(C) } 1 \qquad \textbf{(D) } a+1 \qquad \textbf{(E) } 3</math>
  
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== Solution ==  
+
== Solution 1 ==  
 
We have  
 
We have  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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== Solution 2 ==
 
== Solution 2 ==
−
WLOG, assume <math>a=-1.</math> Then, the given expression simplifies to <math>5</math>:
+
Assume that <math>a=-1.</math> Then, the given expression simplifies to <math>5</math>:
−
<cmath>\left|a-2-\sqrt{(a-1)^2}\right| = \left|-1-2-\sqrt{(-1-1)^2}\right|
+
<cmath>\begin{align*}
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= \left|-1-2-\sqrt{4}\right|
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\left|a-2-\sqrt{(a-1)^2}\right| &= \left|-1-2-\sqrt{(-1-1)^2}\right| \\
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= \left|-1-2-2\right|
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&= \left|-1-2-\sqrt{4}\right| \\
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= 5.</cmath>
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&= \left|-1-2-2\right| \\
−
 
+
&= 5.
 +
\end{align*}</cmath>
 
Then, we test each of the answer choices to see which one is equal to <math>5</math>:
 
Then, we test each of the answer choices to see which one is equal to <math>5</math>:
  
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<math>A:</math> <math>3-2a = 3-2\cdot(-1) = 3+2 = 5.</math>  
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<math>\textbf{(A) } 3-2a = 3-2\cdot(-1) = 3+2 = 5.</math>  
  
−
<math>B:</math> <math>1-a = 1-(-1) = 2 \neq 5.</math>
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<math>\textbf{(B) } 1-a = 1-(-1) = 2 \neq 5.</math>
  
−
<math>C:</math> <math>1 \neq 5.</math>
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<math>\textbf{(C) } 1 \neq 5.</math>
  
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<math>D:</math> <math>a+1 = -1+1 = 0 \neq 5.</math>
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<math>\textbf{(D) } a+1 = -1+1 = 0 \neq 5.</math>
  
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<math>E:</math> <math>3 \neq 5.</math>
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<math>\textbf{(E) } 3 \neq 5.</math>
  
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The only answer choice equal to <math>5</math> for <math>a=-1</math> is <math>A</math>, so the answer is <math>\boxed{\textbf{(A) } 3-2a}.</math>
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The only answer choice equal to <math>5</math> for <math>a=-1</math> is <math>\boxed{\textbf{(A) } 3-2a}.</math>
  
 
-MathWizard09
 
-MathWizard09

Revision as of 02:35, 12 November 2022

Problem

Which expression is equal to \[\left|a-2-\sqrt{(a-1)^2}\right|\] for $a<0?$

$\textbf{(A) } 3-2a \qquad \textbf{(B) } 1-a \qquad \textbf{(C) } 1 \qquad \textbf{(D) } a+1 \qquad \textbf{(E) } 3$

Solution 1

We have \begin{align*} \left|a-2-\sqrt{(a-1)^2}\right| &= \left|a-2-|a-1|\right| \\ &=\left|a-2-(-a+1)\right| \\ &=\left|2a-3\right| \\ &=\boxed{\textbf{(A) } 3-2a}. \end{align*} ~MRENTHUSIASM

Solution 2

Assume that $a=-1.$ Then, the given expression simplifies to $5$: \begin{align*} \left|a-2-\sqrt{(a-1)^2}\right| &= \left|-1-2-\sqrt{(-1-1)^2}\right| \\ &= \left|-1-2-\sqrt{4}\right| \\ &= \left|-1-2-2\right| \\ &= 5. \end{align*} Then, we test each of the answer choices to see which one is equal to $5$:

$\textbf{(A) } 3-2a = 3-2\cdot(-1) = 3+2 = 5.$

$\textbf{(B) } 1-a = 1-(-1) = 2 \neq 5.$

$\textbf{(C) } 1 \neq 5.$

$\textbf{(D) } a+1 = -1+1 = 0 \neq 5.$

$\textbf{(E) } 3 \neq 5.$

The only answer choice equal to $5$ for $a=-1$ is $\boxed{\textbf{(A) } 3-2a}.$

-MathWizard09

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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

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