Difference between revisions of "2005 AMC 12B Problems/Problem 2"

Latest revision as of 12:56, 14 December 2021

The following problem is from both the 2005 AMC 12B #2 and 2005 AMC 10B #2, so both problems redirect to this page.

A positive number $x$ has the property that $x\%$ of $x$ is $4$ . What is $x$ ?

$\textbf{(A) }\ 2 \qquad \textbf{(B) }\ 4 \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ 20 \qquad \textbf{(E) }\ 40$

Since $x\%$ means $0.01x$ , the statement " $x\% \text{ of } x \text{ is 4}$ " can be rewritten as " $0.01x \cdot x = 4$ ":

$0.01x \cdot x=4 \Rightarrow x^2 = 400 \Rightarrow x = \boxed{\textbf{(D) }20}.$

Try the answer choices one by one. Upon examination, it is quite obvious that the answer is $\boxed{\textbf{(D) }20}.$ Very fast.

Solution by franzliszt

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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

@@ Line 5: / Line 5: @@
 <math>
-\mathrm{(A)}\ 2      \qquad
+\textbf{(A) }\ 2      \qquad
-\mathrm{(B)}\ 4      \qquad
+\textbf{(B) }\ 4      \qquad
-\mathrm{(C)}\ 10      \qquad
+\textbf{(C) }\ 10      \qquad
-\mathrm{(D)}\ 20      \qquad
+\textbf{(D) }\ 20      \qquad
-\mathrm{(E)}\ 40
+\textbf{(E) }\ 40
 </math>
 == Solution ==
-Since <math>x \text{%} </math> means <math>0.01x</math>, the statement "<math>x \text{ % of } x \text{ is 4}</math>" can be rewritten as "<math>0.01x \cdot x = 4</math>":
+===Solution 1===
-<math>0.01x \cdot x=4 \Rightarrow x^2 = 400 \Rightarrow x = \boxed{20}.</math>
+Since <math>x\%</math> means <math>0.01x</math>, the statement "<math>x\% \text{ of } x \text{ is 4}</math>" can be rewritten as "<math>0.01x \cdot x = 4</math>":
+<math>0.01x \cdot x=4 \Rightarrow x^2 = 400 \Rightarrow x = \boxed{\textbf{(D) }20}.</math>
+===Solution 2===
+Try the answer choices one by one. Upon examination, it is quite obvious that the answer is <math>\boxed{\textbf{(D) }20}.</math> Very fast.
+Solution by franzliszt
 == See also ==