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

(Solution)
(Solution 2)
 
(10 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 +
{{Duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #2]] and [[2005 AMC 10B Problems|2005 AMC 10B #2]]}}
 +
 
== Problem ==
 
== Problem ==
 
A positive number <math>x</math> has the property that <math>x\%</math> of <math>x</math> is <math>4</math>.  What is <math>x</math>?
 
A positive number <math>x</math> has the property that <math>x\%</math> of <math>x</math> is <math>4</math>.  What is <math>x</math>?
  
 
<math>
 
<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>
 
</math>
  
 
== Solution ==
 
== Solution ==
  
<math>0.01x \cdot x=4 \Rightarrow x^2 = 400 \Rightarrow x = \boxed{20}.</math>
+
===Solution 1===
 +
 
 +
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 ==
 
== See also ==
* [[2005 AMC 12B Problems]]
+
{{AMC10 box|year=2005|ab=B|num-b=1|num-a=3}}
 +
{{AMC12 box|year=2005|ab=B|num-b=1|num-a=3}}
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

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.

Problem

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

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

Solution

Solution 1

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

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

Solution 2

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

See also

2005 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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