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

m (Solution)
(Solution)
(2 intermediate revisions by one other user not shown)
Line 14: Line 14:
 
== Solution ==
 
== 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{\text{(D)}20}.</math>
 +
 
 +
===Solution 2===
 +
 
 +
Use the answer choices. Upon examination, it is quite obvious that the answer is <math>\boxed{\text{(D)}20}.</math> Very fast.
 +
 
 +
 
 +
Solution by franzliszt
  
 
== See also ==
 
== See also ==

Revision as of 19:43, 12 July 2020

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 $x$?

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

Solution

Solution 1

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

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

Solution 2

Use the answer choices. Upon examination, it is quite obvious that the answer is $\boxed{\text{(D)}20}.$ 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