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

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

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

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

@@ Line 1: / Line 1: @@
+{{Duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #2]] and [[2005 AMC 10B Problems|2005 AMC 10B #2]]}}
 == Problem ==
-The equations <math>2x+7=3</math> and <math>bx-10=-2</math> have the same solution for <math>x</math>.  What is the value of <math>b</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>
-\mathrm{(A)}\ -8      \qquad
+\mathrm{(A)}\ 2      \qquad
-\mathrm{(B)}\ -4      \qquad
+\mathrm{(B)}\ 4      \qquad
-\mathrm{(C)}\ -2      \qquad
+\mathrm{(C)}\ 10      \qquad
-\mathrm{(D)}\  4      \qquad
+\mathrm{(D)}\ 20      \qquad
-\mathrm{(E)}\  8
+\mathrm{(E)}\ 40
 </math>
 == Solution ==
-<math>2x+7=3 \Rightarrow 2x = -4 \Rightarrow x = -2</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{\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.
-<math>-2b-10=-2 \Rightarrow -2b = 8 \Rightarrow b = \boxed{-4}</math>.
+Solution by franzliszt
 == 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}}

Page

Toolbox

Search