Difference between revisions of "2005 AMC 12B Problems/Problem 2"
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+ | {{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>? | |
<math> | <math> | ||
− | \ | + | \textbf{(A) }\ 2 \qquad |
− | \ | + | \textbf{(B) }\ 4 \qquad |
− | \ | + | \textbf{(C) }\ 10 \qquad |
− | \ | + | \textbf{(D) }\ 20 \qquad |
− | \ | + | \textbf{(E) }\ 40 |
</math> | </math> | ||
== Solution == | == Solution == | ||
+ | |||
+ | ===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 == | ||
− | + | {{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 has the property that of is . What is ?
Solution
Solution 1
Since means , the statement "" can be rewritten as "":
Solution 2
Try the answer choices one by one. Upon examination, it is quite obvious that the answer is Very fast.
Solution by franzliszt
See also
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.