Difference between revisions of "2004 AMC 12A Problems/Problem 9"

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{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #9]] and [[2004 AMC 10A Problems/Problem 11|2004 AMC 10A #11]]}}
 
==Problem==
 
==Problem==
A company sells peanut butter in cylindrical jars.  Marketing research suggests that using wider jars will increase sales.  If the diameter of the jars is increased by <math>25\%</math> without altering the volume, by what percent must the height be decreased?
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A company sells peanut butter in cylindrical jars.  Marketing research suggests that using wider jars will increase sales.  If the [[diameter]] of the jars is increased by <math>25\%</math> without altering the [[volume]], by what percent must the height be decreased?
  
 
<math> \mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 25 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ } 60  </math>
 
<math> \mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 25 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ } 60  </math>
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== See also ==
 
== See also ==
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{{AMC12 box|year=2004|ab=A|num-b=8|num-a=10}}
 
{{AMC10 box|year=2004|ab=A|num-b=10|num-a=12}}
 
{{AMC10 box|year=2004|ab=A|num-b=10|num-a=12}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 19:11, 3 December 2007

The following problem is from both the 2004 AMC 12A #9 and 2004 AMC 10A #11, so both problems redirect to this page.

Problem

A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume, by what percent must the height be decreased?

$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 25 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ } 60$

Solution

When the diameter is increased by $25\%$, is is increased by $\frac54$, so the area of the base is increased by $\left(\frac54\right)^2=\frac{25}{16}$.

To keep the volume the same, the height must be $\frac{1}{\frac{25}{16}}=\frac{16}{25}$ of the original height, which is a $36\%$ reduction $\Rightarrow\mathrm{(C)}$.

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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
2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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