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

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When the diameter is increased by <math>25\%</math>, it is increased by <math>\frac54</math>, so the area of the base is increased by <math>\left(\frac54\right)^2=\frac{25}{16}</math>.
 
When the diameter is increased by <math>25\%</math>, it is increased by <math>\frac54</math>, so the area of the base is increased by <math>\left(\frac54\right)^2=\frac{25}{16}</math>.
  
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To keep the volume the same, the height must be <math>\frac{1}{\frac{25}{16}}=\frac{16}{25}</math> of the original height, which is a <math>36\%</math> reduction <math>\Rightarrow\mathrm{(C)}</math>.
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To keep the volume the same, the height must be <math>\frac{1}{\frac{25}{16}}=\frac{16}{25}</math> of the original height, which is a <math>36\%</math> reduction. <math>\boxed{\mathrm{(C)}\ 36}</math>
  
 
== See also ==
 
== See also ==

Revision as of 23:35, 20 July 2014

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 $25\%$, it 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. $\boxed{\mathrm{(C)}\ 36}$

See also

2004 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

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