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The radius of a right circular cylinder is decreased by 20% and its height is increased by 25%. What is the absolute value of the percent change in the volume of the cylinder?

Guest Jun 6, 2018
 #1
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Hello, Guest!

 

A formula that will be important for this problem is the volume formula for a cylinder. It is the following:

 

\(V_{\text{cylinder}}=\pi r^2h\)

 

Let's compute the volume of the original right circular cylinder:

 

\(V_{\text{original}}=\pi r_{\text{old}}^2 h_{\text{old}}\) The original volume is the one where radius and height remained unchanged.
   

 

Now, let's consider a volume wherein the variables are tweaked somewhat. 

 

\(r_{\text{new}}=r_{\text{old}}-20\%*r_{\text{old}}\) As a decimal, 20%=0.2
\(r_{\text{new}}=r_{\text{old}}-0.2r_{\text{old}}=0.8r_{\text{old}}\) Now, let's find how the height was affected.
\(h_{\text{new}}=h_{\text{old}}+25\%*h_{old}\) As a decimal, 25%=0.25
\(h_{\text{new}}=h_{\text{old}}+0.25h_{\text{old}}=1.25h_{\text{old}}\) Now, we have tweaked both variables to fit the description in the original problem.
   

 

Now, let's find the volume of the new right cylinder:

 

\(V_{\text{new}}=\pi r_{\text{new}}^2h_{\text{new}}\) Plug in the known values for the radius and height.
\(V_{\text{new}}=\pi (0.8r_{\text{old}})^2*1.25h_{\text{old}}\) The only thing left to do is simplify.
\(V_{\text{new}}=\pi *0.64r_{\text{old}}^2*1.25h_{\text{old}}\)  
\(V_{\text{new}}=0.8\pi r_{\text{old}}^2h_{\text{old}}\)  
   

 

Now the only thing left to do is to calculate the percent change. I want you to try to do that. See what you can do. Check in with me if you would like. 

TheXSquaredFactor  Jun 6, 2018
 #2
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the answer is 80%

Guest Jun 6, 2018
 #3
avatar+2068 
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You are correct in stating that the new volume of the cylinder is 80% of the original right cylinder. Although this is a true observation, it does not answer what the value of the percent is. 

 

However, what 80% tells you is that this is certainly a percent decrease. 

 

\(V_{\text{new}}=V_{old}-xV_{old}\) I am subtraction a portion of the old volume to see what the decimal will be. Use substitution here. 
\(V_{\text{new}}=0.8\pi r_{\text{old}}^2h_{\text{old}}\\ V_{\text{original}}=\pi r_{\text{old}}^2 h_{\text{old}}\)  
\(0.8\pi r_{\text{old}}^2 h_{\text{old}}=\pi r_{\text{old}}^2 h_{\text{old}}-x\pi r_{\text{old}}^2 h_{\text{old}}\) Now, solve. I will factor. 
\(0.8\pi r_{\text{old}}^2 h_{\text{old}}=\pi r_{\text{old}}^2 h_{\text{old}}(1-x)\) I will divide both sides by \(\pi r^2 h\). This get's rid of a lot of the variables. 
\(0.8=1-x\)  
\(-0.2=-x\)  
\(x=0.2\Rightarrow 20\%\text{ change}\) The question asks for the percent change, so I convert from a decimal to a percent. 
   
TheXSquaredFactor  Jun 6, 2018

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