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A cylindrical quarter has a \(\frac{15}{16}\) inch diameter and a \(\frac{1}{16}\) inch height. What would be the number of inches in the height of a coin whose volume is exactly four times that of the given quarter and whose diameter equals \(1 \frac{1}{8}\) inches? Express your answer as a common fraction.

 Jan 1, 2019
 #1
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The old radius is  ..... (1/2)(15/16) = 15/32 inches

The old volume is  .....    pi  * ( 15/32)^2 * (1/16)  in ^3

And 4 times this volume is   pi * (15/32)^2 * (1/16) * 4   in^3  =  pi * (15/32)^2 *(1/4) in^3

The  new diameter is   1 + 1/8  in   =  9/8 inches.....so....the new radius is 1/2 of this = (9/8) (1/2) = 9/16 inches

 

So we have that

 

pi * (15/32)^2 * (1/4)  =  pi * (9/16)^2  * height            [ divide out pi ]

 

(15/32)^2 / 4   =  (9/16)^2 * height            [  divide both sides by (9/16)^2 ]

 

(15/32)^2 / [ 4 * (9/16)^2 ]  =  height   =    25 /144  in

 

 

cool cool cool

 Jan 1, 2019

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