+0  
 
0
445
1
avatar

A cylindrical quarter has a 15/16 inch diameter and a 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 1/8 inches? Express your answer as a common fraction.

 Jun 5, 2018
 #1
avatar+20217 
0

Volume of the quarter = pi r^2 x h

     = pi * (15/16 * 1/2)^2 * 1/16

 

FOUR times this volume is  (the new coin volume)

    4 * pi * (15/16 * 1/2)^2 * 1/16

 

the new coin volume = pi r^2 h    also

 

pi * (9/8 * 1/2)^2  h   =   4 * pi * (15/16 * 1/2)^2 *  1/16     solve for the new coin's h

(9/16)^2  h   =  1/4 (15/32)^2

 

h=  25/144

 Jun 5, 2018
edited by ElectricPavlov  Jun 5, 2018

27 Online Users

avatar
avatar