Q:

An hourglass consists of two sets of congruent composite figures on either end. Each composite figure is made up of a cone and a cylinder, as shown below:Hourglass with sand measuring 47 millimeters high© 2011 Jupiterimages CorporationEach cone of the hourglass has a height of 12 millimeters. The total height of the sand within the top portion of the hourglass is 47 millimeters. The radius of both the cylinder and cone is 4 millimeters. Sand drips from the top of the hourglass to the bottom at a rate of 10π cubic millimeters per second. How many seconds will it take until all of the sand has dripped to the bottom of the hourglass? A(6.4) B(62.4) C(8.5) D(56.0)

Accepted Solution

A:
The volume of a cylinder is given by the formula V=πr²h². In this instance, r is 4 and the height is 47-12 (the height of the cone at the bottom is 12 mm and the sand goes up to 47 mm on the top portion of the hourglass, including both the cone and cylinder) or 35 mm.
V=π(4²)(35)=560π mm³.
The volume of a cone is given by the formula V=1/3πr²h².  In this instance, r is 4 and the height is 12 mm.
V=1/3π(4²)(12)=π(1/3)(12)(4²)=π(4)(4²)=64π mm³.  
This gives a total volume of 560π+64π=624π mm³ of sand.
Since the sand goes down to the bottom at a rate of 10π mm³/second, it will take 624π/10π=62.4 seconds for the sand to all drain out.