## Water Transfer

Two cylindrical water tanks stand side by side. One has a radius of $4$ meters and contains water to a depth of $12.5$ meters. The other has a radius of $3$ meters and is empty. Water is pumped from the first tank to the second tank at a rate of $10$ cubic meters per minute. How long to the nearest tenth of a minute, must the pump run before the depth of the water is the same in both tanks?
Source: NCTM Mathematics Teacher, January 2006

SOLUTION
Let $h$ represent the same depth of water in each tank. The sum of the volumes in each tank equals the original volume in the first tank.
$\pi 4^2 h+\pi 3^2 h=\pi 4^2\times 12.5$
Simplify
$h(4^2+3^2)=4^2\times 12.5$
$h=8$
Volume that is transferred to the second tank
$\pi3^2\times 8=\pi 72$
Time it takes to transfer the water
$\pi 72/10=22.6$ minutes

Answer: $22.6$ minutes