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

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About mvtrinh

Retired high school math teacher.
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