Question

A 2000 gallon fish aquarium can be filled by water flowing at a constant rate in 10 hours. When a decorativve rock is placed in the aquarium, it can be filled in 9.5 hours. Find the volume of the rock in cubic feet

Answer 1

Answer:

The volume of the rock is 13.368 cubic feet.

Step-by-step explanation:

The volume of the rock ($V_{r}$), measured in gallons, is equal to the volume of the fish aquarium without the decorative rock ($V'$) minus the volume of the fish aquarium with the decorative rock ($V''$), both measured in gallons.

Since the water flow is at constant rate, the volume of the rock is expressed by the following equation:

$V_{r} = V' - V''$

$V_{r} = \dot V \cdot (\Delta t_{1}-\Delta t_{2})$ (1)

Where:

$\dot V$ - Flow rate, measured in gallons per hour.

$\Delta t_{1}$ - Filling time of the fish aquarium without the decorative rock, measured in hours.

$\Delta t_{2}$ - Filling time of the fish aquarium with the decorative rock, measured in hours.

And the flow rate is:

$\dot V = \frac{2000\,gal}{10\,h}$

$\dot V = 200\,\frac{gal}{h}$

The flow rate is 200 gallons per hour.

If we know that $\dot V = 200\,\frac{gal}{h}$, $\Delta t_{1} = 10\,h$ and $\Delta t_{2} = 9.5\,h$, then the volume of the rock is:

$V_{r} = \left(200\,\frac{gal}{h}\right)\cdot (10\,h-9.5\,h)$

$V_{r} = 100\,gal$

$V_{r} = 13.368\,ft^{3}$

The volume of the rock is 13.368 cubic feet.