Question

A water tank can hold 4320 gallons of water. Due to a leak, 3/8 of the tank's capacity is lost. Later, 2/5 of the remaining capacity is filled with water. Then, 0.25 of the current water level is pumped out for maintenance. If 1/3 of the remaining water is contaminated and needs to be drained, what fraction and decimal representation of the original tank capacity remains?

Answer 1

Let's work through the problem step-by-step.

1. **Initial Capacity of the Tank**:
The water tank has a total capacity of 4320 gallons.

2. **Water Lost Due to Leak**:
The amount of water lost due to the leak is \( \frac{3}{8} \) of the tank's capacity:
\[
\text{Water lost} = \frac{3}{8} \times 4320 = 1620 \text{ gallons}
\]

3. **Remaining Capacity After Leak**:
After the leak, the remaining capacity of the tank is:
\[
\text{Remaining capacity} = 4320 - 1620 = 2700 \text{ gallons}
\]

4. **Water Filled After Leak**:
Now, \( \frac{2}{5} \) of the remaining capacity is filled with water:
\[
\text{Water filled} = \frac{2}{5} \times 2700 = 1080 \text{ gallons}
\]

5. **Current Water Level**:
The current water level in the tank is now:
\[
\text{Current water level} = 1080 \text{ gallons}
\]

6. **Water Pumped Out for Maintenance**:
Next, we pump out \( 0.25 \) (or \( \frac{1}{4} \)) of the current water level:
\[
\text{Water pumped out} = \frac{1}{4} \times 1080 = 270 \text{ gallons}
\]

7. **Water Left After Pumping Out**:
The amount of water remaining in the tank after pumping out is:
\[
\text{Water remaining} = 1080 - 270 = 810 \text{ gallons}
\]

8. **Water Contaminated and Drained**:
Of the remaining water, \( \frac{1}{3} \) is contaminated and needs to be drained:
\[
\text{Contaminated water} = \frac{1}{3} \times 810 = 270 \text{ gallons}
\]

9. **Final Amount of Water in the Tank**:
Thus, the final amount of water remaining in the tank after draining the contaminated water is:
\[
\text{Final water level} = 810 - 270 = 540 \text{ gallons}
\]

10. **Calculating the Fraction and Decimal Representation of the Original Tank Capacity**:
The fraction of the original tank capacity that remains is:
\[
\text{Fraction remaining} = \frac{540}{4320}
\]
To simplify:
\[
\frac{540 \div 540}{4320 \div 540} = \frac{1}{8}
\]

The decimal representation of \( \frac{1}{8} \) is:
\[
0.125
\]

So, the fraction of the original tank capacity remaining is \( \frac{1}{8} \) and in decimal form, it is \( 0.125 \).