Answer :
To determine which tank is leaking faster, we need to find and compare the leak rates for both tanks. The leak rate is defined as the volume of water leaked per unit time.
### Step-by-Step Solution
1. Find the leak rate for Tank A:
- Volume of water leaked from Tank A: [tex]\(\frac{1}{16}\)[/tex] gallon.
- Time taken for Tank A to leak: [tex]\(\frac{1}{12}\)[/tex] minute.
The leak rate [tex]\( R_A \)[/tex] for Tank A can be calculated by dividing the volume by the time:
[tex]\[
R_A = \frac{\frac{1}{16} \text{ gallon}}{\frac{1}{12} \text{ minute}} = \left(\frac{1}{16}\right) \div \left(\frac{1}{12}\right)
\][/tex]
Dividing by a fraction is the same as multiplying by its reciprocal:
[tex]\[
R_A = \left(\frac{1}{16}\right) \times \left(\frac{12}{1}\right) = \frac{12}{16} = 0.75 \text{ gallons per minute}
\][/tex]
2. Find the leak rate for Tank B:
- Volume of water leaked from Tank B: [tex]\(\frac{3}{80}\)[/tex] gallon.
- Time taken for Tank B to leak: [tex]\(\frac{1}{30}\)[/tex] minute.
The leak rate [tex]\( R_B \)[/tex] for Tank B can be calculated similarly by dividing the volume by the time:
[tex]\[
R_B = \frac{\frac{3}{80} \text{ gallon}}{\frac{1}{30} \text{ minute}} = \left(\frac{3}{80}\right) \div \left(\frac{1}{30}\right)
\][/tex]
Again, dividing by a fraction is the same as multiplying by its reciprocal:
[tex]\[
R_B = \left(\frac{3}{80}\right) \times \left(\frac{30}{1}\right) = \frac{90}{80} = 1.125 \text{ gallons per minute}
\][/tex]
3. Compare the leak rates:
[tex]\[
R_A = 0.75 \text{ gallons per minute}
\][/tex]
[tex]\[
R_B = 1.125 \text{ gallons per minute}
\][/tex]
Since [tex]\(1.125 > 0.75\)[/tex], the leak rate of Tank B is higher than that of Tank A.
### Conclusion
Tank B is leaking faster with a rate of [tex]\(1.125\)[/tex] gallons per minute compared to Tank A, which is leaking at a rate of [tex]\(0.75\)[/tex] gallons per minute.
### Step-by-Step Solution
1. Find the leak rate for Tank A:
- Volume of water leaked from Tank A: [tex]\(\frac{1}{16}\)[/tex] gallon.
- Time taken for Tank A to leak: [tex]\(\frac{1}{12}\)[/tex] minute.
The leak rate [tex]\( R_A \)[/tex] for Tank A can be calculated by dividing the volume by the time:
[tex]\[
R_A = \frac{\frac{1}{16} \text{ gallon}}{\frac{1}{12} \text{ minute}} = \left(\frac{1}{16}\right) \div \left(\frac{1}{12}\right)
\][/tex]
Dividing by a fraction is the same as multiplying by its reciprocal:
[tex]\[
R_A = \left(\frac{1}{16}\right) \times \left(\frac{12}{1}\right) = \frac{12}{16} = 0.75 \text{ gallons per minute}
\][/tex]
2. Find the leak rate for Tank B:
- Volume of water leaked from Tank B: [tex]\(\frac{3}{80}\)[/tex] gallon.
- Time taken for Tank B to leak: [tex]\(\frac{1}{30}\)[/tex] minute.
The leak rate [tex]\( R_B \)[/tex] for Tank B can be calculated similarly by dividing the volume by the time:
[tex]\[
R_B = \frac{\frac{3}{80} \text{ gallon}}{\frac{1}{30} \text{ minute}} = \left(\frac{3}{80}\right) \div \left(\frac{1}{30}\right)
\][/tex]
Again, dividing by a fraction is the same as multiplying by its reciprocal:
[tex]\[
R_B = \left(\frac{3}{80}\right) \times \left(\frac{30}{1}\right) = \frac{90}{80} = 1.125 \text{ gallons per minute}
\][/tex]
3. Compare the leak rates:
[tex]\[
R_A = 0.75 \text{ gallons per minute}
\][/tex]
[tex]\[
R_B = 1.125 \text{ gallons per minute}
\][/tex]
Since [tex]\(1.125 > 0.75\)[/tex], the leak rate of Tank B is higher than that of Tank A.
### Conclusion
Tank B is leaking faster with a rate of [tex]\(1.125\)[/tex] gallons per minute compared to Tank A, which is leaking at a rate of [tex]\(0.75\)[/tex] gallons per minute.
