Answer :
To determine which pool is leaking faster, we need to calculate the leakage rate of each pool in gallons per minute. Here's a step-by-step breakdown:
Step 1: Determine the leakage rate for Pool A
- Pool A leaks [tex]\( \frac{2}{3} \)[/tex] gallon in 15 minutes.
- To find the leakage rate in gallons per minute, divide the amount leaked by the time:
[tex]\[
\text{Rate for Pool A} = \frac{2/3 \text{ gallon}}{15 \text{ minutes}} = \frac{2}{3} \div 15
\][/tex]
- This calculation gives us approximately 0.0444 gallons per minute.
Step 2: Determine the leakage rate for Pool B
- Pool B leaks [tex]\( \frac{3}{4} \)[/tex] gallon in 20 minutes.
- Similarly, divide the amount leaked by the time to find the leakage rate:
[tex]\[
\text{Rate for Pool B} = \frac{3/4 \text{ gallon}}{20 \text{ minutes}} = \frac{3}{4} \div 20
\][/tex]
- This calculation gives us approximately 0.0375 gallons per minute.
Step 3: Compare the leakage rates
- Pool A's rate of 0.0444 gallons per minute is greater than Pool B's rate of 0.0375 gallons per minute.
- Therefore, Pool A is leaking faster.
Step 4: Solve the complex fractions
- Rewrite each complex fraction as a division problem:
[tex]\( \frac{2}{3} \div \frac{1}{4} \)[/tex] becomes:
[tex]\[
\frac{2}{3} \times \frac{4}{1} = \frac{8}{3} \approx 2.67
\][/tex]
[tex]\( \frac{3}{4} \div \frac{1}{3} \)[/tex] becomes:
[tex]\[
\frac{3}{4} \times \frac{3}{1} = \frac{9}{4} = 2.25
\][/tex]
Conclusion
Pool A is leaking faster. The division of [tex]\(\frac{2}{3} \div \frac{1}{4}\)[/tex] results in approximately 2.67, and [tex]\(\frac{3}{4} \div \frac{1}{3}\)[/tex] results in 2.25.
Step 1: Determine the leakage rate for Pool A
- Pool A leaks [tex]\( \frac{2}{3} \)[/tex] gallon in 15 minutes.
- To find the leakage rate in gallons per minute, divide the amount leaked by the time:
[tex]\[
\text{Rate for Pool A} = \frac{2/3 \text{ gallon}}{15 \text{ minutes}} = \frac{2}{3} \div 15
\][/tex]
- This calculation gives us approximately 0.0444 gallons per minute.
Step 2: Determine the leakage rate for Pool B
- Pool B leaks [tex]\( \frac{3}{4} \)[/tex] gallon in 20 minutes.
- Similarly, divide the amount leaked by the time to find the leakage rate:
[tex]\[
\text{Rate for Pool B} = \frac{3/4 \text{ gallon}}{20 \text{ minutes}} = \frac{3}{4} \div 20
\][/tex]
- This calculation gives us approximately 0.0375 gallons per minute.
Step 3: Compare the leakage rates
- Pool A's rate of 0.0444 gallons per minute is greater than Pool B's rate of 0.0375 gallons per minute.
- Therefore, Pool A is leaking faster.
Step 4: Solve the complex fractions
- Rewrite each complex fraction as a division problem:
[tex]\( \frac{2}{3} \div \frac{1}{4} \)[/tex] becomes:
[tex]\[
\frac{2}{3} \times \frac{4}{1} = \frac{8}{3} \approx 2.67
\][/tex]
[tex]\( \frac{3}{4} \div \frac{1}{3} \)[/tex] becomes:
[tex]\[
\frac{3}{4} \times \frac{3}{1} = \frac{9}{4} = 2.25
\][/tex]
Conclusion
Pool A is leaking faster. The division of [tex]\(\frac{2}{3} \div \frac{1}{4}\)[/tex] results in approximately 2.67, and [tex]\(\frac{3}{4} \div \frac{1}{3}\)[/tex] results in 2.25.
