To find the change in water level per hour if Mr. Gill used the stronger pump, we need to multiply the current drainage rate by the factor of 2 1/2 (which is 2.5).

Current drainage rate:

$-2 \frac{1}{4} \text{ inches per hour} = -\frac{9}{4} \text{ inches per hour}$

Stronger pump rate (2 1/2 times faster):

Multiply the current rate by 2.5: $\text{New rate} = -\frac{9}{4} \times 2.5$

First, convert 2.5 to a fraction: $2.5 = \frac{5}{2}$

Now multiply: $\text{New rate} = -\frac{9}{4} \times \frac{5}{2} = -\frac{45}{8} \text{ inches per hour}$

Conclusion:

The change in water level would be $-\frac{45}{8}$ inches per hour, or approximately $-5.625$ inches per hour if Mr. Gill used the stronger pump.

Would you like further explanation or any additional details?

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Here are 5 related questions:

1. How would the rate change if the pump drained 3 times as fast?
2. What would the total water level change be after 4 hours with the stronger pump?
3. How do you multiply mixed numbers and fractions efficiently?
4. How much faster is 2 1/2 times compared to 1 1/4 times?
5. What if the original rate were different, like $-3 \frac{1}{2}$ inches per hour?

Tip: When multiplying mixed numbers, always convert them to improper fractions first for easier calculation.