QuestionMr. Gill's pump drains water at $-2 \frac{1}{4}$ inches/hour. A stronger pump drains $2 \frac{1}{2}$ times faster. Find the rate.

Studdy Solution

STEP 1

Assumptions1. The current pump changes the water level by $- \frac{1}{4}$ inches per hour. . A stronger pump would drain the pool $\frac{1}{}$ times as fast.
3. The negative sign indicates that the water level is decreasing.

STEP 2

First, we need to convert the mixed number $-2 \frac{1}{4}$ to an improper fraction to make the calculation easier.
$-2 \frac{1}{4} = -\frac{9}{4}$

STEP 3

Now, we need to multiply the rate at which the current pump is draining the pool by the factor by which the stronger pump is faster. This will give us the rate at which the stronger pump would drain the pool.
$Rate_{stronger\, pump} = Rate_{current\, pump} \\times Speed\, factor$

STEP 4

Now, plug in the given values for the rate of the current pump and the speed factor to calculate the rate of the stronger pump.
$Rate_{stronger\, pump} = -\frac{9}{4} \\times2 \frac{1}{2}$

STEP 5

Convert the mixed number $2 \frac{1}{2}$ to an improper fraction to make the calculation easier.
$2 \frac{1}{2} = \frac{5}{2}$ $Rate_{stronger\, pump} = -\frac{9}{4} \\times \frac{5}{2}$

STEP 6

Multiply the fractions to find the rate of the stronger pump.
$Rate_{stronger\, pump} = -\frac{9}{4} \\times \frac{5}{2} = -\frac{45}{8}$

STEP 7

Convert the improper fraction $-\frac{45}{}$ to a mixed number to make it easier to understand.
$-\frac{45}{} = -5 \frac{5}{}$ The stronger pump would change the water level by $-5 \frac{5}{}$ inches per hour. So, the answer is (A) $-5 \frac{5}{}$ inches.

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