tab B alone takes 2 minutes longer than tap A to fill one tub. Working together, they can fill the tub 4 times in 5 minutes. How long does it take tap A (alone) to fill one tub? ((to 2 decimal places if necessary))

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- May 24th 2008, 09:36 AM #1

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- May 24th 2008, 09:53 AM #2

- May 24th 2008, 10:01 AM #3

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- May 24th 2008, 10:19 AM #4
Hi, finch41. As galactus explained, $\displaystyle a$ is the amount of time it takes tap A to fill 1 tub. That means tap A fills 1 tub per $\displaystyle a$ minutes, so the rate is $\displaystyle \frac1a\text{ tubs/min}$. Similarly, tap B takes $\displaystyle a + 2$ minutes (2 minutes more than tap A) to fill the tub, so it has a fill rate of 1 tub per $\displaystyle a + 2$ minutes, or $\displaystyle \frac1{a + 2}\text{ tubs/min}$. But, when both taps are on, they fill 4 tubs every 5 minutes, so we have a rate of $\displaystyle \frac45\text{ tubs/min}$. This gives us the equation,

$\displaystyle \text{rate of tap A}\;+\;\text{rate of tap B}\;=\;\text{combined rate}\Rightarrow\frac1a + \frac1{a + 2} = \frac45$

Solve for $\displaystyle a$.