Two pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?

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- Jun 10th 2007, 04:04 AMblueridgeTwo Pumps, One Tank
Two pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?

- Jun 10th 2007, 04:10 AMQuick
I'll start you off...

Define your variables: I'd call the**b**igger pump $\displaystyle b$ and the**s**maller pump $\displaystyle s$

$\displaystyle b=$the number of tanks per hour the bigger pump can empty

$\displaystyle s=$the number of tanks per hour the smaller pump can empty

Both of those numbers will be fractions.

Now, since you only wanted a hint, I'm only going to give you the first equation and you'll have to find the rest...

Quote:

Two pumps of different sizes working together can empty a fuel tank is 5 hours.

Do you need any more help? - Jun 10th 2007, 04:12 AMblueridgeYes
If you can set up the other equation, I can take it from there.

- Jun 10th 2007, 04:22 AMQuick
The next equation is somewhat weird.

If $\displaystyle b=\frac{\text{tanks}}{\text{hours}}$

than $\displaystyle \frac{1}{b}=\frac{\text{hours}}{\text{tanks}}$

So in fact:

$\displaystyle \frac{1}{b}=$the number of hours to empty a tank

So we know that: $\displaystyle \frac{1}{b}-\frac{1}{s}=4$ - Jun 10th 2007, 05:54 AMblueridgetell me...
I am dealing with two equations in two unknowns?

- Jun 10th 2007, 06:59 AMCaptainBlack
- Jun 10th 2007, 07:39 AMSoroban
Hello, blueridge!

Here's another approach . . .

Quote:

Two pumps of different sizes working together can empty a fuel tank is 5 hours.

The larger pump can empty this tank in 4 hours less than the smaller one.

If the larger one is out of order, how long will it take the smaller one to do the job alone?

. . In one hour, they can do $\displaystyle \frac{1}{5}$ of the job. .**[1]**

The smaller pump can do the job in $\displaystyle x$ hours. .[Note that: .$\displaystyle x > 4$.]

. . In one hour, it can do $\displaystyle \frac{1}{x}$ of the job.

The larger pump takes 4 hours less; it takes $\displaystyle x - 4$ hours.

. . In one hour, it can do $\displaystyle \frac{1}{x-4}$ of the job.

Together, in one hour, they can do: .$\displaystyle \frac{1}{x} + \frac{1}{x-4}$ of the job. .**[2]**

But [1] and [2] describe the same thing:

. . the fraction of the job done in one hour.

There is our equaton! . . . .$\displaystyle \boxed{\frac{1}{x} + \frac{1}{x-4} \:=\:\frac{1}{5}}$

Multiply by the common denominator: $\displaystyle 5x(x - 4)$

. . $\displaystyle 5(x - 4) + 5x \:=\:x(x-4)$

. . which simplifies to the quadratic: .$\displaystyle x^2 - 14x + 20 \:=\:0$

The Quadratic Formula gives us: .$\displaystyle x \;=\;\frac{14\pm\sqrt{116}}{2} \;=\;7 \pm\sqrt{29} \;\approx\;\{1.6,\:12.4\}$

Since $\displaystyle x > 4$. the solution is: .$\displaystyle x = 12.4$

Therefore, the smaller pump will take about 12.4 hours working alone.

- Jun 10th 2007, 11:14 AMblueridgetell me...
Soroban,

I thank you for sharing yet a more simplistic avenue to understanding this question.