Two Pumps, One Tank

**blueridge** · June 10th 2007, 04:04 AM

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?

**Quick** · June 10th 2007, 04:10 AM

Originally Posted by blueridge

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?

I'll start you off...

Define your variables: I'd call the bigger pump $b$ and the smaller pump $s$

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

$s=$ 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...

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

$b+s=\frac{1\text{tank}}{5\text{hours}}$

Do you need any more help?

**blueridge** · June 10th 2007, 04:12 AM

If you can set up the other equation, I can take it from there.

**Quick** · June 10th 2007, 04:22 AM

Originally Posted by blueridge

If you can set up the other equation, I can take it from there.

Originally Posted by blueridge

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?

The next equation is somewhat weird.

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

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

So in fact:

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

So we know that: $\frac{1}{b}-\frac{1}{s}=4$

**blueridge** · June 10th 2007, 05:54 AM

I am dealing with two equations in two unknowns?

**CaptainBlack** · June 10th 2007, 06:59 AM

Originally Posted by blueridge

I am dealing with two equations in two unknowns?

Yes, solve for $b$ and $s$ , and the required answer is $s$

RonL

**Soroban** · June 10th 2007, 07:39 AM

Hello, blueridge!

Here's another approach . . .

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?

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

The smaller pump can do the job in $x$ hours. .[Note that: . $x > 4$ .]
. . In one hour, it can do $\frac{1}{x}$ of the job.

The larger pump takes 4 hours less; it takes $x - 4$ hours.
. . In one hour, it can do $\frac{1}{x-4}$ of the job.
Together, in one hour, they can do: . $\frac{1}{x} + \frac{1}{x-4}$ 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! . . . . $\boxed{\frac{1}{x} + \frac{1}{x-4} \:=\:\frac{1}{5}}$

Multiply by the common denominator: $5x(x - 4)$
. . $5(x - 4) + 5x \:=\:x(x-4)$

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

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

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

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

**blueridge** · June 10th 2007, 11:14 AM

Soroban,

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

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