1. ## Emptying a tank

Hello everyone! Boy these problems are so interesting, but I'm struggling to get the equation setup correct. Either way I find these problems so tasty.

A water tank can be emptied by using one pump for 5 hours. A second, smaller pump can empty the tank in 8 hours. If the larger pump is started at 1:00 p.m. at what time should the smaller pump be started so that the tank will be emptied at 5:00 p.m. ?

So 5 hours passes.

In one hour the bigger pump empties 1/5 of the tank

In one hour the smaller pump empties 1/8 of the tank

Let x be the amount of hours.

So I thought: $\displaystyle \frac{1}{5}x\,+\,\frac{1}{8}(5\,-\,x)\,=\,\frac{1}{5}$

But that make no sense. Can someone show me how to get the equation?

2. The larger pump empties $\displaystyle \frac{4}{5}$ of the tank in those $\displaystyle 4$ hours so... The smaller tank must empty the remaining $\displaystyle \frac{1}{5}$ of the tank.

So how long does it take to remove that much of the tank with the smaller pump? It is $\displaystyle \frac{8}{5}$ hours because. At a rate of $\displaystyle \frac{1}{8}$ per hour, the equation is:

$\displaystyle \frac{1}{8} x = \frac{1}{5}$

So that's 1.6 hours and we find we must start the smaller pump at $\displaystyle \boxed{\text{3:24 pm}}$

3. How do you know the tanks empties 4/5 in 4 hours? I'm totally lost.

4. Jonboy:

Here's another way to look at it.

Pump A can empty 1/5 of the tank in 1 hour, because it can empty the whole thing in 5 hours. See?.

Therefore, it starts at 1:00 pm and empties $\displaystyle \frac{1}{5}x$ of the tank by x hours past 1:00 pm.

After the other pump kicks in x hours after 1:00 pm, the both of them pump

$\displaystyle \frac{1}{5}+\frac{1}{8}=\frac{13}{40}$ of the tank per hour.

The remaining time is 4-x hours.

So, we have $\displaystyle \frac{1}{5}x+\frac{13}{40}(4-x)=1$

The 1 on the right side is the entire tank.

Solve for x and we get:

$\displaystyle x=\frac{12}{5}$ hours past 1:00 pm.

That's 144 minutes past 1:00 pm or 3:24 pm

Just as Aradesh said. Except, he/she subtracted 96 minutes from 5:00 pm. Same thing.

Did that explain it a little better?.

5. Originally Posted by Jonboy
Hello everyone! Boy these problems are so interesting, but I'm struggling to get the equation setup correct. Either way I find these problems so tasty.

A water tank can be emptied by using one pump for 5 hours. A second, smaller pump can empty the tank in 8 hours. If the larger pump is started at 1:00 p.m. at what time should the smaller pump be started so that the tank will be emptied at 5:00 p.m. ?

So 5 hours passes.

In one hour the bigger pump empties 1/5 of the tank

In one hour the smaller pump empties 1/8 of the tank

Let x be the amount of hours.

So I thought: $\displaystyle \frac{1}{5}x\,+\,\frac{1}{8}(5\,-\,x)\,=\,\frac{1}{5}$

But that make no sense. Can someone show me how to get the equation?

The larger pump empties 1/5 of a tank per hour, the smaller 1/8 of a tank per hour.

The larger pump runs for 4 hours so empties 4/5 of the tank, leaving 1/5
to be emptied by the smaller pump. So the smaller pump will take (1/5)(8)
hours to do its share, or 1.6 hours or 1 hour and 36 minutes, so it will have
to start pumping at 3:24.

RonL

6. Originally Posted by galactus

Did that explain it a little better?.
Yes thanks