Thread: 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:

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

The larger pump empties of the tank in those hours so... The smaller tank must empty the remaining of the tank.

So how long does it take to remove that much of the tank with the smaller pump? It is hours because. At a rate of per hour, the equation is:



So that's 1.6 hours and we find we must start the smaller pump at

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

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 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

of the tank per hour.

The remaining time is 4-x hours.

So, we have

The 1 on the right side is the entire tank.

Solve for x and we get:

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?.

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:

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

Originally Posted by galactus

Did that explain it a little better?.

Yes thanks