Emptying a tank

J

jonboy

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Jun 8, 2006
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I need help setting up the equation here... that's it.

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. ?

I'm kinda lost... any tips?
 
jonboy said:
I need help setting up the equation here... that's it.

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. ?

I'm kinda lost... any tips?
Click to expand...

You know that the larger pump can empty the tank in 5 hours. So, in 1 hour that pump can empty 1/5 of the tank.

If the larger pump is started at 1 pm, and the tank is to be emptied at 5 pm, then that pump will work for 5 - 1, or 4 hours, and will empty 4/5 of the tank.

The smaller pump can empty the tank in 8 hours. In 1 hour, it empties 1/8 of the tank. If the smaller pump works for x hours, it will empty x/8 of the tank.

amount emptied by large pump + amount emptied by small tank = whole job

4/5 + x/8 = 1

Now, solve for x....
 
And when you find the value of x, that will be the amount of time BEFORE 5 pm at which you need to start the smaller pump. So, you should be able to determine when to start the smaller pump.
 
Thanks Mrspi! You explained this very clearly and I wish you was my teacher. ;)
 
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