Fractional equation? 'Pipes A, B fill tank in 8, 12 hrs...'

[dunit0001]

Pipes A and B can fill a storage tank in 8 hours and 12 hours, respectively. With the tank empty, pipe A was turned on at noon, and then pipe B was turned on at 1:30pm. At what time was the tank full?

How do I solve this??

I know that A fills 1/8 in one hour and B fills 1/12 in one hour...

3/2 is the time A has been turned on already...

 

[soroban]

Re: Fractional equation???

Hello, dunit0001!

Pipes A and B can fill a storage tank in 8 hours and 12 hours, respectively.
With the tank empty, pipe A was turned on at noon, and then pipe B was turned on at 1:30pm.
At what time was the tank full?

This is the way I baby-talk my way through these "Work Problems".

Pipe A fills the tank in 8 hours.
. . In one hour, it fills \(\displaystyle \frac{1}{8}\) of the tank.
. . In \(\displaystyle x\) hours, it fills \(\displaystyle \frac{x}{8}\) of the tank.

Pipe B fills the tank is 12 hours.
. . In one hour, it fills \(\displaystyle \frac{1}{12}\) of the tank.
. . In \(\displaystyle x\) hours, it fills \(\displaystyle \frac{x}{12}\) of the tank.


Pipe A has a 1½-hour headstart.
It has already filled: \(\displaystyle \:\frac{3}{2}\cdot\frac{1}{8} \:=\:\frac{3}{16}\) of the tank.


In the next \(\displaystyle x\) hours:
. . Pipe A will fill \(\displaystyle \frac{x}{8}\) of the tank.
. . Pipe B will fill \(\displaystyle \frac{x}{12}\) of the tank.

Together, they fill: \(\displaystyle \:\frac{x}{8}\,+\,\frac{x}{12}\:=\:\frac{5}{24}x\) of the tank.
. . And they must fill the remaining \(\displaystyle \frac{13}{16}\) of the tank.


There is our equation! . . . \(\displaystyle \L\frac{5}{24}x \:=\:\frac{13}{16}\)

. . . Go for it!