Filling A tank

[karipeters]

Filling a tank. Two pipes are connected to the same tank. Working together, they can fill the tank in 4 hr. The larger pipe, working alone, can fill the tank in 6 hr less time than the smaller one. How long would the smaller one take, working alone, to fill the tank?

I have:
t(t-6)/(2t-6)
t(t-6/(2t-6=4
t^2-6t=8t-48
t^2-14t+48=0
(t-8)(t-6)=0
It would take the smaller pipe 8 hours to fill the tank.

Is this correct? It's suppose to be in quadratic form, but I'm lost on how those work.

 

[karipeters]

"No.
Sorry, but you need serious classroom help..."

Well I already figured that it was wrong, which is why I was asking for help here. Actually I'm pulling a B in the class, I'm just stuck on this one for the moment. So any help would be greatly appreciated.

 

[soroban]

Hello, karipeters!

Two pipes are connected to the same tank.
Working together, they can fill the tank in 4 hours.
The larger pipe, working alone, can fill the tank in 6 hr less time than the smaller one.
How long would the smaller one take, working alone, to fill the tank?

This is the way I talk my way through these "work" problems.

Let \(\displaystyle L\) = hours for the large pipe to fill the tank alone.
Let \(\displaystyle S\) = hours for the small pipe to fill the tank alone.

In one hour, the large pipe can fill \(\displaystyle \dfrac{1}{L}\) of the tank.
. . In 4 hours, the large pipe can fill \(\displaystyle \dfrac{4}{L}\) of the tank.

In one hour, the small pipe can fill \(\displaystyle \dfrac{1}{S}\) of the tank.
. . In 4 hours, the small pipe can fill \(\displaystyle \dfrac{4}{S}\) of the tank.

Working together for 4 hours, they will fill the entire tank: .\(\displaystyle \dfrac{4}{L} + \dfrac{4}{S} \:=\:1\) .[1]

We are told that: .\(\displaystyle L \:=\:S - 6\) .[2]

Substitute [2] into [1]: .\(\displaystyle \dfrac{4}{S - 6} + \dfrac{4}{S} \:=\:1\) . . . There!

Multiply by \(\displaystyle S(S-6)\!:\;4S + 4(S-6) \:=\:S(S-6) \quad\Rightarrow\quad S^2 - 14S + 24 \:=\:0\)

Factor:. \(\displaystyle (S-2)(S-12) \:=\:0 \quad\Rightarrow\quad \begin{Bmatrix} \rlap{/////}S = 2 \\ S = 12\end{Bmatrix}\)

The small pipe will take 12 hours to fill the tank alone.