Two hoses, a and b, are used to fill a fish tank with water. hose a puts water into the tank twice as fast as hose b. if both hoses are used, the tank is filled five minutes faster than if just hose a is used. how many minutes would it take for hose b to fill the tank on its own?

Let $A$ be the rate that hose $A$ pumps water. From the given information, hose $B$ pumps water at a rate of $.5A$. Then, let $V$ equal the volume of the tank, and $t$ equal the time for hose $A$ to fill the tank on its own. The volume filled will be equal to the rate multiplied by the time, so we can set up two equations: \begin{align*}

V&=t(A)\\

V&=(t-5)(A+.5A)

\end{align*}The first represents the tank filled by just hose $A$, and the second represents the tank filled 5 minutes faster by both hoses. Setting the two equations equal to each other, we can solve for $t$ as shown: \begin{align*}

(t-5)(A+.5A)&=t(A)\\

\Rightarrow\qquad (t-5)(1.5A)&=tA\\

\Rightarrow\qquad 1.5(t-5)&=t\\

\Rightarrow\qquad 1.5t-7.5&=t\\

\Rightarrow\qquad .5t&=7.5\\

\Rightarrow\qquad t&=15

\end{align*}So, it would take 15 minutes for hose $A$ to fill the tank alone. Since hose $B$ is half as fast, it would take $\boxed{30}$ minutes for hose $B$ to fill the tank on its own.

x+y=90

x=30-y=x

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