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.