Which one is faster?

Let the distance between $A$ and $B$ be $d$ , the river flow, Harry's swimming and rolling speeds be $v_0$ , $v_1$ and $v_2$ respectively, and the time for Harry to row from $A$ to $B$ be $t$ . The we have:

$\begin{cases} 40(v_1+v_0) = d & \Rightarrow v_1+v_0 = \dfrac{d}{40} &...(1) \\ 45(v_1-v_0) = d & \Rightarrow v_1-v_0 = \dfrac{d}{45} &...(2) \\ 15(v_2-v_0) = d & \Rightarrow v_2-v_0 = \dfrac{d}{15} &...(3) \\ \space \space t\space (v_2+v_0) = d & &...(4) \end{cases}$

$\begin{cases} (1)-(2): & 2v_0 = \left(\dfrac{1}{40} - \dfrac{1}{45} \right) d & \Rightarrow v_0 = \dfrac{d}{720} \\ (3): & v_2 - \dfrac{d}{720} = \dfrac{d}{15} & \Rightarrow v_2 = \dfrac{49d}{720} \\ (4): & t \left(\dfrac{49d}{720} + \dfrac{d}{720} \right) = d & \Rightarrow t = \dfrac{720}{50} = 14.4 = 14:24 \end{cases}$

Therefore, the required answer is $\boxed{1424}$ .

Let the distance from $A$ to $B$ is $1$ , and let $x,y,z$ be the the speeds of the current and Harry's swimming and rowing, respectively. Then

$y+x=\frac{1}{45}=\frac{9}{360}$ and $y-x=\frac{1}{45}=\frac{8}{360}$ and $z-x=\frac{1}{15}=\frac{24}{360}$

so

$z+x=(y+x)-(y-x)+(z-x)=\frac{9-8+24}{360}=\frac{25}{360}$

so it take harry $\frac{360}{24}=14.4min$ to row from $A$ to $B$

which mean $14min$ and $24seconds$