In short if pipe A is much shorter than B, and both A and B have the same size and outlet pressure (aka back pressure), then pipe A and B will have identical pressure drop and pipe A will have a higher flow rate.
Given a fixed inlet and outlet pressure, the pressure drop is also fixed ($\Delta P = P_{inlet} - P_{outlet}$) and therefore the flow in the two pipes would have to adjust to balance this constraint. It is assumed that the inlet pressure is the same as both pipes branch of the same line at the same point in a Tee configuration. It is assumed that the back pressure on pipes A and B is the same in absence of better information.
Pressure drop along a pipe is dependent on several factors and may be calculated using the equation below:
$$
\Delta P = f \frac{L}{D} \frac{\rho v^2}{2}
$$
Where:
$P$ is the pressure drop.
$f$ is the friction factor.
$L$ is the pipe length.
$D$ is the pipe internal diameter.
$\rho$ is the fluid density.
$v$ is the fluid velocity.
The fluid properties will be identical in each branch as they are fed from the same main line and therefore we do not need to consider the effect of $\rho$.
If pipes A and B are of identical construction (same size and schedule) they will have the same internal diameter and therefore will do not need to consider the effect of $D$.
The pipe friction factor is a complex function of the Reynolds number and the pipe wall roughness. Assuming the pipes are constructed from the same material and have similar histories (i.e. manufactured and installed at the same time) the wall roughness of the two pipes should be similar. Therefore we would only need to consider the effect of Reynolds number (more precisely the velocity) on the friction factor. However due to its lesser effect on the pressure drop compared to fluid velocity and pipe length and the complexity of its relationship we will not ignore the friction factor moving forward.
At this point we can see pressure drop is primarily a function of the fluid velocity and the pipe length.
$$
\Delta P \propto L v^2
$$
Given the assumption that that $\Delta P$ is fixed, if we reduce $L$, $v$ will be increased to maintain the 'fixed' pressure drop. Similarly if we increase $L$, $v$ will be reduced.