What happens exactly, depends on what happens downstream of pipes A and B.
Suppose both pipes flow into the open air, then the pressure at both outlets has to be the same, e.g, at ambient pressure $p_0$ (ignore altitude etc effects). The pressure at the branch is some pressure $p_1$, which is the same for both pipes. In other words
$$\Delta p_A = \Delta p_B $$.
This pressure drop, follows from, e.g. the Hagen-Poiseuille equation,
$$\Delta p \propto L Q, $$
with $L$ the length of the pipe, and $Q$ the flow rate. The scaling can be exactly derived for laminar flow, but for turbulent flow (which you probably have), it relation is similar, with a different proportionality. In other words, the lengths and flowrates in both pipes, are relates as follows.
$$\frac{L_A}{L_B}=\frac{Q_B}{Q_A}$$
Example: If pipe A is three times longer than pipe B, than three times more liquid will flow through pipe B. Thus $75 m^3/h$ through B, and only $25 m^3/h$ through pipe A.
The default Bernoulli equation for incompressible fluids does not consider losses, however you can add correction to it, which leads to the following equation:
$$
p_1+\frac{1}{2}{\rho}v_1^2+{\rho}gz_1=p_2+\frac{1}{2}{\rho}v_2^2+{\rho}gz_2+\left(f\frac{L}{D}+{\sum}K\right)\frac{1}{2}{\rho}v_2^2,
$$
where $p_i$ are pressures, $\rho$ is the density of the fluid, $v_i$ is the average flow velocity, $g$ the acceleration due to gravity, $z_i$ the height, $f$ the friction factor, $L$ the length of the pipe, $D$ the diameter of the pipe and $K$ additional friction losses (such as in- or outlets, bends in the pipe, ect.).
(1) This gives the pressure difference between $p_1$ and $p_2$, so 30 psi or 207 kPa.
(3) This is the diameter of the pipe, so 0.01384 m in SI units.
(2) From this the volume flow rate can be determined and when combined with the diameter of the pipe, then the velocity can be determined, this assumes that the pipe has a constant diameter. Thus $v = v_1 = v_2 = \frac{4V}{\pi D^2t} = 0.7397\ \frac{m}{s}$.
The density of water at 16°C is equal to 999.2 kg/m${}^3$. According to your profile you are from Massachusetts, so for the acceleration due to gravity I will use 9.804 m/s${}^2$. I am not sure how to interpret the height difference, I will assume that you measured the pressure difference between the faucet and atmospheric pressure, such that this already incorporates the hydrostatic pressure and thus both $z_1$ and $z_2$ can be set to zero.
This allows us to rewrite the first equation to,
$$
\Delta p = \frac{1}{2}{\rho}v^2\left(f\frac{L}{D}+{\sum}K\right).
$$
To determine $f$ we first need to know the Reynolds number (and possibly the relative roughness of the pipes if the flow is turbulent). The Reynolds number can be calculated as follows,
$$
Re = \frac{vD}{\nu},
$$
where $\nu$ is equal to the kinematic viscosity of the fluid. The kinematic viscosity of water at 16°C is roughly 1.12 10$^{-6}\ m^2/s$. This leads to a Reynolds number of 9141, which is definitely well within the turbulent regime, which would require relative roughness. For this I will assume a value between 0.001 and 0.01, which lead to a friction factor of 0.035 to 0.043. I will assume that the length of the pipe is roughly equal to the sum of the vertical and horizontal distance, since pipes usually are incorporated into the walls and floors and do not travel diagonally that often. Thus $L$ would be roughly equal to 45 feet or 13.7 m. Combining all this information allows use to guess the sum of loss factors,
$$
\sum K = \frac{2\Delta p}{\rho v^2} - f\frac{L}{D} = 719 \pm 4.
$$
To get an idea of how much this is, every connection between two pipes segments adds 0.08, a right angle outlet (such as a faucet) adds 1, a 90° bend adds 0.75 to 1.4 (depending on the radius) and a completely open faucet can add 0.15 to 10 (depending on the type). Without knowing more exact numbers for the length, the amount of bends, it will be hard to tell how clean the pipes are. For instance the length will probably much longer since there will also be pipes before the main faucet.
Best Answer
For an incompressible (or essentially incompressible) fluid, pressure is determined only to within an arbitrary constant. The only way to establish an absolute pressure is to match a boundary condition, which is not available in this problem specification.