Step 1: Express specific gravity using the reciprocal density ratio:
Instead of dividing oil density by water density directly, first find the reciprocal, $\dfrac{1}{SG} = \dfrac{\rho_{water}}{\rho_{oil}} = \dfrac{62.4}{50}$. Computing this gives $\dfrac{62.4}{50} = 1.248$.
Step 2: Substitute this reciprocal directly into the API gravity formula:
The API gravity formula can be rewritten as ${}^{\circ}API = 141.5 \times \left(\dfrac{1}{SG}\right) - 131.5$, which avoids computing SG as a decimal first and reduces rounding error. Substituting the reciprocal value found in Step 1: ${}^{\circ}API = 141.5 \times 1.248 - 131.5$.
Step 3: Perform the multiplication:
$141.5 \times 1.248 = 176.592$. This is the value of $141.5/SG$ computed to more decimal precision than the direct division method.
Step 4: Subtract to find the API gravity:
${}^{\circ}API = 176.592 - 131.5 = 45.092$, which rounds to $45.1$ to one decimal place, matching the result obtained from the direct division method.
Step 5: Confirm physical reasonableness:
Since the given oil density of 50 lbm/ft$^3$ is well below the water density of 62.4 lbm/ft$^3$, the oil is light and buoyant relative to water, and an API gravity in the mid forties is typical for a light or condensate like crude oil, confirming the answer is physically sensible.
Final Answer:
\[\boxed{45.1 \ ^{\circ}API}\]