What is the sum of $7\frac{1}{16}$ and $7\frac{1}{48}$?

Let the given mixed fractions be separated into their integer and fractional parts:
$$\text{Sum}=\left(7+\frac{1}{16}\right)+\left(7+\frac{1}{48}\right)$$

Group the integer parts together and the fractional parts together:

$$\text{Sum}=(7+7)+\left(\frac{1}{16}+\frac{1}{48}\right)$$

$$\text{Sum}=14+\left(\frac{1}{16}+\frac{1}{48}\right)$$

Find a common denominator for the fractions. The Least Common Multiple (LCM) of $16$ and $48$ is $48$:

$$\frac{1}{16}=\frac{1\times 3}{16\times 3}=\frac{3}{48}$$

Now combine the fractions:

$$\frac{3}{48}+\frac{1}{48}=\frac{4}{48}$$

Reduce the fraction component to its simplest form:

$$\frac{4}{48}=\frac{1}{12}$$

Combining the integer part and the simplified fractional part gives:

$$14+\frac{1}{12}=14\frac{1}{12}$$

Looking closely at the unsimplified options provided in the problem statement:

$$14+\frac{4}{48}=14\frac{4}{48}$$

This exactly matches Option A.