An element loses 31/32 of its original atoms in a certain time. If its half-life is 7 years, how long did it take for this amount of decay?

The problem states that the element will lose $\frac{31}{32}$ of its original atoms.
The fraction of atoms remaining is:

$$1 - \frac{31}{32} = \frac{32}{32} - \frac{31}{32} = \frac{1}{32}$$

The formula for the fraction of a radioactive substance remaining after $n$ half-lives is:

$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$$

Where $\frac{N}{N_0}$ is the fraction remaining, and $n$ is the number of half-lives.

Substitute the fraction remaining into the formula:

$$\frac{1}{32} = \left(\frac{1}{2}\right)^n$$

We know that $2^5 = 32$. So, $\frac{1}{32} = \left(\frac{1}{2}\right)^5$.
Therefore, $n = 5$.
This means 5 half-lives have passed.

The half-life ($T_{1/2}$) of the element is given as 7 years.
The total time ($T$) is the number of half-lives multiplied by the half-life period:

$$T = n \times T_{1/2}$$ $$T = 5 \times 7 \text{ years}$$ $$T = 35 \text{ years}$$

The final answer is $\text{D}$.