If something loses $p\%$, you multiply by $1 - p/100$. If it keeps $p\%$, you multiply by $p/100$. If it gains $p\%$, you multiply by $1 + p/100$. Decide which of these applies at each step.

Start from your assumed original (for example, $100$). First apply the $20\%$ loss, then the $5\%$ discard on the new amount, and finally the $12\%$ increase on the latest amount. Be careful not to apply any percentage to the original unless the problem says so.

After you finish all three steps using an original of $100$ units, whatever final number you get is directly the percentage of the original amount.

In Desmos, type the expression 100*0.8*0.95*1.12 and look at the numerical output. That value represents the final weight of the beans (out of the original 100 units), so read it as the final percentage of the original weight.

Let the original weight of the unroasted beans be $100$ units.

Translate each percentage change into a multiplier:

• Losing $20\%$ means keeping $80\%$, so multiply by $0.8$.
• Discarding $5\%$ means keeping $95\%$, so multiply by $0.95$.
• Absorbing water equal to $12\%$ of the current weight means the new weight is $112\%$ of the old weight, so multiply by $1.12$.

We will apply these multipliers in sequence to $100$.

Start from $100$ units.

After roasting (lose $20\%$):

After discarding $5\%$ of the roasted beans (keep $95\%$ of $80$):

So, after roasting and discarding imperfect beans, $76$ units remain.

Now apply the $12\%$ water absorption to the $76$ units. Absorbing water equal to $12\%$ of the current weight means multiplying by $1.12$:

We started with $100$ units, so the final usable coffee is $85.12$ units out of $100$, which means the final weight is $85.12\%$ of the original. Thus, $w = 85.12$. (Choice C.)