Answer:
B.
Step-by-step explanation:
Let's write an equation that expresses the relationship.
Let s denote the amount of spiders.
We know that every time Alex <em>catches</em> one spider, the <em>remaining </em>population <em>doubles</em>.
So, if s represent the population of spiders, if Alex <em>catches</em> 1, then the population will become (s-1).
However, for every catch, the population <em>doubles</em>, so, the new population will be 2(s-1).
Therefore, we can write the following equation:

Where s_n represents the new population and s_o represents the original population.
We know that there are now 194 spiders in Alex's house. We want to find the number of spiders in his house before he caught the last <em>2</em> spiders. So, let's solve for s_o <em>twice</em>.
Substitute 194 for s_n. Solve for s_o:

Divide by 2:

Therefore, before catching the <em>last</em> spider, there were 98 spiders.
So, to find the number of spiders before the last <em>2</em>, we will solve for s_0 again. This time, we will use 98 for s_n. So:

Solve for s_o:

Therefore, there were 50 spiders in Alex's house before he caught the last 2 spiders given that the current population is 194 spiders.
Our answer is B.
Edit: Typo