In: Statistics and Probability
I think there is a problem in your question. Just think about it, if 2 people are sitting per table and you still need 4 more tables to seat everyone. And if 4 sit in each table then you still need 4 more. This looks iffy. However, I will give you the way to solve the problem.
Let denote the number of students. and the number of tables in the lab.
If she assigns 2 people per table then she needs 4 more tables to occupy all students. So, if she seats 2 students each in tables then she can seat all students. Therefore:
Similarly if she assigns 4 people per table then she needs 4 more tables to occupy all students. So again if she seats 4 students in y+4 tables then she can seat all students. Therefore:
So, this is the way to solve it but the question here is not correct that is why I got n=0.
Please let me know any doubts you have in the comments.