This is a inequality question. There are two separate inequalities and the values n can take are the solutions of n that overlap between the 2 inequalities. First inequality: 3n< 12 therefore n <4 . Second inequality : 6n > n2 + 5 then n2 - 6n + 5 < 0 therefore (n-5)(n-1) < 0.For the second inequality we can then use the graph.= of the quadratic. The part below 0 is between 5 and 1 so for this inequality the solution is 1< n < 5 and combining the two solutions is 1< n<4. This is the answer to the question.If student interested in maths at a level or found it too easy: potential extension looking at 1/n < 5 and approaches that can be taken