George the chimp found a large guava tree full of ripe fruit and decided to celebrate this find with his friends. He gave half of the total bunch and one extra to his closest pal. He then gives half of the remaining guavas and one extra to his 2nd closest pal and carries on in this manner. The tree is out of guavas after treating 5 pals. How many guavas did the tree have when George initially found it?
If we continue with the logic of (x/2)+1 the ans should be 62 but I am confused in the language of the question.
If there are $x$ guavas at any time, then you are correct that George gives $\frac{x}{2} + 1$ guavas to his friend, which leaves $x - \left(\frac{x}{2} + 1 \right) = \frac{x}{2} - 1$ guavas left.
You can check that $62$ is the correct answer. After giving the guavas to his 1st closest friend, there are $\frac{62}{2} - 1 = 30$ guavas left. Repeating the process gives $14, 6, 2, 0$. Since he has $0$ guavas after treating his 5th closest friend, your answer is correct.
As to where $62$ came from, work backwards. If there are $y$ guavas at one stage, $y = \frac{x}{2} - 1 \Rightarrow 2(y + 1) = x$, so there were $2(y+1)$ guavas at the previous stage. Substitute $y = 0$ as there are $0$ guavas at the end, then do it $4$ more times ($5$ in total) to arrive at $62$.
The recurrence can be described as $a_{n+1} = 2(a_n + 1)$, with initial condition $a_0 = 0$. $a_5 = 62$ for this question, and in general, $a_{n} = 2(2^n - 1)$.