Question

At a business meeting every person shakes each other's hand once if they were 91 handshakes in total the number of people at the meet is

Answer 1

This is basically asking you to find the number of vertices $n$ of a complete graph that has 91 edges. (I've attached an example of one where $n=7$, taken from Wiki "complete graph" article)

Let's say you have $n$ unconnected vertices, each labelled with a number $v_1,v_2,\ldots,v_n$ (just for the purpose of tracking which ones you've already taken into account). Draw edges connecting $v_1$ to all the other vertices. There are $n-1$ possible edges you can draw.

Now starting from $v_2$, there's already an edge between it and $v_1$, which means there are $n-2$ other possible edges that you can draw.

Next, from $v_3$, you can only draw $n-3$ new edges because $v_3$ is already connected to $v_1$ and $v_2$.

Continuing in this pattern, you find that the second-to-last edge, $v_{n-1}$ can only be connected once to $v_n$, and finally arriving at $v_n$ you find that all possible connections have been exhausted.

The takeaway here is that the total number of possible connections is the sum

$S=(n-1)+(n-2)+(n-3)+\cdots+1+0$

Writing the terms in reverse order, you have

$S=0+1+2+\cdots+(n-2)+(n-1)$

So now if you add up the first, second, ..., terms of the two equivalent sums, you have

$2S=(n-1+0)+(n-2+1)+(n-3+2)+\cdots+(1+n-2)+(0+n-1)$
$2S=(n-1)+(n-1)+(n-1)+\cdots+(n-1)+(n-1)$

In other words, you're adding up $n-1$, $n$ times (since each $n-1$ corresponds to a distinct vertex), so you have

$2S=n(n-1)$

Dividing by 2, you end up with

$S=\dfrac{n(n-1)}2$

Now, since you know that 91 handshakes took place in total, you have

$91=\dfrac{n(n-1)}2\implies 182=n(n-1)\implies n=-14,13$

Obviously, we omit the negative solution, so there were 13 people at the meeting.