No Honor Among Thieves

If each (of the remaining 3/5 of the group) thief received an extra 4 coins after killing 2/5 of the original group, this means that originally, each of the thieves must have had 6 coins , since:

$\text { dead : alive } = \frac {2}{5} : \frac {3}{5} = 2 : 3$

$\text {original number of coins } = \frac {4}{ \frac {2}{3} } = 6$

This also means, that after receiving the extra 4 coins, each living thief had 6 + 4 = 10 coins (including their leader).

After killing the leader, his 10 coins were received by 10 ÷ 2 = 5 thieves, each having now 10 + 2 = 12 coins.

Hence, the total number of coins:

$5 × 12 = \boxed {60}$

( It also follows, that there were 60 ÷ 6 = 10 thieves in the original group.)

Let $x$ be the number of gold coins, and $n$ be the number of thieves. Then we solve the system of equations

$\dfrac{x}{n}+4=\dfrac{x}{\frac{3}{5}n}$

$\dfrac{x}{n}+4+2=\dfrac{x}{\frac{3}{5}n-1}$

to get $x=60$ and $n=10$

Let $S$ the total number of gold coins in the bag; $x$ be the number of thieves in the group; and $q$ be the original quotient.

Thus, initially, $S = xq$ .

Then, after the first killing, $S = \dfrac{3x}{5}(q+4)$ .

And after the second kill, $S = (\dfrac{3x}{5} - 1)(q+6)$ .

We can see that $S$ never changes as the thieves wouldn't leave any gold coin behind, and so equating the first two equations, we can solve for $q$ :

$S = xq = \dfrac{3x}{5}(q+4)$

$\dfrac{2xq}{5} = \dfrac{12x}{5}$ .

$q = 6$

Therefore, $S = 6x = (\dfrac{3x}{5} - 1)(6+6)$

$x = \dfrac{6x}{5} - 2$

$x = 10$

As a result, $S = \boxed{60}$ .

Checking the answers, first, there were initially $10$ thieves getting a share of $6$ gold coins each.

Then $\dfrac{3}{5}$ of the group or $6$ thieves remained and got a share of $10$ gold coins each (4 more added).

Finally, one more was killed, and so $5$ thieves got a share of $12$ gold coins each (2 more added).