Transfer Game

Let the initial and end amounts $A$ and $B$ have be $a_0$ and $b_0$ , and $a_1$ and $b_1$ . Then $a_0+b_0 = a_1+b_1 = 320$ .

Since $\dfrac {b_1}{a_1} =\dfrac 32$ , $\implies a_1 + \dfrac 32 a_1 = 320 \implies a_1 = 128$ and $b_1 = 192$ .

Then we have:

$\begin{cases} b_1 = 0.8(0.2a_0+b_0) & \implies 0.16 a_0 + 0.8 b_0 = 192 & ...(1) \\ a_1 = 0.8a_0 + 0.2(0.2a_0+b_0) & \implies 0.84a+0.2b_0 = 128 & ...(2) \end{cases}$

From $4\times(2) - (1): \ 3.2a_0 = 320 \implies a_0 = \boxed{100}$ .