Lets compare the Fluids!

Conservation of energy $\tfrac12\rho v^2 = \rho gy$ gives for the volume outflow rate of liquid $\frac{dV}{dt} = -av = -a \sqrt{2gy}.$ Since $dV = A(y)\:dy$ , where $A(y)$ is the cross-sectional area at height $y$ , we have $\frac{dt}{dy} = A(y)\frac{dt}{dV} = -\frac{A(y)}{a\sqrt{2gy}} = -k\frac{A(y)/\pi}{\sqrt{y}}$ for a constant $k$ .

The cross-sectional areas at height $y$ for both containers is $A_1(y)/\pi = R^2 - (R - y)^2 = 2Ry - y^2;\ \ \ A_2(y) /\pi = R^2 - y^2.$ Thus $t_1(y) = -k\int (2R\sqrt y - y\sqrt y)\:dy = t_{1,\text{empty}} - k (\tfrac43 Ry\sqrt y - \tfrac25 y^2\sqrt y);$ $t_2(y) = -k\int (R^2/\sqrt y - y\sqrt y)\:dy = t_{2,\text{empty}} - k (2 R^2\sqrt y - \tfrac25 y^2\sqrt y).$ Finally, setting $t(R) = 0$ , $t_{1,\text{empty}} = k (\tfrac43 - \tfrac 25)R^2\sqrt R;\ \ \ t_{2,\text{empty}} = k (2 - \tfrac 25)R^2\sqrt R.$ The desired traction is $\frac{t_{1,\text{empty}}}{t_{2,\text{empty}}} = \frac{\tfrac43 - \tfrac 25}{2 - \tfrac 25} = \frac{14/15}{24/15} = \frac{7}{12}.$ Therefore we submit the answer $7 + 12 = \boxed{19}$ .