A bob of mass $$m$$ is suspended by a light string of length $$L$$. It is imparted a minimum horizontal velocity at the lowest point $$A$$ such that it just completes half circle reaching the top most position $$B$$. The ratio of kinetic energies $$\frac{(K.E.)_A}{(K.E.)_B}$$ is:

Condition at the Topmost Point ($$B$$)

For the bob to just complete the circle, the tension in the string at point B must be zero. The centripetal force at this point is provided entirely by the gravitational force:

$$mg = \frac{mv_B^2}{L}$$

$$v_B^2 = gL$$

Kinetic Energy at Point $$B$$

The kinetic energy at the highest point is: 

$$(K.E.)_B=\frac{1}{2}mv_B^2=\frac{1}{2}mgL$$

Conservation of Mechanical Energy

$$(K.E.)_A = (K.E.)_B + (P.E.)_B$$

$$(K.E.)_A = \frac{1}{2}mgL + mg(2L)$$

$$(K.E.)_A = \frac{5}{2}mgL$$

The ratio is $$5 : 1$$.