Let the original program be of x steps, and total time taken be t. Time per step = t/x
Now, E1 : 5% of x has been enchanced by 10x(time per step reduced 10 times) (x/20) *(t/10x) + (19x/20)*t/x = 0.955t
Similarly, for E2 : (x/10)*t/5x + 9x/10*t/x = 0.92t
As, overall time taken by E2 is less than E1 so E2 is better enhancement than E1.