Question 1166543
Let l, w be the length and width of the lawn. Let s = the strip width. 
The length of the rectangle remaining to mow is l - 2s, since it's reduced by s on either side
Similarly, the width remaining to mow is w - 2s
The new area is 16% of the original area: 
A1 = 0.16*l*w = 0.16*100*25 = 400 = (l-2s)(w-2s)
(100-2s)(25-2s) = 400
2500 - 200s - 50s + 4s^2 - 400 = 0
4s^2 - 250s + 2100 = 0
2s^2 - 125s + 1050 = 0
This can be factored as: (2s-105)(s-10) = 0
The first solution gives a strip wider than the width of the lawn, so we take 
the 2nd: s=10
A 10 ft wide strip will leave 16% of the original area
Check: (100-20)(25-20) = 80*5 = 400