I did this a little differently.
We know that \(5r_1=12\), where \(r_1\) equals the hours each machine takes to clean their respective section of the lot.
Solving for \(r_1\) yields \(r_1=2.4\)
Using the same approach for 6 machines yields:
\(6r_2=12\)
\(=>r_2=2\)
I then did a percent change:
\(\frac{2.4-2}{2.4}\approx{0.17}\)
That is we can expect a 17% change in our total time. Which yields,
\(12*0.17\)\(\approx{2}\).
Thus, \(12-2=10\). The answer is E.
This may not prove to be the most time efficient way, but I sometimes find it helpful in rate/work problems to think of % change to help visualize what the problem is asking. I also find that it's not too hard to estimate these numbers to get a more educated guess. When I initially did this problem I actually came up with a 20% change because I rounded to make the math easier. When using 20%, you get an answer of 9.6 hours total which for me was enough to know the answer was E.