(I) Substitute values for variables..
Let s = 100 and x=10..
Time taken while going 100 at 10 feet per minute= \(\frac{100}{10}=10\) minutes
Time taken on return 1.1*100 at 10 meter per minute or @\(\frac{10}{0.3}\) feet per minute= \(110/\frac{10}{0.3}=\frac{33}{10}=3.3\) minutes

Change or decrease = \(\frac{10-3.3}{10}*100=67\)%

(II) Algebra..
Time taken while going s feet at x feet per minute= \(\frac{s}{x}\) minutes
Time taken on return 1.1*s at x meter per minute or @\(\frac{x}{0.3}\) feet per minute= \(1.1s/\frac{x}{0.3}=\frac{0.33s}{x}\) minutes

Change or decrease = \((\frac{s}{x}-\frac{0.33s}{x})/\frac{s}{x}*100=(1-0.33)*100=67\)%

C

let distance s feet = 10feet = 3 mtr
speed = 10 feet per minute ; total time ; 3/10 ; .3 min
new distance ; 3*1.1 ; 3.3 mtrs
speed ; 10/.3 feet per min; time taken ; 3.3/33.33 = .099 min
% ∆ in time = .099-.3/.3 ; (67%)

IMO C

"Tom ended up walking 10% more" could refer to time or distance, as it is new route...

It says walking 10% more than he usually does, and hence it refers to distance.

If it was, taking 10% more than he usually does, then would have referred to time. (In which case the answer would simply have been 10% increase)

Hope this helps

Arun Kumar
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