If speed increases by 33.33%, what is the percent reduction in time taken to travel the same distance

We know, T=D/S ;
As speed increases by 33.33%, new time taken say T1 = D/(1+1/3)S = D/(4/3)S
i.e T1= 3D/4S
Reduction in time taken=[ {(D/S)-(3D/4S)} / (D/S) ] * 100
=25%

33% is (1/3) so dec in time will be (1/(1+3)) which will be 1/4. Hence 25%

take a example if speed was 100kmph and took 100 hour hour distance becomes t*d =10000km,now distance doesn't change hence 10000=133.33*x(133.33 is new speed adding 33.33to it) , then x comes out to be 75. now difference is 100-75=25 change in time .

assume distance is 60 and speed is 60 ....so speed increase by 33.33% new speed 80
60 km with speed 80 ...time =45 min
time reduction%=(60-45)/60*100=25

25%

25% reduction

25

25%

How? Please can you provide me with an easy solution

25 percent