Question 993999
Let distance = d and time taken =t
so speed = d/t
New time = (3/4)t
So new speed = d/ (3t/4)
=4d/3t
increase in speed = 4d/3t -d/t 
= d/3t
So, percentage increase in speed
={{{(d/3t)/(d/t) *100%}}}
= 33.33%