Maths 百分法

88.42.
The answer is: C. [100k/(100 + k)]%

Originally, let s and t be the speed and time respectively.
Distance = s × t

When the speed is increased by k%, let t1 be the time.
Distance = [(1 + k%) × s] × t1

The distance between P and Q is constant. Then,
s × t = [(1 + k)% × s] × t1
t = t1(1 + k%)
t1 = t/(1 + k%)


The time is reduced by
= [(t - t1) / t] × 100%
= {[t - (t/(1 + k%))] / t} × 100%
= {1 - [1/(1 + k%)]} × 100%
= {[(1 + k%)/(1 + k%)] - [1/(1 + k%)]} × 100%
= [(1 + k% - 1)/(1 + k%)] × 100%
= [k%/(100 + k)%] × 100%
= [100k/(100 + k)]%