Question

A man covers a distance by $3/4$ of its original speed and reach by $30$ min. late. If he walks by increasing $40$ % of original speed how much time before he reach?

Answer 1

Let orignal speed be $=v$ , original time $=t$, distance$=d$

Given :

$\frac{3}{4}\times v=\frac{d}{t+30}d=\left(\frac{3}{4}v\right)(t+3)$

Distance is always same

$d=v\times t$

Equating the two distances

$vt=\frac{3}{4}v\times (t+30)(\frac{4}{5}-1)t=30t=90$

Hence, the correct answer is $90$ minutes.