lean_workbook_plus_48646

Let the speed of the train be $v$ .\nLet the distance be $d$ .\nLet the original time be $t$ .\nIf the train hadn't broken down, then $vt=d$ ...........(1)\nBut when the train suffers a breakdown.\nThen\n $t'=t+3$\nSo, $\frac{3}{2}v+\frac{4}{5}v(t+1)=d$ ...........(2) (Because for the first 1.5 hrs it travels with speed $v$ , then 0.5 hrs for repairs, this gives $t+3-1.5-0.5=t+1$ hrs in which it travels at speed $\frac{4v}{5}$ ).\n\nCombining equation (1) and (2) and solving, we get\n $\frac{d}{v}=\frac{23}{2}$ .