Question

# A fast train takes 3 hours less than a slow train for a journey of ( 600 mathrm{km} ). If the speed of the slow train is ( 10 mathrm{km} / mathrm{h} ) less than that of the fast train, find the speeds of the two trains.

Solution

Lit spect of slow tran ( m=w mathrm{km} / mathrm{hr} )

( Rightarrow ) specd of fast train ( =(v+10) mathrm{km} / mathrm{m} )

( 3+ ) Time taken by fast hain ( = ) set time takou by slow thain

( Rightarrow quad 3+frac{600}{v+10}=frac{600}{v} )

( Rightarrow frac{600}{14}-frac{600}{v+10}=3 )

( Rightarrow frac{600(v+10-v)}{v(v+10)}=3 )

( Rightarrow quad v^{2}+10 v=2000 )

( Rightarrow quad v^{2}+10 v-2000=0 )

2) ( quad v^{2}+50 v-40 u-2000=0 )

( Rightarrow(u+50)(v-40)=0 )

( Rightarrow quad v=40 )

( (because v eq-50) )

Speed

of slow frain ( =40 mathrm{km} / mathrm{h} ) sbeed of fast train = 50 kmlh.