Solution (By Examveda Team)
$$\eqalign{
& {\text{Relative speed of the trains }} \cr
& {\text{ = }}\left( {\frac{{100 + 200}}{{2 \times 60}}} \right){\text{m/sec}} \cr
& {\text{ = }}\left( {\frac{5}{2}} \right){\text{m/sec}} \cr
& {\text{Speed of train B}} \cr
& {\text{ = 120 kmph}} \cr
& = \left( {120 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr
& {\text{ = }}\left( {\frac{{100}}{3}} \right){\text{m/sec}} \cr
& {\text{Let the speed of second train be }}x{\text{ m/sec}} \cr
& {\text{Then, }} \frac{{100}}{3} - x = \frac{5}{2} \cr
& \Rightarrow x = \left( {\frac{{100}}{3} - \frac{5}{2}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{185}}{6}} \right){\text{m/sec}} \cr
& \therefore {\text{Speed of second train}} \cr
& {\text{ = }}\left( {\frac{{185}}{6} \times \frac{{18}}{5}} \right){\text{ kmph}} \cr
& {\text{ = 111 kmph}} \cr} $$
Speed = distance / time
s1 -s2 = x+y/t
120-s2 = 100+200/ 2×60
120-s2 =300/120
120-s2 = 5/2
120-s2 = 5/2 × 18/5
120-s2 =9
s2 = 120-9
s2 =111 kmph is answer
Ans should be 111
How can the speed of train C be greater than the B's as B crosses C? The equation will be, 100/3-x=5/2
Hence, x=111
Correct it.
Thanks