time taken by a after crssing c is 12 hrs and speed of a is 12....
so dist frm c-->b is 12*48=576


now let speed of b be "sb".... so dist cover by b after crossing c ie CA is 3*"Sb"


now we know that both trains meet at C.......this implies that both take equal time to reach C from start.........

hence.... we cn equate the equation of time taken by both to rech c......

=> 576/"sb"=3*"Sb"/48


which give "sb"=96........

i hope i hv explained enuff.........bt if still prb post back a rep.....willl def repp

Time taken by T1 and T2 to reach C from their respective starting point is SAME

Time taken by T1 to reach C from A = Time taken by T2 to reach C from B

lets take speed of T1=“x” and T2=”y”

Train T1 : takes 12 hours to reach B from C so the distance is : 12x
Train T2: takes 3 hours to reach A from C so the distance is : 3y

So T1 speed will be : 3y/x ( Distance of AC/speed of T1)
So T2 speed will be : 12x/y ( Distance of AC/speed of T2)

So 3y/x = 12x/y
3y^2 = 12x^2
y^2 = (12*48*48)/3
y = 96.

Let distance from A to B = d
Let distance from A to C = c

Therefore: distance from C to B = d-c

A<---c--->C<---(d-c)--->B
<----------d------------->

So:

let them meet at C after traveling for a time t each: (we assume that they left the stations at the same time and i think that would be part of the question anyway)

T1=> covers distance c in time t
T1=> covers distance d-c in time t

therefore equating for t:

time = distance/speed

so:

t = c/S1 = (d-c)/S2 (S1 = speed of T1 and S2 - speed of T2)
=>c/48 = (d-c)/S2( since T1 = 48 given)----(1)

Now:

T1 covers rest of distance (d-c) in 12 hrs
T2 covers rest of distance (c) in 3 hrs

for T1
=> speed = distance/time
=>48=(d-c)/12
=>(d-c)=(48)(12)-----(2)

for T2
=> speed = distance/time
=> S2=c/3
=> c=3S2-----(3)

Use (2) and (3) in (1)

=>3S2/48 = (48)(12)/S2
=>(S2)(S2)=(48)(48)(12)/3=(48)(48)(4)
=>S2=(48)(2)=96

Hence D

Ta Da!!!!!!!!!