two trains T1 and T2 start from point A and B respectively and meet at a point C . After passing each other at point C , T1 took 12 hrs an T2 took 3 hrs to reach B and A respectively . If speed of Train T1 is 48kmph , then what is the speed of train T2(kmph)?
a) 50
b) 144
c) 24
d) 96
e) 168
OA D
time dis
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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
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
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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.
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!!!!!!!!!
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!!!!!!!!!