a train leaves station a at 5 a m and reaches station b at 9am on the same day another

Let the distance between X and Y is d km
Then, speed of A is d/4 km/hr and that of B is 2d/7 km/hr.

X ------------------- Y (XY = d km)
Relative speed =(d/4+2d/7)=15d/28 km/hr

Now distance between these trains at 7 am =d−d/2=d/2km

Hence, time = (d/2)(15d/28) = 14/15×60 = 56 min
Hence both of them meet at 7:56 am
5 years agoHelpfull: Yes(4) No(0)
Let the distance between X and Y is d km
Then, speed of A is
d
4
km/hr and that of B is
2d
7
km/hr.

X ------------------- Y (XY = d km)
Relative speed =(
d
4
+
2d
7
)=
15d
28
km/hr

Now distance between these trains at 7 am =d−
d
2
=
d
2
km

Hence, time =
(
d
2
)
(
15d
28
)
=
14
15
×60=56 min
Hence both of them meet at 7:56 am
5 years agoHelpfull: Yes(1) No(0)
M takes 4 hrs to go from A to B
and
N takes 3.5 hrs to from B to A
also
Train N leaves B, 2 hrs after M leaves A
:
It will easier if we choose a distance value for A to B
I chose 280 mi, both 4 and 3.5 will divide into it evenly
Find the speed of both trains from this
280/4 = 70 mph speed of M
280/3.5 = 80 mph speed of N
:
When they cross the total distance traveled by both trains = 280 mi
Let t = time (hrs) of train M
then
(t-2) = time of train N
:
Dist = speed * time
70t + 80(t-2) = 280
70t + 180t - 160 = 280
150t = 280 + 160
t = 440%2F150
t = 2140%2F150 = 214%2F15 hrs travel time of M when the cross
:
Change 14/15 hrs to min: 14/15 * 60 = 56 min
:
2 hr 56 min from 5 am = 7:56 am time when they cross.
7 years agoHelpfull: Yes(0) No(0)

a train leaves station a at 5 a.m and reaches station b at 9am on the same day.another train leaves station b at 7 a.m and reaches station a at 10:30am on the same day.the time at which the two trains cross each other is?