<PRE><B>Question 684391</B>
 1) Two trains X and Y enter at the same time a tunnel 1 Km long at the opposite ends.
 They meet in 45 seconds, after X has traveled 200 meters more than Y.
:
Find the distance each train has traveled when they meet
let a = distance traveled by x when they meet
then
(a-200) = distance traveled by y when they meet
therefore
a + (a-200) = 1000 meters
2a = 1000 + 200
2a = 1200
a = 600 meters traveled by x, and 400 meters traveled by y
then
600/45 = 13.33 m/sec is the speed of x
and
400/45 = 8.89 m/sec is the speed of y
: 
They cross each other in 4.5 seconds
Find the total Length of the two trains
L = 4.5(13.33+8.89)
L = 100 meters
:
 and 30 seconds later X is completely out of the tunnel.
total time for x to enter and clear the tunnel
45 + 4.5 + 30 = 79.5 seconds
Find how far the train X travels in 79.5 sec at 13.33 m/sec
79.5*13.333 = 1060 meters
the tunnel is 1000 meters, therefore the train is 60 meters long
:
Find the length of y
100 - 60 = 40 meters is the length of train y
:
find the length and speed of each of the trains.
X is 60 meters long
Y = 40 meters long
and
X = {{{(13.333*3600)/1000}}} = 48 km/hr
Y = {{{(8.889*3600)/1000}}} = 32 km/hr
:
:
2) A train after traveling 50 Km. Meets with an accident and then proceeds at ž th of its former speed and arrives at its destination 35 minutes late.
let s = train&#39;s normal speed
then
.75s = train&#39;s speed after the the accident
Let d = distance of the whole trip
then
(d-50) = distance at a speed of .75s
Write a time equation
{{{(d-50)/(.75s)}}} - {{{(d-50)/s}}} = {{{35/60}}}
multiply by 60s so get rid of the denominators, resulting in:
80(d-50) - 60(d-50) = 35s
80d - 4000 - 60d + 3000 = 35s
20d - 35s - 1000 = 0
20d - 35s = 1000
then we are dealing with a distance of (d-74) 
 Had the accident happened 24 Km. Further on, it would have reached the destination only 23 minutes late.
{{{(d-74)/(.75s)}}} - {{{(d-74)/s}}} = {{{23/60}}}
multiply by 60s so get rid of the denominators, resulting in:
80(d-74) - 60(d-74) = 23s
80d - 5920 - 60d + 4440 = 23s
20d - 23s - 1480 = 0
20d - 23s = 1480
:
Use elimination here
20d - 23s = 1480
20d - 35s = 1000
------------------subtraction eliminates d, find s
12s = 480
s = 480/12
s = 40 km/hr is the normal speed (30 km/hr is the slower speed)
:
Find the speed of the train and the distance.
20d - 23(40) = 1480
20d - 920 = 1480
20d = 1480 + 920
20d = 2400
d = 2400/20
d = 120 km
:
:
Check that in the other equation
20(120) - 35(40) = 
2400 - 1400 = 1000; confirms our solution of: 
speed: 40 km/hr, normal speed
dist: 120 km </PRE>