SOLUTION: 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 travelled 200 meters more than Y. They cross each other

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Question 684391: 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 travelled 200 meters more than Y. They cross each other in 4.5 seconds and 30 seconds later X is completely out of the tunnel. find the length and speed of each of the trains.
2) A train after travelling 50 Km. Meets with an accident and then proceeds at ¾ th of its former speed and arrives at its destination 35 minutes late. Had the accident happened 24 Km. Further on, it would have reached the destination only 23 minutes late. Find the speed of the train and the distance.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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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 = %2813.333%2A3600%29%2F1000 = 48 km/hr
Y = %288.889%2A3600%29%2F1000 = 32 km/hr
:
:
2) A train after traveling 50 Km. Meets with an accident and then proceeds at ¾ th of its former speed and arrives at its destination 35 minutes late.
let s = train's normal speed
then
.75s = train'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
%28d-50%29%2F%28.75s%29 - %28d-50%29%2Fs = 35%2F60
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.
%28d-74%29%2F%28.75s%29 - %28d-74%29%2Fs = 23%2F60
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