SOLUTION: Two trains leave two cities that are 310 km apart, traveling toward each other on parallel tracks. One train travels at 70 kph and the other travels at 90 kph. If the slower train

Question 820133: Two trains leave two cities that are 310 km apart, traveling toward each other on parallel tracks. One train travels at 70 kph and the other travels at 90 kph. If the slower train leaves at 9:40 a.m. and the faster train leaves at 11:00 a.m., at what time will they pass each other? Round the answer to the nearest minute.
Found 2 solutions by mananth, josmiceli:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
slower train 9:40 AM Faster train 11:00 AM
difference = 1:20 =1 1/3 = 4/3 hours
In 4/3 hours slower train covers 4/3 * 70 = 280/3 .
At 11:00AM they are at distance of 310-280/3 = 650/3
speed = 70+90 = 160 km/h
D= 650/3
time = d/r
=(650/3)/160 =1.35 hours
They will meet at 12:21 PM

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
The time of the slower train's head start is
11:00 AM minus 9:40 AM = 1 hour 20 minutes
------------
Let = the slower train's head start in km

km
--------------------
Start a stopwatch when the faster train leaves
Let = the time on the stopwatch when
the trains meet
Let = the distance that the faster train travels
until they meet
---------------
Faster train's equation:
(1)
Slower train's equation:
(2)
Substitute (1) into (2)
(2)
(2)
(2) hrs
This is the time after 11:00 AM when they meet

--------------------
They will meet at about 12:21 PM
-----------
check:
(1)
(1)
(1)
and
(2)
(2)
(2)
(2)
(2)
error may be due to rounding off

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