Train A leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Train B, which is making the trip from Boston to New York at a constant speed. If Train B left Boston at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in New York?
(1) Train B arrived in New York before Train A arrived in Boston.
(2) The distance between New York and Boston is greater than 140 miles
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gmatnmein2010
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Please search before you postgmatnmein2010 wrote:Train A leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Train B, which is making the trip from Boston to New York at a constant speed. If Train B left Boston at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in New York?
(1) Train B arrived in New York before Train A arrived in Boston.
(2) The distance between New York and Boston is greater than 140 miles
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I hate to resurrect this thread, but I was asked about this problem today and realized there wasn't an expert answer.
How I'd do it:
Let b = Train B's rate. We're going to solve this problem entirely in terms of this variable.
When the two trains meet, Train B has traveled for (1/6) of an hour at b mph, so Train B's distance = b/6.
After this, Train A has b/6 miles left to travel, and Train B has 100 miles left to travel. Since the trains have traveled for 70 minutes already, they have 50 minutes left to travel, or 5/6 of an hour.
Train B's remaining D = 100 and its R = b, so its T = 100/b.
Train A's remaining D = b/6 and its R = 100, so its T = b/600. But we also know that its remaining T can be written as 5/6 - 100/b, or Total T Left For Both Trains - B's Remaining T.
Since b/600 and 5/6 - 100/b represent the same thing, we can say that
b/600 = 5/6 - 100/b
b = 500 - 60000/b
b² = 500b - 60000
b² - 500b + 60000 = 0
and this quadratic is easy to factor:
(b - 200) * (b - 300) = 0
So B's rate is either 200mph or 300mph.
Remember that the distance for the trip = b/6 + 100 miles, since B travels b/6 before the meeting and 100 miles after it.
With that in mind, if B's rate = 200, then the distance = 200/6 + 100 and B's time for the trip = 2/3 of an hour. If this is the case, B travels 30 minutes after the meeting.
If B's rate = 300, then the distance = 300/6 + 100 and B's time for the trip = 1/2 an hour. If this is the case, B travels 20 minutes after the meeting.
Notice that we STILL HAVEN'T used the statements! We can do virtually all the work without them.
S1 tells us that B travels for less than 25 minutes after the meeting. Only b = 300 from our two solutions above fits, so this is sufficient.
S2 tells us that B travels for less than 26 minutes after the meeting. (If the total distance = 140 miles, then A will take a total of 84 minutes to make the trip, meaning A will travel for 24 minutes after the meeting. But the distance is greater than that, so A will travel > 24 minutes, forcing B to travel < 26 minutes.) Again, only b = 300 fits, and this is also sufficient.
How I'd do it:
Let b = Train B's rate. We're going to solve this problem entirely in terms of this variable.
When the two trains meet, Train B has traveled for (1/6) of an hour at b mph, so Train B's distance = b/6.
After this, Train A has b/6 miles left to travel, and Train B has 100 miles left to travel. Since the trains have traveled for 70 minutes already, they have 50 minutes left to travel, or 5/6 of an hour.
Train B's remaining D = 100 and its R = b, so its T = 100/b.
Train A's remaining D = b/6 and its R = 100, so its T = b/600. But we also know that its remaining T can be written as 5/6 - 100/b, or Total T Left For Both Trains - B's Remaining T.
Since b/600 and 5/6 - 100/b represent the same thing, we can say that
b/600 = 5/6 - 100/b
b = 500 - 60000/b
b² = 500b - 60000
b² - 500b + 60000 = 0
and this quadratic is easy to factor:
(b - 200) * (b - 300) = 0
So B's rate is either 200mph or 300mph.
Remember that the distance for the trip = b/6 + 100 miles, since B travels b/6 before the meeting and 100 miles after it.
With that in mind, if B's rate = 200, then the distance = 200/6 + 100 and B's time for the trip = 2/3 of an hour. If this is the case, B travels 30 minutes after the meeting.
If B's rate = 300, then the distance = 300/6 + 100 and B's time for the trip = 1/2 an hour. If this is the case, B travels 20 minutes after the meeting.
Notice that we STILL HAVEN'T used the statements! We can do virtually all the work without them.
S1 tells us that B travels for less than 25 minutes after the meeting. Only b = 300 from our two solutions above fits, so this is sufficient.
S2 tells us that B travels for less than 26 minutes after the meeting. (If the total distance = 140 miles, then A will take a total of 84 minutes to make the trip, meaning A will travel for 24 minutes after the meeting. But the distance is greater than that, so A will travel > 24 minutes, forcing B to travel < 26 minutes.) Again, only b = 300 fits, and this is also sufficient.
