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Train A leaves New York for Boston at 3 PM and travels at

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Train A leaves New York for Boston at 3 PM and travels at [#permalink] New post 21 Jul 2010, 23:49
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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.


Please help me know the difficulty level of this question. I was not able to solve it in even 5 mins :(
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Re: Train A leaves New York [#permalink] New post 22 Jul 2010, 02:34
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rohitgoel15 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.

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

Please help me know the difficulty level of this question. I was not able to solve it in even 5 mins :(


Let:
d be the distance between cities;
x be the rate of Train B.

"An hour later (so at 4:00PM), Train A passes Train B" --> before they pass each other A traveled 1 hour (4PM-3PM) and B traveled 1/6 hours (4PM-3:50PM).

"Combined travel time of the two trains is 2 hours" --> d/100(time to cover all distance for train A)+d/x(time to cover all distance for train B)=2 --> \frac{d}{100}+\frac{d}{x}=2;

As before they pass A traveled 100 miles (1 hour at 100 miles per hour), then distance to cover for B after they pass is this 100 miles and as B traveled x*1/6 miles before they pass (1/6 hour at x miles per hour), then distance to cover for A after they pass is this x*1/6 miles --> 100+\frac{x}{6}=d;

So, we have:
\frac{d}{100}+\frac{d}{x}=2 and 100+\frac{x}{6}=d.

Solving for d and x
d=150 and x=300;
OR:
d=\frac{800}{6}\approx{133.3} and x=200.

(1) Says that train B arrived before A.
If x=200 A arrives at 4:20, B at 4:30, not good;
If x=300 A arrives at 4:30, B at 4:20, OK.
Sufficient

(2) Says that d>140 --> d=150 --> x=300, arrival time for B 4:20. Sufficient

Answer D.

P.S. This is definitely a hard (700+) question.

Hope it's clear.
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Re: Train A leaves New York [#permalink] New post 23 Jul 2010, 05:18
Nice explanation. Thanks Bunuel!
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Re: Train A leaves New York [#permalink] New post 06 Mar 2011, 12:36
very good problem,

thank you Bunuel!
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Re: Train A leaves New York [#permalink] New post 06 Mar 2011, 18:36
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Interesting question. I would like to share my thoughts on it. The first thing I notice is that the statements do not provide any concrete data. I cannot solve anything using them so most probably I will be able to get an answer from the data in the question stem but I will get multiple possible answers. The statements will probably help me choose one of them. (all a speculation based on the statements. The answer may be E) I know a quadratic gives me multiple answers.

Attachment:
Ques2.jpg
Ques2.jpg [ 5.7 KiB | Viewed 16667 times ]

The diagram above incorporates the data given in the question stem. Let x be the distance from meeting point to Boston.

Speed of train A = 100 mph
Speed of train B = x/(10 min) = 6x mph (converted min to hr)
Total time taken by both is 2 hrs. Already accounted for is 1hr + (1/6) hr
The remaining (5/6) hrs is the time needed by both together to reach their respective destinations.

Time taken by train A to reach B + time taken by train B to reach NY = 5/6
x/100 + 100/6x = 5/6
3x^2 - 250x + 5000 = 0 (Painful part of the question)
x = 50, 33.33

(1) Train B arrived in New York before Train A arrived in Boston.
If x = 50, time taken by train A to reach B = 1/2 hr, time taken by train B to reach NY = 1/3 hr
If x = 33.33, time taken by train A to reach B = 1/3 hr, time taken by train B to reach NY = 1/2 hr
Since train B arrived first, x must be 50 and B must have arrived at 4:20. Sufficient.

(2) The distance between New York and Boston is greater than 140 miles.
x must be 50 to make total distance more than 140. Time taken by train B must be 1/3 hr and it must have arrived at 4:20. Sufficient.
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Re: Train A leaves New York [#permalink] New post 06 Mar 2011, 20:24
Is there a intuitive way to solve this rather than with equations, quadratics. I guesses for a sec that A is very sufficient information. Since only two variables are unknown - distance covered by B before it intersects A and the speed of train B. Two variables and two equations will be fine.

But then I looked at B and suddenly the difficulty jumped up. A or D appear likely. I thought train B is super fast so increasing the distance will make even more likely that train B arrives before A. This is the same as stem 1)

hence I guessed D.

VeritasPrepKarishma wrote:
Interesting question. I would like to share my thoughts on it. The first thing I notice is that the statements do not provide any concrete data. I cannot solve anything using them so most probably I will be able to get an answer from the data in the question stem but I will get multiple possible answers. The statements will probably help me choose one of them. (all a speculation based on the statements. The answer may be E) I know a quadratic gives me multiple answers.

Attachment:
Ques2.jpg

The diagram above incorporates the data given in the question stem. Let x be the distance from meeting point to Boston.

Speed of train A = 100 mph
Speed of train B = x/(10 min) = 6x mph (converted min to hr)
Total time taken by both is 2 hrs. Already accounted for is 1hr + (1/6) hr
The remaining (5/6) hrs is the time needed by both together to reach their respective destinations.

Time taken by train A to reach B + time taken by train B to reach NY = 5/6
x/100 + 100/6x = 5/6
3x^2 - 250x + 5000 = 0 (Painful part of the question)
x = 50, 33.33

(1) Train B arrived in New York before Train A arrived in Boston.
If x = 50, time taken by train A to reach B = 1/2 hr, time taken by train B to reach NY = 1/3 hr
If x = 33.33, time taken by train A to reach B = 1/3 hr, time taken by train B to reach NY = 1/2 hr
Since train B arrived first, x must be 50 and B must have arrived at 4:20. Sufficient.

(2) The distance between New York and Boston is greater than 140 miles.
x must be 50 to make total distance more than 140. Time taken by train B must be 1/3 hr and it must have arrived at 4:20. Sufficient.
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Re: Train A leaves New York [#permalink] New post 06 Mar 2011, 22:47
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Phew!!! Difficult one for me; took me more than 3 minutes just to formulate the equation and more than 3 minutes to solve and arrive at conclusion.

Sol:
The total time is 2 hours

"A" traveled 100 miles in 1 hour when it met train B, which by then would have traveled 10 mins or 1/6 hours.

Let's take distance traveled by B in 10 minutes or 1/6 hours to be "x" miles. So; train A travels 100 miles + x miles and B travels x miles+ 100 miles

Now; let's just talk about time

A traveled 100 miles in 1 hour
A would have traveled x miles in x/100 hour

B traveled x miles in 1/6 hour
B would have traveled 100 miles in 100/(6x) hour

Total time combined is 2; Thus;

1+ x/100 + 1/6 + 100/(6x) = 2 ---> This is the equation

Solving the above; we get
3x^2-100x-150x+5000=0
(x-50)(3x-100)=0

x could be 50 miles
or
x could be 100/3 miles approx 33 miles

1.
It says B arrived at NY before A arrived at Boston.

Say x=50
B spent 10 minutes to travel x miles or 50 miles
B will spend 20 minutes to travel remaining 100 miles

A spent x/100 hour to travel x miles means; 1/2 hour

As we can see after A and B met; B traveled 20 minutes and A 30 minutes.
This satisfies the statement 1 for x=50

Let's check x=33 as well
B spent 10 minutes to travel x miles or 33 miles
B will spend approx 30 minutes to travel 3 times the distance (100=3*33), which is remaining 100 miles.

A spent x/100 hour to travel x miles means; 33/100 hour approx 1/3 hours; 20 minutes approx

As we can see after A and B met; B traveled 30 minutes and A 20 minutes.
This will make statement 1 false. Thus x can't be 33.

We found unique solution for x=50.
Thus we know; train B arrived New York 30 minutes after it started. i.e. at 4:20PM

Sufficient.

2.
This one is easy;
It says the distance > 140 miles
if x=33
Distance = 100+x = 133 <140
x=33 can't be true

if x=50
Distance = 100+x = 150 >140
x=50 is true

Sufficient.

Ans: "D"
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Re: Train A leaves New York [#permalink] New post 07 Mar 2011, 11:14
I have one question to all of you ! Why havent you taken into consideration relative speed concept ????

if you take that the equations will be completely different
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Re: Train A leaves New York [#permalink] New post 07 Mar 2011, 17:05
This is not true. Well relative speed is not an elixir. Its derivative and based on usual speed distance formula. It niether changes the variables not the relationships between the variables - so equations cant be different.

Am I am missing something?
lastattack wrote:
I have one question to all of you ! Why havent you taken into consideration relative speed concept ????

if you take that the equations will be completely different
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Re: Train A leaves New York [#permalink] New post 07 Mar 2011, 18:36
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lastattack wrote:
I have one question to all of you ! Why havent you taken into consideration relative speed concept ????

if you take that the equations will be completely different


Relative speed concept has its uses. This question is not one of them. We use it when 2 people cover some distance together in the same time... Here we already know that they meet at 4:00 when A has traveled 100 miles. After that we know that they take a total of 50 mins to reach their respective destinations independently.

Give the equations you have in mind... we can tell you what works and what doesn't and why...
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Re: Train A leaves New York [#permalink] New post 07 Mar 2011, 18:39
folks one thing.. wht are the chances that this might turn up in the real GMAT? chances of solving this in less than 2 mins are next to none..
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Re: Train A leaves New York [#permalink] New post 07 Mar 2011, 18:40
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gmat1220 wrote:
This is not true. Well relative speed is not an elixir. Its derivative and based on usual speed distance formula. It niether changes the variables not the relationships between the variables - so equations cant be different.

Am I am missing something?


That's right. Relative speed is just speed of one relative to the other... It doesn't matter from whose perspective you see, the answer would never be different. It does not change the relation between the variables.
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Re: Train A leaves New York [#permalink] New post 07 Mar 2011, 18:44
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gladvijay wrote:
folks one thing.. wht are the chances that this might turn up in the real GMAT? chances of solving this in less than 2 mins are next to none..


such a question is not unfathomable... it is based on logic and sound interpretation... it is a higher level question for sure but the questions at this level are challenging... I would expect a more straight forward quadratic to save time but otherwise the question is fine... also remember, if you reach a level where you get such a question, you would have solved the really easy ones fairly quickly.. so you would actually have 3-4 mins to invest in such a question which is more than sufficient time... try using diagrams.. they help you grasp the concepts quickly...
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Re: Train A leaves New York [#permalink] New post 07 Mar 2011, 18:52
VeritasPrepKarishma wrote:
if you reach a level where you get such a question, you would have solved the really easy ones fairly quickly.. so you would actually have 3-4 mins to invest in such a question which is more than sufficient time... try using diagrams.. they help you grasp the concepts quickly...


This makes sense, thank you!
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Re: Train A leaves New York [#permalink] New post 06 Jun 2011, 11:21
CAN THE STEM STATE SOMTHING ELSE THAN THE STATMENTS?
ACCORDING TO THE STEM: D=130
ACCORDING TO STATMENT B D>140



Let:
be the distance between cities;
be the rate of Train B.

"An hour later (so at 4:00PM), Train A passes Train B" --> before they pass each other A traveled 1 hour (4PM-3PM) and B traveled 1/6 hours (4PM-3:50PM).

"Combined travel time of the two trains is 2 hours" --> d/100(time to cover all distance for train A)+d/x(time to cover all distance for train B)=2 --> ;

As before they pass A traveled 100 miles (1 hour at 100 miles per hour), then distance to cover for B after they pass is this 100 miles and as B traveled x*1/6 miles before they pass (1/6 hour at x miles per hour), then distance to cover for A after they pass is this x*1/6 miles --> ;

So, we have:
and .

Solving for and
and ;
OR:
and .

(1) Says that train B arrived before A.
If A arrives at 4:20, B at 4:30, not good;
If A arrives at 4:30, B at 4:20, OK.
Sufficient

(2) Says that --> --> , arrival time for B 4:20. Sufficient
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Re: Train A leaves New York [#permalink] New post 06 Jun 2011, 11:31
BUNUEL HAS THE BEST ABILITY TO SIMPLIFY THE HARDEST OF THEM ALL
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Re: Train A leaves New York [#permalink] New post 14 Jun 2011, 22:03
a (d-100)/(1/6) > 100
Speed b > speed a

d > 116.
so taking d = 117 there will be fixed time for B to cover the distance considering
ta+tb = 2hrs.

SUfficient.

b essentially gives the same d>140 and ta+tb = 2 will give a fixed value for tb.

Thus D it is.
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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink] New post 16 Dec 2011, 10:38
Clean bowled till I read Bunuel's explanation. I would have guessed and moved on. Quite difficult for me. :(
Bunuel, thanks to you, I am at least able to understand the solutions.
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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink] New post 27 Jun 2013, 00:02
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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink] New post 30 Jul 2013, 10:36
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?

As an aside, trying to visualize this problem using real world examples isn't a good idea. I travel the NEC (North East Corridor - rail line between NYC and Boston) frequently. The actual train is much slower and travels a greater distance than the train in this problem meaning any visualization I did was grossly incorrect!

Train A leaves NYC for Boston, Train B leaves Boston for NYC.
Train A covers 100 miles in one hour
Train A passes B at 4PM after covering 100 miles
Train B leaves Boston 10 minutes before it passes A.

So, we are given that the total time for both trains is two hours. This means the time of A and the time of B added together = 2.

Time = Distance/Speed
Speed = Distance/Time

d/100 + d/x = 2
This represents the total time it took for A and B to pass one another and get to their respective destinations.

We know that when A and B passed one another, A traveled for 1 hour and B traveled for 1/6th of an hour (it left at 3:50 and passed A at 4:00) which means that the trains combined traveled for another 5/6ths of an hour)

Speed (B): = Distance/Time
Speed (B): = x/(1/6) (1/6) is the time in minutes (10 minutes), converted to hours, that the train covered from 3:50 to 4:00 when it passed A.
Speed (B): = (x/1) / (1/6) = (x/1) * (6/1) = 6x

Now we need to see how long it took each train (combined) to reach it's intended destination. The combined time of both trains must = 5/6ths

When B passed A, A had traveled 100 miles. Therefore, B had 100 miles to travel from that point (where the passed) to NYC.

T = D/S

t1 + t2 = 5/6

x/100 + 100/6x = 5/6

Could someone explain to me why we use the time it took from A to cross paths with B (x/100) in the formula trying to attain the time for the last 5/6ths of an hour of the journey?

3x^2 - 250x + 5000 = 0
x = 50, 33.33

(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.
Re: Train A leaves New York for Boston at 3 PM and travels at   [#permalink] 30 Jul 2013, 10:36
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