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Train A leaves New York for Boston at 3 PM and travels at [#permalink]
 22 Jul 2010, 00:49
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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]
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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 [ 5.7 KiB | Viewed 4068 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]
22 Jul 2010, 03:34
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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 xd=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]
06 Mar 2011, 23:47
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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)=0x 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]
23 Jul 2010, 06:18
Nice explanation. Thanks Bunuel!
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Re: Train A leaves New York [#permalink]
06 Mar 2011, 13:36
very good problem, thank you Bunuel!
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Re: Train A leaves New York [#permalink]
06 Mar 2011, 21:24
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]
07 Mar 2011, 12:14
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]
07 Mar 2011, 18:05
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]
07 Mar 2011, 19:36
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]
07 Mar 2011, 19:39
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]
07 Mar 2011, 19:40
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]
07 Mar 2011, 19:44
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]
07 Mar 2011, 19:52
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]
06 Jun 2011, 12:21
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]
06 Jun 2011, 12:31
BUNUEL HAS THE BEST ABILITY TO SIMPLIFY THE HARDEST OF THEM ALL
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Re: Train A leaves New York [#permalink]
14 Jun 2011, 23:03
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]
16 Dec 2011, 11:38
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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Train A leaves New York for Boston at 3 PM and travels at [#permalink]
26 Feb 2013, 13:54
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.
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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink]
26 Feb 2013, 13:59
jaskarannagra 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. Merging similar topics. Please refer to the solutions above.
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Re: Train A leaves New York for Boston at 3 PM and travels at
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