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Re: Train A leaves New York for Boston at 3 PM and travels at
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01 Aug 2013, 08:32
I modeled my approach after Bunuel's. x would be the speed traveled for train B. VeritasPrepKarishma wrote: WholeLottaLove 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?
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 ........................................ (I) 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 ................................. (II)
I am a little confused here: what is x? Is it the speed of train B (as used in equation I) or is it the distance traveled by B in 10 mins (as used in equation II)? Also, trying to visualize is a great idea but don't try to find a parallel in the real world (and even if you do find one, the numbers may not match). Our questions are inspired from real world but they do not accurately depict real world. Think of New York as a station N and Boston as a station B. A train starts from N when the clock shows 3 and so on...



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Re: Train A leaves New York for Boston at 3 PM and travels at
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01 Aug 2013, 09:22
Here is my revised solution, Karishma. I hope it makes a bit more sense!
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?
t1+t2 = total
time = distance/speed
(d/100) + (d/x) = 2 Where x = speed of train B and 2 = combined time of 2 hours.
We know that when A and B passed A was traveling for exactly one hour and B was traveling for exactly 1/6th of an hour (6 ten minute intervals passed and 1 ten minute interval passed when A and B passed one another) we can solve for 'd' in the original equation by finding d for train B: 100 (the miles train B has to cover when it passes A) + x*(1/6) which represents the distance B had already traveled as x is the unknown rate and 1/6 is the fraction of time it was traveling compared to A. So, the distance B has to travel is 100 + (x/6) = d. Now we can plug in and solve.
d = 150 and x = 300 OR d = 133 and x = 200
Using the total time formula [(d/100) + (d/x) = 2] we can plug in and see what makes sense.
As an aside, I see how x*(1/6) = x/6 but If I were using the formula distance = speed / time how would I know to put 6 in as the time?
(1) Train B arrived in New York before Train A arrived in Boston.
I.) d = 150 and x = 300 OR II.) d = 133 and x = 200
The time it takes for A to get to B is d/100. The time it takes for B to get to A is d/x. Keep in mind that A leaves at 3 PM and B leaves at 3:50 PM
Using data set I. Keep in mind that d/100 and d/x represents the time taken by train A and B respectively:
A: (d/100) ==> 150/100 ==> 1.5 hours (i.e. 90 minutes) which means that A arrived at B by 4:30 B: (d/x) ==> 150.300 ==> .5 hours (i.e. 30 minutes) which means that B left at 3:50 and arrived at A by 4:30.
This is only possible with one set of data. (i.e. both sets of data don't allow for the same train to arrive earlier than the other one) SUFFICIENT
(2) The distance between New York and Boston is greater than 140 miles.
Remember, we have two sets of data:
d = 150 and x = 300 OR d = 133 and x = 200
2) tells us that the distance from NYC to Boston is greater than 140. Only one set of data (d=150 and x=300) falls withing the range of d>140. SUFFICIENT
(D)



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Re: Train A leaves New York for Boston at 3 PM and travels at
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01 Aug 2013, 22:09
WholeLottaLove wrote: As an aside, I see how x*(1/6) = x/6 but If I were using the formula distance = speed / time how would I know to put 6 in as the time?
Note that the formula is Distance = Speed*Time and you are using it when you say x*(1/6) = Speed*Time The rest is fine.
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Re: Train A leaves New York for Boston at 3 PM and travels at
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10 Nov 2013, 08:18
Bunuel wrote: 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 (4PM3PM) and B traveled 1/6 hours (4PM3:50PM). "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. I'm trying to really nail down these DS questions; I got almost all of them wrong on my test so I have to retake it. I'm trying to look conceptually at what the problem is giving you for information. Before I would have looked at this problem and just assumed it was C or E, because there are those two variables (d & x) involved. But by looking at the fact that they tell you A traveled 100mph, and in one hour it met B, I realized that it must have gone exactly 100 miles, and met B at 4:00, which meant that B had only been traveling for 10 minutes. I was able to set up the equations in my head based on that, and then realize what information I would need to solve them. I find it a lot easier that way; rather than looking at what a problem gives you and THEN figuring out if you need it or not, I'm trying to figure out what I need before I even look at the options, that seems to be helping me greatly.



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Re: Train A leaves New York for Boston at 3 PM and travels at
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31 Dec 2013, 13:20
I did all the logic that the question required, but obviously doing that in less than 2 (even 3 minutes) seems imporbable... The aproach of most of the explanations was the same as I did, but it took me far more than 3 minutes, especially that the answers are specific (we need to make ALL the maths)! Nevertheless, really good question!
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Re: Train A leaves New York for Boston at 3 PM and travels at
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23 Jan 2014, 07:00
I think its a Yes/No question and we dont need to solve the entire mathematical problem.We just need to know that whether the statements are sufficient or not.



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Re: Train A leaves New York for Boston at 3 PM and travels at
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02 Feb 2014, 21:24
Hi,
Thanks Bunuel. Can you tell me a quick way to solve the 2 equations for d and x ?
\(\frac{d}{100}+\frac{d}{x}=2\) \(100+\frac{x}{6}=d\)



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Re: Train A leaves New York for Boston at 3 PM and travels at
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04 Feb 2014, 01:40



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Re: Train A leaves New York for Boston at 3 PM and travels at
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09 Jun 2014, 11:40
D two statements are sufficient,I have just started taking GMAT prep a few days.Is explanation of data is neccesary in GMAT exams?.and I have solved this problem in about 3 minutes



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Re: Train A leaves New York for Boston at 3 PM and travels at
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09 Jun 2014, 12:26



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Re: Train A leaves New York for Boston at 3 PM and travels at
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09 Jun 2014, 19:10
foisal wrote: D two statements are sufficient,I have just started taking GMAT prep a few days.Is explanation of data is neccesary in GMAT exams?.and I have solved this problem in about 3 minutes You only have to select the right answer. No explanation is needed.
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Re: Train A leaves New York for Boston at 3 PM and travels at
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11 Jun 2014, 08:49
another way of solving Assuming D is the distance between NY and Boston x hrs is the additional time taken by Train A to reach Boston, then Total time taken By train A =1+x By train B = 1x (as total time taken is 2). At 4 PM Train A covers 100m and has (D100) pending. At 4 PM Train B covers (D100) and has 100m pending. In 10 mins Train B Covered (D100) In 1 Hour it covers 6* (D100) = 6D600 So. speed of Train A is = \(\frac{D}{(1+x)}\)=100 > 1 Speed of Train B is = \(\frac{D}{(1x)}\)= 6D600 > 2 From eq. 1 Replacing value of D=100 (1+x) in eq. 2 We get quad. eq. with 2 solutions. x=0.5 ====> D=150 x=0.3333 ====> D=133.33 (1) Says that train B arrived before A. If x=0.333 A arrives at 4:20, B at 4:30, not good; If x=0.5 A arrives at 4:30, B at 4:20, OK. Sufficient (2) Says that D>140 > D=150 >x=0.5 , arrival time for B 4:20. Sufficient Answer D.
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Re: Train A leaves New York for Boston at 3 PM and travels at
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25 Feb 2015, 19:20
VeritasPrepKarishma wrote: 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... Hi, I tried solving this problem with the below formula, which covers the concept of combination of speed when 2 objects are travelling towards each other. Total distance travelled till they met / Sum of speeds = Total time travelled till they met This should give the combination of time over a set distance. Let x be the distance travelled by B till it met A So, 100 + x / 100 + 6x = 7/6 (1hr 10 mins) Now, 6(100 + x) = 7(100 + 6x) >> This doesnt give the desired outcome.. Can you tell me where did i go wrong with the above formula? The idea was to find a suitable value for x and then fit it alongwith the information provided. Please assist.



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Re: Train A leaves New York for Boston at 3 PM and travels at
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02 Jun 2015, 04:25
Bunuel wrote: batman08 wrote: Hi,
Thanks Bunuel. Can you tell me a quick way to solve the 2 equations for d and x ?
(i) \(\frac{d}{100}+\frac{d}{x}=2\)
(ii) \(100+\frac{x}{6}=d\) No super fast way. Just substitute value of d from (ii) in (i) and solve quadratics for x. Hi, if you havd substituted (i) into (ii) instead this would have created a quadratic equation to hard to solve (in such a short time frame). But making that mistake and then doing the other substitution would have eaten up precious minutes. How did you decide which equation to sub into which?



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Re: Train A leaves New York for Boston at 3 PM and travels at
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02 Jun 2015, 05:05
zumiker wrote: Bunuel wrote: batman08 wrote: Hi,
Thanks Bunuel. Can you tell me a quick way to solve the 2 equations for d and x ?
(i) \(\frac{d}{100}+\frac{d}{x}=2\)
(ii) \(100+\frac{x}{6}=d\) No super fast way. Just substitute value of d from (ii) in (i) and solve quadratics for x. Hi, if you havd substituted (i) into (ii) instead this would have created a quadratic equation to hard to solve (in such a short time frame). But making that mistake and then doing the other substitution would have eaten up precious minutes. How did you decide which equation to sub into which? Point 1: As per the explanation you are referring to, x is rate of Train B and to answer the questions you need the rate of train B only therefore you identify that you have to eliminate unwanted variable d and find value of x Hence, you substitute the value of d from equation II to equation I Point 2: You don't have to solve the quadratic equation here. Since it's a Data sufficiency question so you only have to find out whether the information is sufficient to answer the question or not hence you just make the equation, Identify that there will a unique solution and call is SUFFICIENT.
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Re: Train A leaves New York for Boston at 3 PM and travels at
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18 Aug 2015, 03:03
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. Please help me know the difficulty level of this question. I was not able to solve it in even 5 mins It takes a lot of time. However, I explain how I worked out. at 3.50 PM A covered 500/6 miles. Before it meets B it has to cover 100/6 miles and it takes 10 min to meet B. let 'd' be the distance covered by B in this 10 min. Therefore, B's speed is 6d mph. (100/6d)+(d/100)=50/60 (it is given that the total time taken is 2 hours; when they meet they had already taken 70 min) solving, d=33.33 miles or 50 miles. It seems statement 2 alone is sufficient to answer this question.



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Re: Train A leaves New York for Boston at 3 PM and travels at
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04 Sep 2015, 08:39
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. Please help me know the difficulty level of this question. I was not able to solve it in even 5 mins Let x be the speed of train B. Let M be the point where Train A and B meet. Train A travels 100 miles in 1 hr before it meets B at point M. Time taken by train B to travel 100 miles will be 100/xTrain B travels for 10 mins or 1/6 hrs before it meets train A at point M. So the distance covered by train B in 10 mins will be x/6 and total distance will be 100 + x/6 Time taken by train A to travel x/6 distance will be x/600It is given that total time for travel for both trains is 2 hours. So, 1 + \(\frac{x}{600}\) +\(\frac{1}{6}\) + \(\frac{100}{x}\) = 2 or \(\frac{x}{600}\) + \(\frac{100}{x}\) = \(\frac{5}{6}\) or \(x^2 + 60000 = 500x\) or \(x^2  500x + 60000 =0\) or \((x300)(x200) = 0\) or \(x= 300, 200\) Now, the distance between point M and boston when x= 200 is 200/6 miles and when x = 300 is 300/6 or 50 milesTotal travel time taken by train A and train B, when distance between point M and boston is 200/6 miles is Train A: 1+(\(\frac{200}{6}\) * \(\frac{1}{100}\)) = 1 + \(\frac{1}{3}\) = 1 hour and 20 minutes. time of arrival at boston is 4:20 pmTrain B: \(\frac{1}{6}\) + \(\frac{100}{200}\) = \(\frac{2}{3}\) = 40 minutes time of arrival at new york is 4:30 pm Total travel time taken by train A and train B, when distance between point M and boston is 50 miles is Train A: 1+ (50*\(\frac{1}{100}\)) = 1+ \(\frac{1}{2}\) = 1 hour and 30 minutes time of arrival at boston is 4:30 pm Train B: \(\frac{1}{6}\) + \(\frac{100}{300}\) = \(\frac{1}{2}\) = 30 minutes time of arrival at new york is 4:20 pmStatement 1: Train B arrived in New York before Train A arrived in Boston. This is true when speed of Train B is 300 miles per hour. Train B arrives new york at 4:20 pm SUFFICIENT Statement 2: The distance between New York and Boston is greater than 140 miles. This is possible only when the distance between point P and boston is 50 miles. So, at 300 miles/hr train B arrives new york at 4:20 pm SUFFICIENT Answer: D



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Re: Train A leaves New York for Boston at 3 PM and travels at
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12 Oct 2015, 11:23
I was trying to find a pure geometric solution to this problem. I could't. However, I think I came up with an alternative solution to this problem: Since the velocities are constants, the trajectories of the trains are represented by straight lines in a graph Distance x Time. So, this problem can be represented accordingly with the figure attached. By simmilarity of triangles we have: A \(\triangle ABD\)\(\approx\)\(\triangle GFD\) \(>\) \(\frac{\frac{1}{6}}{t_2}=\frac{d}{100}\) \(>\)\(t_2=\frac{100}{6d}\); B \(\triangle BCD\)\(\approx\)\(\triangle FED\) \(>\) \(\frac{t_1}{1}=\frac{d}{100}\) \(>\)\(t_1=\frac{d}{100}\) Now, the problem says that the total time traveled by the trains is 2 h, or \((t_1+1)+(t_2+\frac{1}{6})=2\) \(>\) C \(t_1+t_2=\frac{5}{6}\). Clearly, this will lead a second degree equation in d by substituting A and B in C: \(\frac{d}{100}+\frac{100}{6d}=\frac{5}{6}\); or by multiplying the equation by 6d: D \(\frac{6d^2}{100}5d+100=0\) It is easy to calculate the discriminant of equation D: \(\triangle=254*\frac{6}{100}*100=1\). So \(d=\frac{5+1}{2*\frac{6}{100}}=50\) or \(d=\frac{51}{2*\frac{6}{100}}=\frac{100}{3}\). We have two possible values for the distance between New York and Boston. One is 150 miles other is 133 miles. Also, this leads to two pairs of values for \((t_1,t_2)\): If d=50 miles then \((t_1,t_2)=(0.50,0.33)\) or If d=33 miles then \((t_1,t_2)=(0.33,0.50)\) (1) This statement allows us to decide between the two situations above. Since the trains encounter in point D, if \(t_2<t_1\) then train B arrives at NY earlier than A arrives in Boston (just like the figure represents). Sufficient. (2) Again, this statement allows us to decide between the two situations above (only if d=50 miles). Sufficient. Letter D.
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Re: Train A leaves New York for Boston at 3 PM and travels at
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24 Oct 2015, 12:16
Bunuel wrote: 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 (4PM3PM) and B traveled 1/6 hours (4PM3:50PM). "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. BunuelCan you explain why do we need statements for this question? As we can determine everything from the stem. Train A = 100 mph; Train B = 600mph (100 miles in 1/6 of an hour). We can then determine the total distance by using d/100 + d/600 = 2; Then we can determine the speed of train B. I chose D because the statements were irrelevant(or so I think).



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Re: Train A leaves New York for Boston at 3 PM and travels at
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13 May 2016, 20:44
I assumed that train B travelled 100 miles in 10 minutes(started at 3:50 and met Train A at 4:00 pm), as that is when both trains met. Since B travelled 100 miles in 1/6 of an hour, it must be going at 600 mph. Am I making an incorrect assumption here?




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