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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]

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13 May 2016, 21:44
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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Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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13 May 2016, 22:28
anandmehrotra wrote:
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?

Let the distance be D..
the two meet each other after A has travelled for 1 hour that is 100*1 = 100miles...
now it meets the other train coming from other side..
the other train has travelled for 10 min as you have correctly stated but it does not travel 100 miles..
IT travels D-100 miles.. and Now you have to work further
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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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13 May 2016, 22:46
chetan2u wrote:
anandmehrotra wrote:
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?

Let the distance be D..
the two meet each other after B has travelled for 1 hour that is 100*1 = 100miles...
now it meets the other train coming from other side..
the other train has travelled for 10 min as you have correctly stated but it does not travel 100 miles..
IT travels D-100 miles.. and Now you have to work further

Thanks, that make sense. Just a minor correction:It was Train A not B that travelled for 1 hour

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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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23 Sep 2016, 10:41
Let the distance travelled by Train B when both the trains meet at 4 pm be x miles.
Therefore, speed of train B = x miles / 10 minutes
= 6x miles per hour

At 4 pm, Train A has travelled for 60 minutes, and Train B has travelled for 10 minutes.
Thus total time travelled = 70 minutes.
Remaining total time for both trains to cover their respective distance = 2hours - 70 minutes
= 50 minutes
= \frac{5}{6} hours

Now, train A has travelled 100 miles in 1 hour, and train B has travelled x miles so far.
Total distance between New York and Boston = (100 + x) miles
Remaining distance to be covered by train A = x miles, and remaining distance to be covered by train B = 100 miles.

Time taken to do so, by train A = \frac{x}{100} hours
Time taken to do so, by train B = \frac{100}{6x} hours

By definition, \frac{x}{100} + \frac{100}{6x} = \frac{5}{6}
Solving, we get $$6x^2 -500x + 10000 = 0$$

x = \frac{100}{3} or 50

Then, as other users have mentioned, each statement can be shown to be sufficient to answer the question.

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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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23 Feb 2017, 05:56
Prompt analysis
Let the distance be x, speed of train B be v, t be the time taken after departure to meet train B
x = v/6 +100*1
x/v +x/100 =2
Solving these two equations we get
x =150 , v = 300
And
x =133.33 , v = 200

Superset
The arrival time will be a positive real number based on the selection of one case.

Transaltion
In order to find the value, we need
1# another equation to select one case
2# one information to select one case

Statement analysis

St 1: for case 1, train A arrives at 4:30PM and train b arrives at 4;20pm
For case 2, train A arrives at 4:20 PM and train B arrives at 4:10PM INSUFFICIENT
St 2: Case 2 is rejected. HEnce train B arrives at 4:20PM. Option C

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Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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22 Mar 2017, 23:23
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 (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

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

Hope it's clear.

The algebra is tough for me.

$$\frac{d}{100}+\frac{d}{x}=2$$

$$dx + 100d = 200x$$

$$100d = x(200-d)$$ -----(1)

$$100 + \frac{x}{6}=d$$

$$600 + x = 6d$$

$$x = 6d - 600$$ -----(2)

Substitute (1) into (2)

$$6d^2 - 1700d + 120000 = 0$$

$$(d - 150)(6d - 800)=0$$

How to identify $$(d - 150)(6d - 800)=0$$ in a faster way? The algebra is complicated.

Dear Bunuel, There is two different solutions for Statement 1. Could you help to elaborate why it is sufficient for statement 1? Thank you.

(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
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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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22 Mar 2017, 23:43
ziyuen wrote:
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 (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

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

Hope it's clear.

The algebra is tough for me.

$$\frac{d}{100}+\frac{d}{x}=2$$

$$dx + 100d = 200x$$

$$100d = x(200-d)$$ -----(1)

$$100 + \frac{x}{6}=d$$

$$600 + x = 6d$$

$$x = 6d - 600$$ -----(2)

Substitute (1) into (2)

$$6d^2 - 1700d + 120000 = 0$$

$$(d - 150)(6d - 800)=0$$

How to identify $$(d - 150)(6d - 800)=0$$ in a faster way? The algebra is complicated.

Dear Bunuel, There is two different solutions for Statement 1. Could you help to elaborate why it is sufficient for statement 1? Thank you.

(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

Because of the first statement which says that that train B arrived before A. Only if x = 300, train B arrives before A but of x = 200 A arrives before B.

Hope it's clear.
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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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22 Mar 2017, 23:44
ByjusGMATapp wrote:
Prompt analysis
Let the distance be x, speed of train B be v, t be the time taken after departure to meet train B
x = v/6 +100*1
x/v +x/100 =2
Solving these two equations we get
x =150 , v = 300
And
x =133.33 , v = 200

Superset
The arrival time will be a positive real number based on the selection of one case.

Transaltion
In order to find the value, we need
1# another equation to select one case
2# one information to select one case

Statement analysis

St 1: for case 1, train A arrives at 4:30PM and train b arrives at 4;20pm
For case 2, train A arrives at 4:20 PM and train B arrives at 4:10PM INSUFFICIENT
St 2: Case 2 is rejected. HEnce train B arrives at 4:20PM. Option C

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Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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09 Aug 2017, 21:11
Let the distance between New York and Boston be d miles. Let the speed of train B be s2

d/100 + d/s2 = 2. – (1)

Also, by 3:50, train A would have traveled (5/6)*100. So distance traveled between 3:50 to 4:00 is
d- (5/6) *100 = (100+s2) *1/6 -- (2)

From (1) and (2) we know we have two values of d and s2.
At this point we can guess especially if short of time, that statements (1) and (2) try to eliminate one of these values and hence D.
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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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12 Aug 2017, 12:47
An easy way to look at the same problem especially for those who are not fond of math

Assuming total distance to be 150 miles

(For the graph please open the doc file attached with this post)

From first Statement
Since train B takes 10 minutes to cover 50 miles, it will take 20 minutes to cover remaining 100 miles. Thus total time taken by B is 30 minutes. Also train A takes 1hr (60 mins) to cover 100 miles, so it will take 30 minutes to cover remaining 50 miles. Thus total time taken by A is 90 minutes. Thus total time taken by both trains = 120 mins (2hrs).
Also train B is reaching earlier than train A
Assuming any other number like 140 or 125 for distance between two stations will either not lead to total time as 2 hours or will breach the condition of train B reaching
From second Statement
Since distance is greater than 140, 150 suites the situation.
Thus we can find the time at which train B reaches New York station from both statements alone.
Attachments

File comment: An easy way to look at the same problem especially for those who are not fond of math
A train leaves from NY to Boston.docx [12.1 KiB]

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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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22 Sep 2017, 21:04
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.

Hi VeritasPrepKarishma. Thanks for the insightful wallkthrough. One question though. How do you solve 3x^2 - 250x + 5000 = 0.
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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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25 Sep 2017, 21:26
streetking wrote:
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.

Hi VeritasPrepKarishma. Thanks for the insightful wallkthrough. One question though. How do you solve 3x^2 - 250x + 5000 = 0.

$$3x^2 - 250x + 5000 = 0$$

$$3x^2 -150x - 100x + 5000 = 0$$

$$3x(x - 50) - 100(x - 50) = 0$$

$$(x - 50)*(3x - 100) = 0$$

x = 50, 100/3

Here is a post on how to solve hard quadratic equations:
https://www.veritasprep.com/blog/2013/1 ... equations/
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Re: Train A leaves New York for Boston at 3 PM and travels at   [#permalink] 25 Sep 2017, 21:26

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