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

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

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30 May 2007, 21:16
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Difficulty:

95% (hard)

Question Stats:

26% (02:30) correct 74% (02:38) wrong based on 178 sessions

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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: train-a-leaves-new-york-for-boston-at-3-pm-and-travels-at-97694.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Jun 2014, 02:41, edited 2 times in total.

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30 May 2007, 22:08
No!! think more...

This is tough question.

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30 May 2007, 23:57
C

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31 May 2007, 00:36
vidyasagar wrote:
E.

E.

Neither (1) and (2) provides informative figure for calculation...

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31 May 2007, 02:05
Come on guyzz!!

Think Hard!! This is really nasty question.

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31 May 2007, 02:40
johnycute wrote:
Come on guyzz!!

Think Hard!! This is really nasty question.

Hell... It first appeared to me an easy question. Seems not. How would we realize a TRICKY question?

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31 May 2007, 03:15
johnycute 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.

Train B has already travelled for 10 min
Train A - for 1 hour.
The total combined time for both will be 2 hrs. Hence 50 min of combined time left for the trains to reach each its own destination.
T2(A) + T2(B) = 50/60
Distance1(A)=Distance2(B)=1hour*100mph=100m - distance covered by A and which must be yet covered by B.
We need to find out time which will be consumed by B to travel 100 m. We need to know the speed of B.
Let's take (1) - Train B arrived in New York before Train A arrived in Boston.
It gives us nothing - INSUFF
Let's take (2) - The distance between New York and Boston is greater than 140 miles.
Distance1(B)>140-100 i.e. B has already travelled for more than 40 miles
It took him 10/60 hours as he left at 3.50 PM and met A at 4.00PM
hence Speed(B) > 240 mph
T2(B) < 100/240 i.e. time left for B to arrive to New York is less than 25 minutes.
It can be 24, 23, 22 minutes and so on....
Let's take now (1) and (2) together.
As T(A) + T(B) = 50/60
and T(B) < 25/60
then T(A) is more or equal to 25 and less than 50
We know that Train B arrived in New York before Train A arrived in Boston. I.e. T(B) is less than T(A).

I don't know
I am getting E here.
Could you please point out fault in my logic?

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31 May 2007, 22:59
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johnycute 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.

The question is very tough and it took me lot of time to solve it....

Lets first rephrase the question.

Consider for train B:
Rate=r; Time=t; Distance=d
ForTrain A:
Rate=100mph;Time=2-t;Dist=d

Now from this information we know 2 equations.
d=100(2-t) ----------------(A)

d=rt-------------------------(B)

We also know that when the trains pass each other (going in opposite directions), Train A has been traveling for 1 hour and Train B has been traveling for 10 minutes or 1/6 of an hour.

This means that Train A has traveled 100 miles and Train B has traveled r/6 miles.
Thus, the total distance from New York to Boston can be expressed using the equation

d=100+r/6------------------(C)

If we compare A and B;

t=200/100+r----------------(D)

Now compare equation B and C and put the value of "t" from equation D.

r(200/100+r) = 100 + r/6

(r-300)(r-200)=0

Thus, the rate, r, of Train B is either 300 mph or 200 mph.

If Train B has a rate of 300 mph. It travels 50 miles in 1/6 hour, at which point it meets Train A which has already traveled 100 miles. Therefore, the total distance from Boston to New York must be 150 miles. Thus, Train B's total traveling time was 1/2 hour, and Train A's total traveling time was 1 1/2 hours. Train B arrived in New York at 4:20 PM and Train A arrived in Boston at 4:30 PM.

If Train B has a rate of 200 mph. It travels 33 1/3 miles in 1/6 hour, at which point it meets Train A which has already traveled 100 miles. Therefore, the total distance from Boston to New York must be 133 1/3 miles. Thus Train B's total traveling time was 2/3 hour, and Train A's total traveling time was 1 1/3 hours. Train B arrived in New York at 4:30 PM and Train A arrived in Boston at 4:20 PM.

Statement (1) tells us that Train B arrived in New York before Train A arrived in Boston. From this, we know that Scenario 1 must have occurred and Train B arrived in New York at 4:20 PM. We have sufficient information to answer the question.

Statement (2) tells us that the distance between New York and Boston is greater than 140 miles. This means that Scenario 2 is not possible so Scenario 1 must have occurred: Train B arrived in New York at 4:20 PM. Again, we have sufficient information

The correct answer is (D): Each statement ALONE is sufficient

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31 May 2007, 23:43
Johny, you are really cute

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01 Jun 2007, 02:50
good question. But solvable in 3 minutes.

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22 Sep 2009, 16:48
This is the most difficult DS RTD question I remember seeing.

It requires solving, and calculations are long.
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Please kudos if my post helps.

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04 Oct 2009, 17:32
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Make it simple:

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

Hope it's clear.
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05 Oct 2009, 02:04
Nicely explained Bunuel!

I think it is one of the most hardest DS questions I have ever faced!!! Even after the explanation it took me a lot of time to solve by myself.

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

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07 Jan 2014, 20:52
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Re: Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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08 Jan 2014, 07:24

these questions can really make you feel stupid...

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

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08 Jan 2014, 07:28
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masokaz wrote:

these questions can really make you feel stupid...

This is a very tough problem. So, don't be discouraged if you solved it wrong.

You can check more solutions of the problem here: train-a-leaves-new-york-for-boston-at-3-pm-and-travels-at-97694.html

Hope this helps.
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28 May 2014, 00:12
Bunuel wrote:
Make it simple:

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

Hi Bunnel

I don't understand why do we have to add $$100+\frac{x}{6}=d$$ ?

Thanks

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02 Jun 2014, 02:41
pretzel wrote:
Bunuel wrote:
Make it simple:

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

Hi Bunnel

I don't understand why do we have to add $$100+\frac{x}{6}=d$$ ?

Thanks

When the trains pass each other A covered 100 miles and B covered x/6 miles. Together they covered the whole distance of d miles, no?

{NY}----(100 miles)----{meeting point}--(x/6 miles)--{Boston}

OPEN DISCUSSION OF THIS QUESTION IS HERE: train-a-leaves-new-york-for-boston-at-3-pm-and-travels-at-97694.html
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Re: DS- Time/Distance   [#permalink] 02 Jun 2014, 02:41
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