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

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?

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/x
Train 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/600
It 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 \((x-300)(x-200) = 0\)

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 miles
Total 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 pm

Train 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 pm

Statement 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

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: \(\triangle=25-4*\frac{6}{100}*100=1\). So \(d=\frac{5+1}{2*\frac{6}{100}}=50\) or \(d=\frac{5-1}{2*\frac{6}{100}}=\frac{100}{3}\).

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.

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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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.

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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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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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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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.

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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/1 ... equations/
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amazing question... not sure if this can be done within 2 minutes if it pops up in the actual exam

nice problem... took 5 minutes to solve...

Where r the multiple choice answers?

This is a data sufficiency question. Options for DS questions are always the same.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

I suggest you to go through the following post ALL YOU NEED FOR QUANT.

Hope this helps.
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Sorry to be so remedial but can someone walk me through why 2 is sufficient, specifically X must be 50?

Correct option D

information which we have :
1. Train A - Starts 3:00 PM from NY to Boston @ 100mph
2. Train B - Starts 3:50 PM from Boston to NY @ constant speed
3. Train A and Train B cross each other at 4:00 PM
(Train A : 60 Mins starting from NY and Train B : 10 Mins starting from Bostan)
4. The combined travel time of the two trains is 2 hours

To get to know, what time did Train B arrive in New York?,
We either need distance between 2 station or Time refrence, which one is early.

(1) Train B arrived in New York before Train A arrived in Boston - Sufficient
- Combine travel : 2 hrs,
In next 1 hr 4 PM, both train crossed each other,
Train A in 60 Mins and Train B in 10 Mins,
It means Train B has more speed than Train A
Distance covered by Train B = 6 times Train A.
So, if Train A reaches Bostan in 2 hrs jouney, Train B will reach in 20 Mins, i.e 3:50 Pm + 20 Mins = 4:15 Pm

(2) The distance between New York and Boston is greater than 140 miles
if distance between NY and Boston is 160 Miles, with (Train A : Train B) speed ratio (1x : 6x)
x+6x =140, (x = 20) i.e train A covers 20 miles, then Train B covers 120 miles
(time = Distance/Speed) = (120/6) = 20 Mins
Train B started at 3:50 PM + 20 mins = 4:15 PM

Option D

in this kind of question, it would be better if we picture its concept by drawing lines that explain both the time train A & B need to get Boston & New York and the distance between Boston & New York

train A New York ----------1hr, 100miles----------------**(4:00pm)------x hr, 100x miles------>> Boston

speed: 100 miles/ 1hr

3:00pm set out from New York to Boston



train B New York <<----100/y hr, 100 miles-----------**(4:00pm)------1/6 hr, y/6 miles--------- Boston

speed: y miles/ 1hr

3:50pm set out from Boston to New York


here we assume that the speed of train B is y miles/ 1hr, and, after the two trains pass each other in 4:00pm, train A travels another x hr to reach to the destination Boston, and from the statement given in the question
we know that, since the total travel time of the two trains is 2 hr, there comes the equation of( 1 + x + 100/y + 1/6 = 2 ), also the distance of train A still need to travel through to get to New York after two trains passed each other is 100x miles, this will equal to the distance of train B have already traveled before two trains passed each other, thus comes another equation( 100x = y/6 ), two unknown and two equations so we could solve this unknown of x = 1/2 or 1/3 and y = 300 or 200, with my logic of solving illustatrate as below

1 + x + 100/y + 1/6 = 2 ......(1)
100x = y/6 ......(2)

arrange the numbers in (1) to the left side we get x + 100/y = 5/6 .....(3)and multiply by 6 to equation (2) to cancel the denomiator we get y = 600x ....(4), take the unknown y in equation(4) to (3) we get x + 100/600x =5/6, multiply two sides by x it becomes x^2 + 1/6 = 5/6x, so its (x-2)(x-3)=0, thus there comes two scenarios x = 1/2 or 1/3 and y = 300 or 200

SCENARIO(1)
if Vb = y =200, the distance from New York to Boston is 133.3 miles,
train A spends 1 1/3hr to reach Boston while train B spends 100/200 + 1/6 = 4/6hr to reach New York

SCENRIO(2)
if Vb = y =300, the distance from New York to Boston is 150 miles,
train A spends 1 1/2hr to reach Boston while train B spends 100/300 + 1/6 = 3/6hr to reach New York

here we go back to the question, in proposition (1) it says train B travel faster than train A, thus only scenario (2) fits, sufficient to get us to the answer, and in proposition (2) it requires that the distance they travel will be more than 140 miles, it is also only scenario (2) fits, sufficient