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

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
\(3x^2-100x-150x+5000=0\)
\((x-50)(3x-100)=0\)

x 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"

Nice explanation. Thanks Bunuel!

very good problem,

thank you Bunuel!

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.

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

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?

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...

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..

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.

Check this video for when to use relative speed: https://youtu.be/wrYxeZ2WsEM

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...

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

BUNUEL HAS THE BEST ABILITY TO SIMPLIFY THE HARDEST OF THEM ALL

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.

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.

Hi,

Thanks Bunuel.
Can you tell me a quick way to solve the 2 equations for d and x ?

\(\frac{d}{100}+\frac{d}{x}=2\)
\(100+\frac{x}{6}=d\)

No super fast way. Just substitute value of d from (ii) in (i) and solve quadratics for x.

Hi Bunnel

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

Thanks