By the combined travel time, it means t1 (Total travel time taken by Train A ) + t2 (Total Travel time for Train B).

Is the OA D ?

Hi,
Lets denote New York by 'N', Boston by 'B' and the meeting point by 'P'

Let the speed of train B be 'v'
Let A take 't1' to reach from P to B
Let B take 't2' to reach from P to N
So, NP = 100.1 = 100 miles, NP = v.t2 => v.t2 = 100
PB = 100.t1 = v(1/6), since train B takes 10 mins from B to P
Total time is 2 hours, that means 1+(1/6)+t1+t2 = 2 =>t1+t2 = 5/6
From the 3 equations we get (v/600) + (100/v) = 5/6 =>v^2+60000=500v =>(v-200)(v-300) =0 =>v= 200 or 300.
case-1:if v=200, t1=v/600 = 1/3, t2 = 1/2
case-2:if v=300, t1 = v/600 =1/2, t2 = 1/3
From(1): t2<t1
case - 2
Sufficient

From(2):100t1>40 =>t1>2/5.
case-2
Sufficient

Hence, D

Cheers!

So,

MI3 wrote:Q. 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.

Barely got a headway, but i'll try.
Let the total Distance = D
Speed of B = B
Speed of A = 100

Now D = 100 + B/6
D= (600 + B)/6 (equation 1)

Given total time is 2 hours
Therefore => 2 = (D/100) + (D/B)=> D(B+100)/100B = 2
Therefore =>D= 200B/(B+100) (equation 2)
If we equate equations 1 & 2, we get a quadratic B^2 + 700B + 6*10^4 = 0

Hence we are bound to get the value of B by solving the quadratic.
Once we get the value of B, we can determine the value of D & subsequently the time take by train B to cover diatance D.

To pick up an example.

Assume D = 150
Therefore time taken by A = 90
This means B covered 50 miles in 10 mins as B started at 3:50 to meet A at 4pm.
This means time taken by B = 30
Total time taken for combined travel = 120 mins = 2hrs.

Having said this, i think we can solve this problem without either of the stems.

Wonder what others have to say.