Bunuel wrote:

Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

(1) Train B arrived at Newcastle before train A arrived at Birmingham.

(2) The distance between Newcastle and Birmingham is greater than 140 km.

Kudos for a correct solution.

Let the total distance between Birmingham and Newcastle be D, and let x be the rate of Train B.

A travels at 100km/hr so total travel time of A is D/100 hours. B travels at x km/hr so total travel time of B is D/x hours.

\(\frac{D}{100} + \frac{D}{x} = 2\)

During the first hour, A travels 100 km. B travels x kilometers/hour * (1/6 hours) = x/6 kilometers. The total distance traveled by either A or B is then

\(100 + \frac{x}{6} = D.\)

This is a very nasty quadratic to solve which is much harder than typical GMAT quadratics.

The solutions to (D,x) are (133.3, 200) and (150,300).

(1) Train B arrived at Newcastle before train A arrived at Birmingham.If (D,x) = (133.3, 200) then Train B traveled 133.3 km at 200km/hr, so B traveled for 2/3 hours, or 40 minutes and arrived at 4:30. A traveled 133.33 km at 100km/hr, so A traveled for 1.33 hours, or 80 minutes, and arrived at 4:20.

If (D,x) = (150,300) then Train B traveled 150km at 300 km/hr, so B traveled for 30 minutes and arrived at 4:20. A traveled for 1.5 hours and arrived at 4:30.

If B has to arrive before A, then only (150,300) fits. Sufficient.

(2) The distance between Newcastle and Birmingham is greater than 140 km.This says that D>140. Only (D,x) = (150,300) fits. Sufficient.

Therefore, either option is sufficient. Answer: D