Constant speed is about ratios.
Train A leaves New York for Boston at 3 p.m. 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 p.m. and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in New York?
To translate the information in the text into a diagram and later set up an appropriate equation, I will use the following guidelines:
Now for the diagram.
The passing juncture at the top of the hour (how considerate!) splits the journey into an NYC section and a BOS section, plied by both trains at different but constant speeds. The 60 and 10 minutes are immediate from the text. For the two missing times, I used the NYC section for train B as unknown. The combined travel time of 120 minutes links the fourth time to the three others.
We can now use the proportionality of time and distance if speed is constant, which yields \(60:(50-T) = 100:D = T:10\). Alternatively, you can argue that the different speed is equivalent to a factor applied to the times (e.g., doubling speed will halve times) and conclude that \[ \frac{60}{T} = \frac{50 -T}{10}\,. \] Both proportions are equivalent to \(T^2 -50T +600 = 0\) with solutions \(T_1 = 20\) and \(T_2 = 30\). The two possible scenarios are summarized in the following table.
Arrival NYC | Arrival BOS | NYC -> BOS |
---|---|---|
4:20 p.m. | 4:30 p.m. | 150 miles |
4:30 p.m. | 4:20 p.m. | 133 miles |
The two premises refute, separately, precisely one of the scenarios and determine the arrival in NYC. The correct answer choice to the data sufficiency question is (D).