IMAT 2012 Q33 [Train Tunnels]

A regular train service operates between Jayford and Kayton, a 16 km journey which takes 19 minutes. The trains travel at a constant speed of 60 km per hour in both directions except through a tunnel, where they are limited to 20 km per hour. Trains travelling towards Kayton enter the tunnel 4 km after setting off from Jayford.

How long is the tunnel?

A. 4.5 km
B. 2.5 km
C. 1.5 km
D. 3.5 km
E. 0.5 km

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This entire question can be solved by a single equation, but to construct it we must understand that:

If the entire train was to theoretically travel at 60km/h, the trip would take 16 minutes.

But we know that for a segment of the trip, which we can call x, the train traveled 3 times slower. From 60km every 60 minutes into 20 km every 60 minutes.

We must remove that distance x from the theoretical 16 minutes, and add it back at the rate at which it actually goes- which we know is 3 times slower.

The equation therefore will look as so:

\frac{16 km}{60km/60min} - \frac{x km}{60km/60min} + \frac{x km}{20km/60min} = 19 km

\frac{16}{1} - \frac{x}{1} + \frac{x}{1/3} = 19 km

\frac{16}{1} - \frac{x}{1} + \frac{3x}{1} = 19 km

16-x+3x= 19 km

16+2x= 19 km

2x= 3km

x= 2/3 km
x= 1.5 km

And our answer is C.