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
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= 3/2 km
x= 1.5 km
And our answer is C.