IMAT 2015 Q16 [Hovercraft Ferries]

There are two services between Sandpoint and Genville: a fast hovercraft service which takes 50 minutes and a slow ferry which takes 1 hour 40 minutes. They both stop in port for 20 minutes at each end before starting the return journey. They start out from Sandpoint together at the same time each morning.

For how long has the slow ferry been sailing before it meets the hovercraft coming back? (Answer to the nearest minute if necessary.)

A. 1 hour and 20 minutes
B. 1 hour and 7 minutes
C. 1 hour and 30 minutes
D. 1 hour and 25 minutes
E. 1 hour and 10 minutes

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Hello! First, let’s say that the whole path has X meters (S → G)

Part 1:

Hovercraft
. 50 min
. X m

Ferry
. 50 min
. X/2 m or 50X/100 m

Part 2:

Hovercraft
. 70 min (50 + 20 min stopped)
. X m

Ferry
. 70 min
. 70X/100 m (because if in 100 min it walks X m, then in 70 it walks 70X/100 m)

Part 3 (we’re gonna test where the Ferry and the Hovercraft will meet):

The Ferry is at 70X/100 m in 70 min, so that means that every 10 min it walks 10X/100 m.

In 10 min, the Ferry will be at 80X/100 m, which is the same for 4/5 of X. Surprisingly, the Hovercraft (which is at G) will arrive at 80X/100 m in 10 min too (in the opposite direction), because if it can complete the path in 50min, in 10 min it will complete 1/5 of the path.

Part 4 (clarifying):

The Ferry will be at 4/5 of X in 10 min (from 70X/100 to 80X/100) and the Hovercraft will also be at 4/5 of X in 10 min (since it walks 1/5 in 10 from G → S in the opposite direction of the Ferry). Therefore, the Ferry, until that point, took:

50min + 20min + 10min = 80min = 1h20min

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