Which one of the following must be true about Freedom's schedule of voyages?

on September 25 at 07:35PM

Explain

Please explain why the correct answer is D

1 Reply

Irina on September 25 at 08:25PM

@delossantosjackie7,

The game requires us to determine a schedule of voyages for weeks 1-7, each week a ship visits one of four destinations - G J M T.

___ ___ ___ ___ ___ ___ ___
1. 2. 3. 4. 5. 6. 7

The following rules apply:

(1) J will not be a destination in week 4
(2) T will be a destination in week 7.

___ ___ ___ ___ ___ ___ T
1. 2. 3. 4. 5. 6. 7
~J
(3) There will be exactly two voyages to M and at least one voyage to G must occur in between them.

This rule tells us that there are exactly two M voyages but they are not in consecutive weeks, so we cannot have MM string anywhere in the schedule. There must be at least 1 G voyage between two M ones- M ...G...M

(4) G will be its destination in the week preceding any voyage to J.

This is an interesting rule, and it tells us that anytime there is J on the schedule, G must be visited the week before - GJ combination. It does not mean though that G must always be followed by J though, so we can have G on the schedule without J following it, but we can never have J without G preceding it.

J -> G the week before
~G -> ~J the week after

Let's think what it means for the overall schedule. We have 6 open slots left (T is in #7). We know that exactly two of them are occupied by M and at least one by G voyage that happens between two M voyages - M G M. Now we have 3 open slots left, at least one of them must be J because we have to visit every destination at least once. Let's say J is a part of M G J M sequence. Then we have only two open slots left, and if we put another J voyage on the schedule it must be preceded by G, so at most we could have another GJ combination in these remaining two open slots.

In this scenario, the overall schedule could be: M G J M G J T. Since there is not enough open weeks on the schedule to fit any more than 2 G J combinations - if we have 3 G J combinations, we no longer have space for two Ms - we can thus conclude that at most we can have two J voyages as (D) correctly concludes.

Let me know if this makes sense and if you have any further questions.