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Vinny Gambini
Vinny Gambini
 
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Joined: November 19th, 2014
 
 
 

How I solved "A Fishy Game" (hard)

by tpcupid Fri Nov 28, 2014 2:07 pm

What does the Question ask?: first off this particular game seems to be a complex order game, and what it asks us is to determine which fish (out of 6 different types from 3 different regions) to feature in a market from Monday- Saturday. Another important limitor is that the fish scheduled each day must be either farm-raised or wild-caught (which I indicate as f and w on my diagram below)

The Variables, Conditions/ Rules, and Diagram:

Variables
Ea: HJ
Al: OP
Ca: RS

Conditions (~: this symbol signifies a condition that cannot be)
*remember to check your rules
1. R(f) and S(f)
2. O/P --> Thrs
3. w --> Thrs
4. H/J --> Tues
5. ~ff (i.e. ~RS)
6. ~OP/PO (therefore O/P cannot be on Wed or Fri)

Answering the questions:

Q1: plug in the rules and cancel out the wrong answers. C is the only one that satisfies all the rules

Q2: so you know from the conditions that R and S must be farm raised so therefore other possible choises would be H/J, since one of them already occupies a place on Tuesday. O/P are not likely options because they can only occupy the Monday, Thurs, or Sat slots and since ~ff and ~OP, they are quite limited in where they can go. We also know that from the rule ~ff, in order for there to be 3 f's the diagram should look something like this (Mon: f Tues: w Wed: f Thrs: w Fri: w/f Sat: f/w). Therefore O/P cannot be one of the f.

Now C is the best answer because it is the only answer that must be true since we know that a farm raised animal must go on the Monday slot. A and B are attractive answers as well but they are not necessary conditions because R/S can fluctuate between Mon and Wed.

Q3: the condition for this question is that P(f) and P -->Sat. So the first thing we must do is plug this condition into our diagram and determine some of the other limitations to include. So because P is farm raised and in the Saturday slot, O must go on Thrs. Also the Fri slot must be a wild caught fish since ~ff, and also based on the original rules Thursday is also a wild caught fish. Therefore the rest of the diagram should look like this (1. Mon: f Tues: w Wed:f ). The full diagram would be this (Mon: R/S [f] Tues: H/J [w] Wed: S/R [f] Thrs: O [w] Fri: J/H [w] Sat: P [f]). Based on this new diagram C is the only answer that does not have to be true.

Q4: the condition is that R-->Sat and to get the correct answer we have to eliminate anything that MUST be true. after plugging this condition into our original diagram the best answer to fulfill this question is E. Remember that because of the original rules R must be farm raised. Here's what the diagram should look like (Mon: P/O [f] Tues: H/J [w] Wed: S [f] Thrs: O/P [w] Fri: J/H [w] Sat: R [f])

Q5: one quick way to get to the answer is first eliminating any answer that includes Tuesday since only H/J can go there. That eliminates choices C and D. Then we have to recall P's restrictions (O/P--> Thrs; Thrs--> w; ~OP/PO). From these conditions and the original rules that we are given, we know that the only days P could occupy are Mon, Thrs, and Sat, and this so happens to be one of the answer choices (E), so therefore E is the correct answer.

Q6: I found this question difficult and I was stuck between C and D, but I reasoned that C was the most essential. Based on our diagram and the restrictions of ~ff to ensure that this rule does not get violated H/J in the Tuesday slot must be w. and this has always been true in all the times I restructured the diagram. That is why I chose this answer, but I'm sure a better explanation exists.