Can someone please tell me if I diagrammed this game right and how you get answers 16 & 17
S-->W
/W-->/S
U-->S
/S-->/U
/Y-->R
/R-->Y
W-->/T or /R
T & R-->/W
so then I made this chain:
U-->S-->W--> /T or /Y-->Y
Contrapositive:
/Y-->T & R-->/W-->/S-->/U ( the Y pointing to the R)
Dec 2004 LG # 3 Photographs Forum
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Re: Dec 2004 LG # 3 Photographs
Just took this PT today. Let me go grab my PT booklet, and I will help you out.roballen wrote:Can someone please tell me if I diagrammed this game right and how you get answers 16 & 17
S-->W
/W-->/S
U-->S
/S-->/U
/Y-->R
/R-->Y
W-->/T or /R
T & R-->/W
so then I made this chain:
U-->S-->W--> /T or /Y-->Y
Contrapositive:
/Y-->T & R-->/W-->/S-->/U ( the Y pointing to the R)
Edit:
Okay, rule diagrams:
S --> W (if S is chosen, W must be chosen)
U --> S (if U is chosen, S must be chosen)
W -->
We can combine the first two rules to one super rule, U --> S --> W So we know everytime U is chosen, S and W must be chosen. Everytime S is chosen, W must be chosen. Also note, that due to the fourth rule, if W is chosen, T AND R must NOT be chosen. So we also must note that when U or S are chosen, [W must be chosen] and T AND R must NOT be chosen.
Contra-positives (I don't personally write them out but it might be helpful to you):
T or R -->
Okay, now that we have our base set up, let's look at the questions that were giving you trouble.
#16:
If U and Z are selected, then how many other friends MUST also appear. So we're looking for what MUST be true.
Let's see what happens when U and Z are selected. Because of our super-rule (U --> S --> W), we know that when U is selected, S and W MUST be selected. We know because of rule #4 that if W is selected, T AND R CANNOT be selected. So at this point we got:
Selected- U Z S W
Not selected- T R
So now we only have Y left to figure out. Well based on our contrapositive of rule #3, (
So at this point (the end) we have:
Selected- U Z S W Y
Not selected- T R
But remember to answer exactly what the question is asking. In this case, it is in ADDITION to U and Z how many friends MUST appear in the photograph. So we know that the friends that MUST appear besides U and Z are: S, W, and Y. So the credited answer is B) Three.
#17.
We're looking for a could be true EXCEPT (translate that to can't be true) answer that functions on having EXACTLY 3 friends selected.
This one is easier answered by going through each answer choice and thinking (or writing down) the consequences of selecting the people each answer choice have selected.
A) S and Z both are selected. Well if S is selected, we know from the rules that W must also be selected. And if W is selected, T AND R can NOT be selected. And if R is NOT selected, Y MUST be selected. This leaves us with 4 selections that must occur, S, Z, W, Y; this violates the question conditions that EXACTLY 3 people are selected. We don't have to go any further, answer choice A can't be true, and is thus correct. You can keep going and repeat the same process for B-E if you wanted to check your answer, but in the interest of time i'd move on immediately after finding A is the credited answer.
Hope this helped, and please feel free to pm me if you have any further questions.