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

- Knock
**Posts:**5152**Joined:**Wed Jun 10, 2009 3:09 pm

### Re: Dec 2004 LG # 3 Photographs

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)

Just took this PT today. Let me go grab my PT booklet, and I will help you out.

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.

Return to “LSAT Prep and Discussion Forum”

### Who is online

Users browsing this forum: Baidu [Spider], Yahoo [Bot] and 2 guests