Q11

[atyanez]

I'm totally stuck on #11, can you please help

Here's the diagram I get after using the rule from the question stem and the initial setup:

There must be 2 O's, 1 G, 2 R's, 1 Y

In Window 1: GPO
Window 2: YPR
Window 3: ROP?

Window 3 is kind of open, but none of this corresponds to the AC's?

[seanays]

The way I reasoned it...

IF Y -> NOT G AND NOT O , the contrapositive is IF G OR O -> NOT Y.
If you combine that with IF NOT P -> O, you get IF NOT P -> NOT Y or the contrapositive of IF Y -> P

Since the first window has G AND P, it cannot contain Y, but because each color must be used at least once then one of the other two windows must have Y. Whichever one that is, will also have P. Since O must be used once more than G and G cannot be in the window with Y and it cannot be in the third window with O because then it would be exactly equal to O, then G must only appear in the first window. However, O must appear twice and since it cannot be in the second window with Y it must appear with G AND P. Since the mate for the lonely O cannot be Y,G, or itself, we are left with P or R. At this point there are two possibilities: The two Rs appear with GP or YP and you have O AND P or the two Rs appear with Y AND O. Since O AND R is not one of the answers listed, the next option is O AND P which is an answer. Conversely, we can see that (B) and (C) cannot be the answer because we require the second O to be in that group. (D) and (E) are missing purple.

So the final grouping is:

1) G P O R
2) Y P R
3) O P

If anyone has a more concise answer, I'd also like to see it.

[noah]

What a tough question!

Here's what I did. First, understand the new rule! If there are more Os than G, we must have 2 Os and 1 G (we can't have 3 Os, since that would mean there are no Ys).

So, now we can set up this:

O O Y
1 2 3

(the window numbers irrelevant here and throughout the game, but useful for discussion)

As others have noted, with Y, we must have P, so that gives us

P
O O Y
1 2 3

And now, the crucial question: Who's left?! We need to place the GP chunk, as well as the two Rs. The GP chunk can't go with Y, so that gives us:

(R, R)

G
P _ P
O O Y
1 2 3

(With some open slots above each window - we can put the Rs anywhere, and . . . I say too much!)

There are no restrictions on the Rs, so let's look at the answer choices from (E) upwards.

(E) No way we can have G without P.

(D) No way we can have G without P.

(C) No way we can have G without O! (That G is quite needy!)

(B) No way we can have G without O.

(A) It doesn't seem right - where's that extra Y coming from? But, it turns out that adding a Y to window 2 is fine. GP stays in window 1, and the Rs go in windows 1 and 3.

Voila?

[yoohoo081]

G
P _ P
O O Y
1 2 3

(With some open slots above each window - we can put the Rs anywhere, and . . . I say too much!)

There are no restrictions on the Rs, so let's look at the answer choices from (E) upwards.

(E) No way we can have G without P.

(D) No way we can have G without P.

(C) No way we can have G without O! (That G is quite needy!)

(B) No way we can have G without O.

(A) It doesn't seem right - where's that extra Y coming from? But, it turns out that adding a Y to window 2 is fine. GP stays in window 1, and the Rs go in windows 1 and 3.

Voila?

I'm so confused.. How can C be possible? G HAS to be paired with P since there can only be one G. However, P and O cannot be together.

Did I do it wrong if I take contrapositve of the condition that is given??? ~p-> o (~o->p)

[noah]

I'm so confused.. How can C be possible?

I'm so confused! The answer is not (C)!

[jh2352]

I'm so confused.. How can C be possible?

I'm so confused! The answer is not (C)!

The best way for me was to look at the answers and make eliminations.

First off, we know that there must be O:2 and G:1

With this in mind compare to the answers like so:

A) 1. O P This gives us some options. So we know that there must be another group with OG because of the
rule involving Y (Y-> ~O and Y -> ~G) so we now know that Y must have its own group and there must be a group of OG. Filling this is would look like this...

1. O P
2. O G P (remember one group needs GP)
3. Y

Now we can place the two R's anywhere without restriction. One of which should help fill up group three.

So now you have:

1. O P
2. O G P R
3. Y R

If you place that second R in group two as I did above then you now have group one with O and P only.

This leads you to (A) The Correct Answer!!