Q10

[ShehryarB30]

Could you pls explain this

[ohthatpatrick]

Four of these must be false, so on our first pass, we should be asking ourselves, "Why WOULDN'T this be possible?"

(A) Why can't G go Mon? Because G appears only once, in a QG chunk, so the earliest spot for G is Tues.

(B) Why can't N be Tues? For the same reason that L couldn't be Tues on Q9:
The QG has to go on Mon/Tues or Tues/Wed.
Tues will always be Q or G.

(C) If P were on Wed (it's always on Thurs), then P would be the double, but the double can't go consecutively.

(D) Since G is part of a QG chunk, if G were on Fri then Q would be on Thurs, but that's where P lives.

(E) Apparently this could be true. We could trust our work and pick it, or we could verify it works by writing a scenario such as this: Q G N P L Q

[ConorD250]

Four of these must be false, so on our first pass, we should be asking ourselves, "Why WOULDN'T this be possible?"

(A) Why can't G go Mon? Because G appears only once, in a QG chunk, so the earliest spot for G is Tues.

(B) Why can't N be Tues? For the same reason that L couldn't be Tues on Q9:
The QG has to go on Mon/Tues or Tues/Wed.
Tues will always be Q or G.

(C) If P were on Wed (it's always on Thurs), then P would be the double, but the double can't go consecutively.

(D) Since G is part of a QG chunk, if G were on Fri then Q would be on Thurs, but that's where P lives.

(E) Apparently this could be true. We could trust our work and pick it, or we could verify it works by writing a scenario such as this: Q G N P L Q

This is a reply to the incorrect game. There are no chunks/days of the week in the relative ordering game for PT83 Game 2.

[Laura Damone]

Oh dear, sorry about that!

For Q 10, if S is 1st, we're invoking one option for rule 3: S and T are both before H. Because H is before R, this, in turn, invokes the contrapositive of rule 2: T before R guarantees T before G. In order for T to be before H, R, and G, T must be 2nd. That is why E is the correct answer.

If you framed this game around the options for rule 3, you will have already pretty much created the frame described above: S and T before H, then R, with G coming sometime after T. The condition provided by Q10 (S in 1) can only be satisfied in this frame. Placing S in 1 forces T into 2, with H -- R , G in a cloud over 3, 4 and 5.

Hope this helps!