The rules tell us that library E gets its delivery after library D, and that library E is not the last library to receive its delivery. The latest day library E could get its delivery is therefore day , which means that day  is the latest library D could get its delivery. 

Consequently, the delivery to library D CANNOT be scheduled for day .

If B is first and A is fourth, then H and D must be walked second and third.  The reason for this is that C, G, and F must be three of the four dogs walked after A. E would also have to be walked after A and would have to be closer to G than is allowed.

Lambast must precede Grand Slam. Titrus must precede Lambast. Titrus and Piston are side by side, so Piston must precede Grand Slam. Fury precedes Piston, which means it necessarily has to precede Grand Slam. Thus, we know 4 cars out of a possible 5 must precede Grand Slam. If there is one space between Grand Slam and Moxie, then Moxie must precede Grand Slam as well. Therefore, we must have the following:

F < TP or PT < M < L < G

Moxie is fourth, Lambast is fifth, and Grand Slam is sixth. We also know Fury must be first. However, there is no set order for T and P. They have to be in the 2 and 3 garages, but in either order.

Thus, the garages for F, M, L, and G are known. The garages for T and P have been narrowed down to two, but are not known specifically. The answer is 4.

Will can not be the shortest or tallest. If he is shorter than Dan, than he is also shorter than Corrin. This also means he must be shorter than Jonathan, who is taller than the girls. Thus, Will can only be the 5th tallest, in between Theresa (the shortest), and Dan in 4th tallest.

The question gives you the information that Sue must give her speech immediately before Quinn gives hers. You know that Sue cannot give her speech fifth, so you need to check if she can give her speech first, second, third, and/or fourth.

She can give it first. If she gives it first, Quinn will give hers second. The first global rule tells you if Sue gives hers first, then Mort gives his fifth. This still leaves room for Othello to go before Paul, third and fourth, to follow the fourth global rule: Sue, Quinn, Othello, Paul, Mort.

She cannot give it second. This would require Quinn to give her speech third. This would then only leave room for Othello and Paul to go fourth and fifth. This forces Mort to go first, which breaks the third global rule.

She can give it third. Quinn would give her speech fourth, leaving room for Othello and Paul to give their speeches first and second, respectively. Mort would give his speech fifth, following all global rules: Othello, Paul, Sue, Quinn, Mort.

She can give it fourth. Quin would give her speech fifth. Othello and Paul would give theirs first and second, respectively, so that Mort doesn't have to give his speech first. Mort would give his third: Othello, Paul, Mort, Sue, Quinn.

So Sue can give her speech first, third, and fourth.

For this problem, you are looking for the outcome that must always be true.

If Paul gives his speech fifth, you know from the fourth global rule that Othello must give his speech fourth:

___, ___, ___, Othello, Paul

You know that Mort cannot give his speech first from the third global rule. You also know Sue cannot give her speech first from the first global rule because this would require Mort to give his speech fifth, which cannot happen here.

This means that only Quinn can give her speech first. Mort and Sue give their speeches second and third, in either order. It's important to not misinterpret the second global rule and assume that since Quinn gives her speech first, Mort must give his speech second - only the reverse is true.

Quinn, Mort/Sue, Sue/Mort, Othello, Paul

You can eliminate the answer choice of Quinn giving her speech third, since that can never happen. You can also eliminate the three answer choices that identify what can be true, but are not always true: Mort giving his speech third, Sue giving her speech second, and Sue giving her speech third.

The correct answer is Quinn gives her speech first. This will always be true under these conditions.

For this problem you are looking for the outcome that must always be true.

If Quinn gives her speech third, that leaves you only two options for when Othello and Paul can give theirs, to follow the fourth global rule - either first and second or fourth and fifth.

Othello and Paul must give theirs first and second, though, because Mort and Sue cannot give their speech first. Mort cannot give his speech first because of the third global rule. Sue cannot give her speech first because the first global rule would require Mort to give his fifth, which would leave no room for Othello to go immediately before Paul.

This leaves Sue and Mort to give their speeches fourth and fifth in either order:

Othello, Paul, Quinn, Sue/Mort, Mort/Sue

You can eliminate the three answer choices that can never be true here: Mort giving his speech before Paul, Mort giving his speech before Quinn, and Sue giving her speech before Paul.

You can also eliminate the answer choice that can sometimes be true, but is not always necessarily true: Sue gives her speech before Mort.

The correct answer is Paul gives his speech before Sue gives hers. This will always occur given these conditions.

If C is invited to speak, then we know that C and X will be speaking on consecutive nights. Since C and A are invited to speak, then we know that Y is NOT invited to speak, which means that B IS invited (because at least one of the two mystery writers must speak. L always speaks.

Thus, we know our diagram will be some sequence of CX, A, B, and L. Since L and B are both invited, neither may speak first or last. Thus, a possible diagram may be something like C, X, L, B, A.  L and B will always be surrounded on either side by the groups CX and A. Thus, there is no combination in which the A and CX blocks touch, and no consecutive grouping of those letters may take place.

 "<" indicate the order of names relative to each other. 1st performed to last performed is respective to left to right.

The correct answer is (F, D, A, B, J, K, G) as it does not disobey any of the rules.

                                - Incorrect answer explaination -

-If you chose  (A, B, F, D, K, G, J), you should refer to the 1st rule concerning G, J < G and K < G. This incorrect answer switches the rule regrading J < G.

-If you chose (A, B, F, K, D, J, G), you should refer to the 4th rule, As these two do NOT perform directly before or after each other.

-If you chose (K, A, B, F, D, J, G), you should refer to the 3rd rule where you will find A < K is disobeyed.

- If you chose (A, F, D, J, K, B, G), you should refer to the 2nd rule where you'll find B < J is disobeyed. 

This question requires that we combine all the rules together. By combining the sequencing and disregarding the FD non-specific consecutive order (they can be switched back and forth between FD and DF) for the moment, we can extrapolate that A must occupy one of the first 3 spots. (See sequence just below) This immidiately kicks the answer choices 4, 5, and 6 out the window.

A < B < J < G and also A < K

We are then left to decide whether A can occupy spot 2 or 3. We must now work with rule 4: FD or DF. Because FD must be consecutive and all other letters force A to perform earlier, A must occupy spot 1 or 3. If it occupies spot 2, it forces FD to split up or disobeys a sequencing order.

Example of incorrect answers:

F, A, D, B, J, K, G    - sperated FD makes this incorrect. D and F are interchangable here. 

B, A, DF, J, K, G      - Here, B < A, which disobeys rule 3 which states, A < K and B.

Hence, A can occupy spot 1 or 3 and since 1 is not an option, 3 is the correct answer.

If you think about DF as a unit and this system as having 6 spots instead of seven, it helps to visualize how consecutive characters function as a unit in this question.