The key insight here is that Anchorage and Boston must form a block one time. Boston cannot be in Month 4 or Month 7 by rule. It cannot be Month 5 based on the setup of this question. Additionally, Boston cannot be Month 6 because Anchorage has to come immediately before it, and Denver occupies the Month 5. Finally, Boston cannot be in Month 1 as A must come before it. So Boston can only be in  Month 2 or 3.

Once this is established, we see there are possibilities for Chicago to be 1st, 4th or 6th only. None of the other cities can be a work site in the months listed.

When the dime is second and the colonial and territorial are next to each other, the gold and the quarter cannot be separated by exactly one coin.

When the colonial is fourth, we know the location of four coins.  Since gold must be after the colonial and exactly one place removed from the quarter, gold is fifth and the quarter is third.  Since the gold coin is fifth, the dime must be second.  Either the bullion or the territorial is first--the other is sixth.

When the territorial is fifth, the dime must be second. We do not know the exact location of the remaining coins, but we do know there are only two possibilities for each of them.  The quarter and gold coin both must be either fourth or sixth.  The bullion and colonial both must be either first or third.

In a question like this, it is always useful to make use of information we have learned from correctly answering previous questions. The answer from the first question for this game is very useful.

Here were are looking for the car model that must always be put on sale during the second week. This means that car model will appear in week 2 for all possible correct outcomes, including the one we identified as the correct answer for the first question in this logic game. In that possible outcome, Q and M fall under week 2 - so it is either Q or M that must always appear in week 2. We can immediately eliminate the 3 other answer choices.

To determine which of the two it is, you can draw up a different possible outcome to see which one can possibly go elsewhere. If a car model can successfully go during a different week, you can eliminate that answer choice.

If you try to place M elsewhere, you'll see it is able to go outside of week 2. M cannot go in week 1, because there's no way to put either Q or V with it (which is required according to the second global rule). Q cannot go there, since there's no room for N to go before it, and V cannot go there since V is in week 3, and thus cannot be duplicated.

However, M can go with V in week 3. You can use the rules to fill out the following possible out come:

Week 1: R, N; Week 2: P, Q; Week 3: V, M; Week 4: R, Y

Since M can go somewhere other than Week 2, it does not necessarily always go in Week 2 - so it is not the correct answer.

This means that Q is the correct answer. You can verify this by trying to fit Q in any of the other 3 weeks - which will not allow for a valid sequence.

In this problem, you can disprove incorrect answers by showing that the pair can possibly go together in the same week. A sure way of doing this is creating hypothetical outcomes to show these pairings. Outcomes you have determined from earlier problems in this set can be of help as well.

However, if you pick up on the limitations one rule presents, this problem moves much quicker.

One of the answer choices lists Q and V. Based off the second global rule, you know that M must be paired with Q or V. We also know that V goes on sale during week 3 based off of the fourth global rule. So if Q and V were to go together, it must happen during week 3. The third global rule also tells us that since these two are in the third week, they are not put on sale twice - in other words, neither Q nor V can show up elsewhere in this set-up to then pair up with M. So this set-up cannot work. Thus, Q and V cannot be put on sale during the same week - and that is your correct answer.

You can double check this by creating possible outcomes that pair up each of the other options.

For a car model to be put on sale twice, it must show up one of those two times during week 4, and it cannot show up during week 3.

One answer choice immediately breaks those guidelines. If M was to be put on sale twice, it would have to be paired up once with Q and once with V - based off of the second global rule. However, V is always put on sale during week 3 - so if M was paired with V, it would also have to go on sale during week 3, thus breaking the third global rule.

So M can never be put on sale twice.

You can verify this answer by drawing up possible outcomes that show each of the other four being put on sale twice.

From the stem, you know the cat is on shelf 5, which puts it immediately after the lion on shelf 4:

__, __, __, lion, cat, __, __

A big help here is recognizing that the monkey-bear pair can only go on shelves 1-2, 2-3, or 6-7.

This means that the monkey can never go on shelf 3, and thus can never be consecutive with the lion, eliminating "monkey, lion."

You can eliminate "bear, dog" because for this to work, the dog must go after the bear, since the monkey goes before the bear. This would only work with shelves 1-2-3, but this would fail because it would be impossible to keep the frog ahead of the dog.

You can eliminate "bear, penguin" because whether you put penguin on shelf 1 or 7, there is no way for the bear to be next to it and keep the monkey-bear pair together.

You can eliminate "dog, monkey" because you run into the same problem as "bear, dog" - only the order would be dog-monkey-bear, leaving no way for the frog to stay ahead of the dog.

This leaves you with the correct answer "penguin, dog" - and an allowable sequence: frog, monkey, bear, lion, cat, dog, penguin.

Here you are looking for the greatest number of shelves that can be put in between the frog and the dog. A natural place to start is to put the frog on shelf 1, since it must always go before the dog.

The farthest the dog might go is to shelf 7. However, this would mean breaking the global rule that the penguin has to go on shelf 1 or 7. So you can put the dog on shelf 6, and the penguin on shelf 7. You need to make sure a sequence can be made, following all rules, to verify this. You already know the lion must go on the fourth shelf from the fourth global rule:

frog, __, __, lion, __, dog, penguin

You can complete this sequence by placing the monkey-bear pair on shelves 2-3, and the cat on shelf 5:

frog, monkey, bear, lion, cat, dog, penguin

This puts four shelves in between the frog and the dog (monkey, bear, lion, cat). You can verify this is the maximum by sending the dog to the seventh shelf, moving the frog the second shelf, and moving the penguin to the first shelf:

penguin, frog, cat, lion, monkey, bear, dog

Still, the greatest number of shelves between them is four.

Here we are looking for the answer that is always necessarily true. The four incorrect answers are all sometimes true, but not always true.

You know the lion must go on shelf 4, and this question stem tells you the cat is to go on shelf 2. You can tackle this problem by determining how many possible outcomes can work with these local rules and the global rules. If you place the penguin on shelf 7, there is only one possible outcome that keeps the monkey-pair together and the frog ahead of the dog:

frog, cat, dog, lion, monkey, bear, penguin

If you place the penguin on shelf 1, there are two possible outcomes to keep the monkey-pair together and the frog ahead of the dog:

penguin, cat, frog, lion, monkey, bear, dog

penguin, cat, frog, lion, dog, monkey, bear

The second and third possible outcomes show the dog does not always have to go on a lower-numbered shelf than the penguin, so you can eliminate that answer.

The first possible outcome shows the lion does not always have to go on a lower-numbered shelf than the dog, so you can eliminate that answer.

The first and third possible outcomes show the monkey does not always have to go on a lower-numbered shelf than the dog, so you can eliminate that answer.

The first possible outcome shows the penguin does not always have to go on a lower-numbered shelf than the bear, so you can eliminate that answer.

All three possible outcomes show that the frog must go on a lower-numbered shelf than the lion, and that is your correct answer.