Main Diagram
Rule 4 eliminates B. Gleeson cannot visit Sydney
I’ve added two other things:
• G must go in the other city, since that’s the only way to have two people in each group without I (G + F/H).
• In the second scenario, rule 3 applies. Since G is in Manila, then H is in Tokyo.
That’s pretty much it for the scenarios. Remember that you have to place both F and H at least once in each scenario, as well. (But not together)
Main Diagram
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Question 17
For acceptable order questions, go through the rules and use them to eliminate answers one by one.
Note that I use the rules themselves. I don’t use my diagrams for these questions. Reading the rules again for this question will help you memorize them, and it’s also more efficient.
Rule 1 eliminates A. Ibanez must visit exactly two cities.
Rule 2 eliminates E. Fan and Haley cannot visit the same city as each other.
Rule 3 eliminates D. Manila is visited by Gleeson, so Tokyo must be visited by Haley.
Rule 4 eliminates B. Gleeson cannot visit Sydney.
C is CORRECT. It violates no rules.
M S I
G I G
T
M S I
I H G G
T
M S I
G I G
T
M S I
I H G G
T
I = 2 2 F H
G
MT
H3
1
G I
4
I G
Question 18
To answer a “completely determines” question, you should look at the scenarios + the rules and think about what makes something else happen. That
“something else” is the key, since determining everything requires a chain reaction that forces 3-5 things to occur.
Here’s are the scenarios and rules for reference:
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We can more or less ignore A and C for now. F and H aren’t totally interchangeable, but they practically are, so these are very nearly the same answer.
Interchangeable answers can’t be right. (While rule 3 does place H in Tokyo, H in Tokyo isn’t a sufficient condition for anything else.)
D and E have the same issue. They’re the same, except one is F and one is H. Interchangeable answers can’t both be right. (And, since G or H in Tokyo isn’t a sufficient condition for anything, these answers don’t force anything else to happen.) So let’s try B. Placing G in two cities forces G into Manila and Tokyo, since G can’t go in Sydney (rule 4)
This in turn triggers rule 3 (G Manila ➞ H Tokyo).
So, this is a promising answer, because it forces another thing to happen. Let’s draw it:
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Next, we know Ibanez has to go in two spots. I placed one Ibanez in Sydney. There are only two cities with space open: Manila and Sydney. So Ibanez must go in both:
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We have to place each person at least once, and we haven’t placed F. So, we have to put F in Sydney:
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Everything is determined. B is CORRECT.
Question 19
We saw in the setup that Ibanez must visit Sydney.
D is CORRECT.
Let’s recap why this is true:
1. Every city needs two people 2. Only G, I, F and H are available.
3. But, F and H can’t go together, so only G, I and one of F/H can go
4. G can’t go in Sydney
5. That leaves only I + one of F/H to go there
M S I
G I G
T
M S I
I H G G
T
I = 2 2 F H
G
MT
H3
1
G I
4
I G
M S I
G H G I G
T
M
S I F G H G I G
T
Question 20
Before doing a general could be true, it’s good to skim all the answers and see which are plausible and which aren’t. I find narrowing it down to a couple of candidates in order to brute force only those helps clear the mind.
D and E are clearly wrong. F and H can’t go together, and every product manager must go at least once. So, neither F nor H can go everywhere (because then the other would have nowhere to go) If you got question 19 right, you’ll recognize C is wrong. Ibanez goes exactly twice (rule 1), and as we saw in the setup, they must go in Sydney (because G can’t go there, so I + F/H must). Therefore Ibanez can’t also go in Manila and Tokyo: that would be three cities.
That leaves A and B. Now that there are only two contenders, we can try drawing both.
A places F and I in Manila:
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We know that G must go in Tokyo, since it can’t go in Sydney. Then we can put either F or H in the other spots. Here’s one example:
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So, this works. A is CORRECT.
Let’s try B, just to be sure. It places G and I in Tokyo:
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Manila is the big space left to fill. We have already placed both I’s, so the only managers who can go in Manila are G + F/H. And here’s the problem: if we place G in Manila, then we’d have to put H in Tokyo (rule 3). But, Tokyo is full. So, B doesn’t work.
Question 21
This question places G and H together. G can only go in Manila or Tokyo (rule 4), so we can try both scenarios:
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The first scenario triggered rule 3: G in Manila ➞ H in Tokyo
It looks like H is in Tokyo in either scenario. So, D is CORRECT.
There’s no need to do any more diagramming. We would merely be drawing out “could be true”
scenarios, which is a great waste of time on “must be true” questions you have already solved. You should focus on certainties.
M S I
F I G
T
M
S I F G H F I G
T
M S I
G I G
T
M S
H G H G
T
M S
G H
T
Question 22
This question places Ibanez in Tokyo. You should then consider the next most restricted factor: G. G must go either in Manila or Tokyo:
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In the left hand scenario, Ibanez and G fill up Tokyo.
This means that G can’t go in Manila, since that would trigger rule 3 (H in Tokyo). This can’t happen, since Tokyo is full.
Ibanez also can’t go in Manila, since they have already gone twice. That leaves just H and F, but they can’t go together (rule 2).
So the left hand scenario actually doesn’t work!
That leaves us in the right hand scenario. G in Manila has triggered rule 3, so H must go in Tokyo.
Since we have placed our two Ibanez and can’t place G in Sydney, this means both open spaces will be filled with F or H (and at least one must be F, because F hasn’t gone yet.)
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A is CORRECT, because it’s possible. B through D contradict the diagram. E is a trap answer. It seems like it would work, but if you filled both open spots with H then there would be no room for F. Every product manager must go at least once.
Question 23
On a rule substitution question, the correct answer must:
• Forbid everything normally forbidden
• Allow everything normally allowed
A lot of people miss this second quality. It’s very useful. It means that if one of the answers contradicts a scenario that would normally be allowed, then that answer is wrong.
Likewise, if an answer makes something happen that didn’t happen under the normal rules, then it is wrong.
A is wrong because normally G and Ibanez can go together. Here’s an example; it’s a scenario I made for question 20. It also disproves D:
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This scenario proves B is too restrictive (I made this diagram just for this question):
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C is a trickier answer. It’s true according to the rules, but it’s not restrictive enough. It allows F and H to be together elsewhere. This scenario is possible with C:
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So we can eliminate four answers by recycling one scenario from question 20, and then making two quick custom scenarios. Not bad!
H
That leaves us with E, which is CORRECT. I covered why in the setup, but I’ll repeat it here.
• Each city must have two managers. F/H can’t go together.
• The three options for each city are G, I and one of F/H
• So, if you are missing one of G or I, you must have the other in order to fill the two spots.
If you are missing both G and I, you’d end up with F + H, which violates rule 2.
Ok, but why does this substitute for the rule?
Because it ensures every city is filled with at least one of G or I. There are then no open spaces for F + H to go together. For example:
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As per E, every space has either I or G. Do you see any place to put F and H together? I don’t. So E indirectly replaces the rule.
It’s best to prove the right answer. But, if you were short on time, you could also have merely eliminated the first four answers as I did with scenarios above.