Logic Game Challenge #36: The Trainee Game (Easy)

[ohthatpatrick]

Our latest logic game challenge, The Trainee Game, is live! Post your answer explanations (or questions) here!

[ashabodhi]

Since there are only three possibilities for management position, it will be easy to set up frames.

1. M: W
A: S/Z, TU
R: S/Z
X and Y can be in A or R

2. M: X
A: S/Z, W/TU
R: S/Z, W/TU
Y can be in A or R

3. M: Y
A: S/Z, W/TU
R: S/Z, W/TU
X can be in A or R
With these frames it will be easy to answer all the questions.

[ohthatpatrick]

Great work!

Indeed, those 3 frames are a great way of narrowing down the possibilities of this game.

To make sure everyone follows your logic, let me spell out a couple parts of it.

The "if and only if" rule, (also called a bi-conditional statement), could be symbolized as

Wa <---> Tr

"Contrapositives" of a bi-conditional statement are different from usual contrapositives in the sense that you don't really need to flip the order of parts (since the arrows go both ways).

~Wa <---> ~Tr

I find that an easier way to think about bi-conditional statements is, "either BOTH these things happen or NEITHER of these things happen".

In the frame in which we have W assigned to Mgmt., we know that NEITHER thing is happening (i.e., T must go to Accounting).

In the other two frames, W and TU are forced to go into Accounting or Research. If BOTH things happen, W goes to A and TU goes to R. If NEITHER thing happens, then W must go to R, and TU must go to A.

So the poster cleverly notated that as W/TU in one group and TU/W in the other.

Similarly, S and Z have to be split up between A and R, so again we use placeholders to notate that.

And it's a great idea to remind yourself which elements are still floating about, waiting to be placed. Since Mgmt. only gets one element, and there are no quantity min/max issues to worry about for Acctng. or Research, the remaining elements can fall into those two slots however they'd like.

For those of you who haven't yet tried the game (or those who did it without frames), see if the frames really do make the questions easy for you. If so, try writing an explanation for each question, describing how you could use the frames to get the answer.

If you're not too keen on frames, let us know how you solved the questions otherwise.

[ptraye]

See below.

[ptraye]

STUWXYZ

M: Management
R: Research
A: Accounting

Distribution:
3-3-1 or 4-2-1

Rules
1) T, U must be together
2) S, Z cannot be together
3) Wa <--> Tr
4) M: W, X or Y

Inference:
1) Because S and Z must be in separate division and neither can be in M, S and Z will always be in either R or A.

2) Because of the biconditional in rule #3, whenever W is in A, T will be in R, and whenever T is in A, W will be in R. Additionally, if W is in M, T can be in R or A.

Basic Diagrams:
3-3-1

Possibility #1
M: X/Y
R: T, U, S/Z
A: W, S/Z, X/Y

Possibility #2
M: X/Y
R: W, S/Z, X/Y
A: T, U, S/Z

Making more examples is not necessary. At this point, I went to the questions. Just remember, there is the possibility of a 4-2-1 distribution as well.


Question 1
Rule #1 eliminates answer choice B.
Rule #2 eliminates answer choice D.
Rule #3 eliminates answer choices C and E. C is eliminated because the biconditional statement works both ways: whenever T is in R, W must be in A. In (C), W is in M.

Question 2
Answer choice C must be false. U and W cannot both be in accounting because T must be in the same division as U. And, the biconditional in rule #3 says that W and T cannot both be placed in the accounting division together.

Question 3
4-2-1
M: X/Y
R: W, S/Z
A: X/Y, T, U, S/Z

In both T and X are in accounting, 4 trainees can be in the accounting division.

Question 4
4-2-1
If fewer are in accounting than research, that mean 2 trainees must be in accounting, 4 trainees must be in research, and 1 trainee must be in management.

This problem is similar to question #3.

M: X/Y
R: X/Y, S/Z, T, U
A: W, S/Z

Because T must pair with U, T cannot be in accounting, with only two positions. This is because S and Z must be in separate divisions, and W and T cannot be in the same division.

Question 5
For question #5, I checked my previous diagrams against the answer choices.

From Possibility #1, I saw answer choice A and B are possible. From Possibility #2, I saw answer choice D and E are possible.

The problem with answer choice C is that the biconditional from rule #3 does not allow W into M. The biconditional is broken:
Rule #3: Wa <--> Tr
Answer choice C
M: W
R: T, U, Z
A: S, X, Y

Question 6
As we saw in the diagram from question #3, whenever 4 trainees are in the accounting division, X and Y must either be in M or A.

4-2-1
M: X/Y
R: W, S/Z
A: X/Y, T, U, S/Z

Answer choice E is correct.

[huongsn86]

Draw a chart to lay out like this
A | R | M

RULES:
s, t, u, w, x, y
1. TU (meaning together)
2. S ~ Z (meaning not together)
3. Wa(accounting) <-> TUr(research)
4. M: W, X, or Y

Infer:
TU can only be in A or R
S can only be in A or R
when TU is in research W will be in accounting, which leaves only X or Y in management
Z is a floater which means it can be plugged into A or R depending on S, X and Y are floaters as well

Possibilities:
I would draw columns with A R and M on top and the possible people below it
A: W, S/Z, X/y
R: T, U, S/Z
M: x/y

A: T,U, s/z
R: x/y, s/z, w
M: x/y

So just know that in either A or R there can not be less than 3 in there because of the S not in the same as Z and the extra floater of X, Y, and sometimes W

Q 1:
- B, E, D all are eliminated because of rules

Its either A or C but requires you to plug it in. Then with A if T is in research, W must be in A. So C works best

Q 2:
process of elimination

Q3:
Just by the rules, you can plug in the floaters and notice that TUSX or TUSY or TUXZ or TUZY can be together

Q4:
since we know that manangerment needs 1, you know it has to be a 4-2 ratio to make Accounting less, what better way than to put the TU together in a pair because that already adds two to one group, which makes TU go into R so that W can be alone

Q5:
I looked at my drwn out chart plugged in the conditional rule and saw that certain ones didnt work

Q6:
just looking back at my chart, I knew that it had to have 4 accounting trainees and so i used my own graph from question 3.

[kathrynlrosenberg]

Posted this in its own thread by accident! Whoops!

Question 1
Correct Answer: A

This is a straightfoward orientation question. Trake the constraints one at a time and apply to the answer choices to find the correct one. Starting with the first constraint, "T and U are assigned to the same division," scan the answer choices for the one which violates this rule. In letter (B), we see that T is assigned to accounting and and U is assigned to research. That won't work. The second constraint, "S and Z are assigned to different divisions" eliminates choice (D), since they are both assigned to Research in that choice. The third constraint, "W will be assigned to accounting if and only if T is assigned to research" is a bi-conditional statement. (W will only be assigned to accounting when T is assigned to research. If T is assigned to something else, W won't be assigned to accounting.) We're looking for the answer choice that either has W assigned to accounting and T is not assigned to research, or has T assigned to research and W not assigned to accounting. Scanning the choices, we see that choice (C) has T assigned to research, but W is not assigned to accounting. Further, choice (E) allows W to be assigned to accounting WITH T, but we know that cannot happen; if W is assigned to accounting, T is assigned to research. Choice (E) is eliminated leaving us with letter (A), without having to explore the final constraint. (None violate it, anyway.)

Question 2:
Correct Answer: C

For this question, we're looking for the answer that must be false, meaning that the rest of our answer choices could be true.

Letter A: Can T and Z both be assigned to accounting? Possibly. All we know is that T must be with U, and that Z must not be with S. S nor U have additional constraints, so nothing further could eliminate T and Z from appearing together in accounting. We might want to check out the other constraint that features T: that W will be assigned to accounting if and only if T is assigned to research. Well, T isn't assigned to research, so W can't be assigned to accounting, but W could easily be assigned to research or management. Cross A out.

Letter B: Can U and S both be assigned to accounting? Let's see. If we do that, then we must include T in accounting, since U and T must be together. Then, we must not include Z in accounting since Z and S cannot be together. There's only one more option for Z since it cannot be in management: Z must be in research. W, X, and Y are left unassigned. W cannot be in accounting because that would force T into research, but we know T is already in accounting. Therefore, W must be in research or management. It doesn't matter which, both will work, as will X and Y in any other divisions as long as one of W, X, or Y fills the necessary management spot. Choice B is plausible.

Letter C: Can U and W both be assigned to accounting? If U is assigned accounting, then T must also be. If W is assigned accounting - wait! If W is assigned to accounting, according to our constraints, T must be in research. This is an impossible choice. Letter C must be false, so it is our answer.

Question 3
Correct Answer: D

If both T and X are assigned to accounting, then U must also be part of accounting always, since T must appear with U. These three letters, in any team arrangement, must be included. This leaves a combination of the other four letters to figure out with these three. Starting with S: If S is part of accounting, Z is not. This also means if Z is part of accounting, S is not. We've got two combinations already (TUXS, TUXZ). Cross out choice A.

This has an additional implication. Our combinations MUST include either an S or a Z. Neither of them can be in management, leaving accounting and research as their only two options. Our combinations now must add letters to either TUXS or TUXZ.

Next, let's look at W. If W is in accounting, T must be in research, but this question explicitly states that T is part of accounting. W must not be in accounting for this to be true. W cannot be part of our combinations.

Can Y be part of another combination? If Y is included with TUXS or TUXZ, then W can fill the management position and all our letters are accounted for. This leaves us TUXSY and TUXZY as two more combinations. Added with the first two we found, this gives us four possible combos - choice (D).

Question 4
Correct Answer: D

All the answer choices in this questions must be true, except one, which could be false. Our correct answer will be the choice which is definitely or possibly false.

This question gives us a clue about the numbers involved in this game. We know that only one person fills the management spot, with the other six filling some combination of accounting and research. Of these six positions left up in the air, we know now it cannot be split evenly three and three. Either research has four or five (it cannot have six - S and Z have to be in separate divisions, so we know accounting has at least one space filled). Our constraints don't give us many specifics about the exact assignments of our trainees - only constraint #4 tells us that only three people can possibly be assigned to management. Scanning the answer choices, (D) should jump out. Let's see what happens if we place either X or Y in accounting.

Let's first try X in accounting. If X is there, then there is at most one other space for someone else in accounting. We know that S and Z must not be together, and they must either be part of accounting or research. If we place S OR Z in accounting, can we place the others? We know that T and U must be in research together, since they are linked and there's no more room for them in accounting. Yet, when we place T in research, we must also place W in accounting, and there's no room for that, either. Therefore, we don't have to continue and try Y in accounting or X and Y together since this statement must NOT be true, it could be false.

Question 5
Correct Answer: C

As we check out our answers, notice that all five choices include three trainees. We can set our sketch so that we know we must have three spots in accounting, leaving three spots in research and one in management. We are searching for the combination of three which cannot be the full list of trainees in accounting. The right choice won't work out for some reason specified by our rules.

Can S, T, U be the three in accounting? (It's the wrong answer if it can work.) If we place STU in accounting, then we know Z must be in research. W is the only letter left with a real constraint, since W, X, and Y remain and they can all fill the management spot. W is only an issue if it is assigned to accounting, which it is not, or if T is assigned to research, which it is not. (A) can work, meaning it is incorrect.

Can T, U, Z work? Again, we don't have to worry about the W and T relationship here because W can't be in accounting and T can't be in research. Quickly, we'll find this works because S can be placed in research and again we're left with the three who can be part of management. (B) works, so we cross it out.

Can S, X, and Y work? Already, we'll notice that this choice includes two of the three that could be part of management. This leaves W as the management position filler, leaving T, U, and Z in research. But, when T is in research, W must be in accounting, therefore this answer cannot work. (C) is correct.

Question 6
Correct Answer: E

This question gives another number clue, except this time, there are four people in accounting. This leaves two in research and one in management. Scanning our answer choices, we should know by now to check out answers that include X and Y, since they're two of our three that are the only ones capable of filling the management role. They've presented a problem all game, so we should start there.

We are looking for an answer that must be true. All other answers could be true or could be false, but the correct one will definitely always be true. Since all the answers could potentially be true, we need to find the one that, if disregarded, would disallow us from making a proper assignment - if it must be true, then we literally won't be able to solve this if we don't use that information.

So, let's see what happens if we disregard (E) and assign X and Y to the SAME division. Putting X and Y both in accounting means we must put W in management. We have two spots left in accounting and two spots left in research. We know S and Z can't be together, so we can split them up arbitrarily, leaving T and U as our remaining letters. T and U must always be together though, so this won't work. X and Y must be separate for this to be possible, so (E) is our answer.

What if we didn't realize how valuable the X and Y are? Quickly reviewing the other answer choices:

A) We've seen plenty of other times in this game that S can be assigned to both accounting or research. If S is research, then Z is in accounting, leaving the rest of the board wide open. Even with our tricky W in accounting if and only if T is in research will work out here. We can just put W in management and the problem is solved. Keeping T and U together, we can place those in accounting, and fill in the rest with X or Y.
B) This literally says the same thing. S and Z are interchangeable. They work fine here, as long as they are not together or part of management. Z does not have to be assigned to accounting.
C) We've seen that Y is essentially a floater if it is not in management. If T and U are both in accounting, then Y can easily be in management. They do not have to be together for this to work.
D) W and S can definitely be assigned to the same division. If W and S are both together in research, then T and U can be placed together in accounting with Z. Then, either Y or X can fill the management spot and the other can round out the accounting four. This works if violated, so it is wrong.


- Sketch: For this problem, you can use an open board. It's definitely an assignment question since we're not ordering anything and it cannot be binary grouping since there are three groups! It is not quite a closed board because we don't know what combination of people will compose the accounting and research divisions. We do know that each person will be assigned a division, so we can create a board of three spaces wide and seven spaces tall, with M, A, and R underneath. Off the bat, we can cross out six of the seven spaces in the M column, since management can only have one person. Underneath the M column, we can write the letters W, X, Y to show that these are the only viable options in this column. We can take a space off the top of both accounting and research since we know a spot must be given to management and at most, either of these can have six trainees. Next to our sketch, we can write vertically "TU" to show these will always appear together, and "SZ" crossed out since these will never appear together. Finally, we can write out our conditional statements: W(accounting) --> T(research) and T(research) --> W (accounting), with the contrapositives: - W(accounting) --> -T(research) and -T(research) --> -W(accounting). This will remind us of these rules. To save time, you might also elect to simply write W(accounting) <-> T(research) and -W(accounting) <-> -T(research) to show that this is a biconditional statement and works both ways.

[tomada36]

I'm lost on Q4. With any luck, someone will show me the error of my ways...

As (D) is the correct answer, doesn't this mean that both X and Y could be assigned to accounting? If so, how could both X and Y be assigned to accounting while adhering to the condition that there are fewer people assigned to accounting than to research? After all, either S or Z must be assigned to accounting, since they cannot be in the same division. This would man that there'd be three people assigned to accounting and three people assigned to research, which violates that condition.

Can someone kindly point out my fallacy?

[tomada36]

Please help me understand your reasoning for Question 4.
You seem to be saying that X and Y are not in the same division, but doesn't answer (D) infer that X, Y can be in the same division (i.e. "Either X or Y but not both can be in the same division" is not necessarily true). Wouldn't this mean that a scenario exists in which X and Y are in the same division?

STUWXYZ

M: Management
R: Research
A: Accounting

Distribution:
3-3-1 or 4-2-1

Rules
1) T, U must be together
2) S, Z cannot be together
3) Wa <--> Tr
4) M: W, X or Y

Inference:
1) Because S and Z must be in separate division and neither can be in M, S and Z will always be in either R or A.

2) Because of the biconditional in rule #3, whenever W is in A, T will be in R, and whenever T is in A, W will be in R. Additionally, if W is in M, T can be in R or A.

Basic Diagrams:
3-3-1

Possibility #1
M: X/Y
R: T, U, S/Z
A: W, S/Z, X/Y

Possibility #2
M: X/Y
R: W, S/Z, X/Y
A: T, U, S/Z

Making more examples is not necessary. At this point, I went to the questions. Just remember, there is the possibility of a 4-2-1 distribution as well.


Question 1
Rule #1 eliminates answer choice B.
Rule #2 eliminates answer choice D.
Rule #3 eliminates answer choices C and E. C is eliminated because the biconditional statement works both ways: whenever T is in R, W must be in A. In (C), W is in M.

Question 2
Answer choice C must be false. U and W cannot both be in accounting because T must be in the same division as U. And, the biconditional in rule #3 says that W and T cannot both be placed in the accounting division together.

Question 3
4-2-1
M: X/Y
R: W, S/Z
A: X/Y, T, U, S/Z

In both T and X are in accounting, 4 trainees can be in the accounting division.

Question 4
4-2-1
If fewer are in accounting than research, that mean 2 trainees must be in accounting, 4 trainees must be in research, and 1 trainee must be in management.

This problem is similar to question #3.

M: X/Y
R: X/Y, S/Z, T, U
A: W, S/Z

Because T must pair with U, T cannot be in accounting, with only two positions. This is because S and Z must be in separate divisions, and W and T cannot be in the same division.

Question 5
For question #5, I checked my previous diagrams against the answer choices.

From Possibility #1, I saw answer choice A and B are possible. From Possibility #2, I saw answer choice D and E are possible.

The problem with answer choice C is that the biconditional from rule #3 does not allow W into M. The biconditional is broken:
Rule #3: Wa <--> Tr
Answer choice C
M: W
R: T, U, Z
A: S, X, Y

Question 6
As we saw in the diagram from question #3, whenever 4 trainees are in the accounting division, X and Y must either be in M or A.

4-2-1
M: X/Y
R: W, S/Z
A: X/Y, T, U, S/Z

Answer choice E is correct.

[RobW512]

Since there are only three possibilities for management position, it will be easy to set up frames.

1. M: W
A: S/Z, TU
R: S/Z
X and Y can be in A or R

2. M: X
A: S/Z, W/TU
R: S/Z, W/TU
Y can be in A or R

3. M: Y
A: S/Z, W/TU
R: S/Z, W/TU
X can be in A or R
With these frames it will be easy to answer all the questions.

Does this not make question 3 incorrect? Scenario 1 says W can be management, S/Z, TU and then X and Y can be in A or R. so that means to me we can have M- W, A- S/Z, TXUW, R- S/Z? Answer should be 5 possible in A.

[tomasl359]

Hi everyone. Really interesting game. I will try it :)

[smith8878]

Thanks for sharing this excellent blog which is really cool to read.
Vex 4

[henkaim886]

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[JANIEW730]

This is a very interesting game. I have a lot of entertainment. Let's develop more good games!
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