Challenge #36 (HARD)

[zdarakjian]

In answering question 3 of Challenge #36 (Hard): (Trainee Game)

The correct answer is given as D (no more than 2 trainees on project O).

But why can't T, U and also W be assigned to project O? What rule would that violate?

Thank you.

Zareh

zdarakjian@gmail.com

[ohthatpatrick]

Sorry for the delay. We've been consumed with trying to get Test 88 explanations up.

Here in our initial diagram:

M: s, t, u, w
N: __ __ / /
O: u __ __ /
P: __ __ / /
R: w, t, s/u, /

Your confusion with Q3 may stem more from initial deductions.

Since R has exactly 3, but lacks either S or U, we know it has
w, t, and exactly one of s/u

Since M has more than anyone, and R has 3, M needs to have all four.

When Q3 tells us that T is in exactly three, we already know that two of those have to be M and R.
We also know that if you're in N, then you're in 4 places. So T can't be in N.
The 2nd rule says that everyone works on N or P (but not both). So if T can't be in N, then it must be in P.

Hence, we know T's three groups: M, P, R

T is not in N or O.
We also know that S is not in O.

If T and S can't be in O, then the most that O could have would be U and W.
Hence, (D) must be true.

For the example you were thinking about, with T, U, and W all working in O, it would break different rules depending on how you wrote out the rest of your scenario. Write out a "complete" scenario that includes T, U, and W working in O and check that complete scenario against all the rules, and you'd always be sure to be breaking a rule (or breaking the constraint of the question that T is in exactly 3).

Hope this helps.