Q21

[zhangstagangsta]

Could someone tell us how to answer maximum/minimum questions in these conditional games? What is the best strategy?

[dan]

Good question.

Some Logic Games questions will ask you for the "maximum number of ways" that something could happen, or the "minimum number of ways" that something could happen.

As a general rule, you want to tackle these questions as follows:

1. MAXIMUM questions: start with the answer choice that represents the largest number and work your way down. Try to make the scenario "work" with the biggest number possible. If it does, you've got your answer. If it doesn't, move to the next biggest number.

2. MINIMUM questions: start with the answer choice that represents the smallest number and work your way up. Try to make the scenario "work" with the smallest number possible. If it does, you've got your answer. If it doesn't, move to the next smallest number.

Let's try it for #21 on this game (using the diagram that Aileen posted above). Since we're looking for the minimum number that could be at S, let's start with the smallest answer, (A) zero.

(A) Can we make this work? Well, if we're going to have zero at S, that means we'd have all six at R. Can we have all six at R? No. If even one doctor at R forces another over to S, we know we can't have all six at R. For instance, When P is at R, both O and K are at S. We should be able to see this quickly (or any of the other doctors at R that force others to S). Eliminate (A). Now move to the next smallest answer.

(B) Can we have just one at S? Let's try. We know the one letter at S can't be O, b/c O also forces N into S. So, O can't be at S; it must be at R. But O at R forces both P and J to S. In summary, when O is at S, we get both O and N at S. Doesn't work. But when O is at R, we get both P and J at S. Doesn't work. There's no way to get just one doctor at S. Eliminate (B). Now move to the next smallest answer.

(C) Can we have just two at S? Let's try. The most strategic way to approach this is to try putting letters at R that don't force many things into S. For example, if we put L into R, that doesn't force anything into S. We'll start there:

R: L
S: nothing

Next, if we put N into R, that only forces O into R:

R: L, N, O
S:

But, O at R forces J and P over to S. We can't forget about this:

R: L, N, O
S: J, P

... and P at S forces L over to R, which we already had. Also notice that J at S forces K into R!

R: L, N, O, K
S: J, P

This works. Only two doctors at S, so (C) is the answer.

This happens to be a pretty tough minimum question, but it's made easier if you start with the smallest possibility and work your way up.

[qccgraphix]

Good question.

Some Logic Games questions will ask you for the "maximum number of ways" that something could happen, or the "minimum number of ways" that something could happen.

As a general rule, you want to tackle these questions as follows:

1. MAXIMUM questions: start with the answer choice that represents the largest number and work your way down. Try to make the scenario "work" with the biggest number possible. If it does, you've got your answer. If it doesn't, move to the next biggest number.

2. MINIMUM questions: start with the answer choice that represents the smallest number and work your way up. Try to make the scenario "work" with the smallest number possible. If it does, you've got your answer. If it doesn't, move to the next smallest number.

Let's try it for #21 on this game (using the diagram that Aileen posted above). Since we're looking for the minimum number that could be at S, let's start with the smallest answer, (A) zero.

(A) Can we make this work? Well, if we're going to have zero at S, that means we'd have all six at R. Can we have all six at R? No. If even one doctor at R forces another over to S, we know we can't have all six at R. For instance, When P is at R, both O and K are at S. We should be able to see this quickly (or any of the other doctors at R that force others to S). Eliminate (A). Now move to the next smallest answer.

(B) Can we have just one at S? Let's try. We know the one letter at S can't be O, b/c O also forces N into S. So, O can't be at S; it must be at R. But O at R forces both P and J to S. In summary, when O is at S, we get both O and N at S. Doesn't work. But when O is at R, we get both P and J at S. Doesn't work. There's no way to get just one doctor at S. Eliminate (B). Now move to the next smallest answer.

(C) Can we have just two at S? Let's try. The most strategic way to approach this is to try putting letters at R that don't force many things into S. For example, if we put L into R, that doesn't force anything into S. We'll start there:

R: L
S: nothing

Next, if we put N into R, that only forces O into R:

R: L, N, O
S:

But, O at R forces J and P over to S. We can't forget about this:

R: L, N, O
S: J, P

... and P at S forces L over to R, which we already had. Also notice that J at S forces K into R!

R: L, N, O, K
S: J, P

This works. Only two doctors at S, so (C) is the answer.

This happens to be a pretty tough minimum question, but it's made easier if you start with the smallest possibility and work your way up.

Thanks Dan. That was a great explanation.