Q23

[opulence2001]

Could you please explain question 23? it was the only question I got wrong for this game because I had no idea how to approach this question. They ask...if Barb selects at least one fish then which one of the following is the min and max possible numbers respectively of different species?

I was so thrown off by this question because I didn't understand what they were asking , but when I returned to it I deduced that if Barb chooses K, J, or L she need not choose anything else so the minimum species is 1. But max I don't get. The answer is B, 1 and 5, but I can't figure out which species would force to Barb to choose four more.

Am I even approaching this question correctly? Please advise.

[ManhattanPrepLSAT1]

Not exactly, but you're not far off either.

There is no constraint limiting the minimum number of fish that Barbara could select. Only rules that move from negative to positive affect the minimum number that can be selected. So if a rule said

~F ---> G

then you would know that at least F or G must be selected and then try and determine whether F or G require other elements to be selected.

For a maximum work in reverse. Only rules that move from positive to negative affect the maximum number that can be selected. So if the rule said

F ---> ~G

then you would know that at most one of those was selected - meaning at least one was not selected. Then determine whether not selecting G or not selecting F require other elements to also not be selected.

Since this game has no constraints that move from negative to positive the minimum is 1 (only because the question stem says that Barbara selects at least one fish; if this had not been there, the correct minimum would be zero).

There are two positive to negative rules.

K ---> ~O
M ---> ~N

Kick out the O and the M and nothing else is required to be not selected. Thus the maximum could be everything beside O and M -- that would be J, K, L, N, P (5).

Does that answer your question?

[monicaiannacone]

Thanks for the explanation on #23. But I'm a little confused. For the rules that go from positive to negative for this problem, why aren't there four rules:

1) k -> ~O (given constraint)
2) O -> ~K (contrapositive)

3) M -> ~N (given constraint)
4) N -> ~M (contrapositive)

In this case, are we only considerign the given rules (and not their contrapostives) since only one or the other is IN?

The process just confused me, Is there another way to quickly answer this question without trial and error (and aside from the given answer)?

Thanks!!

[timmydoeslsat]

Matt did list just the given rules and not their contrapositives.

You can immediately look at a rule and see that it forces at least one out:

A ---> ~B

This conditional structure forces at least one of the two variables to be out. Never could both variables be included.

So with this in mind, we will determine what is the max # of different species.

We would obviously have the random variables of J and L included.

We would want O to be included, as if we do not, it would take down so many variables.

And we know that if we select O we will need to select P. We also know that by selecting O to be in, it would force K out due to the first rule.

So right now we have this:

J L O P.................K

We only have 2 left to place: M N

There is a rule governing it:

M ---> ~N

It could be true that both M and N are out. However, since we want to determine the max that could be in, we would want to have one of those guys be in. They both could not be in, but one could.

That means that 5 is the max.

[chibunna.chimezie]

I approached this question a little differently given the question stem. Since the question says AT LEAST ONE, I looked for the one species she could select and would bring in the max amount of species. Knowing that the Max would consist of the two wildcard variables of J AND L, I look at M and N. From the rules I know M and N can not both be in but either M or N brings in O.

So i chose M as the Min(1) and Max accounts for 0(2), J, L, P; Because choosing 0 brings in 2 O's, then i count 1 j, l, p for a total of 5

Is this a flawed way to look at this..? I understand the way Matt went about it but I decided to take what the question stem gave in saying "at least one"

Thanks a lot

[chibunna.chimezie]

I approached this question a little differently given the question stem. Since the question says AT LEAST ONE, I looked for the one species she could select and would bring in the max amount of species. Knowing that the Max would consist of the two wildcard variables of J AND L, I look at M and N. From the rules I know M and N can not both be in but either M or N brings in O.

So i chose M as the Min(1) and Max accounts for 0(2), J, L, P; Because choosing 0 brings in 2 O's, then i count 1 j, l, p for a total of 5

Is this a flawed way to look at this..? I understand the way Matt went about it but I decided to take what the question stem gave in saying "at least one"

Thanks a lot

[shaynfernandez]

There are two positive to negative rules.

K ---> ~O
M ---> ~N

Kick out the O and the M and nothing else is required to be not selected. Thus the maximum could be everything beside O and M -- that would be J, K, L, N, P (5).

Does that answer your question?

How can that be correct, rule 4 says if she selects at least one N then she selects at least one O"

You listed N without including O which violates the rule.

[natalie.seni]

Not exactly, but you're not far off either.

Kick out the O and the M and nothing else is required to be not selected. Thus the maximum could be everything beside O and M -- that would be J, K, L, N, P (5).

Does that answer your question?

I have the same question as the poster above me. If there's N, then, as the fourth rule stipulates, doesn't there also have to be O?