Bird-watchers explore a forest to see which of the following six kinds of birds -- grosbeak, harrier, jay, martin, shrike, wren -- it contains. The findings are consistent with the following conditions:If harriers are in the forest, then grosbeaks are not.If jays, martins, or both are in the forest, then so are harriers If wrens are in the forest, then so are grosbeaksIf jays are not in the forest, then shrikes areIf jays are not in the forest, then which one of the following must be false?
Answer(s): D
Rejecting J means (Rule 4) selecting S for sure. Therefore, option [Neither martins nor shrikes...] is false: At least S must be present in the forest. Simple as that. A variety of birds and combinations can be present but no others need be, so none of the other choices must be true.
Bird-watchers explore a forest to see which of the following six kinds of birds -- grosbeak, harrier, jay, martin, shrike, wren -- it contains. The findings are consistent with the following conditions:If harriers are in the forest, then grosbeaks are not.If jays, martins, or both are in the forest, then so are harriers If wrens are in the forest, then so are grosbeaksIf jays are not in the forest, then shrikes areWhich one of the following is the maximum number of the six kinds of birds the forest could contain?
Answer(s): C
To maximize a total selected we need to minimize a total rejected, and here we should start with the most limiting rule, which is probably Rule 4. That rule allows us to choose both J and S with impunity, so let's do so.Choosing J requires choosing H (Rule 2) and therefore rejecting G (Rule 1), and with G rejected, W must be rejected as well (Rule 3). M is left and can be chosen with impunity, so we have worked out the maximum as 4 = H, J, M, and S.An alternative approach would have been to start with the other limiting rule, Rule 1. The most we can pick of the G / H pair is one, so we can work out each possibility. If it's G, then we can pick three birds at most (G, S, and W), but if it's H, then we can pick the four already discussed.
Bird-watchers explore a forest to see which of the following six kinds of birds -- grosbeak, harrier, jay, martin, shrike, wren -- it contains. The findings are consistent with the following conditions:If harriers are in the forest, then grosbeaks are not.If jays, martins, or both are in the forest, then so are harriers If wrens are in the forest, then so are grosbeaksIf jays are not in the forest, then shrikes areWhich one of the following pairs of birds CANNOT be among those birds contained in the forest?
Answer(s): A
Use previous work. We saw, chosen together in our work for questions 7 and 9, "H, J, M, S," which itself demonstrates that the pairs in options [jays, shrikes], [jays, martins], and [shrikes, martins] can in fact be chosen together. Option [shrikes, wrens] is a possible pair. But option [jays, wrens] is impossible, a fact demonstrable through trial. Selecting J requires H (Rule 2), and selecting W requires G (Rule 3), and voila. We have violated Rule 1.
Bird-watchers explore a forest to see which of the following six kinds of birds -- grosbeak, harrier, jay, martin, shrike, wren -- it contains. The findings are consistent with the following conditions:If harriers are in the forest, then grosbeaks are not.If jays, martins, or both are in the forest, then so are harriers If wrens are in the forest, then so are grosbeaksIf jays are not in the forest, then shrikes areIf grosbeaks are in the forest, then which one of the following must be true?
A straightforward chain. Choosing G mandates rejecting H (Rule 1). That in turn mandates rejecting J and M (Rule 2's contrapositive). And once J is gone, S must be chosen (Rule 4). W, the remaining entity, can be chosen but need not be, so neither choice [Wrens are in the forest.] nor [The forest contains both...] must be true. And since the total number of birds here is 2 or 3, options [At most two kinds of...] and [At least three kinds of...] are possible only.
Bird-watchers explore a forest to see which of the following six kinds of birds -- grosbeak, harrier, jay, martin, shrike, wren -- it contains. The findings are consistent with the following conditions:If harriers are in the forest, then grosbeaks are not.If jays, martins, or both are in the forest, then so are harriers If wrens are in the forest, then so are grosbeaksIf jays are not in the forest, then shrikes areSuppose the condition is added that if shrikes are in the forest, then harriers are not. If all other conditions remain in effect, then which one of the following could be true?
Answer(s): B
The immediate impact of the new rule is that S and H are mutually exclusive: "NEVER HS," if you will. The larger impact will likely only be seen by trying out the choices individually.If, in line with option [The forest contains both jays and shrikes...], J and S are selected, a contradiction emerges. The question stem requires us to reject H, but Rule 2 requires us to select H.
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