LSAC LSAC Certifications LSAT-TEST Questions & Answers

• Question 341:

  From among ten stones, a jeweler will select six, one for each of six rings. Of the stones, three -- F, G, and

  H -- are rubies; three -- J, K, and M -- are sapphires; and four -- W, X, Y, and Z -- are topazes. The

  selection of stones must meet the following restrictions:

  At least two of the topazes are selected.

  If exactly two of the sapphires are selected, exactly one of the rubies is selected.

  If W is selected, neither H nor Z is selected.

  If M is selected, W is also selected.

  If J and M are the only sapphires selected, which one of the following could be true?

  A. F and G are both selected.

  B. F and X are both selected.

  C. G and H are both selected.

  D. G and K are both selected.

  E. Y and Z are both selected.

  Correct Answer: B

  Interestingly--if frustratingly--this last question proves to be one of the most concrete, and hence one of the easiest. We hope that you didn't get bogged down on some of the earlier ones and miss out onthis one. In your roster, select J and M, and cross out K, in line with the question stem's if. Rule 4 requires that you select W, and Rule 3 requires that you cross out H and Z. At the same time, Rule 2 (we use it at last!) kicks in--there will be exactly 1 ruby chosen, and hence 3 sapphires. And with Z gone, those 3sapphires are W, X, and Y. So: J, M, W, X, and Y are definitely chosen, and the final stone will be a ruby, either F or G. Option [F and X are both selected.] is therefore the only possibility among the choices. Option [F and G are both selected.] would stick us with a total of 7 stones. And each of the remaining choices mentions a stone that we have definitively rejected: H in option [G and H are both selected.], K in option [G and K are both selected], and Z in option [Y and Z are both selected].

• Question 342:

  From among ten stones, a jeweler will select six, one for each of six rings. Of the stones, three--F, G, and

  H--are rubies; three--J, K, and M--are sapphires; and four--W, X, Y, and Z--are topazes. The selection of

  stones must meet the following restrictions:

  At least two of the topazes are selected.

  If exactly two of the sapphires are selected, exactly one of the rubies is selected.

  If W is selected, neither H nor Z is selected.

  If M is selected, W is also selected.

  Which one of the following must be true?

  A. The selection of stones includes at least one ruby.

  B. The selection of stones includes at most two rubies.

  C. The selection of stones includes either F or Z, or both.

  D. The selection of stones includes either X or Y, or both.

  E. The selection of stones includes either X or Z, or both.

  Correct Answer: D

  We have deduced that at least two topazes must be chosen at all times. And which two? Hard to say; but since we can never choose W and Z together (Rule 3), we will always need to choose at least one of the remaining topazes. That's all that option [The selection of stones includes either X or Y, or both.] is saying. Our work in Step 4, above, demonstrates that neither option [The selection of stones includes at least one ruby.]. nor option [The selection of stones includes at most two rubies.], need be true. Contrary to these both options, we could choose no rubies at all (if we used 3 sapphires and 3 topazes). The remaining choices may require a little trial and error, but the selection of, for instance, "G,J,K,M,W,Y" proves that neither option [The selection of stones includes either F or Z, or both.] nor option [The selection of stones includes either X or Z, or both.] need be true.

• Question 343:

  From among ten stones, a jeweler will select six, one for each of six rings. Of the stones, three -- F, G, and

  H -- are rubies; three -- J, K, and M -- are sapphires; and four -- W, X, Y, and Z -- are topazes. The

  selection of stones must meet the following restrictions:

  At least two of the topazes are selected.

  If exactly two of the sapphires are selected, exactly one of the rubies is selected.

  If W is selected, neither H nor Z is selected.

  If M is selected, W is also selected.

  If exactly two rubies are selected, which one of the following must be true?

  A. H is selected.

  B. J is selected.

  C. Z is selected.

  D. Exactly one sapphire is selected.

  E. Exactly two topazes are selected

  Correct Answer: D

  This is sheer arithmetic. If we choose exactly 2 rubies, we cannot choose 2 topazes -- remember, 2: 2 : 2 is forbidden -- so we need exactly 3 topazes (and have proved thatE.is impossible, incidentally).And 2 rubies + 3 topazes only = 5 stones, so there will be exactly 1 sapphire chosen, choice [Exactly one sapphire...] What that sapphire is, is unclear; it could be J, as option [J is selected.] suggests, but need not. And options [H is selected.]. and [Z is selected.] are much more certain as to the identity of particular stones than we can possibly be, under this question's circumstances.

• Question 344:

  From among ten stones, a jeweler will select six, one for each of six rings. Of the stones, three -- F, G, and

  H -- are rubies; three -- J, K, and M -- are sapphires; and four -- W, X, Y, and Z -- are topazes. The

  selection of stones must meet the following restrictions:

  At least two of the topazes are selected.

  If exactly two of the sapphires are selected, exactly one of the rubies is selected.

  If W is selected, neither H nor Z is selected.

  If M is selected, W is also selected.

  If Z is selected, which one of the following could be true?

  A. All three of the sapphires are selected.

  B. Both J and M are selected.

  C. Both K and M are selected.

  D. None of the rubies is selected.

  E. None of the sapphires is selected.

  Correct Answer: E

  Write out your roster, circle Z on it, and realize that that's a trigger for rejecting W (Rule 3's contrapositive) and thus for rejecting M (Rule 4's contrapositive). That's all we need. With sapphire M gone, clearly options [All three of the sapphires are selected.], [Both J and M are selected.], and [Both K and M are selected.] are impossible. And if option [None of the rubies is selected.] were true -- no rubies -- our remaining 2 sapphires and maximum of 3 topazes would leave us short of our desired total of 6. So [None of the sapphires is selected.] is possible by default. Indeed, all three rubies plus topazes X, Y, Z would constitute an acceptable set of 6.

• Question 345:

  From among ten stones, a jeweler will select six, one for each of six rings. Of the stones, three -- F, G, and

  H -- are rubies; three -- J, K, and M -- are sapphires; and four -- W, X, Y, and Z -- are topazes. The

  selection of stones must meet the following restrictions:

  At least two of the topazes are selected.

  If exactly two of the sapphires are selected, exactly one of the rubies is selected.

  If W is selected, neither H nor Z is selected.

  If M is selected, W is also selected.

  Which one of the following must be true?

  A. G is selected.

  B. J is selected.

  C. X is selected.

  D. Of at least one of the three types of stones, exactly one stone is selected.

  E. Of at least one of the three types of stones, exactly three stones are selected.

  Correct Answer: E

  Those who worked out the number options we laid out above had no trouble with this one. No matter how you slice it, at least one stone type -- and possibly two -- are represented by exactly 3 stones. And even if you didn't work out every possibility, simply seeing that 2 : 2 : 2 was an impossible ratio(because of Rule 2) might have led you to suspect that option [Of at least one ... three stones are selected.] was a true statement. Options [G is selected.], [J is selected.] and [X is selected.] are far from necessarily true--there are lots of possibilities in which G, J, and X respectively are not present. And the possibility of choosing, say, 3 sapphires + 3 topazes means that option [Of at least one ... one stone are selected] need not be true either.

• Question 346:

  From among ten stones, a jeweler will select six, one for each of six rings. Of the stones, three -- F, G, and

  H -- are rubies; three -- J, K, and M -- are sapphires; and four -- W, X, Y, and Z -- are topazes. The

  selection of stones must meet the following restrictions:

  At least two of the topazes are selected.

  If exactly two of the sapphires are selected, exactly one of the rubies is selected.

  If W is selected, neither H nor Z is selected.

  If M is selected, W is also selected.

  Which one of the following could be the selection of stones?

  A. F, G, H, M, X, Y

  B. F, G, J, K, M, W

  C. F, G, J, K, W, X

  D. G, H, J, X, Y, Z

  E. G, H, K, W, X, Z

  Correct Answer: D

  Even if you were blown away by this game's rules, this is a point worth getting, and worth getting quickly. Rule 1's requirement of 2 topazes minimum is violated by option [F, G, J, K, M, W]. Rule 2's 2 sapphires: 1 ruby ratio is violated by option [F, G, J, K, W, X]., which gives us 2 : 2 : 2. Rule 3 is violated by option [G, H, K, W, X, Z], which gives us W, H, and Z all at once. And Rule 4 is violated by option [F, G, H, M, X, Y], which gives us M without W. Option [G, H, J, X, Y, Z] must be correct sight unseen, and perhaps finding it, and then analyzing it, can provide us with some needed insight into the game and the heart to keep going with it.

• Question 347:

  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 grosbeaks If jays are not in the forest, then shrikes are

  Suppose 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?

  A. The forest contains both jays and shrikes.

  B. The forest contains both wrens and shrikes.

  C. The forest contains both martins and shrikes.

  D. Jays are not in the forest, whereas martins are.

  E. Only two of the six kinds of birds are not in the forest.

  Correct Answer: 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.

• Question 348:

  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 grosbeaks If jays are not in the forest, then shrikes are

  If grosbeaks are in the forest, then which one of the following must be true?

  A. Shrikes are in the forest.

  B. Wrens are in the forest.

  C. The forest contains both wrens and shrikes.

  D. At most two kinds of birds are in the forest.

  E. At least three kinds of birds are in the forest.

  Correct Answer: A

  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.

• Question 349:

  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 grosbeaks If jays are not in the forest, then shrikes are

  Which one of the following pairs of birds CANNOT be among those birds contained in the forest?

  A. jays, wrens

  B. jays, shrikes

  C. shrikes, wrens

  D. jays, martins

  E. shrikes, martins

  Correct Answer: 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.

• Question 350:

  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 grosbeaks If jays are not in the forest, then shrikes are

  If jays are not in the forest, then which one of the following must be false?

  A. Martins are in the forest.

  B. Harriers are in the forest.

  C. Neither martins nor harriers are in the forest.

  D. Neither martins nor shrikes are in the forest.

  E. Harriers and shrikes are the only birds in the forest

  Correct Answer: 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.