LSAC LSAT Section 1: Logical Reasoning LSAT SECTION 1: LOGICAL REASONING Dumps in PDF

Free LSAC LSAT SECTION 1: LOGICAL REASONING Real Questions (page: 16)

This morning, a bakery makes exactly one delivery, consisting of exactly six loaves of bread. Each of the loaves is exactly one of three kinds: oatmeal, rye, or wheat, and each is either sliced or unsliced. The loaves that the bakery delivers this morning must be consistent with the following:

Which one of the following statements must be true?

  1. At least one of the loaves is rye.
  2. At least one of the loaves is wheat.
  3. At least one of the loaves is sliced.
  4. No more than four oatmeal loaves are sliced.
  5. No more than four wheat loaves are sliced.

Answer(s): D

Explanation:

This is another one that must be checked choice by choice, but you can (and should!) do so strategically. The best place to start is with the correct choice for Acceptability question; 1
There we saw that we could have all unsliced loaves, and no wheat loaves, which proves that neither option [At least one of the loaves is wheat.] nor option [At least one of the loaves is sliced.] must be true.



This morning, a bakery makes exactly one delivery, consisting of exactly six loaves of bread. Each of the loaves is exactly one of three kinds: oatmeal, rye, or wheat, and each is either sliced or unsliced. The loaves that the bakery delivers this morning must be consistent with the following:

If the bakery delivers exactly four wheat loaves, then the bakery could also deliver

  1. one sliced rye loaf and one unsliced rye loaf
  2. one sliced oatmeal loaf and one unsliced oatmeal loaf
  3. two unsliced rye loaves
  4. two unsliced oatmeal loaves
  5. two sliced oatmeal loaves

Answer(s): B

Explanation:

If four wheat loaves are delivered, then those wheat loaves are sliced (Rule 3). One unsliced oatmeal loaf is always included, and that kills options [one sliced rye loaf and one unsliced rye loaf]., [two unsliced rye loaves], and [two sliced oatmeal loaves]. So what could the sixth loaf be? It can't be an unsliced oatmeal, since that would violate Rule 5. That kills option [two unsliced oatmeal loaves].



The six messages on an answering machine were each left by one of Fleure, Greta, Hildy, Liam, Pasquale, or Theodore, consistent with the following:

At most one person left more than one message.
No person left more than three messages.
If the first message is Hildy's, the last is Pasquale's.
If Greta left any message, Fleure and Pasquale did also.
If Fleure left any message, Pasquale and Theodore did also, all of Pasquale's preceding any of Theodore's. If Pasquale left any message, Hildy and Liam did also, all of Hildy's preceding any of Liam's.

Which one of the following could be a complete and accurate list of the messages left on the answering machine, from first to last?

  1. Fleure's, Pasquale's, Theodore's, Hildy's,Pasquale's, Liam's
  2. Greta's, Pasquale's, Theodore's, Theodore's,Hildy's, Liam's
  3. Hildy's, Hildy's, Hildy's, Liam's, Pasquale's,Theodore's
  4. Pasquale's, Hildy's, Fleure's, Liam's,Theodore's, Theodore's
  5. Pasquale's, Hildy's, Theodore's, Hildy's, Liam's,Liam's

Answer(s): D

Explanation:

This game wasn't easy, but this was the easiest question. Take the rules and use them to eliminate choices.
Rule 1 kills E.. Rule 2 doesn't help, but Rule 3 eliminates [Hildy's, Hildy's, Hildy's, Liam's, Pasquale's,Theodore's]. Rule 4 axes [Greta's, Pasquale's, Theodore's, Theodore's,Hildy's, Liam's]., Rule 5 kills [Fleure's, Pasquale's, Theodore's, Hildy's,Pasquale's, Liam's], and we're down to [Pasquale's, Hildy's, Fleure's, Liam's,Theodore's, Theodore's], the correct answer.



The six messages on an answering machine were each left by one of Fleure, Greta, Hildy, Liam, Pasquale, or Theodore, consistent with the following:

At most one person left more than one message.
No person left more than three messages.
If the first message is Hildy's, the last is Pasquale's.
If Greta left any message, Fleure and Pasquale did also.
If Fleure left any message, Pasquale and Theodore did also, all of Pasquale's preceding any of Theodore's. If Pasquale left any message, Hildy and Liam did also, all of Hildy's preceding any of Liam's.

The first and last messages on the answering machine could be the first and second messages left by which one of the following?

  1. Fleure
  2. Hildy
  3. Liam
  4. Pasquale
  5. Theodore

Answer(s): A



The six messages on an answering machine were each left by one of Fleure, Greta, Hildy, Liam, Pasquale, or Theodore, consistent with the following:

At most one person left more than one message.
No person left more than three messages.
If the first message is Hildy's, the last is Pasquale's.
If Greta left any message, Fleure and Pasquale did also.
If Fleure left any message, Pasquale and Theodore did also, all of Pasquale's preceding any of Theodore's. If Pasquale left any message, Hildy and Liam did also, all of Hildy's preceding any of Liam's.

If Greta left the fifth message, then which one of the following messages CANNOT have been left by Theodore?

  1. the first message
  2. the second message
  3. the third message
  4. the fourth message
  5. the sixth message

Answer(s): A

Explanation:

G leaving the fifth message triggers Rule 4, so we have to hear from F and P, which in turn means that we have to hear from T, and all P's must precede all T's (Rule 5). So we can't put a T in 1, since there would be no way to get a P message before it.



The six messages on an answering machine were each left by one of Fleure, Greta, Hildy, Liam, Pasquale, or Theodore, consistent with the following:

At most one person left more than one message.
No person left more than three messages.
If the first message is Hildy's, the last is Pasquale's.
If Greta left any message, Fleure and Pasquale did also.
If Fleure left any message, Pasquale and Theodore did also, all of Pasquale's preceding any of Theodore's. If Pasquale left any message, Hildy and Liam did also, all of Hildy's preceding any of Liam's

Each of the following must be true EXCEPT:

  1. Liam left at least one message.
  2. Theodore left at least one message.
  3. Hildy left at least one message.
  4. Exactly one person left at least two messages.
  5. At least four people left messages.

Answer(s): D

Explanation:

If you made the Key Deductions we described above (and few did), then you could have quickly eliminated options [Liam left..], [Theodore left...] and [Hildy left...], [At least four...] basically a combination of Rules 1 and 2, is another statement you could have deduced up front. So option [Exactly one person left at least two messages.] is the winner. We could hear once from each of the six people, in which case we'd have no repeat callers.



The six messages on an answering machine were each left by one of Fleure, Greta, Hildy, Liam, Pasquale, or Theodore, consistent with the following:

At most one person left more than one message.
No person left more than three messages.
If the first message is Hildy's, the last is Pasquale's.
If Greta left any message, Fleure and Pasquale did also.
If Fleure left any message, Pasquale and Theodore did also, all of Pasquale's preceding any of Theodore's. If Pasquale left any message, Hildy and Liam did also, all of Hildy's preceding any of Liam's

If the only message Pasquale left is the fifth message, then which one of the following could be true?

  1. Hildy left the first message.
  2. Theodore left exactly two messages.
  3. Liam left exactly two messages.
  4. Liam left the second message.
  5. Fleure left the third and fourth messages.

Answer(s): C

Explanation:

A P in 5 means we'll need to hear from H and L, with all the H's before all the L's. Since P's only message is fifth, we can't have an H in 1, since that would force a P in 6. That kills option [Hildy left the first message.]., and after a little more thought, it kills options [Liam left the second message.] and [Fleure left the third and fourth messages.] as well. If an L is second, then an H is first, and we have the same problem. So [Liam left the second message.] is out. Similarly, if F is third and fourth, then we need a T after we hear from P (Rule 5). So we're looking at this scenario: __ __ F F P T.
But we still need to hear from H and L, in that order, and the only available positions are the first and second positions. We still can't have H in 1, and so [Fleure left the third and fourth messages.] is impossible. Once you have it down to options [Theodore left exactly two messages.] and [Liam left exactly two messages], you only have to check one, although you could guess if you were running short of time. Let's look at option [Theodore left exactly two messages.] first. If we hear from T twice, then we can't put all the T's after all the P's, and so we can't hear from either F or G (contrapositives of Rules 4 and 5). But now we don't have enough messages! T leaves exactly 2, and P,H, and L leave exactly 1. That's only 5 messages, soB.is impossible. So option [Liam left exactly two messages.] is correct. If L left two messages, the sequence F H L L P T works fine.



Exactly five cars ­ Frank's, Marquitta's, Orlando's, Taishah's, and Vinquetta's ­ are washed, each exactly once. The cars are washed one at a time, with each receiving exactly one kind of wash: regular, super, or premium.
The following conditions must apply:

The first car washed does not receive a super wash, though at least one car does. Exactly one car receives a premium wash. The second and third cars washed receive the same kind of wash as each other.
Neither Orlando's nor Taishah's is washed before Vinquetta's.
Marquitta's is washed before Frank's, but after Orlando's. Marquitta's and the car washed immediately before Marquitta's receive regular washes.

Which one of the following could be an accurate list of the cars in the order in which they are washed, matched with type of wash received?

  1. Orlando's: premium; Vinquetta's: regular; Taishah's: regular; Marquitta's: regular; Frank's: super
  2. Vinquetta's: premium; Orlando's: regular; Taishah's: regular; Marquitta's: regular; Frank's: super
  3. Vinquetta's: regular; Marquitta's: regular; Taishah's: regular; Orlando's: super; Frank's: premium
  4. Vinquetta's: super; Orlando's: regular; Marquitta's: regular; Frank's: regular; Taishah's: super
  5. Vinquetta's: premium; Orlando's: regular; Marquitta's: regular; Frank's: regular; Taishah's: regular

Answer(s): B

Explanation:

A standard Acceptability question; we can use either the options listed above or the traditional Method for Acceptability questions. Since we'll be using the options in the rest of the questions, let's use the usual Training Method here for practice. We simply match the rules against the choices, eliminating those that don't conform.
Rule 1 kills [Vinquetta's: super; Orlando's: regular; Marquitta's: regular; Frank's: regular; Taishah's: super]., which gives the first car a super wash, and option [Vinquetta's: premium; Orlando's: regular; Marquitta's:
regular; Frank's: regular; Taishah's: regular], which gives no car a super wash. Option [Vinquetta's: super; Orlando's: regular; Marquitta's: regular; Frank's: regular; Taishah's: super] violates Rule 2 as well, but no other choice conflicts with that one. All the choices conform to Rule 3, but option [Orlando's: premium; Vinquetta's:
regular; Taishah's: regular; Marquitta's: regular; Frank's: super] disobeys Rule 4 by placing O before V. And C.
places M before O, in defiance of Rule 5. Option [Vinquetta's: premium; Orlando's: regular; Taishah's: regular; Marquitta's: regular; Frank's: super] remains and is correct--and, incidentally, corresponds to Option 2 above.



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