LSAT Logic Games : Solving grouping games
Study concepts, example questions & explanations for LSAT Logic Games
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Example Questions
Example Question #667 : Lsat Logic Games
A corporate conference is held with three breakout sessions: motivational interviewing (MI), sales optimizing (SO), and research and development (RD). Each of exactly five persons from one regional office---Nora, Oscar, Paul, Tim, and Vick---attend exactly one breakout session. The following conditions must apply:
Nora and Oscar do not attend the same breakout session.
Exactly two persons from the above-noted regional office attend the sales optimizing (SO) session.
Tim and Paul do not attend the same breakout session.
Neither Nora nor Oscar attend a session with Paul.
If Nora or Vick attend the MI session, then the other also attends that same session.
If Vick attends the SO session, then each of the following could be true EXCEPT:
Tim attends the SO session.
Nora attends the SO session
Nora attends the RD session.
Oscar attends the MI session.
Tim attends the MI session.
Tim attends the SO session.
Tim cannot attend the SO session because the SO session only has two slots, and one slot is taken by Vick (by virtue of the question itself), and the other slot must be taken by Nora, Oscar, or Paul. The key deduction in this problem is the fact that Nora, Oscar, or Paul must occupy a single slot in each of the sessions, by virtue of the fact that none of them can attend the same session together.
Example Question #161 : Solving Grouping Games
A corporate conference is held with three breakout sessions: motivational interviewing (MI), sales optimizing (SO), and research and development (RD). Each of exactly five persons from one regional office---Nora, Oscar, Paul, Tim, and Vick---attend exactly one breakout session. The following conditions must apply:
Nora and Oscar do not attend the same breakout session.
Exactly two persons from the above-noted regional office attend the sales optimizing (SO) session.
Tim and Paul do not attend the same breakout session.
Neither Nora nor Oscar attend a session with Paul.
If Nora or Vick attend the MI session, then the other also attends that same session.
Which one of the folloiwng is a pair of persons who could attend the MI session together?
Paul and Tim
Oscar and Tim.
Oscar and Vick
Nora and Tim
Paul and Oscar
Oscar and Tim.
If Nora or Vick are in the MI slot, then both must take up the only two MI slots. That fact eliminates Nora and Tim, and Oscar and Vick, as correct responses. The other two incorrect responses---"Paul and Oscar" and "Paul and Tim"---violate explicit rules. That leaves "Oscar and Tim" as the only viable response.
Example Question #162 : Solving Grouping Games
A corporate conference is held with three breakout sessions: motivational interviewing (MI), sales optimizing (SO), and research and development (RD). Each of exactly five persons from one regional office---Nora, Oscar, Paul, Tim, and Vick---attend exactly one breakout session. The following conditions must apply:
Nora and Oscar do not attend the same breakout session.
Exactly two persons from the above-noted regional office attend the sales optimizing (SO) session.
Tim and Paul do not attend the same breakout session.
Neither Nora nor Oscar attend a session with Paul.
If Nora or Vick attend the MI session, then the other also attends that same session.
Each of the following statments must be false EXCEPT:
Tim and Vick go to the RD session together.
Only Paul goes to the RD session.
Only Tim goes to the RD session.
Only Vick goes to the RD session.
Only Oscar goes to the RD session.
Only Paul goes to the RD session.
Nora, Oscar, or Paul must be included in a slot for the RD session, so three of the responses can be eliminated on that basis alone. If Oscar attends the RD session alone, then Paul and Tim are forced together, in violation of an explicit rule. Thus, it could be true that Paul goes alone to the RD session.
Example Question #163 : Solving Grouping Games
A corporate conference is held with three breakout sessions: motivational interviewing (MI), sales optimizing (SO), and research and development (RD). Each of exactly five persons from one regional office---Nora, Oscar, Paul, Tim, and Vick---attend exactly one breakout session. The following conditions must apply:
Nora and Oscar do not attend the same breakout session.
Exactly two persons from the above-noted regional office attend the sales optimizing (SO) session.
Tim and Paul do not attend the same breakout session.
Neither Nora nor Oscar attend a session with Paul.
If Nora or Vick attend the MI session, then the other also attends that same session.
If Nora attends the SO session, then which one of the following is a complete and accurate list of the persons any one of whom could attend the MI session?
Oscar, Tim
Paul, Vick
Oscar
Paul, Tim, Vick
Oscar, Paul, Tim
Oscar, Paul, Tim
With Nora attending the SO session, Oscar and Paul are eligible for the MI session (and one of them must attend). The credited response is the only one that allows for either Oscar or Paul to attend the MI session.
Example Question #164 : Grouping Games
A corporate conference is held with three breakout sessions: motivational interviewing (MI), sales optimizing (SO), and research and development (RD). Each of exactly five persons from one regional office---Nora, Oscar, Paul, Tim, and Vick---attend exactly one breakout session. The following conditions must apply:
Nora and Oscar do not attend the same breakout session.
Exactly two persons from the above-noted regional office attend the sales optimizing (SO) session.
Tim and Paul do not attend the same breakout session.
Neither Nora nor Oscar attend a session with Paul.
If Nora or Vick attend the MI session, then the other also attends that same session.
If the condition that exactly two persons from the regional office attend the SO session is changed to require that exactly three persons attend the SO session, but all other conditions remain the same, then which one of the following persons must attend a session other than the SO session?
Nora
Tim
Vick
Paul
Oscar
Paul
We now know that one person attends MI and one person attends RD. Three persons, by virtue of this problem, attend SO. Because Nora, Oscar, and Paul must occupy slots across the three sessions, Vick and Tim must be the persons attending the SO session, along with one other person. But Paul cannot be that other person because Paul cannot attend a session with Tim.
Example Question #165 : Grouping Games
A zoo has four exhibition regions---labelled R, T, V, and W---and each region consists of a vegetarian wing and a carnivore wing. The following conditions apply:
Each exhibition wing houses only mammals or amphibians.
Three of the wings have mammals.
R's vegetarian wing and T's vegetarian wing house amphibians.
If an exhibition region has mammals in one of its wings, then the other wing in that exhibition region has amphibians.
If mammals are housed in V's carnivore wing, then W's vegetarian wing has mammals, too.
If amphibians are assigned to V's carnivore wing and V's vegetarian wing, then which one of the following could be two other wings that also house amphibians?
T's carnivore wing and W's carnivore wing.
R's carnivore wing and W's carnivore wing.
R's vegetarian wing and T's carnivore wing.
T's vegetarian wing and W's carnivore wing.
R's carnivore wing and T's vegetarian wing.
T's vegetarian wing and W's carnivore wing.
The key insight needed to answer this question is that only one region can be exclusively female. This insight can be derived from the fact that only three wings house mammals. That means five wings must house amphibians, which necessitates one region to be exclusively amphibian. The problem establishes that region V is exclusively amphibian. Therefore, region W must house amphibians in one of its wings. Only the credited response has amphibians in region W while also not forcing another region to be exclusively amphibian.
Example Question #164 : Solving Grouping Games
A zoo has four exhibition regions---labelled R, T, V, and W---and each region consists of a vegetarian wing and a carnivore wing. The following conditions apply:
Each exhibition wing houses only mammals or amphibians.
Three of the wings have mammals.
R's vegetarian wing and T's vegetarian wing house amphibians.
If an exhibition region has mammals in one of its wings, then the other wing in that exhibition region has amphibians.
If mammals are housed in V's carnivore wing, then W's vegetarian wing has mammals, too.
It CANNOT be true that amphibians are assigned to both:
R's carnivore wing and W's carnivore wing.
T's carnivore wing and W's carnivore wing.
R's carnivore wing and V's vegetarian wing.
R's carnivore wing and T's carnivore wing.
V's vegetarian wing and W's carnivore wing.
R's carnivore wing and T's carnivore wing.
Because only one region can be exclusively amphibian, the credited response forces two regions to be exclusively amphibian---in this instance, amphibians being housed in R's carnivore wing and T's carnivore wing.
Example Question #165 : Solving Grouping Games
A zoo has four exhibition regions---labelled R, T, V, and W---and each region consists of a vegetarian wing and a carnivore wing. The following conditions apply:
Each exhibition wing houses only mammals or amphibians.
Three of the wings have mammals.
R's vegetarian wing and T's vegetarian wing house amphibians.
If an exhibition region has mammals in one of its wings, then the other wing in that exhibition region has amphibians.
If mammals are housed in V's carnivore wing, then W's vegetarian wing has mammals, too.
If W's vegetarian wing houses amphibians, then amphibians must also be housed in which one of the following?
V's vegetarian wing
V's carnivore wing.
R's carnivore wing
T's carnivore wing
W's carnivore wing
V's carnivore wing.
The last rule states that if mammals are housed in V's carnivore wing, then W's vegetarian wing houses mammals. The contrapositive thus holds that if W's vegetarian wing houses amphibians, then V's carnivore wing houses amphibians. That contrapositive yields the correct response here.
Example Question #168 : Grouping Games
A zoo has four exhibition regions---labelled R, T, V, and W---and each region consists of a vegetarian wing and a carnivore wing. The following conditions apply:
Each exhibition wing houses only mammals or amphibians.
Three of the wings have mammals.
R's vegetarian wing and T's vegetarian wing house amphibians.
If an exhibition region has mammals in one of its wings, then the other wing in that exhibition region has amphibians.
If mammals are housed in V's carnivore wing, then W's vegetarian wing has mammals, too.
If mammals are housed in V's carnivore wing, which one of the following is a complete and accurate list of the wings that CANNOT house mammals?
R's vegetarian wing, T's vegetarian wing
R's vegetarian wing, R's carnivore wing, T's vegetarian wing, V's vegetarian wing, W's carnivore wing
R's vegetarian wing, T's vegetarian wing, V's vegetarian wing
R's vegetarian wing, T's vegetarian wing, V's vegetarian wing, W's carnivore wing.
R's vegetarian wing, T's vegetarian wing, W's carnivore wing
R's vegetarian wing, T's vegetarian wing, V's vegetarian wing, W's carnivore wing.
R and T's vegetarian wings are already designated locations for amphibians, so they are included in any correct response. Because mammals cannot occupy both wings, V's vegetarian wing cannot house mammals, since V's carnivore wing houses mammals. A rule dictates that if mammals are housed in V's carnivore wing, then W's vegetarian wing must house mammals. Therefore, mammals cannot be housed in V's vegetarian wing and W's carnivore wing, in addition to the wings already noted. Hence, the credited response is: R's vegetarian wing, T's vegetarian wing, V's vegetarian wing, W's carnivore wing.
Example Question #166 : Solving Grouping Games
A zoo has four exhibition regions---labelled R, T, V, and W---and each region consists of a vegetarian wing and a carnivore wing. The following conditions apply:
Each exhibition wing houses only mammals or amphibians.
Three of the wings have mammals.
R's vegetarian wing and T's vegetarian wing house amphibians.
If an exhibition region has mammals in one of its wings, then the other wing in that exhibition region has amphibians.
If mammals are housed in V's carnivore wing, then W's vegetarian wing has mammals, too.
If T's carnivore wing houses amphibians, then it could be true that amphibians are housed in both:
R's carnivore wing and W's vegetarian wing
V's carnivore wing and W's carnivore wing
V's vegetarian wing and W's vegetarian wing
R's carnivore wing and W's carnivore wing
V's vegetarian wing and V's carnivore wing
V's carnivore wing and W's carnivore wing
Notice that with amphibians housed in T's carnivore wing, all the other wings will have split between mammals and amphibians (since only one region can have exclusively amphibians). That fact eliminates the option: V's vegetarian wing and V's carnivore wing. Notice further that mammals must be housed in R's carnivore wing, which eliminates two more of the options. V's vegetarian wing and W's vegetarian wing cannot both house amphibians, because that would lead to an additional region being exclusively amphibian, which cannot happen. Thus, the credited response is the only one remaining as viable.
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