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Rahul_Sharma23
01 Jul 2025, 10:35
Kudos
Bookmarks
Abstract and Baroque have to in same Gallery to separate Cubist and Expressionist
Case 1: Gallery C has Folk and Cubist artwork
Case 2: Gallery C has Folk only
Expressionist or Cubist
Case 1: Gallery C has Folk and Cubist artwork
| Gallery A | Gallery B | Gallery C |
| Expressionist or Digital | Abstract | Folk |
| Baroque | Cubist | |
| Expressionist or Digital |
Case 2: Gallery C has Folk only
Expressionist or Cubist
| Gallery A | Gallery B | Gallery C |
| Digital | Abstract | Folk |
| Baroque | ||
| Expressionist or Cubist |
Bunuel
01 Jul 2025, 10:43
Kudos
Bookmarks
First to avoid confusion, Lets call Gallery I, Gallery II and Gallery III.
Types of artwork are A,B,C,D,E and F.
By first rule. We have two possibilities.
I (1) II (3) III (2) or I (2) II (3) III (1)
With number in bracket denoting no of artwork type exhibited.
If we integrate rule 2 into rule 1, First possibility has only one outcome, A,B are in gallery II. Also F is in Gallery III. (from rule 4)
But in Second possibility, two more options arise, A,B is in Gallery II or Gallery I. and Gallery III only has F in both cases. (again from rule 4). So before getting into Rule 3 we have 3 distinct scenarios.
I (1) II (3) III (2) | I(2) II(3) III (1) | I(2) II(3) III (1)
A,B F A,B F A,B F
Now if we fit rule 3 here. Third scenario collapses as all Remaining C,D and E should come in Gallery II which is against Rule 3.
I (1) II (3) III (2) | I(2) II(3) III (1)
A,B F A,B F
But if we take second scenario, Gallery I should have either C,D or E,D (from rule 3). But there is no such option for Gallery I given. So scenario II collapses. Now only one scenario.
I (1) II (3) III (2)
A,B F
If D goes in I it will be lone exhibit. But no such option. So D goes in either II or III. First option is eliminated already as 3 exhibits are in Gallery II.
And there is no option for Gallery III with two exhibits, except, F and C.
So choose that for Gallery III. And since no D only option for Gallery I, Choose E only option for Gallery I.
Types of artwork are A,B,C,D,E and F.
By first rule. We have two possibilities.
I (1) II (3) III (2) or I (2) II (3) III (1)
With number in bracket denoting no of artwork type exhibited.
If we integrate rule 2 into rule 1, First possibility has only one outcome, A,B are in gallery II. Also F is in Gallery III. (from rule 4)
But in Second possibility, two more options arise, A,B is in Gallery II or Gallery I. and Gallery III only has F in both cases. (again from rule 4). So before getting into Rule 3 we have 3 distinct scenarios.
I (1) II (3) III (2) | I(2) II(3) III (1) | I(2) II(3) III (1)
A,B F A,B F A,B F
Now if we fit rule 3 here. Third scenario collapses as all Remaining C,D and E should come in Gallery II which is against Rule 3.
I (1) II (3) III (2) | I(2) II(3) III (1)
A,B F A,B F
But if we take second scenario, Gallery I should have either C,D or E,D (from rule 3). But there is no such option for Gallery I given. So scenario II collapses. Now only one scenario.
I (1) II (3) III (2)
A,B F
If D goes in I it will be lone exhibit. But no such option. So D goes in either II or III. First option is eliminated already as 3 exhibits are in Gallery II.
And there is no option for Gallery III with two exhibits, except, F and C.
So choose that for Gallery III. And since no D only option for Gallery I, Choose E only option for Gallery I.
Bunuel
01 Jul 2025, 10:44
Kudos
Bookmarks
Let GA, GB, Gc be the respective gallery
Let A,B,C,D,E,F be respective artwork
we know A, B in same category and not GC
let's put A,B in GA, then GB->C,D,E which is not possible a/c to rule 3
Hence A,B should be in GB
Using POE(Process Of Elimination)
we know GC has F
opt1: if GC has C,F,D not possible as GB will also have A,B,E and GA will be empty and no gallery should be empty
opt2: if GC has C,F then GA can have E/D which is given in opt5
hence GA->E; GB->A,B,D; GC->C,F
Let A,B,C,D,E,F be respective artwork
we know A, B in same category and not GC
let's put A,B in GA, then GB->C,D,E which is not possible a/c to rule 3
Hence A,B should be in GB
Using POE(Process Of Elimination)
we know GC has F
opt1: if GC has C,F,D not possible as GB will also have A,B,E and GA will be empty and no gallery should be empty
opt2: if GC has C,F then GA can have E/D which is given in opt5
hence GA->E; GB->A,B,D; GC->C,F
Bunuel
