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01 Jul 2025, 08:00
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Gallery A: Expressionist Only
Gallery C: Cubist and Folk only
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
73% (03:11) correct
27%
(03:12)
wrong
based on 586
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A national art exhibition features six different types of artwork: Abstract, Baroque, Cubist, Digital, Expressionist, and Folk. These artworks are divided among three galleries: Gallery A, Gallery B, and Gallery C. Each artwork type is displayed in exactly one gallery, and no gallery is empty.
The following rules apply:
- Exactly three types of artwork are showcased in Gallery B.
- The Baroque and Abstract artworks are displayed together in the same gallery, but not in Gallery C.
- The Cubist artwork is displayed in a different gallery from the Expressionist artwork.
- The Folk artwork is displayed in Gallery C.
Select for Gallery A the set of artwork types displayed in Gallery A, and select for Gallery C the set of artwork types displayed in Gallery C, such that the two selections, from among the given options, would be jointly consistent with the information provided. Make only two selections, one in each column.
| Gallery A | Gallery C | |
| Cubist, Folk, and Digital only | ||
| Cubist and Folk only | ||
| Cubist only | ||
| Baroque and Abstract only | ||
| Expressionist Only | ||
| Folk Only |
ShowHide Answer
Official Answer
Gallery A: Expressionist Only
Gallery C: Cubist and Folk only
01 Jul 2025, 08:30
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Bunuel
GMAT Club Official Explanation:
Abstract, Baroque, Cubist, Digital, Expressionist, and Folk.
Since exactly three types of artwork are in Gallery B, and Baroque and Abstract are together but not in C, then Baroque and Abstract must be in B. If they were in A, then B would need to have the remaining three artworks — Cubist, Digital, and Expressionist — but we know that Cubist and Expressionist cannot be in the same gallery. So, we have the following distribution:
| Gallery A | Gallery B | Gallery C |
| Abstract | Folk | |
| Baroque | ||
| ? |
Now let’s check the options. We know that Folk is in C, so for Gallery C we can consider only the 1st (Cubist, Folk, and Digital only), 2nd (Cubist and Folk only), or 6th (Folk only) options.
The 1st option for Gallery C (Cubist, Folk, and Digital only) is not possible, because B already has three artworks. That would make 6 in total, leaving Gallery A empty — which violates the condition that no gallery is empty.
If we consider the 2nd option for Gallery C (Cubist and Folk only), then Digital and Expressionist remain. One of them can go to Gallery B (to make three artworks there), and the other to Gallery A. Checking the options, we see that the 5th option is Expressionist Only, which fits, while there is no option for Digital Only. So the following is a valid distribution:
| Gallery A | Gallery B | Gallery C |
| Expressionist | Abstract | Folk |
| Baroque | Cubist | |
| Digital |
Now, just to illustrate why the Folk Only option for Gallery C is not a correct answer:
If Gallery C has only Folk, then we are left with Cubist, Digital, and Expressionist. One of these must go to Gallery B (alongside Baroque and Abstract), and Cubist and Expressionist cannot be together. So, for Gallery A, we would need either:
- Cubist and Digital only — no such option
- Expressionist and Digital only — no such option
Thus, the valid allocation only the one shown above:
| Gallery A | Gallery B | Gallery C |
| Expressionist | Abstract | Folk |
| Baroque | Cubist | |
| Digital |
General Discussion
01 Jul 2025, 08:24
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Let GalA = Gallery A, GalB = Gallery B, and GalC = Gallery C
Let Abstract = A, Baroque = B, Cubist = C, Digital = D, Expressionist = E, and Folk = F
Gallery restrictions:
1) No gallery empty
2) GalB have 3 artworks.
Inference: Since we have 6 artworks, between GalA and GalB we will have 1 with only one artwork and the other with 2 artworks.
We know that F is located in GalC. We know that A & B are together, in GalA ou GalB.
Next inference: if A&B are in GalA, then GalC have only F.
We know that F are not in GalA, so we elimate all options with F from GalA. We know that F are in GalC, so we eliminate all options without F from GalC.
Next we know that C and E are not in the same Gallery. If A&B were together in GalA, we would only have F in GalC, and GalB would need to have C and E together, what it not allowed.
So, the only possible option with the constraints is:
Gallery A = Expressionist Only
Gallery C = Cubist and Folk only
(And, therefore, A, B, and D together in GalB).
Let Abstract = A, Baroque = B, Cubist = C, Digital = D, Expressionist = E, and Folk = F
Gallery restrictions:
1) No gallery empty
2) GalB have 3 artworks.
Inference: Since we have 6 artworks, between GalA and GalB we will have 1 with only one artwork and the other with 2 artworks.
We know that F is located in GalC. We know that A & B are together, in GalA ou GalB.
Next inference: if A&B are in GalA, then GalC have only F.
We know that F are not in GalA, so we elimate all options with F from GalA. We know that F are in GalC, so we eliminate all options without F from GalC.
Next we know that C and E are not in the same Gallery. If A&B were together in GalA, we would only have F in GalC, and GalB would need to have C and E together, what it not allowed.
So, the only possible option with the constraints is:
Gallery A = Expressionist Only
Gallery C = Cubist and Folk only
(And, therefore, A, B, and D together in GalB).
