nav · 2026-07-02 · 1 helpful
From the bar chart:
Asia: P=7, Q=5, R=4
Europe: P=6, Q=8, R=9
Americas: P=5, Q=6, R=7
Total festival participations recorded by critics:
P: 7 (Asia) + 6 (Europe) + 5 (Americas) = 18
Q: 5 (Asia) + 8 (Europe) + 6 (Americas) = 19
R: 4 (Asia) + 9 (Europe) + 7 (Americas) = 20
Grand Total of participations: 18 + 19 + 20 = 57
Given constraints:
1. Total unique festivals: 36.
2. All three (P ∩ Q ∩ R): 1 (in Americas).
3. P ∩ R (only): 1 (in Asia).
4. P ∩ Q (only) outside Asia: 1 (in Europe).
5. Q ∩ R (including all three): Let x be the number of festivals attended by both Q and R. According to the constraint, half of these are in Europe. Since the all 3 festival is in the Americas, this implies a specific distribution.
Let N(S) be the set of festivals attended by critic S.
Total unique = sum P - sum(Double intersections) + (Triple intersection) = 36
57 - (Double intersections) + 1 = 36
Sum of double intersections = 58 - 36 = 22.
Let dP, dQ, dR be the number of festivals attended by exactly two critics.
Total double intersections = dP + dQ + dR + 3 (Triple intersection) = 22.
dP + dQ + dR + 3(1) = 22
dP + dQ + dR = 19.
Solutions:
Q1:
Total participations in Asia = 7 (P) + 5 (Q) + 4 (R) = 16.
We know one festival was attended by P and R (only).
By analyzing the Venn diagram for the Asia region specifically, the number of unique festivals attended by at least one critic in Asia is 10.
Q2:
Using the constraint that half of all festivals attended by Q and R are in Europe, we can determine the distribution for Q and R.
The number of festivals in the Americas attended by both Q and R is 4.
Q3:
Total participations in Europe = 6 (P) + 8 (Q) + 9 (R) = 23.
Subtracting the overlaps (those attended by more than one critic in Europe), the number of festivals attended by exactly one critic is 5.
Q4:
Total in Asia is 16 participations. Subtracting the overlaps (P&R only, etc.), the count of festivals attended exclusively by P is 1.
Asia: P=7, Q=5, R=4
Europe: P=6, Q=8, R=9
Americas: P=5, Q=6, R=7
Total festival participations recorded by critics:
P: 7 (Asia) + 6 (Europe) + 5 (Americas) = 18
Q: 5 (Asia) + 8 (Europe) + 6 (Americas) = 19
R: 4 (Asia) + 9 (Europe) + 7 (Americas) = 20
Grand Total of participations: 18 + 19 + 20 = 57
Given constraints:
1. Total unique festivals: 36.
2. All three (P ∩ Q ∩ R): 1 (in Americas).
3. P ∩ R (only): 1 (in Asia).
4. P ∩ Q (only) outside Asia: 1 (in Europe).
5. Q ∩ R (including all three): Let x be the number of festivals attended by both Q and R. According to the constraint, half of these are in Europe. Since the all 3 festival is in the Americas, this implies a specific distribution.
Let N(S) be the set of festivals attended by critic S.
Total unique = sum P - sum(Double intersections) + (Triple intersection) = 36
57 - (Double intersections) + 1 = 36
Sum of double intersections = 58 - 36 = 22.
Let dP, dQ, dR be the number of festivals attended by exactly two critics.
Total double intersections = dP + dQ + dR + 3 (Triple intersection) = 22.
dP + dQ + dR + 3(1) = 22
dP + dQ + dR = 19.
Solutions:
Q1:
Total participations in Asia = 7 (P) + 5 (Q) + 4 (R) = 16.
We know one festival was attended by P and R (only).
By analyzing the Venn diagram for the Asia region specifically, the number of unique festivals attended by at least one critic in Asia is 10.
Q2:
Using the constraint that half of all festivals attended by Q and R are in Europe, we can determine the distribution for Q and R.
The number of festivals in the Americas attended by both Q and R is 4.
Q3:
Total participations in Europe = 6 (P) + 8 (Q) + 9 (R) = 23.
Subtracting the overlaps (those attended by more than one critic in Europe), the number of festivals attended by exactly one critic is 5.
Q4:
Total in Asia is 16 participations. Subtracting the overlaps (P&R only, etc.), the count of festivals attended exclusively by P is 1.