Draw three circles(three charities) overlapping each other. Each circle has 8 members in total of which some of them overlap with other circles(some may be in all three some in just 2 etc...)
Case 1: The area common to all three circles(1 area)will have 4 (since 4 of them serve all three charities)
Case 2: The area thats common to each of the 2 circles (3 areas in total) will have 1 member(since its given each pair of charities have 5 memebers in common. The reason its 1 because 4 already overlap between each pair since they occur in Case 1)
Case 3: Group1 exlcusive, Group2 exlcusive and Group3 exlcusive will be 2 each.
2+2+2+4+1+1+1 = 13
This a three sets problem. Hope this helps!
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Sets Problem
Source: Beat The GMAT — Problem Solving |
Draw three circles(three charities) overlapping each other. Each circle has 8 members in total of which some of them overlap with other circles(some may be in all three some in just 2 etc...)
Case 1: The area common to all three circles(1 area)will have 4 (since 4 of them serve all three charities)
Case 2: The area thats common to each of the 2 circles (3 areas in total) will have 1 member(since its given each pair of charities have 5 memebers in common) each. The reason its 1 because 4 already overlap between each pair since they occur in Case 1
Case 3: Group1 exlcusive, Group2 exlcusive and Group3 exlcusive will be 2 each.
2+2+2+4+1+1+1 = 13
This a three sets problem. Hope this helps!
Case 1: The area common to all three circles(1 area)will have 4 (since 4 of them serve all three charities)
Case 2: The area thats common to each of the 2 circles (3 areas in total) will have 1 member(since its given each pair of charities have 5 memebers in common) each. The reason its 1 because 4 already overlap between each pair since they occur in Case 1
Case 3: Group1 exlcusive, Group2 exlcusive and Group3 exlcusive will be 2 each.
2+2+2+4+1+1+1 = 13
This a three sets problem. Hope this helps!
awesome..I had come across this problem in Kaplan I think. They have given a very crude solution to this. But I like your solution better. Helped me too. Thanks
Its a 3 entity set problem.
where Total (distinct) = A + B + C - (A and B) - (B and C) - (A and C) + (A and B and C)
so, A = B = C = 8 (each has 8 members)
(A and B) + (B and C) + (A and C) = 5 * 3 = 15
( each has 5 common with one of the other)
Common among all 3 = 4
so, Total (distinct) = 8 + 8 + 8 - 15 + 4 => 13
where Total (distinct) = A + B + C - (A and B) - (B and C) - (A and C) + (A and B and C)
so, A = B = C = 8 (each has 8 members)
(A and B) + (B and C) + (A and C) = 5 * 3 = 15
( each has 5 common with one of the other)
Common among all 3 = 4
so, Total (distinct) = 8 + 8 + 8 - 15 + 4 => 13
