The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. Practice Paper 2 Question 4

Three planar regions $$A$$, $$B$$, $$C$$ partially overlap each other, with $$|A| = 90,$$ $$|B| = 90,$$ $$|C| = 60$$ and $$|A \cup B \cup C| = 100,$$ where $$|\cdot|$$ denotes the area. Find the minimum possible $$|A \cap B \cap C|$$.

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Hints

• Hint 1
Try to find the minimum $$|A\cap B|.$$
• Hint 2
... by considering the maximum $$|A\cup B|.$$
• Hint 3
What does that minimal case imply about $$C$$ in relation to $$A$$ and $$B,$$ given the question conditions?
• Hint 4
... more specifically, in relation to $$A\cup B,$$ given $$|A \cup B \cup C| = 100?$$
• Hint 5
Given $$C$$ must be contained within $$A\cup B,$$ what does that imply about $$C$$ in relation to $$A\cap B?$$
• Hint 6
... more specifically, how must $$C$$ be distributed outside of $$A\cap B$$ in order to minimize the full intersection?

Solution

First consider the minimum size of $$A \cap B$$. We know that $$|A \cup B|=|A|+|B|-|A\cap B|$$ and since $$|A \cup B\cup C|=100$$ then $$|A \cup B|\le100$$ so $$|A \cap B|\ge80$$. In this minimal case, $$C$$ must be contained within $$A \cup B,$$ because otherwise $$|A \cup B \cup C| > 100$$. We want to have as much of $$C$$ as possible outside of $$A\cap B,$$ hence the minimal size of the intersection is $$|C| - (|A \cup B| - |A \cap B|)=40$$.

Note: Drawing standard (circular) Venn diagrams for $$A,B,C$$ to meet the above conditions is not possible, but it is possible using other shapes. Give it a try.

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