Here are some tips for solving Venn diagram reasoning questions: Start with intersection: When solving problems with Venn diagrams, always start with the intersection. Set up a two set Venn diagram: Put a box around two sets to show the universal set. Enter numbers: To solve a Venn diagram with three circles, enter the number of items in common to all three sets, then enter the remaining number of items in the overlapping region of each pair of sets. Finally, enter the remaining number of items in each individual set. Use known totals: Use any known totals to find missing numbers. Use... Show more Here are some tips for solving Venn diagram reasoning questions: Start with intersection: When solving problems with Venn diagrams, always start with the intersection. Set up a two set Venn diagram: Put a box around two sets to show the universal set. Enter numbers: To solve a Venn diagram with three circles, enter the number of items in common to all three sets, then enter the remaining number of items in the overlapping region of each pair of sets. Finally, enter the remaining number of items in each individual set. Use known totals: Use any known totals to find missing numbers. Use formulas: Here are some formulas that can help: n(A ∪ B) = n(A) + n(B) – n(A ∩ B) n(A ∩ B) = Φ then n(A ∪ B) = n(A) + n(B) n (A ∪ B ∪ C) = n (A) + n(B) + n(C) – n (A ∩ B) – n (A ∩ C) – n (B ∩ C) Show less
Here are some tips for solving Venn diagram reasoning questions: Start with intersection: When solving problems with Venn diagrams, always start with the intersection. Set up a two set Venn diagram: Put a box around two sets to show the universal set. Enter numbers: To solve a Venn diagram with three circles, enter the number of items in common to all three sets, then enter the remaining number of items in the overlapping region of each pair of sets. Finally, enter the remaining number of items in each individual set. Use known totals: Use any known totals to find missing numbers.
Use formulas: Here are some formulas that can help: n(A ∪ B) = n(A) + n(B) – n(A ∩ B) n(A ∩ B) = Φ then n(A ∪ B) = n(A) + n(B) n (A ∪ B ∪ C) = n (A) + n(B) + n(C) – n (A ∩ B) – n (A ∩ C) – n (B ∩ C)
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