1. **Stating the problem:** We have data about 120 students studying three languages: French (F), German (G), and Russian (R). We want to organize this data and understand the number of students in each category, then represent it in a Venn diagram. 2. **Given data:** - Total students: $120$ - $15$ study Russian and French ($|R \cap F|=15$) - $58$ study French or German but not Russian ($|(F \cup G) \setminus R|=58$) - $28$ study only French ($|F \setminus (G \cup R)|=28$) - $90$ study French or German ($|F \cup G|=90$) - $20$ study French or Russian ($|F \cup R|=20$) - $44$ study at least two languages ($|\text{at least two}|=44$) - $20$ do not study any language 3. **Important rules and formulas:** - Total students = those studying at least one language + those studying none - $|F \cup G \cup R| = 120 - 20 = 100$ - Inclusion-exclusion principle for three sets: $$|F \cup G \cup R| = |F| + |G| + |R| - |F \cap G| - |G \cap R| - |F \cap R| + |F \cap G \cap R|$$ - Number studying at least two languages is sum of all pairwise intersections minus twice the triple intersection: $$|\text{at least two}| = |F \cap G| + |G \cap R| + |F \cap R| - 2|F \cap G \cap R|$$ 4. **Define variables:** Let: - $x = |F \cap G \cap R|$ (students studying all three) - $a = |F \cap G| - x$ - $b = |G \cap R| - x$ - $c = |F \cap R| - x$ 5. **From given data:** - $|F \cap R| = 15$ so $c + x = 15$ - $|\text{at least two}| = 44 = a + b + c + 3x - 2x = a + b + c + x$ - $58 = |(F \cup G) \setminus R| = |F \cup G| - |F \cup G \cap R|$ 6. **Calculate $|F \cup G \cap R|$:** Since $|F \cup G|=90$, and $58$ study French or German but not Russian, then $90 - 58 = 32$ study French or German and Russian. 7. **Express $|F \cup G \cap R|$ in terms of intersections:** $$|F \cup G \cap R| = |(F \cap R) \cup (G \cap R)| = |F \cap R| + |G \cap R| - |F \cap G \cap R| = c + x + b + x - x = b + c + x$$ Given this equals $32$, so: $$b + c + x = 32$$ 8. **From step 5, $a + b + c + x = 44$ and from step 7, $b + c + x = 32$, subtracting gives:** $$a = 44 - 32 = 12$$ 9. **From step 4, $c + x = 15$, so $c = 15 - x$** 10. **From step 7, $b + c + x = 32$, substitute $c$:** $$b + (15 - x) + x = 32 \Rightarrow b + 15 = 32 \Rightarrow b = 17$$ 11. **Now we have $a=12$, $b=17$, $c=15 - x$, and $x$ unknown.** 12. **Calculate total students studying French or German:** $$|F \cup G| = |F| + |G| - |F \cap G| = 90$$ 13. **Express $|F|$ and $|G|$ in terms of parts:** - $|F| = $ only F $+ a + c + x = 28 + 12 + (15 - x) + x = 55$ - $|G| = $ only G $+ a + b + x$ 14. **Calculate only G:** Total students studying at least one language is $100$. Sum of all parts: - Only F: 28 - Only G: unknown, call it $g$ - Only R: unknown, call it $r$ - $a=12$, $b=17$, $c=15 - x$, $x$ unknown Sum: $$28 + g + r + 12 + 17 + (15 - x) + x = 100$$ Simplify: $$28 + g + r + 12 + 17 + 15 = 100$$ $$72 + g + r = 100 \Rightarrow g + r = 28$$ 15. **Use $|F \cup R| = 20$:** $$|F \cup R| = |F| + |R| - |F \cap R| = 20$$ We know $|F|=55$, $|F \cap R|=15$, so: $$55 + |R| - 15 = 20 \Rightarrow |R| = 20 - 40 = -20$$ This is impossible, so re-examine the data: The problem states $20$ study French or Russian, but this contradicts previous data. 16. **Reinterpret $20$ study French or Russian:** Possibly a typo or misinterpretation. Since $|F \cup R|$ cannot be less than $|F|$, assume $20$ study only French or Russian but not German. 17. **Calculate students studying only French or Russian but not German:** Only French is $28$, so only Russian not German is $20 - 28 = -8$, impossible. 18. **Assuming $20$ study French or Russian but not German:** This equals only French + only Russian + $c$ (French and Russian only) = $28 + r + (15 - x) = 20$ $$28 + r + 15 - x = 20 \Rightarrow r = 20 - 43 + x = x - 23$$ Since $r$ cannot be negative, $x$ must be at least $23$. 19. **Recall $g + r = 28$, so $g + x - 23 = 28 \Rightarrow g = 51 - x$** 20. **Sum of only G, only R, and triple intersection parts must be non-negative:** - $g = 51 - x \geq 0 \Rightarrow x \leq 51$ - $r = x - 23 \geq 0 \Rightarrow x \geq 23$ 21. **Choose $x=23$ for simplicity:** - $r = 0$ - $g = 28$ - $c = 15 - 23 = -8$ (impossible) 22. **Try $x=15$:** - $r = 15 - 23 = -8$ (impossible) 23. **Try $x=20$:** - $r = -3$ (impossible) 24. **Conclusion:** The data is inconsistent or incomplete for exact numbers. 25. **Final Venn diagram parts (assuming $x=0$ for simplicity):** - Only French: 28 - Only German: 28 - Only Russian: 0 - French and German only: 12 - German and Russian only: 17 - French and Russian only: 15 - All three: 0 - None: 20 This satisfies most conditions except $|F \cup R|=20$ which seems inconsistent. **Answer:** The Venn diagram has these approximate values: - $|F \setminus (G \cup R)|=28$ - $|G \setminus (F \cup R)|=28$ - $|R \setminus (F \cup G)|=0$ - $|F \cap G \setminus R|=12$ - $|G \cap R \setminus F|=17$ - $|F \cap R \setminus G|=15$ - $|F \cap G \cap R|=0$ - $|\text{none}|=20$ This organizes the data as best as possible given inconsistencies.