1. **Stating the problem:** We analyze the breakfast habits of 15 classmates divided into categories: no breakfast, only sweet, only savory, and both sweet and savory. 2. **Given data:** - 1 of 15 has no breakfast. - Ratio no breakfast : breakfast = 1 : 15 (smallest ratio). - Ratio both sweet and savory : only sweet = 6 : 8. - Of those who eat breakfast, 1 in 5 eats both sweet and savory. - 1 in 4 classmates eats only sweet. 3. **Interpreting the ratios and counts:** - Total classmates = 15. - No breakfast = 1 (given). - Breakfast eaters = 15 - 1 = 14. 4. **From the ratio both sweet and savory : only sweet = 6 : 8, simplify:** $$\frac{6}{8} = \frac{3}{4}$$ 5. **Let the number of both sweet and savory be $x$, then only sweet is $\frac{4}{3}x$.** 6. **From the statement 1 in 5 breakfast eaters eat both sweet and savory:** $$x = \frac{1}{5} \times 14 = 2.8$$ 7. **Calculate only sweet:** $$\text{only sweet} = \frac{4}{3} \times 2.8 = \frac{4}{3} \times 2.8 = 3.733...$$ 8. **Given 1 in 4 classmates eat only sweet:** $$\frac{1}{4} \times 15 = 3.75$$ 9. **The calculated only sweet (3.733...) is close to 3.75, consistent with the data.** 10. **Calculate only savory:** $$\text{only savory} = \text{breakfast eaters} - (\text{only sweet} + \text{both sweet and savory}) = 14 - (3.75 + 2.8) = 14 - 6.55 = 7.45$$ 11. **Summary:** - No breakfast: 1 - Only sweet: approx 3.75 - Both sweet and savory: approx 2.8 - Only savory: approx 7.45 12. **Check totals:** $$1 + 3.75 + 2.8 + 7.45 = 15$$ **Final answer:** - Geen ontbijt: 1 - Alleen zoet: 4 (rounded) - Zowel zoet als hartig: 3 (rounded) - Alleen hartig: 7 (rounded)