1. **State the problem:** We have 100 apples. Among them, 20 have worms, 15 have bruises, and 10 have both worms and bruises. We want to find how many apples have neither worms nor bruises and thus can be sold. 2. **Identify the sets:** Let $W$ be the set of apples with worms, $B$ be the set of apples with bruises. 3. **Given values:** - Total apples, $|U| = 100$ - Apples with worms, $|W| = 20$ - Apples with bruises, $|B| = 15$ - Apples with both worms and bruises, $|W \cap B| = 10$ 4. **Use the principle of inclusion-exclusion to find apples with worms or bruises:** $$|W \cup B| = |W| + |B| - |W \cap B| = 20 + 15 - 10 = 25$$ 5. **Calculate apples with neither worms nor bruises:** $$|\text{neither}| = |U| - |W \cup B| = 100 - 25 = 75$$ 6. **Conclusion:** There are 75 apples that have neither worms nor bruises and can be sold.