1. **Problem statement:** You took $\frac{1}{3}$ of the chocolates from a box, then your brother took $\frac{3}{5}$ of the remaining chocolates. After that, 20 chocolates were left in the box. We need to find the original number of chocolates in the box. 2. **Define variables:** Let the original number of chocolates be $x$. 3. **Step 1:** You took $\frac{1}{3}$ of $x$, so the number of chocolates you took is $\frac{1}{3}x$. 4. **Step 2:** The remaining chocolates after you took yours is $x - \frac{1}{3}x = \frac{2}{3}x$. 5. **Step 3:** Your brother took $\frac{3}{5}$ of the remaining chocolates, so he took $\frac{3}{5} \times \frac{2}{3}x = \frac{2}{5}x$ chocolates. 6. **Step 4:** The chocolates left after your brother took his share is the remaining chocolates minus what he took: $$\frac{2}{3}x - \frac{2}{5}x = \left(\frac{10}{15} - \frac{6}{15}\right)x = \frac{4}{15}x$$ 7. **Step 5:** We know this leftover amount is 20 chocolates, so: $$\frac{4}{15}x = 20$$ 8. **Step 6:** Solve for $x$: $$x = 20 \times \frac{15}{4} = 20 \times 3.75 = 75$$ **Final answer:** The box originally contained $\boxed{75}$ chocolates.