1. **State the problem:** A drink-seller had 520 cups in total: small (S), medium (M), and large (L). She used 34 small cups and 20% of medium cups. Then she increased large cups by 50%. After these changes, the ratio of remaining small, medium, and large cups is 9:4:6. 2. **Define variables and write equations:** Let the initial numbers be $S$, $M$, and $L$. We know: $$S + M + L = 520$$ 3. **Calculate remaining cups:** - Small cups left: $S - 34$ - Medium cups used: 20% of $M$ means used $0.2M$, so left $M - 0.2M = 0.8M$ - Large cups increased by 50%, so new large cups: $L + 0.5L = 1.5L$ 4. **Use the ratio of remaining cups:** The ratio of remaining small : medium : large is 9 : 4 : 6. So, $$\frac{S - 34}{9} = \frac{0.8M}{4} = \frac{1.5L}{6} = k \text{ (some constant)}$$ 5. **Express each in terms of $k$:** $$S - 34 = 9k$$ $$0.8M = 4k \implies M = \frac{4k}{0.8} = 5k$$ $$1.5L = 6k \implies L = \frac{6k}{1.5} = 4k$$ 6. **Substitute $S$, $M$, $L$ into total cups equation:** $$S + M + L = 520$$ $$ (9k + 34) + 5k + 4k = 520$$ $$9k + 34 + 5k + 4k = 520$$ $$18k + 34 = 520$$ $$18k = 520 - 34 = 486$$ $$k = \frac{486}{18} = 27$$ 7. **Find $S$, $M$, $L$:** $$S = 9k + 34 = 9 \times 27 + 34 = 243 + 34 = 277$$ $$M = 5k = 5 \times 27 = 135$$ $$L = 4k = 4 \times 27 = 108$$ 8. **Check total:** $$277 + 135 + 108 = 520$$ correct. 9. **Answer part (a): Fraction of medium cups left:** Medium cups left = $0.8M = 0.8 \times 135 = 108$ Fraction left = $\frac{108}{135} = \frac{4}{5}$ 10. **Answer part (b): Total cups in the end:** Total cups left = small left + medium left + large left Small left = $S - 34 = 277 - 34 = 243$ Medium left = $108$ Large left = $1.5L = 1.5 \times 108 = 162$ Total = $243 + 108 + 162 = 513$ **Final answers:** (a) Fraction of medium cups left = $\frac{4}{5}$ (b) Total number of cups in the end = 513