1. **State the problem:** Three housewives A, B, and C jointly purchased a basket of oranges costing 80. We need to find how much each paid. 2. **Define variables:** Let $A$, $B$, and $C$ be the amounts paid by housewives A, B, and C respectively. 3. **Write down the given equations:** - Total cost: $$A + B + C = 80$$ - One-half of A plus one-fifth of B plus one-tenth of C equals 30: $$\frac{1}{2}A + \frac{1}{5}B + \frac{1}{10}C = 30$$ - A plus one-eighth of B minus one-quarter of C equals 50: $$A + \frac{1}{8}B - \frac{1}{4}C = 50$$ 4. **Solve the system of equations:** From the first equation: $$C = 80 - A - B$$ Substitute $C$ into the second and third equations: Second equation: $$\frac{1}{2}A + \frac{1}{5}B + \frac{1}{10}(80 - A - B) = 30$$ Multiply both sides by 10 to clear denominators: $$5A + 2B + 80 - A - B = 300$$ Simplify: $$4A + B + 80 = 300$$ $$4A + B = 220$$ Third equation: $$A + \frac{1}{8}B - \frac{1}{4}(80 - A - B) = 50$$ Multiply both sides by 8: $$8A + B - 2(80 - A - B) = 400$$ Simplify inside parentheses: $$8A + B - 160 + 2A + 2B = 400$$ Combine like terms: $$10A + 3B - 160 = 400$$ $$10A + 3B = 560$$ 5. **Solve the two equations:** $$4A + B = 220$$ $$10A + 3B = 560$$ Multiply the first equation by 3: $$12A + 3B = 660$$ Subtract the second equation from this: $$(12A + 3B) - (10A + 3B) = 660 - 560$$ $$2A = 100$$ $$A = 50$$ Substitute $A=50$ into $4A + B = 220$: $$4(50) + B = 220$$ $$200 + B = 220$$ $$B = 20$$ Find $C$: $$C = 80 - A - B = 80 - 50 - 20 = 10$$ 6. **Answer:** Housewife A paid 50, B paid 20, and C paid 10.