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 A, B, and C respectively. 3. **Write the equations from the problem:** - Total cost: $$a + b + c = 80$$ - Half of A's payment plus one-fifth of B's plus one-tenth of C's equals 30: $$\frac{1}{2}a + \frac{1}{5}b + \frac{1}{10}c = 30$$ - A's payment plus one-eighth of B's minus one-quarter of C's equals 50: $$a + \frac{1}{8}b - \frac{1}{4}c = 50$$ 4. **Solve the system:** 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:** A paid 50, B paid 20, and C paid 10.