1. **State the problem:** We have two boxes, A and B, with a total mass of sand 250g. Box A gave Box B $\frac{1}{4}$ of its sand. Box B gave Box A $\frac{3}{7}$ of its sand. We want to find the initial mass of sand in Box A. 2. **Define variables:** Let the initial mass of sand in Box A be $x$ grams. Then the initial mass in Box B is $250 - x$ grams. 3. **Write the expressions for the sand after exchange:** - Box A gives $\frac{1}{4}x$ to Box B. - Box B gives $\frac{3}{7}(250 - x)$ to Box A. 4. **Write the equation for the final mass in Box A:** $$\text{Final mass in A} = x - \frac{1}{4}x + \frac{3}{7}(250 - x)$$ 5. **Simplify the expression:** $$x - \frac{1}{4}x = \frac{3}{4}x$$ So, $$\text{Final mass in A} = \frac{3}{4}x + \frac{3}{7}(250 - x)$$ 6. **Since the total mass is conserved, the final mass in A plus final mass in B equals 250g.** But we only need to find $x$, so we use the fact that the final mass in A equals the initial mass in A after exchange (since the problem implies the exchange is simultaneous and total mass is constant). 7. **Set up the equation:** The final mass in A is the same as the initial mass in A (since the problem asks for initial mass, we solve for $x$): $$x = \frac{3}{4}x + \frac{3}{7}(250 - x)$$ 8. **Multiply both sides to clear denominators:** Multiply both sides by 28 (LCM of 4 and 7): $$28x = 28 \times \frac{3}{4}x + 28 \times \frac{3}{7}(250 - x)$$ Calculate each term: $$28x = 7 \times 3x + 4 \times 3 (250 - x)$$ $$28x = 21x + 12(250 - x)$$ 9. **Expand and simplify:** $$28x = 21x + 3000 - 12x$$ $$28x = (21x - 12x) + 3000$$ $$28x = 9x + 3000$$ 10. **Isolate $x$:** $$28x - 9x = 3000$$ $$19x = 3000$$ 11. **Solve for $x$:** $$x = \frac{3000}{19} \approx 157.89$$ **Answer:** The initial mass of sand in Box A is approximately 157.89 grams.