1. **State the problem:** We have product M made by mixing chemicals X and Y in ratio 5:4. Chemical X is made from raw materials A and B in ratio 1:3. Chemical Y is made from raw materials B and C in ratio 2:1. We mix 864 units of product M with water to get a final mixture where raw material B concentration is 50%. We need to find how much water was added. 2. **Express the composition of X and Y:** - Chemical X ratio A:B = 1:3 means total parts = 4. So fraction of B in X = $\frac{3}{4}$. - Chemical Y ratio B:C = 2:1 means total parts = 3. So fraction of B in Y = $\frac{2}{3}$. 3. **Express the composition of product M:** - Product M ratio X:Y = 5:4 means total parts = 9. - Fraction of X in M = $\frac{5}{9}$, fraction of Y in M = $\frac{4}{9}$. 4. **Calculate fraction of B in product M:** $$\text{Fraction of B in M} = \left(\frac{5}{9}\right)\left(\frac{3}{4}\right) + \left(\frac{4}{9}\right)\left(\frac{2}{3}\right)$$ Calculate each term: $$\left(\frac{5}{9}\right)\left(\frac{3}{4}\right) = \frac{15}{36} = \frac{5}{12}$$ $$\left(\frac{4}{9}\right)\left(\frac{2}{3}\right) = \frac{8}{27}$$ Find common denominator 108: $$\frac{5}{12} = \frac{45}{108}, \quad \frac{8}{27} = \frac{32}{108}$$ Sum: $$\frac{45}{108} + \frac{32}{108} = \frac{77}{108}$$ So fraction of B in product M is $\frac{77}{108}$. 5. **Calculate amount of B in 864 units of product M:** $$864 \times \frac{77}{108} = 864 \times \frac{77}{108} = 8 \times 77 = 616$$ 6. **Final mixture has 50% B concentration:** Let water added be $w$ units. Total mixture = $864 + w$ units. Amount of B remains 616 units (water has no B). So concentration equation: $$\frac{616}{864 + w} = 0.5$$ 7. **Solve for $w$:** $$616 = 0.5 (864 + w)$$ $$616 = 432 + 0.5 w$$ $$616 - 432 = 0.5 w$$ $$184 = 0.5 w$$ $$w = \frac{184}{0.5} = 368$$ **Answer:** 368 units of water were added.