1. **Problem Statement:** We have two alloys A and B with Aluminum (Al) and Nickel (Ni) in ratios 4:3 and 3:5 respectively. We mix these alloys to form a new alloy where Al and Ni are in ratio 1:1. We need to find the ratio of alloy A to alloy B in the new alloy. 2. **Given Ratios:** - Alloy A: Al : Ni = 4 : 3 - Alloy B: Al : Ni = 3 : 5 - New alloy: Al : Ni = 1 : 1 3. **Let the amounts of alloys A and B mixed be $x$ and $y$ respectively.** 4. **Calculate the amount of Aluminum and Nickel in the mixture:** - Aluminum from A = $4x/(4+3) = \frac{4x}{7}$ - Nickel from A = $3x/7$ - Aluminum from B = $3y/(3+5) = \frac{3y}{8}$ - Nickel from B = $5y/8$ 5. **Total Aluminum and Nickel in the new alloy:** - Aluminum = $\frac{4x}{7} + \frac{3y}{8}$ - Nickel = $\frac{3x}{7} + \frac{5y}{8}$ 6. **Since the new alloy has Al : Ni = 1 : 1, set the ratio equal:** $$\frac{\frac{4x}{7} + \frac{3y}{8}}{\frac{3x}{7} + \frac{5y}{8}} = 1$$ 7. **Cross multiply and simplify:** $$\frac{4x}{7} + \frac{3y}{8} = \frac{3x}{7} + \frac{5y}{8}$$ 8. **Bring like terms together:** $$\frac{4x}{7} - \frac{3x}{7} = \frac{5y}{8} - \frac{3y}{8}$$ $$\frac{x}{7} = \frac{2y}{8}$$ 9. **Simplify:** $$\frac{x}{7} = \frac{y}{4}$$ 10. **Cross multiply:** $$4x = 7y$$ 11. **Therefore, the ratio of alloy A to alloy B is:** $$x : y = 7 : 4$$ **Final answer:** The ratio of alloy A to alloy B in the new alloy is 7 : 4.