Two alloys both are made up of copper and tin. The ratio of the copper and tin in the first alloy is 1 : 3 and in the second alloy is 2 : 5. In what ratio should the two alloys be mixed to obtain a new alloy in which the ratio of tin and copper be 8 : 3 ?

Solution(By Examveda Team)

Given:
⇒ Let the alloys be mixed in the ratio of $$x$$ : $$y$$ (Assumption)
⇒ In 1st alloy, Copper = $$\frac{x}{4}$$
⇒ In 1st alloy, Tin = $$\frac{3x}{4}$$
⇒ In 2nd alloy, Copper = $$\frac{2y}{7}$$
⇒ In 2nd alloy, Tin = $$\frac{5y}{7}$$

To find:
⇒ The ratio in which 2 alloys must be mixed to get a new alloy with a ratio of copper and tin be 3 : 8 =?

Now we have,
$$\eqalign{ & \left( {\frac{x}{4} + \frac{{2y}}{7}} \right) : \left( {\frac{{3x}}{4} + \frac{{5y}}{7}} \right) = 3 : 8 \cr & \Rightarrow \frac{{\left( {\frac{x}{4} + \frac{{2y}}{7}} \right)}}{{ \left( {\frac{{3x}}{4} + \frac{{5y}}{7}} \right) }} = \frac{3}{8} \cr & \Rightarrow \frac{{\frac{{7x + 14y}}{{28}}}}{{ \frac{{21x + 20y}}{{28}} }} = \frac{3}{8} \cr & \Rightarrow \frac{{7x + 14y}}{{21x + 20y}} = \frac{3}{8} \cr & \Rightarrow 56x + 64y = 63x + 60y \cr & \Rightarrow 64y - 60y = 63x - 56x \cr & \Rightarrow 4y = 7x \cr & \therefore \frac{x}{y} = \frac{4}{7} \cr} $$

⇒ Ratio in which 2 alloys must be mixed to get a new alloy with a ratio of copper and tin be 3 : 8 = 4 : 7