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 ?

Answer: Option B

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,
$$\begin{array}{rl}& \left(\frac{x}{4}+\frac{2y}{7}\right):\left(\frac{3x}{4}+\frac{5y}{7}\right)=3:8\\ & \Rightarrow \frac{\left(\frac{x}{4}+\frac{2y}{7}\right)}{\left(\frac{3x}{4}+\frac{5y}{7}\right)}=\frac{3}{8}\\ & \Rightarrow \frac{\frac{7x+14y}{28}}{\frac{21x+20y}{28}}=\frac{3}{8}\\ & \Rightarrow \frac{7x+14y}{21x+20y}=\frac{3}{8}\\ & \Rightarrow 56x+64y=63x+60y\\ & \Rightarrow 64y-60y=63x-56x\\ & \Rightarrow 4y=7x\\ & \therefore \frac{x}{y}=\frac{4}{7}\end{array}$$

⇒ 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