In an alloy, zinc and copper are in the ratio 1 : 2. In the second alloy, the same elements are in the ratio 2 : 3. If these two alloys be mixed to form a new alloy in which two elements are in the ratio 5 : 8, the ratio of these two alloys in the new alloys is -
Correct Answer: 3 : 10
Let them be mixed in the ratio x : y
Then, in 1st alloy, Zinc = $$\frac{x}{3}$$ and Copper = $$\frac{{2x}}{3}$$
2nd alloy, Zinc = $$\frac{{2y}}{5}$$ and Copper = $$\frac{{3y}}{5}$$
Now, we have
$$\eqalign{ & \frac{x}{3} + \frac{{2y}}{5}:\frac{{2x}}{3} + \frac{{3y}}{5} = 5:8 \cr & {\text{or,}}\frac{{5x + 6y}}{{10x + 9y}} = \frac{5}{8} \cr & {\text{or,}}40x + 48y = 50x + 45y \cr & {\text{or,}}10x = 3y \cr & \therefore \frac{x}{y} = \frac{3}{{10}} \cr} $$
Thus, the required ratio = 3 : 10
Alligation Method :
You must know that we can apply this rule over the fractional value of either zinc or copper.
Let us consider the fractional value of zinc.
![Alligation mcq solution image Alligation mcq solution image](/images/solution-image/arithmetic-ability/alligation/1570951848-59.png)
Therefore, they should be mixed in the ratio
$$\eqalign{ & = \frac{1}{{65}}:\frac{2}{{39}} \cr & {\text{or,}}\frac{1}{{65}} \times \frac{{39}}{2} \cr & = \frac{3}{{10}} \cr & {\text{or,}}\,\,3:10 \cr} $$
Then, in 1st alloy, Zinc = $$\frac{x}{3}$$ and Copper = $$\frac{{2x}}{3}$$
2nd alloy, Zinc = $$\frac{{2y}}{5}$$ and Copper = $$\frac{{3y}}{5}$$
Now, we have
$$\eqalign{ & \frac{x}{3} + \frac{{2y}}{5}:\frac{{2x}}{3} + \frac{{3y}}{5} = 5:8 \cr & {\text{or,}}\frac{{5x + 6y}}{{10x + 9y}} = \frac{5}{8} \cr & {\text{or,}}40x + 48y = 50x + 45y \cr & {\text{or,}}10x = 3y \cr & \therefore \frac{x}{y} = \frac{3}{{10}} \cr} $$
Thus, the required ratio = 3 : 10
Alligation Method :
You must know that we can apply this rule over the fractional value of either zinc or copper.
Let us consider the fractional value of zinc.
![Alligation mcq solution image Alligation mcq solution image](/images/solution-image/arithmetic-ability/alligation/1570951848-59.png)
Therefore, they should be mixed in the ratio
$$\eqalign{ & = \frac{1}{{65}}:\frac{2}{{39}} \cr & {\text{or,}}\frac{1}{{65}} \times \frac{{39}}{2} \cr & = \frac{3}{{10}} \cr & {\text{or,}}\,\,3:10 \cr} $$