Two containers, A and B, contain mixtures of alcohol and water. Container A has alcohol and water in the ratio 4:1, while Container B has them in the ratio 5:2. If 10 liters are drawn from Container A and 14 liters from Container B, and the contents are mixed in a third container, what is the ratio of alcohol to water in the new mixture?

Correct option is A

​Given :

Container A: Alcohol : Water = 4 : 1
Container B: Alcohol : Water = 5 : 2
From A, 10 liters are taken
From B, 14 liters are taken

Formula Used :
​$$\text{Quantity of component} = \text{Total quantity} \times \frac{\text{Part}}{\text{Sum of ratio parts}}$$​​
Solution :

From Container A

Ratio = 4:1 → Total parts = 5

Alcohol from A:
1$$0 \times \frac{4}{5} = 8 \text{ L}$$​​

Water from A:
10$$\times \frac{1}{5} = 2 \text{ L}$$​​
From Container B

Ratio = 5:2 → Total parts = 7

Alcohol from B:
14$$\times \frac{5}{7} = 10 \text{ L}$$​​

Water from B:
14 $$\times \frac{2}{7} = 4 \text{ L}$$​​
Total in New Container

Total alcohol = 8 + 10 = 18 L

Total water = 2 + 4 = 6 L

Required Ratio (Alcohol : Water) :

18 : 6 = 3 : 1