100 mL of alcohol from container A containing 1 L of alcohol is transferred to another container B containing 1 L of water and mixed well. From this, 100 mL is transferred back to container A. The amount of alcohol in container B would be

Correct option is A

Solution:
Initial Setup:
Container A: 1000 mL (1 L) alcohol
Container B: 1000 mL (1 L) water

Step 1: Transfer 100 mL of alcohol from A to B
Container A: 900 mL alcohol
Container B: 100 mL alcohol + 1000 mL water = 1100 mL mixture
Now, the alcohol-to-water ratio in B = 100:1000 = 1:10

Step 2: Transfer 100 mL of mixture from B to A
The total parts = 1 (alcohol) + 10 (water) = 11 parts
So in 100 mL mixture:
Alcohol = $$\frac{1}{11} \times 100 = \frac{100}{11}\ \text{mL}$$

$$\text{Water: }\frac{10}{11} \times 100 = \frac{1000}{11}\ \text{mL}$$

Final Situation:
In Container A:  $$900 + \frac{100}{11} = \frac{9900 + 100}{11} = \frac{10100}{11}\ \text{mL}$$

$$\text{Water: }\frac{1000}{11}\ \text{mL}$$

In Container B:

$$\text{Alcohol:}100 - \frac{100}{11} = \frac{1100 - 100}{11} = \frac{1000}{11}\ \text{mL}$$

​Water: $$\text{}1000 - \frac{1000}{11} = \frac{11000 - 1000}{11} = \frac{10000}{11}\ \text{mL}$$

$$\textbf{Conclusion:}\\\ \\\text{Alcohol in B} = \frac{1000}{11}\ \text{mL}\quad \text{and} \quad\text{Water in A} = \frac{1000}{11}\ \text{mL}$$

Therefore, both are equal.​​​