Two vessels A and B contains acid of different concentration. The capacity of A and B are 6 lt and 3 lt respectively. We take out same quantity from both vessels and put into one another, now the concentration in both vessel are same. Find the taken out quantity.

Options

A. 2 litres
B. 1 litres
C. `1(1)/(2)` litres
D. `(3)/(4)` litres

Answer

To solve the problem step by step, we will denote the quantity taken out from each vessel as \( x \) liters.

### Step 1: Define the concentrations before the exchange
- Let the concentration of acid in vessel A be \( C_A \) and in vessel B be \( C_B \).
- The capacity of vessel A is 6 liters, and the capacity of vessel B is 3 liters.

### Step 2: Assume the initial concentrations
- Assume vessel A contains a certain concentration of acid, say \( C_A = \frac{a}{6} \), where \( a \) is the amount of acid in vessel A.
- Assume vessel B contains \( C_B = \frac{b}{3} \), where \( b = 0 \) since it contains only water (0% acid).

### Step 3: After taking out \( x \) liters from both vessels
- From vessel A, after taking out \( x \) liters, the amount of acid left is \( a - \frac{a}{6} \cdot x \) and the total volume is \( 6 - x \).
- From vessel B, after taking out \( x \) liters, the amount of acid is \( \frac{b}{3} \cdot x \) and the total volume is \( 3 - x \).

### Step 4: Set up the equation for equal concentrations
After exchanging \( x \) liters, the concentration in both vessels becomes equal:
\[ \frac{(a - \frac{a}{6} \cdot x)}{(6 - x)} = \frac{(0 + 0)}{(3 - x)} \]

Since vessel B initially had no acid, the concentration in vessel B after adding \( x \) liters from vessel A will be:
\[ \frac{\frac{a}{6} \cdot x}{3} \]

### Step 5: Cross-multiply to solve for \( x \)
Cross-multiplying gives:
\[ (a - \frac{a}{6} \cdot x) \cdot 3 = (0 + 0) \cdot (6 - x) \]
This simplifies to:
\[ 3a - \frac{3a}{6} \cdot x = 0 \]
Rearranging gives:
\[ 3a = \frac{3a}{6} \cdot x \]
Thus:
\[ x = 2 \text{ liters} \]

### Conclusion
The quantity taken out from each vessel is \( 2 \) liters.