Is prashn ko hum pipe A aur pipe B ki kaam karne ki dar ko samajhkar solve karenge. Aapke diye gaye information ke mutabik:

• Pipe A aur Pipe B dono ko ek saath chalane se tank ko bharne ka samay 3 minute kam ho jata hai.
• Pipe B ko Pipe A aur Pipe B dono ke saath chalane par 64/3 minute zyada samay lagta hai.

Yeh do equations ban rahi hain, jo hum solve karenge:

1. Let the time taken by Pipe A to fill the tank alone be $T_A$ minutes.
2. Let the time taken by Pipe B to fill the tank alone be $T_B$ minutes.

Step 1: First relationship

Jab Pipe A aur Pipe B dono kaam kar rahe hote hain, toh time $T_A$ aur $T_B$ se kam ho jata hai.

Mathematically, $\frac{T_A \cdot T_B}{T_A + T_B} = T_A - 3$ Yeh equation mein yeh dikhata hai ki dono pipes ko saath mein chalane par time $T_A - 3$ minute ho jata hai.

Step 2: Second relationship

Jab pipe B ko alone chalane par, usko 64/3 minute extra time lagta hai. Toh, pipe B ko alone chalane par $T_B$ ka samay $T_A + T_B + \frac{64}{3}$ ke barabar ho jata hai. $T_B = T_A + T_B + \frac{64}{3}$

Hum ab yeh dono equations solve karke $T_A$ aur $T_B$ nikalenge, jo dono pipes ko saath chalane ka samay bataenge.

Let me calculate that for you.Chaliye, hum yeh equations solve karte hain:

Step 1: Relationship between $T_A$ and $T_B$

Aapne pehla equation diya tha: $\frac{T_A \cdot T_B}{T_A + T_B} = T_A - 3$ Isko simplify karte hain: $T_A \cdot T_B = (T_A - 3) \cdot (T_A + T_B)$ $T_A \cdot T_B = T_A^2 + T_A \cdot T_B - 3T_A - 3T_B$ T_A \cdot T_B ko dono taraf se cancel kar denge: $0 = T_A^2 - 3T_A - 3T_B$ $T_A^2 - 3T_A = 3T_B \quad \text{(Equation 1)}$

Step 2: Second relationship

Aapne doosra equation diya tha: $T_B = T_A + T_B + \frac{64}{3}$ Isko simplify karte hain: $0 = T_A + \frac{64}{3}$ $T_A = -\frac{64}{3} \quad \text{(Equation 2)}$ This equation gives a negative time value for $T_A$. This suggests that there might be a mistake in interpreting the problem. Let me go through it again.

Step 3: Let's clarify

Do you want me to try re-explaining?