Two pipes A and B are opened together to fill a tank . Both the pipes fill the tank in time t . If A separately takes 4 min more time than t to fill the tank B takes 64 min more time than t to fill the tank , find the value of t .

Answer

To solve the problem step by step, we will break down the information given and use it to find the value of \( t \).

### Step 1: Understand the problem
We have two pipes, A and B, that fill a tank together in \( t \) minutes. Pipe A takes \( t + 4 \) minutes to fill the tank alone, and Pipe B takes \( t + 64 \) minutes to fill the tank alone. We need to find the value of \( t \).

### Step 2: Set up the equations
Let’s denote the capacity of the tank as \( C \) (in units). The rates of filling for pipes A and B can be expressed as follows:
- Rate of Pipe A = \( \frac{C}{t + 4} \) (units per minute)
- Rate of Pipe B = \( \frac{C}{t + 64} \) (units per minute)

When both pipes are opened together, their combined rate is:
\[ \frac{C}{t + 4} + \frac{C}{t + 64} = \frac{C}{t} \]

### Step 3: Eliminate \( C \)
Since \( C \) is common in all terms, we can cancel it out from the equation:
\[ \frac{1}{t + 4} + \frac{1}{t + 64} = \frac{1}{t} \]

### Step 4: Find a common denominator
The common denominator for the left side is \( (t + 4)(t + 64) \):
\[ \frac{(t + 64) + (t + 4)}{(t + 4)(t + 64)} = \frac{1}{t} \]
This simplifies to:
\[ \frac{2t + 68}{(t + 4)(t + 64)} = \frac{1}{t} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ (2t + 68)t = (t + 4)(t + 64) \]

### Step 6: Expand both sides
Expanding both sides:
Left side:
\[ 2t^2 + 68t \]
Right side:
\[ t^2 + 68t + 256 \]

### Step 7: Set the equation to zero
Setting the equation to zero:
\[ 2t^2 + 68t - (t^2 + 68t + 256) = 0 \]
This simplifies to:
\[ t^2 - 256 = 0 \]

### Step 8: Solve for \( t \)
Factoring gives:
\[ (t - 16)(t + 16) = 0 \]
Thus, \( t = 16 \) or \( t = -16 \). Since time cannot be negative, we take:
\[ t = 16 \text{ minutes} \]

### Final Answer
The value of \( t \) is 16 minutes.

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