To achieve a total fill time of 20 minutes, it is necessary to determine the precise moment pipe B should be deactivated. This involves quantifying the flow rates of each pipe and integrating them with the defined operational duration.
Step 1: Ascertain the fill rate for each pipe.
Pipe A completes the tank in 32 minutes, yielding a rate of 1/32 of the tank per minute.
Pipe B completes the tank in 48 minutes, yielding a rate of 1/48 of the tank per minute.
Step 2: Formulate the equation governing the 20-minute fill cycle.
Let \(x\) represent the duration, in minutes, that pipe B operates before being shut off. During this interval of \(x\) minutes, both pipes A and B function concurrently. Subsequently, for the remaining \(20-x\) minutes, only pipe A continues to operate.
The collective work accomplished by pipes A and B during the initial \(x\) minutes is expressed as:
\(\dfrac{x}{32}+\dfrac{x}{48}\)
The work performed solely by pipe A during the subsequent \(20-x\) minutes is:
\(\dfrac{20-x}{32}\)
The sum of these contributions must equal the completion of one full tank:
\(\dfrac{x}{32}+\dfrac{x}{48}+\dfrac{20-x}{32}=1\)
Step 3: Resolve the equation.
To simplify the equation, a common denominator is required for the fractional terms. The least common multiple of 32 and 48 is 96:
\(\dfrac{3x}{96}+\dfrac{2x}{96}+\dfrac{20-x}{32}=1\)
\((3x+2x)/96+(20-x)/32=1\)
This simplifies to:
\(\dfrac{5x}{96}+\dfrac{20-x}{32}=1\)
Multiplying each term by 96 to eliminate denominators yields:
\(5x+3(20-x)=96\)
Further simplification leads to:
\(5x+60-3x=96\)
\(2x=36\)
\(x=18\)
Conclusion: To ensure the tank is filled precisely within 20 minutes, pipe B must be deactivated after 18 minutes of operation.
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