Question

Two pipes A and B can fill a tank in 32 minutes and 48 minutes respectively. If both the pipes are opened simultaneously, after how much time B should be turned off so that the tank is full in 20 minutes?

• A. 14 minutes
• B. 15 minutes
• C. 16 minutes
• D. 18 minutes

Answer 1

The Correct Option is D

To solve the problem, we need to determine when to turn off pipe B to fill the tank in exactly 20 minutes. The steps involve calculating the rates of the pipes and applying them to the time constraints. 

Step 1: Calculate the rate of each pipe.

Pipe A can fill the tank in 32 minutes, so its rate is 1/32 tank per minute.

Pipe B can fill the tank in 48 minutes, so its rate is 1/48 tank per minute.

Step 2: Set up the equation based on the total operation time of 20 minutes.

Let \(x\) be the time in minutes after which pipe B is turned off. In those \(x\) minutes, both pipes work together. Therefore, for the next \(20-x\) minutes, only pipe A is working.

The equation for the work done by both pipes in the initial \(x\) minutes is:

\(\dfrac{x}{32}+\dfrac{x}{48}\)

The equation for the work done by pipe A in the remaining time \(20-x\) is:

\(\dfrac{20-x}{32}\)

The sum of these equations equals 1 full tank:

\(\dfrac{x}{32}+\dfrac{x}{48}+\dfrac{20-x}{32}=1\)

Step 3: Solve the equation.

To solve this equation, find a common denominator for the fractions. The least common multiple of 32 and 48 is 96:

\(\dfrac{3x}{96}+\dfrac{2x}{96}+\dfrac{20-x}{32}=1\)

\((3x+2x)/96+(20-x)/32=1\)

simplifies to:

\(\dfrac{5x}{96}+\dfrac{20-x}{32}=1\)

Multiply every term by 96 to clear the denominators:

\(5x+3(20-x)=96\)

which simplifies to:

\(5x+60-3x=96\)

\(2x=36\)

\(x=18\)

Conclusion: Pipe B should be turned off after 18 minutes for the tank to be filled in 20 minutes.

Answer 2

Let B be turned off after \( t \) minutes. The amount of water filled by A in 20 minutes:

A's rate \( = \frac{1}{32} \), B's rate \( = \frac{1}{48} \).

Water filled by A in 20 minutes: \( \frac{20}{32} \).

The amount of water filled by B in \( t \) minutes:

Water filled by B: \( \frac{t}{48} \).

The total water filled should equal 1 tank:

\[ \frac{20}{32} + \frac{t}{48} = 1. \]

Simplify:

\[ \frac{5}{8} + \frac{t}{48} = 1 \implies \frac{t}{48} = \frac{3}{8}. \]

\[ t = \frac{3}{8} \times 48 = 18. \]

Thus, pipe B should be turned off after 18 minutes.