Pipes And Cistern All Competence Aptitude

Let's denote:
- Time taken to fill the tank without the leak = \( T_F \) hours.
- Time taken to fill the tank with the leak = \( T_L \) hours.

Given:
- Time taken to fill the tank without the leak = 9 hours.
- Time taken to fill the tank with the leak = 9 + 4 = 13 hours.

Now, let's find the rate at which the leak empties the tank:

The rate at which the tank fills without the leak is \( \dfrac{1}{T_F} \) portion per hour.
The rate at which the tank fills with the leak is \( \dfrac{1}{T_L} \) portion per hour.

When the tank is full, the rate at which the leak empties the tank is equal to the rate at which the tank fills without the leak.

So, we have:

\[ \dfrac{1}{T_F} = \dfrac{1}{T_L}\]

Substituting the given values:

\[ \dfrac{1}{9} = \dfrac{1}{T_L}\]

Solving for \( T_L \):

\[ T_L = \dfrac{9 \times 13}{9 - 13}\]\[ T_L = \dfrac{117}{-4}\]\[ T_L = -29.25 \text{ hours}\]

Since time cannot be negative, we consider the absolute value, and thus, the leak will empty the tank in 29.25 hours.

Therefore, the leak will empty the tank in 29.25 hours. So, the correct answer is 29.25 hrs.

Let's denote:
- Time taken by the pipe to fill the cistern alone = \( T_F \) hours.
- Time taken by the pipe to empty the cistern alone = \( T_E \) hours.

Given:
- Time taken by the pipe to fill the cistern alone = 5 hours.
- Time taken by the pipe to empty the cistern alone = 4 hours.

Now, let's find the rate at which each pipe fills or empties the cistern:

The rate at which the cistern fills with the fill pipe is \( \dfrac{1}{T_F} \) portion per hour.
The rate at which the cistern empties with the empty pipe is \( \dfrac{1}{T_E} \) portion per hour.

When both pipes are open together, their combined rate is the difference of their individual rates:

Combined Rate = Rate of fill pipe - Rate of empty pipe

Substituting the given values:

Combined Rate = \( \dfrac{1}{5} - \dfrac{1}{4} \)

To combine the fractions, we need a common denominator, which is 20:

Combined Rate = \( \dfrac{4}{20} - \dfrac{5}{20} \)

Combined Rate = \( \dfrac{-1}{20} \) portion per hour

Since the combined rate is negative, it means the cistern is being emptied. 

Therefore, when both pipes are open together, the cistern will be emptied in \( \dfrac{20}{1} = 20 \) hours.

Therefore, the correct answer is 20 hours.

Let's denote:
- Time taken by the first pipe to fill the tank alone = \( T_1 = 15 \) hours.
- Time taken by the second pipe to fill the tank alone = \( T_2 = 20 \) hours.
- Time taken by the third pipe to empty the tank alone = \( T_3 = 30 \) hours.

We can use the formula for the rate of filling:

\[ \text{Rate} = \dfrac{1}{\text{Time}}\]

Using this formula, we can find the rates of each pipe:

For the first pipe:\[ \text{Rate}_1 = \dfrac{1}{T_1} = \dfrac{1}{15}\]

For the second pipe:\[ \text{Rate}_2 = \dfrac{1}{T_2} = \dfrac{1}{20}\]

For the third pipe:
Since it empties the tank, its rate is negative:\[ \text{Rate}_3 = -\dfrac{1}{T_3} = -\dfrac{1}{30}\]

When all pipes are opened together, their combined rate is the sum of their individual rates:

\[ \text{Combined Rate} = \text{Rate}_1 + \text{Rate}_2 + \text{Rate}_3\]

Substituting the given values:

\[ \text{Combined Rate} = \dfrac{1}{15} + \dfrac{1}{20} - \dfrac{1}{30}\]

To combine the fractions, we need a common denominator, which is 60:

\[ \text{Combined Rate} = \dfrac{4}{60} + \dfrac{3}{60} - \dfrac{2}{60}\]

\[ \text{Combined Rate} = \dfrac{5}{60}\]

\[ \text{Combined Rate} = \dfrac{1}{12}\]

Therefore, when all pipes are opened together, they will fill the empty tank in \( \dfrac{1}{\dfrac{1}{12}} = 12 \) hours. So, the correct answer is 12 hours.

Let's denote:
- Time taken by pipe A to fill the tank alone = \( T_A = 30 \) minutes.
- Time taken by pipe B to fill the tank alone = \( T_B \).

Given:
- Pipe B can fill the tank 5 times as fast as pipe A. Therefore, \( T_B = \dfrac{T_A}{5} = \dfrac{30}{5} = 6 \) minutes.

Now, let's find the combined rate of both pipes when they are opened together:

The rate at which pipe A fills the tank is \( \dfrac{1}{T_A} = \dfrac{1}{30} \) tank per minute.
The rate at which pipe B fills the tank is \( \dfrac{1}{T_B} = \dfrac{1}{6} \) tank per minute.

When both pipes are open together, their combined rate is the sum of their individual rates:

\[ \text{Combined Rate} = \dfrac{1}{T_A} + \dfrac{1}{T_B} = \dfrac{1}{30} + \dfrac{1}{6}\]

\[ \text{Combined Rate} = \dfrac{1}{30} + \dfrac{5}{30} = \dfrac{6}{30}\]

\[ \text{Combined Rate} = \dfrac{1}{5} \text{ tank per minute}\]

Therefore, when both pipes are open together, the tank will overflow in \( \dfrac{1}{\dfrac{1}{5}} = 5 \) minutes.

So, it will take 5 minutes for the tank to overflow. Thus, the correct answer is 5 minutes.

Let's denote:
- The number of pumps as \( P \).
- The number of hours each pump works per day as \( H \).

Given:
- \( P = 12 \) pumps.
- \( H = 6 \) hours per day.
- Time taken to empty the reservoir = 15 days.

Using the formula: \(\text{Work} = \text{Rate} \times \text{Time}\)

The work done by the pumps is to empty the reservoir.

The work done by the pumps is the same, so:

\[ 12 \times 6 \times 15 = P \times 9 \times 12\]

Solving for \(P\):

\[ P = \dfrac{12 \times 6 \times 15}{9 \times 12}\]\[ P = \dfrac{1080}{108}\]\[ P = 10\]

Therefore, \(P = 10\).

So, 10 pumps working 9 hours a day will empty the reservoir in 12 days. Thus, the correct answer is 10.

Let's denote:
- The rate at which pipe A fills the cistern is \( R_A \).
- The rate at which pipe B fills the cistern is \( R_B \).
- The rate at which the third pipe empties the cistern is \( R_C \).

Given:
- Pipe A fills the cistern in 60 minutes, so its rate is \( R_A = \dfrac{1}{60} \) cistern per minute.
- Pipe B fills the cistern in 75 minutes, so its rate is \( R_B = \dfrac{1}{75} \) cistern per minute.
- When all three pipes are open, the cistern fills in 50 minutes, so their combined rate is \( \dfrac{1}{50} \) cistern per minute.

Now, let's set up the equation based on the rates:

\[ R_A + R_B - R_C = \dfrac{1}{50}\]

Substituting the given values:

\[ \dfrac{1}{60} + \dfrac{1}{75} - R_C = \dfrac{1}{50}\]

To combine the fractions, we need a common denominator, which is 150:

\[ \dfrac{5}{300} + \dfrac{4}{300} - R_C = \dfrac{3}{150}\]

\[ \dfrac{9}{300} - R_C = \dfrac{3}{150}\]

\[ R_C = \dfrac{9}{300} - \dfrac{3}{150}\]

\[ R_C = \dfrac{3}{100} - \dfrac{2}{100}\]

\[ R_C = \dfrac{1}{100}\]

Therefore, the rate at which the third pipe empties the cistern is \( R_C = \dfrac{1}{100} \) cistern per minute.

Now, to find the time taken by the third pipe alone to empty the cistern, we use the formula:

\[ \text{Time} = \dfrac{1}{\text{Rate}}\]

Substituting the rate of the third pipe:

\[ \text{Time} = \dfrac{1}{\dfrac{1}{100}} = 100 \text{ minutes}\]

Therefore, the third pipe alone can empty the cistern in 100 minutes. So, the correct answer is 100 min.

Let’s denote:
- The rate at which pipe A fills the cistern is \(R_A\).
- The rate at which pipe B fills the cistern is \(R_B\).
- The rate at which the third pipe empties the cistern is \(R_C\).

Given:
- Pipe A fills the cistern in 60 minutes, so its rate is \(R_A = \frac{1}{60}\) cistern per minute.
- Pipe B fills the cistern in 75 minutes, so its rate is \(R_B = \frac{1}{75}\) cistern per minute.
- When all three pipes are open, the cistern fills in 50 minutes, so their combined rate is \(\frac{1}{50}\) cistern per minute.

Using the principle that rates are additive when pipes work together, we have:

\[ R_A + R_B - R_C = \frac{1}{50} \]

Substituting the known values:

\[ \frac{1}{60} + \frac{1}{75} - R_C = \frac{1}{50} \]

To combine the fractions, we need a common denominator, which is 300:

\[ \frac{5}{300} + \frac{4}{300} - R_C = \frac{6}{300} \]

Combine the numerators:

\[ \frac{9}{300} - R_C = \frac{6}{300} \]

Solving for \(R_C\):

\[ R_C = \frac{9}{300} - \frac{6}{300} = \frac{3}{300} = \frac{1}{100} \]

So, the third pipe empties the cistern at a rate of \(\frac{1}{100}\) cistern per minute, which means it would take 100 minutes to empty the cistern.

Let's denote:
- The rate at which the first tap fills the tank is \( R_1 \).
- The rate at which the second tap empties the tank is \( R_2 \).

Given:
- The time taken by the first tap to fill the tank alone is 40 minutes, so \( R_1 = \dfrac{1}{40} \) tank per minute.
- The time taken by the second tap to empty the tank alone is 60 minutes, so \( R_2 = -\dfrac{1}{60} \) tank per minute (negative because it empties the tank).

When both taps are open, their combined rate is the sum of their individual rates:

\[ \text{Combined Rate} = R_1 + R_2 = \dfrac{1}{40} - \dfrac{1}{60}\]

\[ \text{Combined Rate} = \dfrac{3}{120} - \dfrac{2}{120} = \dfrac{1}{120} \text{ tank per minute}\]

Therefore, when both taps are open together, they will fill \( \dfrac{1}{120} \) of the tank per minute.

Now, to find the time taken to fill the empty tank, we use the formula:

\[ \text{Time} = \dfrac{1}{\text{Rate}}\]

Substituting the combined rate:

\[ \text{Time} = \dfrac{1}{\dfrac{1}{120}} = 120 \text{ minutes}\]

Converting minutes to hours:

\[ 120 \text{ minutes} = 2 \text{ hours}\]

So, it will take 2 hours for the empty tank to be filled when both taps are open. Thus, the correct answer is 2 hours.

Let's denote:
- The time taken by pipe P alone to fill the cistern is \( T_P = 30 \) minutes.
- The time taken by pipe Q alone to fill the cistern is \( T_Q \).

Given:
- Together, pipes P and Q can fill the cistern in 20 minutes. So, their combined rate is \( \dfrac{1}{20} \) cistern per minute.
- Pipe P alone can fill the cistern in 30 minutes. So, its rate is \( \dfrac{1}{30} \) cistern per minute.

We can use the fact that the combined rate of P and Q is the sum of their individual rates. Therefore:

\[ \dfrac{1}{30} + \dfrac{1}{T_Q} = \dfrac{1}{20}\]

Now, let's solve for \( T_Q \):

\[ \dfrac{1}{T_Q} = \dfrac{1}{20} - \dfrac{1}{30}\]

\[ \dfrac{1}{T_Q} = \dfrac{3}{60} - \dfrac{2}{60}\]

\[ \dfrac{1}{T_Q} = \dfrac{1}{60}\]

Therefore, \( T_Q = 60 \) minutes.

Hence, pipe Q alone can fill the cistern in 60 minutes. So, the correct answer is 60 minutes.

If \( \dfrac{3}{5} \) of a cistern is filled in 1 minute, it means that \( \dfrac{3}{5} \) of the cistern can be filled in 1 minute. 

Let's denote:
- The rate at which the cistern is filled is \( R \) (portion per minute).
- The time needed to fill the remaining \( \dfrac{2}{5} \) of the cistern is \( T \) (minutes).

Given that \( \dfrac{3}{5} \) of the cistern is filled in 1 minute, we can set up the equation:

\[ R \times 1 = \dfrac{3}{5}\]

Solving for \( R \):

\[ R = \dfrac{3}{5}\]

Now, we can find the time needed to fill the remaining \( \dfrac{2}{5} \) of the cistern using the formula:

\[ \text{Time} = \dfrac{\text{Volume}}{\text{Rate}}\]

Substituting the given values:

\[ T = \dfrac{\dfrac{2}{5}}{\dfrac{3}{5}} = \dfrac{2}{3}\]

Converting \( \dfrac{2}{3} \) minutes to seconds:

\[ T = \dfrac{2}{3} \times 60 = 40 \text{ seconds}\]

Therefore, the time needed to fill the rest of the cistern is 40 seconds. Hence, the correct answer is 40 sec.