Pipes And Cistern All Competence Aptitude

Pipe A fills 1/20 of the tank in one minute, and pipe B fills 1/30 of the tank in one minute. 

When both pipes are opened together, their rates of filling add up. So, the combined rate is:

1/20 (from pipe A) + 1/30 (from pipe B) = 3/60 + 2/60 = 5/60 

This simplifies to 1/12, meaning that together they fill 1/12 of the tank in one minute.

Thus, it would take 12 minutes for both pipes A and B to fill the tank together. 

So, the correct answer is:

12 minutes

Let's find out how much of the tank is filled by both pipes A and B working together for 4 minutes.

The rate of filling of pipe A is \( \dfrac{1}{15} \) tank/minute.
The rate of filling of pipe B is \( \dfrac{1}{20} \) tank/minute.

So, in 4 minutes, the combined rate of filling of both pipes A and B is \( \dfrac{1}{15} + \dfrac{1}{20} = \dfrac{4}{60} + \dfrac{3}{60} = \dfrac{7}{60} \) tank/minute.

In 4 minutes, the amount of tank filled by both pipes together is \( 4 \times \dfrac{7}{60} = \dfrac{28}{60} \) of the tank.

Now, after 4 minutes, pipe A is turned off. So, only pipe B continues to work.

The rate of filling of pipe B alone is \( \dfrac{1}{20} \) tank/minute.

The remaining amount of tank to be filled after 4 minutes is \( 1 - \dfrac{28}{60} = \dfrac{32}{60} \) of the tank.

To find out how much time pipe B alone will take to fill this remaining amount, we can use the formula:

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

\[ \text{Time} = \dfrac{\dfrac{32}{60}}{\dfrac{1}{20}} \]

\[ \text{Time} = \dfrac{32}{60} \times 20 \]

\[ \text{Time} = \dfrac{32 \times 20}{60} \]

\[ \text{Time} = \dfrac{640}{60} \text{ minutes} \]

\[ \text{Time} = 10 \dfrac{2}{3} \text{ minutes} \]

Now, let's add the time taken by both pipes together for the initial 4 minutes and the time taken by pipe B alone to fill the remaining amount:

Total time = 4 minutes + \( 10 \dfrac{2}{3} \) minutes

Converting \( \dfrac{2}{3} \) minutes to seconds:\[ \dfrac{2}{3} \times 60 = 40 \] seconds

So, total time = 4 minutes + 10 minutes + 40 seconds

\[ \text{Total time} = 14 \text{ minutes} + 40 \text{ seconds} \]

Therefore, the total time required to fill the tank is 14 minutes and 40 seconds. So, the correct option is 14 min. 40 sec.

If the cistern is normally filled in 8 hours but takes two hours longer to fill due to a leak, it means it takes 8 + 2 = 10 hours to fill the cistern with the leak.

Let's denote the rate at which the cistern fills without the leak as \( F \) (the part filled per hour), and the rate at which the leak empties the cistern as \( L \) (the part emptied per hour).

Given:
- Time taken to fill the cistern without the leak = 8 hours
- Time taken to fill the cistern with the leak = 10 hours

Let's denote the volume of the cistern as \( V \) (the total part).

Using the formula: \[ \text{Time} = \dfrac{\text{Volume}}{\text{Rate}} \]

We can find the rates as follows:
- Without the leak: \( F = \dfrac{V}{8} \)
- With the leak: \( F - L = \dfrac{V}{10} \)

Subtracting the second equation from the first, we get:\[ L = \dfrac{V}{8} - \dfrac{V}{10} \]\[ L = \dfrac{5V - 4V}{40} \]\[ L = \dfrac{V}{40} \]

So, the leak empties \( \dfrac{1}{40} \) of the cistern's volume per hour.

To find the time taken by the leak to empty the cistern when it's full, we use the formula:\[ \text{Time} = \dfrac{\text{Volume}}{\text{Rate}} \]

Substituting the volume of the cistern (\( V \)) and the rate of the leak (\( \dfrac{1}{40} \)) into the formula, we get:\[ \text{Time} = \dfrac{V}{\dfrac{1}{40}} = 40V \]

Since \( V \) represents the volume of the cistern, it will take \( 40 \times 1 \) hours for the leak to empty the cistern when it's full.

Therefore, the leak will empty the cistern in 40 hours. So, the correct option is 40 hrs.

Let's find the rates at which each tap empties the tank per hour.

The first tap can empty the tank in one hour, so its rate is \( \dfrac{1}{1} = 1 \) tank per hour.

The second tap can empty the tank in 30 minutes, so its rate is \( \dfrac{1}{\dfrac{1}{2}} = 2 \) tanks per hour.

Now, when both taps operate simultaneously, their combined rate is the sum of their individual rates.

Combined rate \( = 1 + 2 = 3 \) tanks per hour.

Now, to find out how much time is needed to empty the tank when both taps operate simultaneously, we can use the formula:

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

Since the whole tank needs to be emptied, the amount is 1 tank.

\[ \text{Time} = \dfrac{1 \text{ tank}}{3 \text{ tanks/hour}} \]

\[ \text{Time} = \dfrac{1}{3} \text{ hour} \]

Now, to convert hours to minutes, we multiply by 60:

\[ \text{Time} = \dfrac{1}{3} \times 60 \text{ minutes} \]

\[ \text{Time} = 20 \text{ minutes} \]

Therefore, when both taps operate simultaneously, it will take 20 minutes to empty the tank. So, the correct option is 20 minutes.

Let's denote:
- Rate of Tap A = \( R_A \) (portion per hour)
- Rate of Tap B = \( R_B \) (portion per hour)
- Rate of Tap C = \( R_C \) (portion per hour)

Given:
- Combined rate of all three taps = \( \dfrac{1}{6} \) portion per hour
- Combined rate of Tap A and Tap B (after Tap C is closed) = \( \dfrac{1}{7} \) portion per hour
- Tap C was closed after 2 hours of all taps working together

From the information given, we can set up the following equations:

1. For the combined rates of all three taps working together for 2 hours:\[ 2(R_A + R_B + R_C) = 2 \times \dfrac{1}{6} \]

2. For the combined rates of Tap A and Tap B working together after Tap C is closed:\[ 7(R_A + R_B) = 1 - 2 \times \dfrac{1}{6} \]

We can solve these equations to find the rates of Tap A and Tap B combined, and then find the rate of Tap C.

Let's solve the first equation:\[ 2(R_A + R_B + R_C) = \dfrac{1}{3} \]\[ R_A + R_B + R_C = \dfrac{1}{6} \]\[ R_C = \dfrac{1}{6} - (R_A + R_B) \]

Now, let's solve the second equation:\[ 7(R_A + R_B) = 1 - \dfrac{1}{3} \]\[ 7(R_A + R_B) = \dfrac{2}{3} \]\[ R_A + R_B = \dfrac{2}{21} \]

Now, let's substitute \( R_A + R_B = \dfrac{2}{21} \) into \( R_C = \dfrac{1}{6} - (R_A + R_B) \):\[ R_C = \dfrac{1}{6} - \dfrac{2}{21} \]\[ R_C = \dfrac{7}{42} - \dfrac{4}{42} \]\[ R_C = \dfrac{3}{42} \]\[ R_C = \dfrac{1}{14} \]

So, Tap C's rate is \( \dfrac{1}{14} \) portion per hour.

Therefore, Tap C takes 14 hours alone to fill the tank. So, the correct answer is 14.

Let's denote:
- Rate of Pipe A = \( R_A \) (portion per hour)
- Rate of Pipe B = \( R_B \) (portion per hour)
- Rate of Waste Pipe = \( R_W \) (portion per hour)

Given:
- Pipe A can fill the cistern in 3 hours, so \( R_A = \dfrac{1}{3} \) portion per hour.
- Pipe B can fill the cistern in 4 hours, so \( R_B = \dfrac{1}{4} \) portion per hour.
- Waste Pipe can empty the cistern in 2 hours, so \( R_W = -\dfrac{1}{2} \) portion per hour (negative because it empties the cistern).

The combined rate of filling when all pipes are open is the sum of the individual rates:

Combined Rate = \( R_A + R_B + R_W \)

Substituting the given values:

Combined Rate = \( \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{2} \)

Combining the fractions:

Combined Rate = \( \dfrac{4}{12} + \dfrac{3}{12} - \dfrac{6}{12} \)

Combined Rate = \( \dfrac{1}{12} \) portion per hour

Therefore, the cistern will be filled in 12 hours when all three pipes are kept open. So, the correct answer is 12 hours.

Let's denote:
- Time taken by the slower pipe to fill the tank alone = \( T \) minutes.
- Time taken by the faster pipe to fill the tank alone = \( \dfrac{T}{3} \) minutes.

Given:
- Together, the two pipes can fill the tank in 36 minutes.

We can set up the following equation based on the rates of filling:

\[ \dfrac{1}{T} + \dfrac{1}{\dfrac{T}{3}} = \dfrac{1}{36}\]

Now, let's solve this equation to find the value of \( T \).

\[ \dfrac{1}{T} + \dfrac{3}{T} = \dfrac{1}{36}\]

\[ \dfrac{4}{T} = \dfrac{1}{36}\]

Now, we'll cross multiply:

\[ 4 \times 36 = T\]

\[ T = 144\]

Therefore, the slower pipe alone will be able to fill the tank in 144 minutes. So, the correct answer is 144 min.

Let's denote:
- Rate of the tap that can fill the cistern = \( R_F \) (portion per hour)
- Rate of the tap that can empty the cistern = \( R_E \) (portion per hour)

Given:
- Time taken by the tap to fill the cistern alone = 8 hours, so \( R_F = \dfrac{1}{8} \) portion per hour.
- Time taken by the tap to empty the cistern alone = 16 hours, so \( R_E = -\dfrac{1}{16} \) portion per hour (negative because it empties the cistern).

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

Combined Rate = \( R_F + R_E \)

Substituting the given values:

Combined Rate = \( \dfrac{1}{8} - \dfrac{1}{16} \)

Combining the fractions:

Combined Rate = \( \dfrac{2}{16} - \dfrac{1}{16} \)

Combined Rate = \( \dfrac{1}{16} \) portion per hour

Therefore, when both taps are open, the cistern will be filled in 16 hours. So, the correct answer is 16.

Let's denote:
- Time taken by the first pipe to fill the cistern alone = \( T_1 = 10 \) hours.
- Time taken by the second pipe to fill the cistern alone = \( T_2 = 15 \) 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:\[ \text{Rate}_1 = \dfrac{1}{T_1} = \dfrac{1}{10}\]

For the second pipe:\[ \text{Rate}_2 = \dfrac{1}{T_2} = \dfrac{1}{15}\]

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

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

Substituting the given values:

\[ \text{Combined Rate} = \dfrac{1}{10} + \dfrac{1}{15}\]

To combine the fractions, we need a common denominator, which is 30:

\[ \text{Combined Rate} = \dfrac{3}{30} + \dfrac{2}{30} = \dfrac{5}{30}\]

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

Therefore, the combined rate of the two pipes is \( \dfrac{1}{6} \) portion per hour.

Now, to find the time taken to fill the cistern when both pipes are open, we can use the formula:

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

Substituting the combined rate:

\[ \text{Time} = \dfrac{1}{\dfrac{1}{6}} = 6\]

Therefore, when both pipes are open together, they can fill the cistern in 6 hours. So, the correct answer is 6 hours.

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

Given:
- Pipe A is 3 times faster than pipe B, so the ratio of their rates is \( 3:1 \).
- Pipe A takes 32 minutes less than pipe B to fill the cistern, so \( T_A = T_B - 32 \).

We can set up the following equation based on their rates:

\[ \dfrac{1}{T_A} = 3 \times \dfrac{1}{T_B}\]

Substituting \( T_A = T_B - 32 \):

\[ \dfrac{1}{T_B - 32} = 3 \times \dfrac{1}{T_B}\]

Multiplying both sides by \( T_B (T_B - 32) \):

\[ T_B = 3(T_B - 32)\]

Expanding:

\[ T_B = 3T_B - 96\]

Rearranging:

\[ 2T_B = 96\]

\[ T_B = 48\]

Now, we can find \( T_A \):

\[ T_A = T_B - 32\]\[ T_A = 48 - 32\]\[ T_A = 16\]

Therefore, when both pipes are opened together, they will fill the cistern in \( \dfrac{1}{\dfrac{1}{16} + \dfrac{1}{48}} = \dfrac{48}{4} = 12 \) minutes. So, the correct answer is 12 min.