Pipes And Cistern All Competence Aptitude

If \( \dfrac{1}{3} \) of a tank holds 80 liters of water, then the total capacity of the tank is \( 3 \times 80 = 240 \) liters.

Now, if \( \dfrac{1}{2} \) of the tank holds water, it means \( \dfrac{1}{2} \times 240 = 120 \) liters of water.

Therefore, the quantity of water that \( \dfrac{1}{2} \) of the tank holds is 120 liters. Hence, the correct answer is 120 litres.

If $\frac{3}{4}$ part of a tank is full of water, it means $\frac{3}{4}$ of the tank's capacity is occupied by water.

Let's denote the capacity of the tank as $C$ liters.

Given that $\frac{3}{4}$ part of the tank is full of water, we can set up the equation:

$$\frac{3}{4}\times C=\text{Amount of water in the tank}$$

If 30 liters of water is taken out and the tank becomes empty, the amount of water left in the tank is $0$ liters. So, we have:

$$\frac{3}{4}\times C-30=0$$

Solving for $C$:

$$\frac{3}{4}\times C=30$$

$$C=\frac{30}{\frac{3}{4}}$$

$$C=30\times \frac{4}{3}$$

$$C=40$$

Therefore, the capacity of the tank is 40 liters. So, the correct answer is 40 litres.

Let's first find the rates at which each pipe fills or empties the tank:

- Pipe P fills the tank in 10 hours, so its rate is \( \dfrac{1}{10} \) tank per hour.
- Pipe Q fills the tank in 12 hours, so its rate is \( \dfrac{1}{12} \) tank per hour.
- Pipe C empties the tank in 6 hours, so its rate is \( -\dfrac{1}{6} \) tank per hour (negative because it empties the tank).

When all three pipes are open, their combined rate is the sum of their individual rates:

\[ \text{Combined Rate} = \dfrac{1}{10} + \dfrac{1}{12} - \dfrac{1}{6}\]

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

\[ \text{Combined Rate} = \dfrac{6}{60} + \dfrac{5}{60} - \dfrac{10}{60}\]

\[ \text{Combined Rate} = \dfrac{1}{60} \text{ tank per hour}\]

Now, let's find out how much time it takes to fill one-fourth of the tank, which is \( \dfrac{1}{4} \) of the tank.

Using the formula:

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

Substituting the volume as \( \dfrac{1}{4} \) of the tank and the combined rate:

\[ \text{Time} = \dfrac{\dfrac{1}{4}}{\dfrac{1}{60}} = \dfrac{1}{4} \times 60 = 15 \text{ hours}\]

7 a.m. + 15 hours = 10 p.m.

Therefore, one-fourth of the tank will be filled at 10 p.m.

First, let's find the rate at which each pipe fills the tank:

- Pipe A fills the tank in 6 hours, so its rate is \( \dfrac{1}{6} \) tank per hour.
- Pipe B fills the tank in 9 hours, so its rate is \( \dfrac{1}{9} \) tank per hour.
- Pipe C fills the tank in 12 hours, so its rate is \( \dfrac{1}{12} \) tank per hour.

When pipes B and C are opened together for half an hour, they fill:

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

\[ \text{Volume} = \left(\dfrac{1}{9} + \dfrac{1}{12}\right) \times \dfrac{1}{2}\]

\[ \text{Volume} = \left(\dfrac{4}{36} + \dfrac{3}{36}\right) \times \dfrac{1}{2}\]

\[ \text{Volume} = \left(\dfrac{7}{36}\right) \times \dfrac{1}{2}\]

\[ \text{Volume} = \dfrac{7}{72} \text{ of the tank}\]

After half an hour, the remaining volume of the tank is \( 1 - \dfrac{7}{72} = \dfrac{65}{72} \) of the tank.

Now, when pipes A, B, and C are opened together, their combined rate is:

\[ \text{Combined Rate} = \dfrac{1}{6} + \dfrac{1}{9} + \dfrac{1}{12} = \dfrac{12}{72} + \dfrac{8}{72} + \dfrac{6}{72} = \dfrac{26}{72} \text{ tank per hour}\]

Now, to find the time taken by the three pipes together to fill the remaining part of the tank:

\[ \text{Time} = \dfrac{\text{Volume}}{\text{Rate}} = \dfrac{\dfrac{65}{72}}{\dfrac{26}{72}} = \dfrac{65}{26} = 2.5 \text{ hours}\]

Therefore, the time taken by the three pipes together to fill the remaining part of the tank is \( 2.5 \) hours. So, the correct answer is 2 and 1/2 hours.

Let's first find the rate at which the single tap fills the tank:
- The single tap fills the tank in 6 hours, so its rate is $\frac{1}{6}$ tank per hour.

When half of the tank is filled, it means that $\frac{1}{2}$ of the tank's volume is filled.

Now, let's find the time it takes for the single tap to fill half of the tank:
$$\text{Time}=\frac{\text{Volume}}{\text{Rate}}=\frac{\frac{1}{2}}{\frac{1}{6}}=3\text{ hours}$$

So, it takes 3 hours for the single tap to fill half of the tank.

Now, when three more similar taps are opened, the rate at which the tank is filled becomes four times the rate of the single tap:
$$\text{Total rate}=4\times \frac{1}{6}=\frac{4}{6}=\frac{2}{3}\text{ tank per hour}$$

Now, let's find the time it takes for the four taps together to fill the remaining half of the tank:
$$\text{Time}=\frac{\text{Volume}}{\text{Rate}}=\frac{\frac{1}{2}}{\frac{2}{3}}=\frac{3}{4}\text{ hours}=45\text{ minutes}$$

Therefore, the total time taken to fill the tank completely is $3\text{ hours}+45\text{ minutes}=\text{3 hours 45 minutes}$. 

So, the correct answer is 3 hours 45 minutes.

Let's find the rates at which each tap can fill or empty the cistern:
- The first tap can fill the cistern in 40 minutes, so its rate is \( \dfrac{1}{40} \) cistern per minute.
- The second tap can empty the cistern in 60 minutes, so its rate is \( -\dfrac{1}{60} \) cistern per minute (negative because it empties the cistern).

When both taps are open, the effective rate at which the cistern is filled is the sum of their individual rates:

\[ \text{Effective Rate} = \dfrac{1}{40} - \dfrac{1}{60}\]

To combine the fractions, we need a common denominator, which is 120:

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

Therefore, the cistern will be filled in 120 minutes. So, the correct answer is 120 minutes.

If a cistern is normally filled in 8 hours but takes another 2 hours longer to fill because of a leak, it means that the leak is draining water at a rate that reduces the filling rate of the cistern.

Let's denote the rate at which the cistern is filled normally as \( R_f \) and the rate at which it is filled with the leak as \( R_l \).

Given:
- Normal filling time (\( T_f \)) = 8 hours
- Time taken with the leak (\( T_l \)) = 8 + 2 = 10 hours

Using the formula \(\text{Rate} = \dfrac{1}{\text{Time}}\), we can find the rates of filling:

\[ R_f = \dfrac{1}{8} \text{ cisterns per hour}\]\[ R_l = \dfrac{1}{10} \text{ cisterns per hour}\]

Now, let's find the rate at which the leak empties the cistern:
- Since the leak is causing the cistern to take longer to fill, it's effectively reducing the rate of filling by \( R_f - R_l \):

\[ \text{Leak Rate} = R_f - R_l = \dfrac{1}{8} - \dfrac{1}{10}\]

\[ \text{Leak Rate} = \dfrac{5}{40} - \dfrac{4}{40} = \dfrac{1}{40} \text{ cisterns per hour}\]

Therefore, the leak empties the cistern at a rate of \( \dfrac{1}{40} \) cisterns per hour.

Now, to find the time taken by the leak to empty the full cistern, we use the formula \( \text{Time} = \dfrac{\text{Volume}}{\text{Rate}} \), where \( \text{Volume} = 1 \) cistern:

\[ \text{Time} = \dfrac{1}{\dfrac{1}{40}} = 40 \text{ hours}\]

So, if the cistern is full, the leak will empty it in 40 hours. Therefore, the correct answer is 40 hours.

Let the rate of pipe A be \(a\), pipe B be \(b\), and pipe C be \(c\), such that their combined rate is \(a + b + c\).

Given:
1. \( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{1}{6} \) (Since they can fill the cistern in 6 hours when working together)
2. They work together for 2 hours, so \( (a + b + c) \times 2 \) of the cistern is filled.
3. After 2 hours, only pipes A and B are working, and they take 7 more hours to fill the cistern, so \( (a + b) \times 7 \) of the cistern is filled.

From the first equation, \( \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{6} - \dfrac{1}{c} \).

After 2 hours, \( (a + b + c) \times 2 \) of the cistern is filled, so \( \dfrac{2}{6} = \dfrac{1}{3} \) of the cistern is filled.

After the next 7 hours, \( (a + b) \times 7 \) of the cistern is filled, so \( \dfrac{7}{6} - \dfrac{1}{3} = \dfrac{5}{6} \) of the cistern is filled.

This \( \dfrac{5}{6} \) of the cistern is filled by pipes A and B in 7 hours, so their combined rate is \( \dfrac{5}{6} \div 7 = \dfrac{5}{42} \).

So, \( \dfrac{1}{c} = \dfrac{1}{6} - \dfrac{1}{6} + \dfrac{5}{42} = \dfrac{1}{14} \).

Therefore, the time taken by C alone to fill the cistern is 14 hours.

\[ \text{Pipe filling rate} = \dfrac{1}{9} \;\; \text{(cistern/hour)} \]

\[ \text{Net filling rate with leak} = \dfrac{1}{10} \;\; \text{(cistern/hour)} \]

\[ \text{Leak rate} = \dfrac{1}{9} - \dfrac{1}{10} = \dfrac{10 - 9}{90} = \dfrac{1}{90} \;\; \text{(cistern/hour)} \]

\[ \text{Time taken by leak to empty the full cistern} = \dfrac{1}{\tfrac{1}{90}} = 90 \;\text{hours} \]

Let's denote the rate at which the pipe fills the tank as \( F \) (tanks per hour) and the rate at which the leak empties the tank as \( L \) (tanks per hour).

Given:
1. The pipe can fill the tank in 24 hours, so its filling rate is \( F = \dfrac{1}{24} \) tanks per hour.
2. Due to the leak, the tank is filled in 36 hours, so the effective filling rate is \( \dfrac{1}{36} \) tanks per hour.
3. The difference between the filling rate and the effective filling rate is the rate at which the leak empties the tank: \( F - \dfrac{1}{36} = L \).

Now, let's solve for \( L \):\[ F - \dfrac{1}{36} = L \]\[ \dfrac{1}{24} - \dfrac{1}{36} = L \]\[ \dfrac{3 - 2}{72} = L \]\[ L = \dfrac{1}{72} \] tanks per hour.

Now, if the tank is half full, it means that \( \dfrac{1}{2} \) of the tank's volume is filled. The time taken by the leak to empty the remaining half of the tank would be the same as the time it takes to fill the tank from empty to half-full, which is 36 hours.

Therefore, the leak will take 36 hours to empty the tank when it is half full. Hence, the correct answer is 36 hours.