Pipes And Cistern All Competence Aptitude

Let the rates of pipes A, B, and C be \(a\), \(b\), and \(c\) (in tanks per hour), respectively.

Given:
1. The combined rate of pipes A, B, and C is \(a + b + c\) tanks per hour.
2. They can fill the tank in 6 hours, so the effective filling rate is \(\dfrac{1}{6}\) tanks per hour.
3. After working together for 2 hours, they fill \(2(a + b + c)\) tanks.

So, the remaining part of the tank that can be filled by pipes A and B in 7 hours is \(7(a + b)\) tanks.

Since the total capacity of the tank is 1, we have:

\[2(a + b + c) + 7(a + b) = 1\]

Now, we can find the rates of A and B combined:\[a + b + c = \dfrac{1}{6}\]

\[a + b = \dfrac{1}{6} - c\]

\[7(a + b) = 7(\dfrac{1}{6} - c) = \dfrac{7}{6} - 7c\]

Given that \(2(a + b + c) + 7(a + b) = 1\), we can substitute the values:

\[2(\dfrac{1}{6}) + \dfrac{7}{6} - 7c = 1\]

\[ \dfrac{1}{3} + \dfrac{7}{6} - 7c = 1\]

\[ \dfrac{7}{6} - 7c = 1 - \dfrac{1}{3}\]

\[ \dfrac{7}{6} - 7c = \dfrac{2}{3}\]

Now, solve for \(c\):

\[ 7c = \dfrac{7}{6} - \dfrac{2}{3}\]

\[ 7c = \dfrac{7}{6} - \dfrac{4}{6}\]

\[ 7c = \dfrac{3}{6}\]

\[ 7c = \dfrac{1}{2}\]

\[ c = \dfrac{1}{14}\]

Therefore, the number of hours taken by C alone to fill the tank is 14. Hence, the correct answer is 14 hours.

Let's denote the rate of inflow of the tap as \( F \) (tanks per hour) and the rate of leakage as \( L \) (tanks per hour).

Given:
1. The tap takes 36 hours extra to fill the tank due to the leakage, which means the effective inflow rate is \( F - L \).
2. The leakage is equivalent to half of the inflow, so \( L = \dfrac{1}{2}F \).

So, the effective inflow rate is \( F - \dfrac{1}{2}F = \dfrac{1}{2}F \).

Since the tank can be filled in \( F \) hours normally, and the effective inflow rate is \( \dfrac{1}{2}F \), it will take twice the normal time to fill the tank. Therefore, the normal time taken to fill the tank is half of the time taken with the leakage, which is \( \dfrac{36}{2} = 18 \) hours.

Therefore, the inflow can fill the tank in 18 hours. Hence, the correct answer is 18 hours.

The rate at which water drains through a pipe is directly proportional to the square of its diameter. This means that if the diameter of a pipe is doubled, its drainage rate becomes four times as fast.

Let's denote the time taken by the pipe with diameter \( d \) to drain the tank as \( t \). Since the drainage rate is directly proportional to the square of the diameter, we have:

\[ \dfrac{1}{t} \propto d^2\]

Now, if we double the diameter to \( 2d \), the drainage rate becomes \( 4 \) times as fast. Therefore, the time taken by the pipe with diameter \( 2d \) to drain the tank becomes \( \dfrac{1}{4}t \).

Given that the time taken by the pipe with diameter \( d \) is \( 40 \) minutes, we can find the time taken by the pipe with diameter \( 2d \) as follows:

\[ \dfrac{1}{4}t = \dfrac{1}{4} \times 40 = 10 \text{ minutes}\]

So, the time taken by a pipe of diameter \( 2d \) to drain the same tank is \( 10 \) minutes.

Let's denote the rate at which the slower pipe fills the tank as \( r \) tanks per minute. Since the faster pipe can fill the tank three times as fast, its rate would be \( 3r \) tanks per minute.

Given that together the two pipes can fill the tank in 36 minutes, their combined rate is \( r + 3r = 4r \) tanks per minute.

We can express the combined rate in terms of the time taken to fill the tank:\[ \text{Combined rate} \times \text{Time taken} = \text{Volume of tank}\]\[ 4r \times 36 = 1\]\[ 144r = 1\]\[ r = \dfrac{1}{144}\]

Now, we need to find the time taken for the slower pipe alone to fill the tank, which is \( r \) tanks per minute:\[ \text{Time taken by slower pipe} = \dfrac{1}{r}\]\[ = \dfrac{1}{\dfrac{1}{144}}\]\[ = 144 \text{ minutes}\]

Therefore, the slower pipe alone will be able to fill the tank in 144 minutes.

Let's denote the time taken by the first pipe to empty the tank as \( t_1 \) and the time taken by the second pipe (with twice the diameter) as \( t_2 \).

Given:
- The first pipe can empty the tank in 40 minutes, so \( t_1 = 40 \) minutes.
- Since the second pipe has twice the diameter, its drainage rate is four times as fast. Therefore, \( t_2 = \dfrac{40}{4} = 10 \) minutes.

Now, let's find the combined rate of both pipes. The combined rate is the sum of their individual rates, which is the reciprocal of the time taken by each pipe:\[ \text{Combined rate} = \dfrac{1}{t_1} + \dfrac{1}{t_2}\]

Plugging in the values, we get:\[ \text{Combined rate} = \dfrac{1}{40} + \dfrac{1}{10} = \dfrac{1}{40} + \dfrac{4}{40} = \dfrac{5}{40} = \dfrac{1}{8}\]

So, the combined rate of both pipes is \( \dfrac{1}{8} \) tanks per minute.

Now, let's find the time taken by both pipes together to empty the tank. Using the formula \( \text{Time} = \dfrac{\text{Volume}}{\text{Rate}} \), where the volume of the tank is 1 (assuming it's a unit tank), we get:\[ \text{Time} = \dfrac{1}{\text{Combined rate}} = \dfrac{1}{\dfrac{1}{8}} = 8 \text{ minutes}\]

Therefore, the two pipes together can empty the tank in 8 minutes.

To find the number of litres wasted in 300 days, we first need to find out how many drops are wasted in 300 days.

Given:
- The tap drips at a rate of one drop per second.
- 600 drops make 100 ml.

So, the number of drops wasted in 300 days is calculated as follows:

\[ \text{Number of drops wasted in 300 days} = \text{Number of drops per day} \times \text{Number of days} \]

The number of drops per day can be calculated as:

\[ \text{Number of drops per day} = \text{Number of drops per second} \times \text{Number of seconds in a day} \]

We know that there are \(60 \times 60 \times 24 = 86400\) seconds in a day. 

So, the number of drops wasted in 300 days is:

\[ \text{Number of drops wasted in 300 days} = 1 \times 86400 \times 300 = 25920000 \]

Now, we need to convert the number of drops wasted into litres.

Given that 600 drops make 100 ml, we can calculate how many drops make 1 litre:

\[ \text{Number of drops to make 1 litre} = \dfrac{1000 \times 600}{100} = 6000 \]

Now, to find out how many litres are wasted, we divide the number of drops wasted by the number of drops that make 1 litre:

\[ \text{Number of litres wasted} = \dfrac{25920000}{6000} = 4320 \]

Therefore, the number of litres wasted in 300 days is 4320.

Let's find the rate at which each pipe fills or empties the tank.

Pipe 1 can fill the tank in 15 hours, so its filling rate is \( \dfrac{1}{15} \) tanks per hour.

Pipe 2 can fill the tank in 12 hours, so its filling rate is \( \dfrac{1}{12} \) tanks per hour.

Pipe 3 can empty the tank in 4 hours, so its emptying rate is \( \dfrac{1}{4} \) tanks per hour.

Now, let's find out the cumulative rate of filling or emptying when the pipes are opened in order at 8, 9, and 11 a.m. respectively.

From 8 a.m. to 9 a.m., both Pipe 1 and Pipe 2 are filling the tank. So, the cumulative filling rate during this hour is:

\[ \text{Rate} = \dfrac{1}{15} + \dfrac{1}{12} \]

From 9 a.m. to 11 a.m., only Pipe 2 is filling the tank, so the cumulative filling rate during these 2 hours is:

\[ \text{Rate} = \dfrac{1}{12} \times 2 \]

At 11 a.m., Pipe 3 starts emptying the tank. So, the net rate of filling or emptying from 11 a.m. onwards is:

\[ \text{Rate} = \dfrac{1}{12} \times 2 - \dfrac{1}{4} \]

Now, let's calculate the cumulative filling or emptying rates:

From 8 a.m. to 9 a.m.:\[ \text{Rate} = \dfrac{1}{15} + \dfrac{1}{12} = \dfrac{4}{60} + \dfrac{5}{60} = \dfrac{9}{60} = \dfrac{3}{20} \] tanks per hour.

From 9 a.m. to 11 a.m.:\[ \text{Rate} = \dfrac{1}{12} \times 2 = \dfrac{2}{12} = \dfrac{1}{6} \] tanks per hour.

At 11 a.m. and onwards:\[ \text{Rate} = \dfrac{1}{6} - \dfrac{1}{4} = \dfrac{2}{12} - \dfrac{3}{12} = -\dfrac{1}{12} \] tanks per hour.

Since the net rate is negative, it indicates that the tank is being emptied.

Now, let's calculate the time it takes for the tank to be emptied completely with this negative rate:

\[ \text{Time} = \dfrac{\text{Volume}}{\text{Rate}} = \dfrac{1}{\left(-\dfrac{1}{12}\right)} = 12 \] hours.

So, the tank will be emptied at \( 11 \, \text{a.m.} + 12 \, \text{hours} = 11 \, \text{p.m.} \).

Therefore, the tank will be emptied at 11:40 a.m.

Let's find the rates at which each pipe fills the bucket.

Pipe 1 can fill the bucket in 20 minutes, so its filling rate is \( \dfrac{1}{20} \) buckets per minute.

Pipe 2 can fill the bucket in 25 minutes, so its filling rate is \( \dfrac{1}{25} \) buckets per minute.

When both pipes are opened together for 5 minutes, the total volume of water filled by both pipes is:

\[ \text{Volume filled by both pipes in 5 minutes} = \left( \dfrac{1}{20} + \dfrac{1}{25} \right) \times 5 \]

\[ = \left( \dfrac{5}{100} + \dfrac{4}{100} \right) \times 5 \]

\[ = \left( \dfrac{9}{100} \right) \times 5 \]

\[ = \dfrac{45}{100} \]

\[ = 0.45 \]

So, after 5 minutes, 0.45 of the bucket is filled by both pipes together. This means that the remaining portion of the bucket that needs to be filled by the first pipe alone is \( 1 - 0.45 = 0.55 \) of the bucket.

Now, let's find out the time taken by the first pipe alone to fill this remaining portion of the bucket:

\[ \text{Time taken by first pipe alone} = \dfrac{\text{Volume filled by first pipe alone}}{\text{Rate of first pipe}} \]

\[ = \dfrac{0.55}{\dfrac{1}{20}} \]

\[ = 0.55 \times 20 \]

\[ = 11 \text{ minutes} \]

Therefore, the time taken by the first pipe alone to fill the remaining portion of the bucket is 11 minutes.

Let's denote the rate at which the first pipe fills the tank as \( R_1 \) (tanks per hour) and the rate at which the second pipe empties the tank as \( R_2 \) (tanks per hour).

Given:
- The first pipe can fill the tank in \( x \) hours, so its rate is \( R_1 = \dfrac{1}{x} \) tanks per hour.
- The second pipe can empty the tank in \( y \) hours, so its rate is \( R_2 = \dfrac{1}{y} \) tanks per hour.

When both pipes are open together, the net rate of filling or emptying the tank is the difference between the rate of filling by the first pipe and the rate of emptying by the second pipe.

So, the net rate when both pipes are open is \( R_1 - R_2 = \dfrac{1}{x} - \dfrac{1}{y} \) tanks per hour.

Now, let's find the time taken to fill the tank completely with this net rate:

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

Since we are filling the tank, the volume is positive. So, the time taken to fill the tank will be:

\[ \text{Time} = \dfrac{1}{\left( \dfrac{1}{x} - \dfrac{1}{y} \right)} \]

This can be simplified as:

\[ \text{Time} = \dfrac{xy}{y - x} \] 

Therefore, the tank will be filled in \( \dfrac{xy}{y - x} \) hours.

Let's denote the time after which Y should be closed as \( t \) minutes.

The rate at which pipe X fills the cistern is \( \dfrac{1}{24} \) tanks per minute, and the rate at which pipe Y fills the cistern is \( \dfrac{1}{32} \) tanks per minute.

When both pipes are opened together for \( t \) minutes, they fill a portion of the tank. The portion filled by pipe X in \( t \) minutes is \( \dfrac{t}{24} \) and the portion filled by pipe Y in \( t \) minutes is \( \dfrac{t}{32} \).

So, the total portion filled by both pipes together in \( t \) minutes is \( \dfrac{t}{24} + \dfrac{t}{32} \).

Given that the tank should be full in 18 minutes, the remaining time after \( t \) minutes should be \( 18 - t \) minutes.

The portion that needs to be filled in \( 18 - t \) minutes is \( 1 - \left( \dfrac{t}{24} + \dfrac{t}{32} \right) \).

Setting up the equation:

\[ 1 - \left( \dfrac{t}{24} + \dfrac{t}{32} \right) = \dfrac{18 - t}{24} \]

Solving this equation will give us the value of \( t \). Let's do the math:

\[ 1 - \left( \dfrac{t}{24} + \dfrac{t}{32} \right) = \dfrac{18 - t}{24} \]

\[ 1 - \left( \dfrac{4t + 3t}{96} \right) = \dfrac{18 - t}{24} \]

\[ 1 - \left( \dfrac{7t}{96} \right) = \dfrac{18 - t}{24} \]

\[ 1 - \dfrac{7t}{96} = \dfrac{18 - t}{24} \]

\[ \dfrac{96}{96} - \dfrac{7t}{96} = \dfrac{18 - t}{24} \]

\[ \dfrac{96 - 7t}{96} = \dfrac{18 - t}{24} \]

\[ 24(96 - 7t) = 96(18 - t) \]

\[ 2304 - 168t = 1728 - 96t \]

\[ 2304 - 1728 = 168t - 96t \]

\[ 576 = 72t \]

\[ t = \dfrac{576}{72} \]

\[ t = 8 \]

Therefore, Y should be closed after 8 minutes. So, the correct answer is 8 minutes.