Pipes And Cistern All Competence Aptitude

Let's find the rates at which each pump fills the tank:

- The first pump can fill the tank in 8 hours, so its rate is \( \dfrac{1}{8} \) tanks per hour.
- The second pump can fill the tank in 10 hours, so its rate is \( \dfrac{1}{10} \) tanks per hour.

When both pumps are opened at the same time, their combined rate is the sum of their individual rates:

\[ \text{Combined rate} = \dfrac{1}{8} + \dfrac{1}{10} \]

\[ = \dfrac{5}{40} + \dfrac{4}{40} \]

\[ = \dfrac{9}{40} \]

So, the combined rate of both pumps is \( \dfrac{9}{40} \) tanks per hour.

Now, in 4 hours, the amount of tank filled by both pumps is:

\[ \text{Amount filled} = \text{Rate} \times \text{Time} \]

\[ = \dfrac{9}{40} \times 4 \]

\[ = \dfrac{36}{40} \]

\[ = \dfrac{9}{10} \]

Therefore, the part of the tank that will be filled up in 4 hours is \( \dfrac{9}{10} \).

Given:

Two pipes can fill a cistern in 20 minutes and 30 minutes.

Let the capacity of cistern be ‘100’.

% Cistern is filled by 1^st pipe in one minute = 100/20

⇒ 5%

% Cistern is filled by 2^nd pipe in one minute = 100/30

⇒ 3.33%
% Cistern filled by 1^st and 2^nd pipes in one minute

⇒ 5% + 3.33%

⇒ 8.33%
Cistern is totally filled by 2^nd pipe in last 10 minute.
⇒ 3.33% × 10

⇒ 33.3%

Remaining filled by 1^st pipe and 2^nd pipe

⇒ 100 – 33.3

⇒ 66.6%

Time taken to fill remaining by both pipe

⇒ 66.6/8.33

⇒ 8 minutes

∴ The first pipe must be turned off in 8 minutes.

Let's break down the problem:

Let the efficiency of pipe A be \( a \) (fraction of tank filled per hour),
the efficiency of pipe B be \( b \) (fraction of tank filled per hour), and
the efficiency of pipe C be \( c \) (fraction of tank filled per hour).

Given:
1. Together, pipes A, B, and C can fill the tank in 6 hours.
So, their combined efficiency is \(\dfrac{1}{6}\) of the tank per hour.

2. After working together for 2 hours, C is closed.
So, in those 2 hours, A, B, and C together filled \(\dfrac{2}{6}\) of the tank.

3. After that, A and B fill the tank in 8 hours.
Their combined efficiency is \(\dfrac{1}{8}\) of the tank per hour.

From (2), we know that in the first 2 hours, A, B, and C together filled \(\dfrac{1}{3}\) of the tank, and in the next 8 hours, A and B alone filled the remaining \(\dfrac{2}{3}\) of the tank.

Let's set up the equations based on the efficiencies:

First, let's write the equation for the first 2 hours:\[2(a + b + c) = \dfrac{1}{3}\]

Then, the equation for the next 8 hours:\[8(a + b) = \dfrac{2}{3}\]

From these equations, we can find the values of \(a\), \(b\), and \(c\).

Once we have those, we can find the time it takes for pipe C alone to fill the tank, which would be \(\dfrac{1}{c}\).

Let's solve the equations:

From the first equation:\[2(a + b + c) = \dfrac{1}{3}\]\[a + b + c = \dfrac{1}{6}\]

From the second equation:\[8(a + b) = \dfrac{2}{3}\]\[a + b = \dfrac{1}{12}\]

Subtracting the second equation from the first:\[c = \dfrac{1}{6} - \dfrac{1}{12}\]\[c = \dfrac{1}{12}\]

So, pipe C alone can fill the tank in 12 hours. 

The correct answer is 12.

Let's denote the capacity of the tank as \( V \) gallons.

Pipe A fills the tank in 20 minutes, so its filling rate is \( \dfrac{V}{20} \) per minute.
Pipe B fills the tank in 24 minutes, so its filling rate is \( \dfrac{V}{24} \) per minute.
Pipe C empties the tank at a rate of 3 gallons per minute.

When all three pipes are opened together, they fill the tank in 15 minutes, so their combined filling rate is \( \dfrac{V}{15} \) per minute.

Setting up the equation based on the rates:

\[ \dfrac{V}{20} + \dfrac{V}{24} - 3 = \dfrac{V}{15} \]

Multiplying by the least common multiple of 20, 24, and 15, which is 120, to eliminate the denominators:

\[ 6V + 5V - 360 = 8V \]\[ 11V - 360 = 8V \]\[ 3V = 360 \]\[ V = 120 \]

So, the capacity of the tank is 120 gallons.

Let's denote the capacity of the tank as \( V \) liters.

The leak empties the tank in 6 hours, so its rate is \( \dfrac{V}{6} \) liters per hour.
The inlet pipe fills water at a rate of 4 liters per minute, so its rate is \( 4 \times 60 = 240 \) liters per hour.

When the tank is full, the inlet is opened, and due to the leak, the tank is empty in 8 hours. So, the combined rate of the inlet and the leak is \( \dfrac{V}{8} \) liters per hour.

Setting up the equation based on the rates:

\[ \dfrac{V}{6} - 240 = \dfrac{V}{8} \]

Multiplying both sides by 24 to eliminate the denominators:

\[ 4V - 5760 = 3V \]

\[ V = 5760 \]

So, the capacity of the tank is 5760 liters.

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

The first pipe fills the tank in 4 hours, so its filling rate is \( \dfrac{1}{4} \) tanks per hour.
The second pipe empties the tank in 16 hours, so its emptying rate is \( \dfrac{1}{16} \) tanks per hour.

When both pipes are opened together, their net filling rate is the difference between the filling rate of the first pipe and the emptying rate of the second pipe.

So, the net filling rate when both pipes are opened together is:

\[ \dfrac{1}{4} - \dfrac{1}{16} = \dfrac{4}{16} - \dfrac{1}{16} = \dfrac{3}{16} \]

This means that together, the two pipes fill \( \dfrac{3}{16} \) of the tank per hour.

To find out how long it takes to fill the tank completely, we can set up the equation:

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

Given that the tank is completely filled in \( \dfrac{3}{16} \) of the tank per hour, we can find the time it takes to fill the tank completely:

\[ \text{Time} = \dfrac{1}{\dfrac{3}{16}} = \dfrac{16}{3} \]

This is equivalent to \( 5 \dfrac{1}{3} \) hours.

So, the tank will be filled in \( 5 \dfrac{1}{3} \) hours when both pipes are opened together.

Alright, let's solve it step by step using the approach you provided.

Given:
Pipe A can fill the cistern in 752 minutes.
Pipe B can fill the cistern in 45 minutes.

We find the least common multiple (LCM) of 752 and 45, which is 225.

Using the LCM:
Efficiency of Pipe A = \( \dfrac{225}{37.5} = 6 \) (since 752 minutes = 225/37.5 minutes)
Efficiency of Pipe B = \( \dfrac{225}{45} = 5 \)

Let \( x \) be the time in minutes for which both pipes are opened together.

According to the question:

\[ ( \text{Efficiency of A} + \text{Efficiency of B}) \times x + (\text{Efficiency of A}) \times (30 - x) = 225\]

Plugging in the values:

\[ (6 + 5) \times x + 6 \times (30 - x) = 225\]

Simplify the equation:

\[ 11x + 180 - 6x = 225\]\[ 5x = 225 - 180\]\[ 5x = 45\]\[ x = 9\]

So, the required value of \( x \) is 9 minutes.

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

Pipe A can fill the tank in 2 hours, so its filling rate is \( \dfrac{1}{2} \) tank per hour.
Pipe B can fill the tank in 6 hours, so its filling rate is \( \dfrac{1}{6} \) tank per hour.

Now, let's find out how much of the tank pipe A fills from 10 A.M. to 11 A.M., which is 1 hour:

\[ \text{Amount filled by pipe A} = \text{Rate of pipe A} \times \text{Time} \]\[ = \dfrac{1}{2} \times 1 \]\[ = \dfrac{1}{2} \]

So, from 10 A.M. to 11 A.M., pipe A fills \( \dfrac{1}{2} \) of the tank.

Now, at 11 A.M., pipe B is opened. Together, pipes A and B will fill the remaining \( 1 - \dfrac{1}{2} = \dfrac{1}{2} \) of the tank.

The combined rate of pipes A and B is:

\[ \text{Combined rate} = \text{Rate of pipe A} + \text{Rate of pipe B} \]\[ = \dfrac{1}{2} + \dfrac{1}{6} \]\[ = \dfrac{3}{6} + \dfrac{1}{6} \]\[ = \dfrac{4}{6} \]\[ = \dfrac{2}{3} \]

So, the combined rate of pipes A and B is \( \dfrac{2}{3} \) tank per hour.

Now, let's find out how long it takes for pipes A and B together to fill \( \dfrac{1}{2} \) of the tank:

\[ \text{Time} = \dfrac{\text{Amount of work}}{\text{Rate of work}} \]\[ = \dfrac{\dfrac{1}{2}}{\dfrac{2}{3}} \]\[ = \dfrac{1}{2} \times \dfrac{3}{2} \]\[ = \dfrac{3}{4} \]

So, it takes \( \dfrac{3}{4} \) of an hour for pipes A and B together to fill \( \dfrac{1}{2} \) of the tank.

Adding this time to 11 A.M. gives us the time when the tank will be filled:

\[ \text{Time} = 11 \text{ A.M.} + \dfrac{3}{4} \text{ hours} \]\[ = 11 \text{ A.M.} + 45 \text{ minutes} \]\[ = 11 \text{ A.M.} + 0.75 \text{ hours} \]\[ = 11 \text{ A.M.} + 45 \text{ minutes} \]

So, the tank will be filled at 11:45 A.M.

Let's find the rates at which each tap fills the tank:

Tap A fills the tank in 4 hours, so its filling rate is \( \dfrac{1}{4} \) tank per hour.
Tap B fills the tank in 6 hours, so its filling rate is \( \dfrac{1}{6} \) tank per hour.
Tap C fills the tank in 12 hours, so its filling rate is \( \dfrac{1}{12} \) tank per hour.

When all three taps are opened together, their combined rate is the sum of their individual rates.

So, the combined rate when all three taps are opened together is:

\[ \text{Combined rate} = \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{12} \]

\[ = \dfrac{3}{12} + \dfrac{2}{12} + \dfrac{1}{12} \]

\[ = \dfrac{6}{12} \]

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

This means that together, all three taps fill \( \dfrac{1}{2} \) of the tank per hour.

Now, to find out how long it takes to fill the entire tank, we can use the formula:

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

The amount of work to fill the tank is 1 (the entire tank), and the combined rate of all three taps is \( \dfrac{1}{2} \) tank per hour.

\[ \text{Time} = \dfrac{1}{\dfrac{1}{2}} = 2 \text{ hours} \]

So, when all three taps are opened together, it will take 2 hours to fill the tank. Therefore, the answer is 2 hrs.

Given, A can fill it in 20 minutes and B can empty it in 30 minutes

Pipe A's work for one minute=1/20

Pipe B's work for one minute=1/30

∴ Total work done in two minute \(= \dfrac{1}{{20}} - {\rm{\;}}\dfrac{1}{{30}} = \dfrac{1}{{60}}\)

In every 2 minute, 1/60^th work is done, which means upto 114^th minute, amount of work done will be = 57 × 1/60 = 57/60

Now the remaining work i.e. 3/60 (=1/20^th) will be done by A in next 1 minute.

∴ Total time required = 114 + 1 = 115 minutes