There are three taps of diameter 2 cm, 3 cm and 4 cm, respectively. The ratio of the water flowing through them is equal to the ratio of the square of their diameters. The biggest tap can fill an empty tank alone in 81 min. If all the taps are opened simultaneously, then how long will the tank take (in min) to be filled?
This problem involves understanding how the rate at which water flows through a tap is related to its diameter and then using this information to calculate the time taken to fill a tank when multiple taps are working together.
The question states that the ratio of the water flowing through the taps is equal to the ratio of the square of their diameters. Let \(D_1\), \(D_2\), and \(D_3\) be the diameters of the three taps, and let \(R_1\), \(R_2\), and \(R_3\) be their respective flow rates.
According to the problem, the ratio of flow rates is \(R_1 : R_2 : R_3 = D_1^2 : D_2^2 : D_3^2\).
Let's calculate the squares of the diameters:
So, the ratio of flow rates is \(R_1 : R_2 : R_3 = 4 : 9 : 16\). We can represent the flow rates as \(4k\), \(9k\), and \(16k\) units of volume per minute, where \(k\) is a constant.
The biggest tap has a diameter of 4 cm, and its flow rate is \(16k\). This tap can fill the empty tank alone in 81 minutes.
The total capacity of the tank is the flow rate of the tap multiplied by the time it takes to fill the tank.
Tank Capacity = Flow Rate of Biggest Tap \(\times\) Time taken by Biggest Tap
Tank Capacity = \(16k \times 81\) units.
Tank Capacity = \(1296k\) units.
When all three taps are opened simultaneously, their flow rates add up. The combined flow rate is the sum of the individual flow rates \(R_1\), \(R_2\), and \(R_3\).
Combined Flow Rate = \(R_1 + R_2 + R_3\)
Combined Flow Rate = \(4k + 9k + 16k\)
Combined Flow Rate = \(29k\) units per minute.
To find the time taken for all taps to fill the tank when working together, we divide the total tank capacity by the combined flow rate.
Time Taken = \(\frac{\text{Tank Capacity}}{\text{Combined Flow Rate}}\)
Time Taken = \(\frac{1296k}{29k}\) minutes.
The constant \(k\) cancels out:
Time Taken = \(\frac{1296}{29}\) minutes.
To express the time as a mixed number, we perform the division:
\(1296 \div 29\)
Let's perform the long division:
So, \(1296 \div 29 = 44\) with a remainder of 20.
Therefore, the time taken is \(44 \frac{20}{29}\) minutes.
| Tap Diameter | Diameter Squared | Flow Rate (Ratio) | Flow Rate (with k) |
|---|---|---|---|
| 2 cm | 4 | 4 | \(4k\) |
| 3 cm | 9 | 9 | \(9k\) |
| 4 cm | 16 | 16 | \(16k\) |
| Item | Value |
|---|---|
| Flow Rate of Biggest Tap | \(16k\) |
| Time taken by Biggest Tap | 81 min |
| Tank Capacity | \(16k \times 81 = 1296k\) units |
| Combined Flow Rate | \(4k + 9k + 16k = 29k\) units/min |
| Time Taken Together | \(\frac{1296k}{29k} = \frac{1296}{29}\) min |
| Time Taken Together (Mixed Number) | \(44 \frac{20}{29}\) min |
Thus, if all the taps are opened simultaneously, the tank will take \(44 \frac{20}{29}\) minutes to be filled.
| Concept | Formula/Relationship | Application in Problem |
|---|---|---|
| Flow Rate Ratio | Proportional to Diameter Squared | \(R_1:R_2:R_3 = D_1^2:D_2^2:D_3^2\) |
| Tank Capacity | Flow Rate \(\times\) Time | \(16k \times 81 = 1296k\) |
| Combined Work (Flow) Rate | Sum of Individual Rates | \(4k + 9k + 16k = 29k\) |
| Time Taken Together | Total Work / Combined Rate | \(\frac{1296k}{29k} = \frac{1296}{29}\) |
| Fraction Conversion | Improper to Mixed Number | \(\frac{1296}{29} = 44 \frac{20}{29}\) |
Problems involving taps filling or emptying tanks are a common type of 'Time and Work' problems. Here are some key concepts:
Understanding the relationship between rate, time, and work is crucial for solving these types of problems effectively. Always ensure units are consistent (e.g., all times in minutes, all flow rates per minute).
Pipe A and pipe B running together can fill a cistern in 6 minutes. If B takes 5 minutes more than A to fill it, then the time in which A and B will fill that cistern separately will be, respectively, __________ .
An inlet pipe can fill an empty tank in 140 hours while an outlet pipe drains a completely-filled tank in 63 hours. If 8 inlet pipes and y outlet pipes are opened simultaneously, when the tank is empty, then the tank gets completely filled in 105 hours. Find the value of y.
There are two inlet pipes A and B connected to a tank. A and B can fill the tank in 32 h and 28 h, respectively. If both the pipes are opened alternately for 1 h, starting with A, then in how much time (in hours, to nearest integer) will the tank be filled?
There are two water taps in a tank which can fill the empty tank in 12 hours and 18 hours respectively. It is seen that there is a leakage point at the bottom of the tank which can empty the completely filled tank in 36 hours. If both the water taps are opened at the same time to fill the empty tank and leakage point was repaired after 1 hour, then in how much time the empty tank will be completely filled?
Pipes A and B can fill a tank in 43.2 minutes and 108 minutes, respectively. Pipe C can empty it at 3 litres/minute. When all the three pipes are opened together, they fill the tank in 54 minutes. The capacity (in litres) of the tank is:
Pipes A and B can fill a tank in 10 hours and 40 hours respectively. C is an outlet pipe attached to the tank. If all the three pipes are opened simultaneously, it takes 80 minutes more time than A and B together takes to fill the tank. If A and B kept open for 7 hours and closed and then C opened. How much time will C take to empty the tank :
Two pipes A and B can fill a cistern in \(12\frac{1}{2}\) hours and 25 hours, respectively. The pipes were opened simultaneously, and it was found that, due to leakage in the bottom, it took one hour 40 minutes more to fill the cistern. If the cistern is full, in how much time (in hours) will the leak alone empty 70% of the cistern?
Pipes A, B and C can fill a tank in 20, 30 and 60 hours, respectively. Pipes A, B and C are opened at 7 a.m., 8 a.m., and 9 a.m., respectively, on the same day. When will the tank be full?
Pipes A and B can fill a tank in 16 hours and 24 hours, respectively, and pipe C alone can empty the full tank in x hours. All the pipes were opened together at 10:30 AM, but C was closed at 2:30 PM. If the tank was full at 8:30 PM on the same day, then what is the value of x?
Pipes A and B are filling pipes while pipe C is an emptying pipe. A and B can fill a tank in 72 and 90 minutes respectively. When all the three pipes are opened together, the tank gets filled in 2 hours. A and B are opened together for 12 minutes, then closed and C is opened. The tank will be empty after:
Two pipes A and B can independently fill a tank completely in 20 and 30 minutes respectively. If both the pipes are opened simultaneously, how much time will they take to fill the tank completely?
The compound interest on Rs. 64,000 for 3 years, compounded annually at 7.5% p.a. is
A water tank can be emptied in 40 minutes by a pipe of d 'diameter, so how long will it take for a 2d diameter pipe to be emptied?
A pipe can fill a tank in 4 hours, while a leak which is at one-fourth of the height of the tank from bottom can empty upto that part in 2 hours. If both are operated simultaneously and initially the tank is full, then when it will be one-fourth full?
Two pipes X and Y can fill an empty tank in 16 hours and 20 hours respectively. Pipe Z alone can empty the completely filled tank in 25 hours. Firstly both pipes X and Y are opened and after 6 hours pipe Z is also opened. What will be the total time (in hours) taken to completely fill the tank?