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?
Understanding Tap Flow Rates and Tank Filling Time
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.
Relating Flow Rate to Diameter
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.
- Diameter of Tap 1, \(D_1 = 2\) cm
- Diameter of Tap 2, \(D_2 = 3\) cm
- Diameter of Tap 3, \(D_3 = 4\) cm
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:
- \(D_1^2 = (2 \text{ cm})^2 = 4 \text{ cm}^2\)
- \(D_2^2 = (3 \text{ cm})^2 = 9 \text{ cm}^2\)
- \(D_3^2 = (4 \text{ cm})^2 = 16 \text{ cm}^2\)
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.
Calculating Tank Capacity
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.
Calculating Combined Flow Rate
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.
Calculating Time Taken Together
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.
Converting the Fraction to a Mixed Number
To express the time as a mixed number, we perform the division:
\(1296 \div 29\)
Let's perform the long division:
- \(129 \div 29 = 4\) with a remainder. \(29 \times 4 = 116\). \(129 - 116 = 13\).
- Bring down 6 to get 136.
- \(136 \div 29 = 4\) with a remainder. \(29 \times 4 = 116\). \(136 - 116 = 20\).
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.
Revision Table: Tap and Tank Problem Summary
| 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 | \(\frac{1296}{29} = 44 \frac{20}{29}\) |
Additional Information: Time and Work Problems
Problems involving taps filling or emptying tanks are a common type of 'Time and Work' problems. Here are some key concepts:
- Work Rate: The amount of work done per unit of time. For taps, this is the volume of water flow per minute or hour.
- Relationship between Rate and Time: Work = Rate \(\times\) Time. If the rate increases, the time taken to complete the same work decreases, and vice versa (they are inversely proportional).
- Multiple Workers (Taps): When multiple taps are filling a tank, their rates are added to find the combined filling rate. If there are emptying taps (leaks or drain pipes), their rates are subtracted from the filling rates to find the net rate.
- Efficiency: Often, flow rate is directly related to the 'efficiency' of the tap. In this problem, the flow rate is linked to the diameter squared.
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).
Similar Questions
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Important Questions from Pipe and Cistern
A pipe can fill a cistern in 20 minutes where as the cistern when full can be emptied by a leak in 28 minutes. When both are opened, The time taken to fill the cistern is:
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Pipes A and B can fill a tank in 36 minutes and 45 minutes, respectively. Both these pipes were opened simultaneously. After 20 minutes, a leak at the bottom of the tank was spotted which was immediately sealed. The tank was full in another 15 minutes. The leak alone can empty the full tank in:
A cistern has a leak which would empty it in 6 hours. A tap is turned on which admits 10 litres of water per minute into the cistern. When it is full it is now emptied in 10 hours. What is the capacity (in litres) of the cistern?
