There are three taps of diameter $2 ext{ cm}$, $3 ext{ cm}$, and $4 ext{ cm}$, 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 $81$ minutes. If all the taps are opened simultaneously, then how long will the tank take (in minutes) to be filled?

Question

There are three taps of diameter $2 	ext{ cm}$, $3 	ext{ cm}$, and $4 	ext{ cm}$, 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 $81$ minutes. If all the taps are opened simultaneously, then how long will the tank take (in minutes) to be filled?

Accepted Answer

**Solution:**

1. The diameters of the taps are $2 	ext{ cm}$, $3 	ext{ cm}$, and $4 	ext{ cm}$. The flow rates are proportional to the squares of their diameters.

2. Calculate the ratio of flow rates:
   $$2^2 : 3^2 : 4^2 = 4 : 9 : 16$$

3. The biggest tap (diameter 4 cm) fills the tank alone in $81$ minutes.

4. Let the total work (filling the tank) be proportional to the biggest tap's flow rate times time:
   $$	ext{Work} = 16 	imes 81 = 1296 	ext{ units}$$

5. Total efficiency of all taps combined:
   $$4 + 9 + 16 = 29 	ext{ units}$$

6. Time taken when all taps work together:
   $$	ext{Time} = \frac{	ext{Total Work}}{	ext{Total Efficiency}} = \frac{1296}{29} = \frac{44 	imes 29 + 20}{29} \approx 44.69 	ext{ minutes}$$

7. Simplifying the fraction exactly:
   $$\frac{1296}{29} = \frac{44 	imes 29 + 20}{29} = 44 + \frac{20}{29} \approx 44.69$$

Hence, the time taken to fill the tank when all taps are opened simultaneously is approximately $\frac{44}{29}$ minutes.

**Answer:** $\frac{44}{29}$ minutes

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