Question : 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?
Option 1: $34 \frac{20}{29}$
Option 2: $64 \frac{20}{29}$
Option 3: $54 \frac{20}{29}$
Option 4: $44 \frac{20}{29}$
Correct Answer: $44 \frac{20}{29}$
Solution :
Given: 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.
The ratios of the three pipe's efficiencies = (2)
2
: (3)
2
: (4)
2
= 4 : 9 : 16
The biggest tap can fill an empty tank alone in 81 min.
Let the total work = (16 × 81) units and the total efficiency = (4 + 9 + 16) = 29 units
So, the required time = $\frac{\text{Total work}}{\text{Total efficiency}}$
= $\frac{16 × 81}{29}$
= $\frac{1296}{29}$
= $44\frac{20}{29}$ min
Hence, the correct answer is $44\frac{20}{29}$.
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