Three taps A, B, C fill up a tank independently in 10hr, 20hr, 30hr respectively. Initially the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. What is the minimum number of hours required to fill the tank?

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answer is A.

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Detailed Solution

Concept- Let us assume that the volume of the Tank is V.
Now rate of water flow through A be

, in the same way for B and C be

respectively.
It is said that each pair of taps is open for at least one hour. So, three pairs are open for the first three hours.
Volume filled in first three hours =

The remaining volume of the tank is

Time taken to fill the rest be x hrs.
Taps A and B together fill the tank fastest, so to get minimum time we should assume that the remaining volume of the tank is filled by this pair of taps.
By calculating this we get time taken as 4.22 hrs which can be approximated to 5 hrs.
Total time taken is 8hrs.
Volume of the tank be

.
Time taken by taps A, B, C respectively = 10hrs, 20 hrs, 30 hrs.
Rate of water flow through Tap A =

Rate of water flow through Tap B =

Rate of water flow through Tap C=

Each pair is open for at least one hour,
Volume filled in first three hours =

Time taken to fill the remaining tank be x hrs.
Volume of remaining tank =