The water tank has three losses. The first and second faucets fill the tank in 6 and 10 hours respectively but the third faucet empties in 8 hours.If all three losses start at the same time, how much time will it take to recover?
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12 taps are fitted in a tank in which some are inlet taps and some are outlet taps. Each inlet tap can fill the tank in 30 hours and each outlet tap can empty the tank in 40 hours. Find the number of outlet taps if the whole tank is filled in 6 hours when all the 12 taps are open.
2
8
7
4
5
Answer (Detailed Solution Below)
Option 4 : 4
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Detailed Solution
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Given:
Total number of taps = 12
Time taken by each inlet tap to fill the tank = 30 hours
Time taken by each outlet tap to empty the tank = 40 hours
Total time taken to fill the tank = 6 hours
Concept used:
Total work = Time taken × Efficiency
And, Total efficiency = efficiency of inlet taps – efficiency of outlet taps
Calculation:
Let total number of inlet taps be x
Total number of outlet taps be (12 – x)
L.C.M of 30 and 40 = 120 units = Total work
Total time taken to fill the tank = 6 hours
Efficiency of all 12 taps = 120/6 = 20 units
Efficiency of each inlet tap = 120/30 = 4 units
Now, Efficiency of x inlet taps = 4x
Efficiency of each outlet tap = 120/40 = 3 units
Efficiency of (12 - x) outlet taps = (36 - 3x)
So, Total efficiency of 9 taps = efficiency of inlet taps – efficiency of outlet taps
⇒ Total efficiency of 12 taps = 4x – (36 – 3x)
So, 4x – (36 – 3x) = 20
⇒ 7x – 36 = 20
⇒ 7x = 56
⇒ x = 8
⇒ Outlet taps = (12 – 8) = 4
∴ The number of outlet taps is 4