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A water tank has three taps attached to it at three places namely A, B and C. Of these taps two of them viz. A & B can fill the tank while third tap C can empty the water tank. Time required by A & B to completely fill the tank is 10 minutes and 40 minutes respectively while tap C can empty it in 10 minutes. If all the three taps are kept open successively for 1 minute each, calculate the shortest time in which the entire water tank can be filled?

A water tank has three taps attached to it at three places namely A, B and C. Of these taps two of them viz. A & B can fill the tank while third tap C can empty the water tank. Time required by A & B to completely fill the tank is 10 minutes and 40 minutes respectively while tap C can empty it in 10 minutes. If all the three taps are kept open successively for 1 minute each, calculate the shortest time in which the entire water tank can be filled? Correct Answer 107 minutes

let total Capacity of the tank be 40l.

Then,

⇒ Quantity filled by tap A in 1 min = 4l

⇒ Quantity filled by tap B in 1 min = 1l

⇒ Quantity emptied by tap A in 1 min = 4l

∴ in 3 min Quantity of tank filled = 4 + 1 – 4 = 1l

⇒ 35 litres will be filled in = 35 × 3 = 105 minutes

⇒ As 105th minute is over; C is turned off.

In 106th minute, Tap A is switched on, then

Total Quantity filled = 35 + 4 = 39l

⇒ In 107th minute, Tap B is switched on, then

Total Quantity filled = 39 + 1 = 40

∴ Answer is 107 minutes.

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