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Three pipes A, B, and C fill a cistern in 20 minutes, 30 minutes and 40 minutes respectively. These pipes are opened alternately for one minute but beginning with pipe A. In what time will the cistern be full?


1. 45/2
2. 53/2
3. 47/2
4. 55/2
5. None of these

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Best answer
Correct Answer - Option 4 : 55/2

Given:

Time taken by pipe A = 20 minutes

Time taken by pipe B = 30 minutes

Time taken by pipe C = 40 minutes

Calculation:

Let the total capacity of the tank be LCM (20, 30, 40) = 120 litres

Now,

Tank filled by A in 1 minute = 120/20

⇒ 6 litre/min

Tank filled by B in 1 minute = 120/30

⇒ 4 litre/min

Tank filled by C in 1 minute = 120/40

⇒ 3 litre/min

Tank filled by (A + B + C) in 1 minute = (6 + 4 + 3)

⇒ 13 litre/min

Now,

According to the question,

For first minute pipe, A is opened = 6 litre

For second-minute pipe, B is opened = 4 litre

For third-minute pipe, C is opened = 3 litre

For fourth minute pipe, A is opened = 6 litre

For fifth-minute pipe, B is opened = 4 litre

For sixth-minute pipe, C is opened = 3 litre

Similarly, So…. On

Now,

We conclude that –

3 minutes = 13 litre

× 9                × 9

27 min   =      117 litre

Now,

The remaining work = 120 – 117

⇒ 3 litre

Now,

Again 28th minute A will be opened and fill the remaining tank

⇒ 3/6

⇒ 1/2

Now,

The tank fills in = (27 minute + ½ minute)

⇒ 55/2 minutes

The tank will be filled in 55/2 minutes.

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