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Three pipes A, B & C fill a tank in 3 hours working simultaneously. The pipe C is twice as faster as B and B is twice as fast as A. The time taken by pipe A alone to fill the tank is:

Solution

✅ Correct Option: 4

Three pipes A, B, and C fill a tank in 3 hours working simultaneously. Pipe C is twice as fast as B, and B is twice as fast as A.

Let pipe A take xx hours to fill the tank alone.

Since B is twice as fast as A, pipe B takes x2\frac{x}{2} hours alone.

Since C is twice as fast as B, pipe C takes x4\frac{x}{4} hours alone.


If a pipe fills a tank in nn hours, it fills 1n\frac{1}{n} of the tank in 1 hour.

In 1 hour:

Pipe A fills 1x\frac{1}{x} of the tank

Pipe B fills 1x/2=2x\frac{1}{x/2} = \frac{2}{x} of the tank

Pipe C fills 1x/4=4x\frac{1}{x/4} = \frac{4}{x} of the tank

All three pipes together in 1 hour:

1x+2x+4x\frac{1}{x} + \frac{2}{x} + \frac{4}{x}

=7x= \frac{7}{x} of the tank


All three pipes fill the tank in 3 hours, so the work done in 1 hour multiplied by 3 equals 1 complete tank:

7x×3=1\frac{7}{x} \times 3 = 1

21x=1\frac{21}{x} = 1

x=21x = 21


Therefore, pipe A alone takes 21 hours to fill the tank.

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