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Pipe A takes 10 hours more than Pipe B to fill a tank. Pipe C(which is equally efficient as Pipe A and Pipe B together), fills the same tank in 12 hours. In how many hours will Pipe A, Pipe B and Pipe C fill the tank if Pipe A is accompanied by B and C on alternate hours.

Pipe A takes 10 hours more than Pipe B to fill a tank. Pipe C(which is equally efficient as Pipe A and Pipe B together), fills the same tank in 12 hours. In how many hours will Pipe A, Pipe B and Pipe C fill the tank if Pipe A is accompanied by B and C on alternate hours. Correct Answer 10 hours

Given:

A takes 10 hours more than B 

Pipe C is equally efficient as pipe A and pipe B together 

Pipe C fills the same tank in 12 hours 

Concept used:

Total capacity = LCM

The efficiency of the pipe is inversely proportional to the time taken by pipe

Formula used:

Efficiency = Total capacity/Total time

Calculations:

Let B can fill the tank in x 

So, Time taken by A to fill the tank = x + 10 hours 

The efficiency of A = 1/(x + 10)

The efficiency of B = 1/x

The efficiency of C = 1/12 

Pipe C is equally efficient as pipe A and pipe B together 

So, (1/(x + 10)) + (1/x) = (1/12)

⇒ (x + x + 10)/(x + 10)(x) = 1/12

⇒ 24x + 120 = x2 + 10x

⇒ x2 – 14x – 120 = 0 

⇒ x2 – 20x + 6x – 120 = 0 

⇒ x(x – 20) + 6(x – 20) = 0

⇒ (x – 20)(x + 6) = 0

⇒ x = 20, -6 

Time taken by A = x + 10 = 30 

Time taken by B = x = 20 

Time taken by C = 12 

Total capacity = LCM(30, 20, 12) = 60 

Pipe Time Total capacity  Efficiency 
A 30 60 2
B 20 60 3
C 12 60 5

Total efficiency of A and B = 2 + 3 = 5 

Total efficiency of A and C = 2 + 5 = 7 

Tank fill in 2 hours = 5 + 7 = 12 

This alternate process repeats 5 more time

Time Tank fills
2 × 5 = 10 hours 12 × 5 = 60 

∴ Time taken by A, B, and C to fill the tank alternately in 10 hours 

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