A tank is filled by three pipes with different uniform flow rates. When the first two pipes operate together, they fill the tank in the same time as the third pipe alone. The second pipe can fill the tank 5 hours faster than the first pipe, but 4 hours slower than the third pipe. Find the time (in hours) required by the first pipe alone to fill the tank.

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Accepted Answer

Let the time taken by the first pipe to fill the tank alone be x hours.
Let the time taken by the second pipe be y hours, and by the third pipe be z hours.

Given:
1. The second pipe is 5 hours faster than the first: y = x - 5
2. The second pipe is 4 hours slower than the third: y = z + 4 => z = y - 4

So:
y = x - 5
z = (x - 5) - 4 = x - 9

Let the total work (filling the tank) be 1 unit.

Rates:
- First pipe: 1/x per hour
- Second pipe: 1/y per hour
- Third pipe: 1/z per hour

When first and second pipes work together, their combined rate is 1/x + 1/y.
They fill the tank in the same time as the third pipe alone (z hours):

So,
1/(1/x + 1/y) = z
=> 1/x + 1/y = 1/z

Substitute y and z in terms of x:
1/x + 1/(x-5) = 1/(x-9)

Take LCM:
[(x-5) + x] / [x(x-5)] = 1/(x-9)
=> (2x - 5) / [x(x-5)] = 1/(x-9)

Cross-multiply:
(2x - 5)(x - 9) = x(x - 5)

Expand:
2x^2 - 18x - 5x + 45 = x^2 - 5x
2x^2 - 23x + 45 = x^2 - 5x

Bring all terms to one side:
2x^2 - 23x + 45 - x^2 + 5x = 0
x^2 - 18x + 45 = 0

This is a quadratic equation:
x^2 - 18x + 45 = 0

Solve using quadratic formula:
x = [18 ± sqrt(324 - 180)] / 2
= [18 ± sqrt(144)] / 2
= [18 ± 12] / 2

So, x = (18 + 12)/2 = 30/2 = 15
or x = (18 - 12)/2 = 6/2 = 3

But if x = 3, then y = x - 5 = -2 (not possible, time cannot be negative).
So, x = 15 is the only valid answer.

Therefore, the first pipe alone can fill the tank in 15 hours.

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