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To Fill a Swimming Pool Two Pipes Are Used. If the Pipe of Larger Diameter Used for 4 Hours and the Pipe of Smaller Diameter for 9 Hours, Only Half of the Pool Can Be Filled. - Mathematics

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Question

To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?

Solution

Let the pipe of larger diameter takes x hours.
Then, the pipe of smaller diameter takes + 10 hours to fill the pool.
Now, the part of the pool filled by the larger pipe in 1 hour = $\frac{1}{x}$

and the part of the pool filled by the smaller pipe in 1 hour = $\frac{1}{x + 10}$

If the larger pipe is used for 4 hours and the smaller pipe is used for 9 hours, only half of the pool can be filled,

$\therefore \frac{4}{x} + \frac{9}{x + 10} = \frac{1}{2}$

$\Rightarrow \frac{4(x + 10) + 9x}{x(x + 10)} = \frac{1}{2}$

$\Rightarrow \frac{4x + 40 + 9x}{x^2 + 10x} = \frac{1}{2}$

$\Rightarrow 2(13x + 40) = x^2 + 10x$

$\Rightarrow x^2 + 10x - 26x - 80 = 0$

$\Rightarrow x^2 - 16x - 80 = 0$

$\Rightarrow x^2 - 20x + 4x - 80 = 0$

$\Rightarrow x(x - 20) + 4(x - 20) = 0$

$\Rightarrow (x + 4)(x - 20) = 0$

$\Rightarrow x + 4 = 0\text { or } x - 20 = 0$

$\Rightarrow x = - 4\text { or } x = 20$Since, time cannot be negative.
∴ x = 20
Thus, the pipe of larger diameter takes 20 hours and the pipe of smaller diameter takes (20 + 10) = 30 hours to fill the pool separately.

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