A swimming pool is filled by 3 pipes with uniform flow, The first two pipes operating simultaneously, fill the pool - Maths - Surface Areas and Volumes

Question

A swimming pool is filled by 3 pipes with uniform flow, The
first two pipes operating simultaneously, fill the pool in the same
time during which the pool is filled by the 3rd pipe
alone The 2nd pipe fills the pool 5 hours faster than
the 1st pipe & 4 hours slower than the
3rd pipe Find the time required by each pipe to fill the
pool separately

Gursheen kaur. · Accepted Answer

Let the volume of the pool = V

Let the time taken by second pipe to fill the pool = x hours

time taken by first pipe to fill the pool = (x+5) hours,

and time taken by third pipe to fill the pool = (x-4) hours.

In One hour,

First pipe can fill $\frac{V}{x + 5}$ parts of volume of pool

Second pipe can fill $\frac{V}{x}$ parts of volume of pool

Third pipe can fill $\frac{V}{x - 4}$ parts of volume of pool

Now, It is given that,

Time taken by first and second pipe simultaneously = Time taken by third pipe alone

$\Rightarrow \frac{V}{x + 5} + \frac{V}{x} = \frac{V}{x - 4} \Rightarrow \frac{1}{x + 5} + \frac{1}{x} = \frac{1}{x - 4} \Rightarrow \frac{x + x + 5}{(x + 5) x} = \frac{1}{x - 4} \Rightarrow \frac{2 x + 5}{(x + 5) x} = \frac{1}{x - 4} \Rightarrow (2 x + 5) (x - 4) = (x + 5) x$

Neglect the negative value.

Thus, the time taken by second pipe to fill the pool = x hours=10 hours.

time taken by first pipe to fill the pool = (x+5) hours= 10+5=15 hours.

time taken by third pipe to fill the pool = (x-4) hours=1-4=6 hours.

Neglect the negative value Thus, the time taken by second pipe to fill the pool = x hours=10 hours time taken by first pipe to fill the pool = (x+5) hours= 10+5=15 hours time taken by third pipe to fill the pool = (x-4) hours=1-4=6 hours