Question

A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. Find the time required by each pipe to fill the pool separately.​

Answer 1

$\huge\underline\mathbb {SOLUTION:-}$

• Let x be the number of hours required by the second pipe alone to fill the pool and first pipe (x + 5) hour while third pipe ( x - 4) hour.

$\mathsf {\frac{1}{x + 5} + \frac{1}{x} = \frac{1}{x - 4} }$

$\implies \mathsf {\frac{x + x + 5}{x^2 + 5x} = \frac{1}{x - 4} }$

$\implies \mathsf {x^2 - 8x - 20 = 0}$

$\implies \mathsf {x^2 - 10x + 2x - 20 = 0}$

$\implies \mathsf {x(x - 10) + 2(x - 10) = 0}$

$\implies \mathsf {(x - 10)\:(x + 2) = 0}$

$\underline \mathsf \blue {x\:=\:10\:or\:x\:=\:-2\:(Neglected)}$

Answer 2

Answer:

SOLUTION:−

Let x be the number of hours required by the second pipe alone to fill the pool and first pipe (x + 5) hour while third pipe ( x - 4) hour.

\mathsf {\frac{1}{x + 5} + \frac{1}{x} = \frac{1}{x - 4} }

x+5

1

+

x

1

=

x−4

1

\implies \mathsf {\frac{x + x + 5}{x^2 + 5x} = \frac{1}{x - 4} }⟹

x

2

+5x

x+x+5

=

x−4

1

\implies \mathsf {x^2 - 8x - 20 = 0}⟹x

2

−8x−20=0

\implies \mathsf {x^2 - 10x + 2x - 20 = 0}⟹x

2

−10x+2x−20=0

\implies \mathsf {x(x - 10) + 2(x - 10) = 0}⟹x(x−10)+2(x−10)=0

\implies \mathsf {(x - 10)\:(x + 2) = 0}⟹(x−10)(x+2)=0

\underline \mathsf \blue {x\:=\:10\:or\:x\:=\:-2\:(Neglected)}

Step-by-step explanation:

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