Math, asked by aparna647, 1 year ago

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.​

Answers

Answered by Anonymous
11

\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)}

Answered by Anonymous
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:

hope it will be help you. ......

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