Question

One fill pipe a takes 3 minutes more to fill the cistern than two fill pipes a and b opened together to fill it. second fill pipe b takes 64/3 minutes more to fill cistern than two fill pipes a and b opened together to fill it. when will the cistern be full if both pipes are opened simultaneously.

Answer 1

$\textbf {Hey there, here is the solution.}$

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STEP 1: Define x:

Let the time taken for both the pipe to fill the cistern be x.

⇒ 1 min = 1/x

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STEP 2: Find time needed for Pipe A to fill the cistern:

It takes 3 mins more:

Time needed = x + 3

⇒ 1 min = 1/(x + 3)

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STEP 3: STEP 2: Find time needed for Pipe B to fill the cistern:

It takes 64/3 mins more.

Time needed = x + 64/3

⇒ 1 min = 1 /( x + 64/3)

.

Simplify 1/(x + 64/3)

$\dfrac{1}{x + \frac{64}{3}} = 1 \div ( x + \dfrac{64}{3}) = 1 \div \dfrac{3x + 64}{3} = 1 \times \dfrac{3}{3x + 64} = \dfrac{3}{3x + 64}$

.

⇒ 1 min = 3/(3x + 64)

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STEP 4: STEP 2: Find time needed for both Pipe A and Pipe B to fill the cistern:

1 min = 1/(x + 3) + 3/( 3x + 64) .

.

$\dfrac{1}{x + 3} + \dfrac{3}{3x + 64}$

$= \dfrac{(3x+ 64) + (3x + 9)}{(x+3)(3x + 64)}$

$= \dfrac{3x+ 64 + 3x + 9}{(x+3)(3x + 64)}$

$= \dfrac{6x+ 73}{(x+3)(3x + 64)}$

.

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STEP 5: Form the equation and solve for x:

$\dfrac{6x+ 73}{(x+3)(3x + 64)} = \dfrac{1}{x}$

$x(6x + 73) = (x + 3)(3x + 64)$

$6x^2 + 73x = 3x^2 + 64x + 9x + 192$

$6x^2 + 73x = 3x^2 + 73x + 192$

$3x^2 -192 = 0$

$3x^2 = 192$

$x^2 = 64$

$x = 8$

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STEP 6: STEP 2: Find time needed for Pipe A and Pipe B to fill the cistern:

1 min = 1/x

Substituting x = 8

1 min = 1/8

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1/8 of the cistern = 1 min

8/8 of the cistern = 1 x 8 = 8 min

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Answer: It takes 8 mins for both the pipes to fill up the cistern.

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⭐$\textbf {Cheers}$⭐