Question

Two pipes running together can fill a Cister in 6 minutes. if 1 pipe takes 5 minutes more than the other to fill the Cister. find the time in each pipe fill the Cister

Answer 1

$\huge\textbf\green{Answer :}$

Two pipes are running together.

So,

$\textbf{Let one pipe = x}$

And another pipe take 5 min more than the other. So,

$\textbf{Another pipe = (x + 5) min.}$

We have to find time.

Now

Time taken by one pipe to fill a Cister in 1 min. = $\frac{1}{x}$

And time taken by another pipe to fill a Cister in 1 min. = $\frac{1}{x\:+\:5}$

Both pipes together can fill a Cister in 6 min.

So, Time taken by both pipe to fill the Cister in min. = $\frac{1}{6}$

A.T.Q.

$\frac{1}{x}$ + $\frac{1}{x\:+\:5}$ = $\frac{1}{6}$

$\frac{x\:+\:5\:x}{(x)(x\:+\:5)}$ = $\frac{1}{6}$

$\frac{2x\:+\:5}{{x}^{2}{\:+\:5x}}$ = $6$

$6(2x\:+\:5)$ = ${x}^{2}\:+\:5x$

$12x\:+\:30$ = ${x}^{2}\:+\:5x$

${x}^{2}\:+\:5x\:-\:30\:-\:12x\:=\:0$

${x}^{2}\:-7x\:-\:30\:=\:0$

${x}^{2}\:+\:3x\:-\:10x\:-\:30\:=\:0$

$x(x\:+\:3)\:-10(x\:+\:3)$

$(x\:+\:3)\:(x\:-\:10)$

$x\:=\:-3$

And

$x\:=\:+10$

Time can never be negative. So, negative value will be neglected.

x = +10
$\textbf{Time of one pipe = x = +10 min.}$

And

$\textbf{Time of another pipe = x + 5 = 10 + 5 = 15 min.}$

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Answer 2

Solutions :-

Given :
Two pipes running together can fill a Cister in 6 minutes.
1 pipe takes 5 minutes more than the other to fill the Cister.

Let the time taken by one pipe be x
Time taken by other pipe = x + 5


A/q

=> 1/x + 1/x+5 = 1/6
=> (x+5+x)/x²+5x = 1/6
=> 6(x+5+x) = x² + 5x
=> 12x + 30 = x² + 5x
=> x² - 12x + 5x - 30 = 0
=> x² - 7x - 30 = 0
=> x² - (10x - 3x) - 30 = 0
=> x² - 10x + 3x - 30 = 0
=> x (x - 10) + 3(x - 10) = 0
=> (x - 10) (x + 3) = 0
=> x = 10, - 3 (Time is always taken positively.)
So, x = 10



Hence,
The time taken by one pipe = x = 10 minutes
Time taken by other pipe = x + 5 = 10 + 5 = 15 minutes

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✯ @shivamsinghamrajput ✯