Two pipes running together can fill a cistern in $3\frac{1}{13}$ minutes. If one pipe takes $3$ minutes more than the other to fill it, find the time in which pipe nwould fill the cistern?

Given: Two pipes running together can fill a cistern in $3\frac{1}{13}$ minutes. And one pipe takes $3$ minutes more than the other to fill it.

To do: To find the time in which pipe would fill the cistern.

Solution:

Let the time taken by faster pipe to fill the cistern be $x$ minutes

Therefore, time taken by slower pipe to fill the cistern $=(x+3)$ minutes

Since the faster pipe takes $x$ minutes to fill the cistern.

Portion of cistern filled by the faster pipe in 1 minute $=\frac{1}{x}$

​

Portion of cistern filled by the faster pipe in 1 minute $=\frac{1}{x+3}$

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Portion of cistern filled by the two pipes in 1 minute $=\frac{1}{\frac{40}{13}}$

$=\frac{13}{40}$

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Now according to the question-

$\frac{1}{x}+\frac{1}{x+3}=\frac{13}{40}$

$\Rightarrow \frac{x+x+3}{x( x+3)}=\frac{13}{40}$

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 $\Rightarrow 40(2x+3)=13x(x+3)$

$\Rightarrow 80x+120=13x^2+39x$

$\Rightarrow13x^2−41x−120=0$

Here,

$a=13,\ b=−41,\ c=−120$

Now by quadratic formula-

$x=\frac{−(−41)±\sqrt{(−41)^2−4×(13)×(−120)}}{2×13}$

​

$\Rightarrow x=\frac{41±\sqrt{1681+6240}}{26}$

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$\Rightarrow x=\frac{41±\sqrt{7921}}{26}$

​

$\Rightarrow x=\frac{41±89}{26}$

​

$\because x\
eq−ve$

$\therefore x=\frac{41+89}{26}=5$

Therefore,

Time taken by faster pipe to fill the cistern $=5$ minutes

Time taken by slower pipe to fill the cistern $=(5+3)=8$ minutes

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Updated on: 10-Oct-2022

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