Question

Two pipes running together can fill a cistern in $2\dfrac{8}{{11}}$minutes. If one pipe takes 1 minute more than the other to fill the cistern, find the time in which each pipe would fill the cistern alone.

Answer 1

Hint – In this question let the time taken by the one pipe to fill a cistern be x minutes. The relation between the time taken by pipes to fill the cistern is given, use this to formulate the equation and thus evaluate the variable.

Complete step by step answer:
Let first pipe (A) take x minute to fill a cistern.
Therefore according to question second pipe (B) takes (x + 1) minute to fill a cistern.
Therefore A’s one minute work is $\dfrac{1}{x}$.
And B’s one minute work is $\dfrac{1}{{x + 1}}$
Now it is given that together they fill a cistern in $2\dfrac{8}{{11}}$ minutes.
Now convert this mixed fraction into improper fraction we have,
$ \Rightarrow 2\dfrac{8}{{11}} = \dfrac{{\left( {11 \times 2} \right) + 8}}{{11}} = \dfrac{{30}}{{11}}$ minutes.
Therefore (A + B)’s one minute work = $\dfrac{1}{{\dfrac{{30}}{{11}}}} = \dfrac{{11}}{{30}}$.
Now A’s one minute work + B’s one minute work = (A + B)’s one minute work.
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{x + 1}} = \dfrac{{11}}{{30}}$
Now simplify this equation we have,
$ \Rightarrow 30\left( {x + 1 + x} \right) = 11x\left( {x + 1} \right)$
$ \Rightarrow 30\left( {2x + 1} \right) = 11{x^2} + 11x$
$ \Rightarrow 11{x^2} - 49x - 30 = 0$
Now factorize the equation we have,
$ \Rightarrow 11{x^2} - 55x + 6x - 30 = 0$
$ \Rightarrow 11x\left( {x - 5} \right) + 6\left( {x - 5} \right) = 0$
$ \Rightarrow \left( {11x + 6} \right)\left( {x - 5} \right) = 0$
$ \Rightarrow x = 5,\dfrac{{ - 6}}{{11}}$
But x = $\dfrac{{ - 6}}{{11}}$ cannot possible
Therefore x = 5 is a valid case.
So the first pipe fills a cistern in 5 minutes and the second pipe fills a cistern in (5 + 1) = 6 minutes.
So this is the required answer.

Note – Whenever we face such type of problems the key point is only one pipe’s time is considered as variable as the other pipe’s time was dependent upon the first pipe’s time so eventually if one pipe’s time can be calculated the other can easily be calculated too. These are general unitary method based questions so no defined formula works in such types of problems.