Question

It takes 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for hours and the pipe of smaller diameter for 9 hours, only half the pool is filled. How long would it take for each pipe to fill the pool separately?

Answer 1

Hint – In this question let the amount of time taken by pipe A and pipe B be x hour and y hour respectively to fill the pool completely. Let the work done by pipe A in 1 hour be \[\dfrac{1}{x}\] and the work done by pipe B in one hour be $\dfrac{1}{y}$. Thus use the constraints of equations to formulate two equations in two variables. This will help get the right answer.

Complete step-by-step answer:
Let the larger diameter pipe is A
And the smaller diameter pipe is B.
Let pipe A and pipe B take x hour and y hour respectively to fill the pool completely.
Now it is given that it takes 12 hours to fill a swimming pool using these two pipes.
The amount of pool fill in 1 hour = $\dfrac{1}{{12}}$
Work done by pipe A in 1 hour = \[\dfrac{1}{x}\]
And work done by pipe B in 1 hour = $\dfrac{1}{y}$
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{{12}}$................ (1)
Now it is given that pipe A is used for 4 hours and pipe B is used for 9 hours so that only half the pool is filled.
Work done by pipe A in 4 hour = $\dfrac{4}{x}$
And work done by pipe B in 9 hour = $\dfrac{9}{y}$
Now make a linear equation according to given information we have,
$ \Rightarrow \dfrac{4}{x} + \dfrac{9}{y} = \dfrac{1}{2}$................. (2), (as the pool is half filled)
Now multiply by 4 in first equation and subtract from equation 2 we have,
$ \Rightarrow \dfrac{4}{x} + \dfrac{9}{y} - \left( {\dfrac{4}{x} + \dfrac{4}{y}} \right) = \dfrac{1}{2} - \dfrac{4}{{12}}$
Now simplify this we have,
$ \Rightarrow \dfrac{5}{y} = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{{3 - 2}}{6}$
$ \Rightarrow \dfrac{5}{y} = \dfrac{1}{6}$
$ \Rightarrow y = 30$ hours.
Now substitute this value in equation (1) we have,
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{30}} = \dfrac{1}{{12}}$
$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{12}} - \dfrac{1}{{30}} = \dfrac{{30 - 12}}{{30\left( {12} \right)}} = \dfrac{{18}}{{30\left( {12} \right)}} = \dfrac{3}{{30\left( 2 \right)}} = \dfrac{1}{{20}}$
$ \Rightarrow x = 20$ hours.
So pipe A solemnly takes 20 hours and pipe B solemnly takes 30 hours.
So this is the required answer.

Note – Whenever we face linear equations involving two variables there can be two methods to solve these equations, first one can be method of substitution, in this we take out one variable in terms of other variable and then substitute it in other equation, this helps formulation of an equation which is solemnly in one variable, solve it to get the variable, then use this value obtained to get other variable. The second method can be a method of elimination, in this method the coefficients of a specific variable of both the equations are made the same and then eliminated using basic operations of addition or subtraction. The other variable which is non-eliminated is taken out. The other variable can be taken out using this variable obtained.