Two water taps together can fill a tank in $1\dfrac{7}{8}$ hours. The tap with a longer diameter takes $2$ hours less than the tap with a smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.
Answer
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Hint: Here, we will be going with a unitary method to find the amount of water that is filled in one hour.
Complete step-by-step answer:
Note: Whenever we face such types of problems the key concept that we need to use is first of all let the time taken by each tap to fill the tank be a variable which we need to find and then use unitary method to determine the amount of water filed by each tap in one hour and then use the condition that is given in the question to get the answers.
Complete step-by-step answer:
In the given question it is given that two taps can fill a tank in ${\text{1}}\dfrac{7}{8} = \dfrac{{15}}{8}$ ${\text{hrs}}$
We apply a unitary method for solving the question.
Amount of water filled by two taps in ${\text{1}}$ hour is $\dfrac{1}{{\dfrac{{15}}{8}}} = \dfrac{8}{{15}}$
Now let the time taken by longer diameter tap to fill the tank be $x$ ${\text{hrs}}$
And the time taken by another tap to fill the tank is ${\text{y hrs}}$.
It is given that the longer diameter takes ${\text{2 hour}}$ less than the tap with a smaller one.
,(by applying this condition)
$x = y - 2$
Amount of water filled by longer diameter tap in ${\text{1 hour}} $is $\dfrac{1}{x} = \dfrac{1}{{y - 2}}$
Amount of water filled by a shorter diameter tap in ${\text{1 hour}} $ is $\dfrac{1}{y}$.
Amount of water filled by two taps in ${\text{1 hour}} $ is $\dfrac{1}{{\dfrac{{15}}{8}}} = \dfrac{8}{{15}}$
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{8}{{15}}$
$ \Rightarrow \dfrac{1}{{y - 2}} + \dfrac{1}{y} = \dfrac{8}{{15}}{\text{ }}\because \dfrac{1}{x} = \dfrac{1}{{y - 2}} $
$ \Rightarrow \dfrac{{y + (y - 2)}}{{y(y - 2)}} = \dfrac{8}{{15}} $
$ \Rightarrow 2(15y - 15) = 8y(y - 2) $
$ \Rightarrow 4{y^2} - 23y + 15 = 0 $
$ \Rightarrow 4{y^2} - 3y - 20y + 15 = 0 $
$ \Rightarrow (4y - 3)(y - 5) = 0 $
$ \Rightarrow y = \dfrac{3}{4},5 $
Put the value of $y$ in $\dfrac{1}{x} = \dfrac{1}{{y - 2}}$ to find the value of $x$
$ \Rightarrow x = - \dfrac{5}{4},3{\text{ }}$ But $x = - \dfrac{5}{4}{\text{ }}$ is not possible because time cannot be negative
So $x = 3,y = 5$
The time taken by longer diameter tap to fill the tank be ${\text{3 hrs}}$
The time taken by other tap to fill the tank be ${\text{5 hrs}}$
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