VerifiedHint: We start solving the problem by assigning variables for the time taken to fill the tank by pipe 1 and for the capacity of the tank. Using this we find the time taken to fill the tank by pipe 2. We then find the part of the tank filled by pipes 1 and 2 in one minute. We then use the condition that two pipes running together can fill a tank in \[11\dfrac{1}{9}\] minutes by adding the obtained times. This gives us a quadratic equation which gives us the required time to solve. We then find the product of the times after obtaining them.
Complete answer:
Two pipes can fill the tank together in \[11\dfrac{1}{9}\]min. Now let us consider the time taken by one pipe takes 5 minutes more than the other pipe to fill the tank. Let us assume pipe 1 takes ‘x’ min and using this we get pipe 2 will take (x + 5) min to fill the tank separately. Let us assume that the capacity of the tank be y cubic units.
\[11\dfrac{1}{9}\]min can also be written as = \[\dfrac{11\times 9+1}{9}=\dfrac{100}{9}\]minutes.
We have given that pipe 1 can fill the entire tank of capacity of y in x minutes.
Let us find the capacity of the tank that pipe 1 can fill in one minute. Let us assume that capacity filled in one minute as z.
So, we get $x\times z=y$.
$\Rightarrow z=\dfrac{y}{x}$ -(1).
We have given that pipe 1 can fill the entire tank of capacity of y in (x + 5) minutes.
Let us find the capacity of the tank that pipe 2 can fill in one minute. Let us assume that capacity filled in one minute as p.
So, we get $\left( x+5 \right)\times p=y$.
$\Rightarrow p=\dfrac{y}{x+5}$ -(2).
From equations (1) and (2), we get the capacity of tank that pipe 1 and pipe 2 can fill in one minute = \[\dfrac{y}{x}+\dfrac{y}{x+5}\].
Given that 2 pipes can fill it together in \[\dfrac{100}{9}\] minutes. Thus we can form our equation as,
\[\Rightarrow \dfrac{100}{9}\left( \dfrac{y}{x}+\dfrac{y}{x+5} \right)=y\].
\[\Rightarrow \dfrac{1}{x}+\dfrac{1}{x+5}=\dfrac{9}{100}\].
Now, let us simplify the above expression and get the value of x.
\[\Rightarrow \dfrac{\left( x+5 \right)+x}{x\left( x+5 \right)}=\dfrac{9}{100}\].
Apply cross multiplication property.
\[\Rightarrow 100\left( 2x+5 \right)=9\left( {{x}^{2}}+5x \right)\].
\[\Rightarrow 200x+500=9{{x}^{2}}+45x\].
\[\Rightarrow 9{{x}^{2}}-155x-500=0\].
The above expression is similar to the general quadratic equation, \[a{{x}^{2}}+bx+c=0\].
Here a = 9, b = -155, c = -500.
Let us apply these values in the quadratic formula, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
\[\Rightarrow x=\dfrac{155\pm \sqrt{{{\left( -155 \right)}^{2}}-4\times 9\times \left( -500 \right)}}{2\times 9}\].
\[\Rightarrow x=\dfrac{155\pm \sqrt{24025+18000}}{18}=\dfrac{155\pm \sqrt{42025}}{18}\].
\[\Rightarrow x=\dfrac{155\pm 205}{18}\].
Thus, we get \[x=\dfrac{155+205}{18}\] and \[x=\dfrac{155-205}{18}\].
Thus, we get, x = 20 and \[x=\dfrac{-50}{18}=\dfrac{-25}{9}\].
We know that time cannot be negative, so we neglect the value of \[x=\dfrac{-25}{9}\].
\[\therefore \] We get pipe 1 which takes x = 20 minutes.
\[\therefore \] Pipe 2 would take x + 5 = 20 + 5 = 25 minutes.
Hence pipe 1 will fill the tank in 20 minutes and pipe B will fill the tank in 25 minutes.
\[\therefore \] Product of time = 20 \[\times \] 25 = 500 minutes.
Note: We should not forget to find the product of the time, which gives the final answer of the question. We can also take the capacity of a tank as 1 cubic unit, but it confuses calculations and there will be a high possibility of making mistakes by following this way. We should not make mistakes while solving this problem. Whenever we get this type of problem, we should start solving by assigning the variable to unknown which leads us to the solution.
