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Question

Answers

(A) 25

(B) $\dfrac{{ - 25}}{4}$

(C) 24

(D) $\dfrac{{25}}{4}$

Answer

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Hint- Here, we will be using the concept of how much part of the tank is filled in one hour.

Let us assume that the tap with smaller diameter can separately fill the tank in $x$ hours.

According to the problem statement it is given that the tap with larger diameter can separately fill the tank in $\left( {x - 10} \right)$ hours.

Also, given that both the taps can together fill the tank in $9\dfrac{3}{8} = \dfrac{{\left( {8 \times 9} \right) + 3}}{8} = \dfrac{{75}}{8}$ hours.

So, the tap with a smaller diameter can fill $\dfrac{1}{x}$ part of the tank in 1 hour. Similarly, the tap with a larger diameter can fill $\dfrac{1}{{\left( {x - 10} \right)}}$ part of the tank in 1 hour. Also, both the taps fill $\dfrac{8}{{75}}$ part of the tank in 1 hour.

Then, \[

\dfrac{1}{x} + \dfrac{1}{{\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow \dfrac{{x - 10 + x}}{{x\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow \dfrac{{2x - 10}}{{x\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow 75\left( {2x - 10} \right) = 8x\left( {x - 10} \right) \\

\Rightarrow 150x - 750 = 8{x^2} - 80x \Rightarrow 8{x^2} - 230x + 750 = 0 \Rightarrow 4{x^2} - 115x + 375 = 0 \\

\Rightarrow 4{x^2} - 100x - 15x + 375 = 0 \Rightarrow 4x\left( {x - 25} \right) - 15\left( {x - 25} \right) = 0 \Rightarrow \left( {x - 25} \right)\left( {4x - 15} \right) = 0 \\

\]

Either $x = 25$ or $4x = 15 \Rightarrow x = \dfrac{{15}}{4}$

When $x = 25$, $\left( {x - 10} \right) = 25 - 10 = 15$ and when $x = \dfrac{{15}}{4}$, $\left( {x - 10} \right) = \dfrac{{15}}{4} - 10 = \dfrac{{15 - 40}}{4} = - \dfrac{{25}}{4}$

Since, the time can never be negative so $x = \dfrac{{15}}{4}$ is rejected.

Hence, $x = 25$ and $\left( {x - 10} \right) = 15$

The tap with smaller diameter can separately fill the tank in 25 hours and the tap with larger diameter can separately fill the tank in 15 hours.

Therefore, option A is correct.

Note- In these types of problems, the concept of part of the work done (here it is filling the tank) in one hour is utilized to obtain an equation in one variable so that we can solve for it.

Let us assume that the tap with smaller diameter can separately fill the tank in $x$ hours.

According to the problem statement it is given that the tap with larger diameter can separately fill the tank in $\left( {x - 10} \right)$ hours.

Also, given that both the taps can together fill the tank in $9\dfrac{3}{8} = \dfrac{{\left( {8 \times 9} \right) + 3}}{8} = \dfrac{{75}}{8}$ hours.

So, the tap with a smaller diameter can fill $\dfrac{1}{x}$ part of the tank in 1 hour. Similarly, the tap with a larger diameter can fill $\dfrac{1}{{\left( {x - 10} \right)}}$ part of the tank in 1 hour. Also, both the taps fill $\dfrac{8}{{75}}$ part of the tank in 1 hour.

Then, \[

\dfrac{1}{x} + \dfrac{1}{{\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow \dfrac{{x - 10 + x}}{{x\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow \dfrac{{2x - 10}}{{x\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow 75\left( {2x - 10} \right) = 8x\left( {x - 10} \right) \\

\Rightarrow 150x - 750 = 8{x^2} - 80x \Rightarrow 8{x^2} - 230x + 750 = 0 \Rightarrow 4{x^2} - 115x + 375 = 0 \\

\Rightarrow 4{x^2} - 100x - 15x + 375 = 0 \Rightarrow 4x\left( {x - 25} \right) - 15\left( {x - 25} \right) = 0 \Rightarrow \left( {x - 25} \right)\left( {4x - 15} \right) = 0 \\

\]

Either $x = 25$ or $4x = 15 \Rightarrow x = \dfrac{{15}}{4}$

When $x = 25$, $\left( {x - 10} \right) = 25 - 10 = 15$ and when $x = \dfrac{{15}}{4}$, $\left( {x - 10} \right) = \dfrac{{15}}{4} - 10 = \dfrac{{15 - 40}}{4} = - \dfrac{{25}}{4}$

Since, the time can never be negative so $x = \dfrac{{15}}{4}$ is rejected.

Hence, $x = 25$ and $\left( {x - 10} \right) = 15$

The tap with smaller diameter can separately fill the tank in 25 hours and the tap with larger diameter can separately fill the tank in 15 hours.

Therefore, option A is correct.

Note- In these types of problems, the concept of part of the work done (here it is filling the tank) in one hour is utilized to obtain an equation in one variable so that we can solve for it.

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