Q9.    Two water taps together can fill a tank in 9\frac{3}{8} hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answers (1)
D Divya Prakash Singh

Let the time taken by the smaller pipe to fill the tank be x\ hr.

Then, the time taken by the larger pipe will be: (x-10)\ hr.

The fraction of the tank filled by a smaller pipe in 1 hour:

 = \frac{1}{x}

The fraction of the tank filled by the larger pipe in 1 hour.

= \frac{1}{x-10}
Given that two water taps together can fill a tank in 9\frac{3}{8} = \frac{75}{8} hours.

Therefore,

\Rightarrow \frac{1}{x}+\frac{1}{x-10} = \frac{8}{75}

\Rightarrow \frac{x-10+x}{x(x-10)} = \frac{8}{75}

\Rightarrow \frac{2x-10}{x(x-10)} = \frac{8}{75}

Making it a quadratic equation:

\Rightarrow 150x-750 = 8x^2-80x

\Rightarrow 8x^2-230x+750 = 0

\Rightarrow 8x^2-200x-30x+750 = 0

\Rightarrow 8x(x-25) - 30(x-25) = 0

\Rightarrow (x-25)(8x+30) = 0

Hence the roots are \Rightarrow x = 25,\ \frac{-30}{8}

As the time taken cannot be negative:

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25-10 =15 hours respectively.

 

 

 

 

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