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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.

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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.

 

 

 

 

Posted by

Divya Prakash Singh

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