# 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.

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,

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

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

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

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

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

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

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

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

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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