Two water taps together can fill a tank in $\mathbf{9} \frac{\mathbf{3}}{\mathbf{8}}$ hours.
Question.

Two water taps together can fill a tank in $\mathbf{9} \frac{\mathbf{3}}{\mathbf{8}}$ 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.

Solution:

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

Time taken by the larger pipe = (x – 10) hr

Part of tank filled by smaller pipe in 1 hour $=\frac{\mathbf{1}}{\mathbf{x}}$

Part of tank filled by larger pipe in 1 hour $=\frac{\mathbf{1}}{\mathbf{x}-\mathbf{1 0}}$

It is given that the tank can be filled in $\mathbf{9} \frac{\mathbf{3}}{\mathbf{8}}=\frac{\mathbf{7 5}}{\mathbf{8}}$ hours by both the pipes together.

Therefore,

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

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

$\Rightarrow 15(2 x-10)=8 x^{2}-80 x$

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

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

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

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

i.e., $x=25, \frac{\mathbf{3 0}}{\mathbf{8}}$

Time taken by the smaller pipe cannot be $\frac{\mathbf{3 0}}{\mathbf{8}}=3.75$ hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.

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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