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

(A) Let the time taken by the smaller pipe to fill the tank be $x \text{ hours}$.
Time taken by the larger pipe $= (x - 10) \text{ hours}$.
Part of the tank filled by the smaller pipe in $1 \text{ hour} = \frac{1}{x}$.
Part of the tank filled by the larger pipe in $1 \text{ hour} = \frac{1}{x - 10}$.
It is given that the tank can be filled in $9 \frac{3}{8} = \frac{75}{8} \text{ hours}$ by both pipes together. Therefore,
$\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75}$.
$\frac{x - 10 + x}{x(x - 10)} = \frac{8}{75}$.
$\frac{2x - 10}{x^2 - 10x} = \frac{8}{75}$.
$75(2x - 10) = 8(x^2 - 10x)$.
$150x - 750 = 8x^2 - 80x$.
$8x^2 - 230x + 750 = 0$.
Dividing by $2$,we get $4x^2 - 115x + 375 = 0$.
$4x^2 - 100x - 15x + 375 = 0$.
$4x(x - 25) - 15(x - 25) = 0$.
$(x - 25)(4x - 15) = 0$.
Thus,$x = 25$ or $x = \frac{15}{4} = 3.75$.
If $x = 3.75$,the larger pipe takes $3.75 - 10 = -6.25 \text{ hours}$,which is impossible.
Therefore,the smaller pipe takes $25 \text{ hours}$ and the larger pipe takes $25 - 10 = 15 \text{ hours}$.