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

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

Part of tank filled by smaller pipe in 1 hour = \( \dfrac{1}{x}\)

Part of tank filled by larger pipe in 1 hour = \( \dfrac{1}{x-10}\)

As given, the tank can be filled in

\(9 \dfrac{3}{8}\)= \( \dfrac{75}{8}\) hours by both the pipes together.

Therefore,

= \( \dfrac{1}{x}+ \dfrac{1}{x-10}= \dfrac{8}{75} \)

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

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

= 75(2*x* – 10) = 8*x*^{2} – 80*x*

= 150*x* – 750 = 8*x*^{2} – 80*x*

= 8*x*^{2} – 230*x* +750 = 0

= 8*x*^{2} – 200*x* – 30*x* + 750 = 0

= 8*x*(*x* – 25) -30(*x* – 25) = 0

= (*x* – 25)(8*x* -30) = 0

= *x* = 25,\( \dfrac{30}{8}\)

Time taken by the smaller pipe cannot be \( \dfrac{30}{8}\) = 3.75 hours,

as the time taken by the larger pipe will become 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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