Let the time taken by the smaller pipe to fill the tank =xhr .Time taken by the larger pipe =(x-10) {hr}Part of tank filled by smaller pipe in 1 hour =1 / xPart of tank filled by larger pipe in 1 hour =1 /(x-10)As given, the tank can be filled in 9 \\backslash \\backslash = 75 / 8 hours by both the pipes together.Therefore,​1/x+1/x−10=8/75x−10+x/x(x−10)=8/75⇒2x−10/x(x−10)=8/75⇒75(2x−10)=8x2−80x⇒150x−750=8x2−80x⇒8x2−230x+750=0​ ⇒8x2−200x−30x+750=0 ⇒8x(x−25)−30(x−25)=0 ⇒(x−25)(8x−30)=0 ⇒x=25,30/8