Question -
Answer -
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 = 1/x
Part of tank filled by larger pipe in 1 hour = 1/(x – 10)
As given, the tank can be filled in
= 75/8 hours by both the pipes together.Therefore,
1/x + 1/x-10 = 8/75
x-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
Time taken by the smaller pipe cannot be30/8 = 3.75 hours, as the time taken by the larger pipe will becomenegative, which is logically not possible.
Therefore, time taken individually by thesmaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.