Question

Two water taps together can fill a tank in 9⅜ hrs. The taps of larger diameter takes 10 hrs less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer 1

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 = 1/x

Part of tank filled by larger pipe in 1 hour = 1/x- 10

It is given that the tank can be filled in9⅜ = 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 be 30/8 = 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.

Answer 2

Let the tap of the larger diameter fills the tank alone in (x – 10) hours.
 
In 1 hour, the tap of the  smaller diameter can fill 1/x part of the tank.
 
In 1 hour, the tap of the  larger diameter can fill 1/(x – 10) part of the tank.
 
Two water taps together can fii a tank in 75 / 8 hours.
 
But  in 1 hour the taps fill 8/75 part of the tank.

1 / x  +  1 / (x – 10) = 8 / 75. 
( x – 10 + x ) / x ( x – 10) =  8 / 75. 
2( x – 5) / ( x^2 – 10 x) = 8 / 75. 
4x^2 – 40x = 75x – 375. 
4x^2 – 115x + 375 = 0
 4x^2 – 100x – 15x + 375 = 0 
4x ( x – 25) – 15( x – 25) = 0 
( 4x -15)( x – 25) = 0.
 x = 25, 15/ 4. 

But x = 15 / 4 then x – 10 = -25 /4 which is not possible since time 
But x = 25 then x – 10 = 15.
 Larger diameter =15 hours and
smaller diameter=  25 hours.