A rectangular tank can be filled with water by two pipes in 100/9 minutes. If the larger pipe alone takes 5 minutes less to fill the tank than...

The solution posted above contains some errors and leaves some points explained although the final answer given is correct. For example, it is stated:

... we assume the larger of the pipe alone takes x+5 minutes to fill up the water tank and the smaller alone  in x minutes.

This statement cannot be correct as the larger pipe should take less time to fill the tank as compared to the smaller pipe.

I suggest the following solution.

Let the time taken by the larger pipe to fill the tank be x minutes.

Then capacity of larger pipe to fill the tank in terms of tanks per minute = 1/x.

Now as the smaller pipe takes five minutes longer than the larger pipe the time taken by it to fill the tank = x + 5 minutes.

Then capacity of smaller pipe to fill the tank in terms of tanks per minute = 1/(x +5).

The total capacity of large plus small pipe = 1/x + 1/(x + 5)

= ([x + 5) + x]/[x*(x + 5)] = (2x + 5)/(x^2 + 5x)

The time taken by large plus small pipe to fill the tank =

1/(total capacity of large plus small pipe) = 1/[(2x + 5)/(x^2 + 5x)]

= (x^2 + 5x)/(2x + 5)

This time is given as 100/9 minutes

Therefore: 100/9 = (x^2 + 5x)/(2x + 5)

Therefore: 100*(2x + 5) = 9*(x^2 + 5x)

Therefore: 200x + 500 = 9*x^2 + 45x

Rearranging this equation with all the terms on left hand side we get:

9*x^2 - 155x + 500 = 0

Therefore: 9*x^2 - 180x  + 25x + 500 = 0

Therefore: 9x(x - 20)  + 25*(x + 20) = 0

Therefore: (x - 20)*( 9x + 25) = 0

Therefore x = 20, and -25/9

However the time taken to fill the tank cannot be negative therefore we accept only the first value i.e. x = 20

Therefore large pipe will take 20 minutes to fill the tank.

And small pipe will take 5 minutes more than that i.e. 20 + 5 = 25 minutes.