t1 time the first pipe takes to fill the tank
t2 time the second pipe takes to fill the tank
t3 time the third pipe takes to fill the tank


We know that pipes 1 and 2 operating simultaneously take the same time to fill the tank as pipe 3 operating alone,

so:


1/t1 +1/t2 =1/t3

We also know that pipe 2 fills the tank 5 hours faster than pipe 1: t2= t1 - 5

And pipe 2 is 4 hours slower than the pipe 3: t3= t2-4


So we have the following relationships:

t1

t2= t1-5

t3=t2-4= t1-9


If we try the diferent solutions proposed, we find that Answer C is the correct answer with t1=15; t2=10 and t3= 6


1/15 + 1/10 =5/30 =1/6 (1/t3= 1/t1+1/t2)

Hi Sajjad1994 , Bunuel , and All,

I have managed to solve this problem, but it took me a lot of time and I have solved it via answer options.

Is there any specific method which can save time while solving these problems.

Request you to share that method.

time taken for pipe 1 = x
time taken for pipe 2 = y
time taken for pipe 3 = z

1/x+1/y = 1/z

x = y+5
z= y-4
substituting the value, the equation formed is
y^2-8y-20 = 0

solve for y,
y comes to be 10
x= y+5 , = 15

15hours is the answer