A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:  Choose the correct choice from the following answers
A) 15 hours
B) 20 hours
C) 25hours
D) 28 hours
E) 18 hours
Let the first pipe alone can fill the tank in x hours   Then the second pipe can fill the tank in (x-5) hours   and the third pipe can fill the tank in (x-5)-4 = (x-9) hours   part filled by first pipe and second pipe together in 1 hr = part filled by third pipe in 1 hr   (1/x)+1/(x−5)=1/(x−9)   (x−5)(x−9)+x(x−9)=x(x−5)   x^2−14x+45+x^2−9x=x^2−5x   −14x+45+x^2−9x=−5x   x^2−18x+45=0   (x−15)(x−3)=0   x=15 or 3   We can not take the value of x = 3 because, (x-9) becomes negative which is not possible   because the third pipe can fill the tank in (x-9) hours   Hence, x = 15   15 hoursnswer is 15 hours
<result>A</result>
A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is: