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. Time required by the first pipe to fill the tank is
6 hours
15 hours
10 hours
30 hours
Explanation:
Suppose the first pipe alone can fill the tank in x hours. Then,
second pipe alone can fill the tank in (x−5) hours,
third pipe alone 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
From here, better to find the value of x from the given choices which will be easier. Or we can solve it as follows.
(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
x2−18x+45=0
(x−15)(x−3)=0
x=15 or 3
We can not take the value 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
Suppose the first pipe alone can fill the tank in x hours. Then,
second pipe alone can fill the tank in (x−5) hours,
third pipe alone 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
From here, better to find the value of x from the given choices which will be easier. Or we can solve it as follows.
(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
x2−18x+45=0
(x−15)(x−3)=0
x=15 or 3
We can not take the value 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