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, then the time required by the first pipe to fill the tank alone is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 15 hours
Let the first pipe alone take x hours. Then
\( \displaystyle \text{second pipe: } x-5 \text{ hours} \)
\( \displaystyle \text{third pipe: } (x-5)-4=x-9 \text{ hours} \)
In one hour the pipes fill \( \frac{1}{x} \), \( \frac{1}{x-5} \) and \( \frac{1}{x-9} \) of the tank. The first two together work as fast as the third, so the rates add:
\( \displaystyle \frac{1}{x}+\frac{1}{x-5}=\frac{1}{x-9} \)
\( \displaystyle \frac{2x-5}{x(x-5)}=\frac{1}{x-9} \)
\( \displaystyle (2x-5)(x-9)=x(x-5) \)
\( \displaystyle 2x^2-23x+45=x^2-5x \)
\( \displaystyle x^2-18x+45=0 \)
\( \displaystyle (x-15)(x-3)=0 \)
x = 3 is impossible (the third pipe would take 3 − 9 < 0 hours), so x = 15 hours.
Shortcut: Try the options: x = 15 gives 15, 10 and 6 hours, and \( \frac{1}{15}+\frac{1}{10}=\frac{2+3}{30}=\frac{1}{6} \). It works at once.