Three pipes A, B and C are installed to fill a tank. Pipes A and B opened together can fill the tank in the same time in which C can alone fill the tank. If pipe B can fill the tank 15 minutes faster than pipe A and 5 minutes slower than pipe C, then the time required by pipe A to fill the tank alone is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 30 minutes
Let A alone take x minutes. Then B takes x − 15 minutes, and C, being 5 minutes faster than B, takes x − 20 minutes.
In one minute a pipe fills the reciprocal of its time. A and B together do in one minute what C does alone:
\( \displaystyle \frac{1}{x} + \frac{1}{x - 15} = \frac{1}{x - 20} \)
\( \displaystyle \frac{2x - 15}{x(x - 15)} = \frac{1}{x - 20} \)
\((2x - 15)(x - 20) = x(x - 15)\)
\(2x^2 - 55x + 300 = x^2 - 15x\)
\(x^2 - 40x + 300 = 0\)
\((x - 10)(x - 30) = 0,\quad x = 10 \text{ or } 30\)
x = 10 would give B a time of −5 minutes, which is impossible. So x = 30: A takes 30 minutes (B takes 15, C takes 10).
Check: A and B together fill, per minute,
\( \displaystyle \frac{1}{30} + \frac{1}{15} = \frac{3}{30} = \frac{1}{10} \)
which is the rate of C.