A cistern can be filled by two pipes filling separately in 8 and 10 minutes respectively. Both the pipes are opened together but being clogged for a certain time, only \( \displaystyle\frac{7}{10} \) of the full quantity of water flows through the former and only \( \displaystyle\frac{3}{8} \) through the latter pipe. The obstructions, however, being suddenly removed, the cistern is filled in 2 minutes from that moment. How long was it before the full flow began?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4.4 minutes
In one minute the first pipe fills \( \frac{1}{8} \) and the second \( \frac{1}{10} \) of the cistern.
While clogged, each pipe gives only its stated fraction of its own rate, so the part filled per minute is
\( \displaystyle \frac{7}{10}\cdot\frac{1}{8}+\frac{3}{8}\cdot\frac{1}{10}=\frac{7}{80}+\frac{3}{80} \)
\( \displaystyle =\frac{10}{80}=\frac{1}{8} \)
With full flow for the last 2 minutes:
\( \displaystyle 2\left(\frac{1}{8}+\frac{1}{10}\right)=2\cdot\frac{9}{40}=\frac{9}{20} \)
The rest of the cistern, \( 1-\frac{9}{20}=\frac{11}{20} \), was filled while the pipes were clogged. If that lasted t minutes,
\( \displaystyle \frac{t}{8}=\frac{11}{20} \)
\( \displaystyle t=\frac{88}{20}=4.4\ \text{minutes} \)