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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 710\frac{7}{10} of the full quantity of water flows through the former and only 38\frac{3}{8} through the later 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?

Solution

✅ Correct Option: 3

Let the clogged period be tt minutes.

Work done while clogged =710⋅t8+38⋅t10=7t80+3t80=t8=\frac{7}{10}\cdot\frac{t}{8}+\frac{3}{8}\cdot\frac{t}{10}=\frac{7t}{80}+\frac{3t}{80}=\frac{t}{8}.

After removal, 2 minutes give 2(18+110)=9202\left(\frac18+\frac1{10}\right)=\frac{9}{20}.

So t8=1−920=1120\frac{t}{8}=1-\frac{9}{20}=\frac{11}{20}, giving t=4.4t=4.4 minutes.

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