Two pipes can fill a tank in $20$ min and $30$ min. Both opened together for $5$ min, then the second is…

Two pipes can fill a tank in $20$ min and $30$ min. Both opened together for $5$ min, then the second is closed. The remaining time for the first pipe to fill it alone is:
  1. $\dfrac{25}{3}$ min
  2. $\dfrac{35}{3}$ min
  3. $\dfrac{25}{2}$ min
  4. $15$ min

Solution

In $5$ min, both fill $5\left(\dfrac{1}{20} + \dfrac{1}{30}\right) = 5 \cdot \dfrac{5}{60} = \dfrac{5}{12}$. Remaining $= \dfrac{7}{12}$. First pipe (rate $\dfrac{1}{20}$) takes $\dfrac{7}{12} \div \dfrac{1}{20} = \dfrac{7 \times 20}{12} = \dfrac{35}{3}$ min.

Asked in: IMO

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