Multiple choice

Pipes $A$ and $B$ can fill a tank in 5 and 6 hours respectively. Pipe $C$ can empty it in 12 hours. The tank is half full. All the three pipes are in operation simultaneously. After how much time the tank will be full?

  1. $3\displaystyle\frac { 9 }{ 17 }$ hours
  2. $11$ hours
  3. $2\displaystyle\frac { 8 }{ 11 }$ hours
  4. $1\displaystyle\frac { 13 }{ 17 }|$ hours
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Pipe A fills at 1/5 tank/hr, B fills at 1/6 tank/hr, and C empties at 1/12 tank/hr. Net rate = 1/5 + 1/6 - 1/12 = (12 + 10 - 5) / 60 = 17/60 tank/hr. Since the tank is half full initially, the remaining volume to fill is 1/2 tank. Time required = (1/2) / (17/60) = 30/17 = 1 13/17 hours. Note that option D is formatted with a stray pipe symbol due to OCR.

AI explanation

Since the tank is already half full, only 1/2 of the tank needs to be filled, and the net rate of the three pipes working together is the sum of their individual rates: 1/5 plus 1/6 minus 1/12. Finding a common denominator of 60 gives 12/60 plus 10/60 minus 5/12, which equals 17/60 of the tank per hour. The time required to fill the remaining half is the work divided by the rate, so 1/2 divided by 17/60, which equals 30/34 or 15/17 hours. Converting this to a mixed number gives 1 and 13/17 hours.

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