Question : Three pipes A, B, and C can fill a tank in 6 hours, 9 hours, and 12 hours, respectively. B and C are opened for half an hour, and then pipe A is opened. The time taken by the three pipes together to fill the remaining part of the tank is:
Option 1: $3\ \text{hours}$
Option 2: $2\ \text{hours}$
Option 3: $2\frac{1}{2}\ \text{hours}$
Option 4: $3\frac{1}{2}\ \text{hours}$
Correct Answer: $2\frac{1}{2}\ \text{hours}$
Solution :
Given:
Three pipes A, B, and C can fill a tank in 6 hours, 9 hours, and 12 hours respectively.
So, L.C.M of 6, 9, 12 = Total capacity = 36 units
Efficiency of A = $\frac{36}{6}$ = 6
Efficiency of B = $\frac{36}{9}$ = 4
Efficiency of C = $\frac{36}{12}$ = 3
In half an hour (B + C) must have filled $= \frac{4}{2}+\frac{3}{2} = \frac{7}{2}$ units
Capacity left $ = 36-\frac{7}{2} = \frac{65}{2}$ units
Now all pipes will fill the remaining tank in $\frac{65}{2×(6+4+3)} = \frac{5}{2} = 2\frac{1}{2}$ hours
Hence, the correct answer is $2\frac{1}{2}\ \text{hours}$.
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