Two pipes A and B can fill a tank in (x+4) hours and (x+8) hours, respectively. Both pipes are opened together, and after 4 hours, pipe A is closed. Pipe B continues for another 2 hours to fill the remaining part of the tank. Find the value of x.
Solution: Let the capacity of the tank be 1 unit. Pipe A’s 1 hour work = 1/(x+4) Pipe B’s 1 hour work = 1/(x+8) In 4 hours, work done by both pipes = 4 × (1/(x+4) + 1/(x+8)) Remaining work = 1 - 4 × (1/(x+4) + 1/(x+8)) Remaining work = 2 × 1/(x+8) Solve for x: 1 - 4 × ((x+8 + x+4) / ((x+4)(x+8))) = 2/(x+8) Simplifying: x = 4
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