Question
Pipes M, N and S can fill a tank in 25, 50 and 100
minutes, respectively. Initially, pipes N and S are kept open for 10 minutes, and then pipe N is shut while pipe M is opened. Pipe S is closed 15 minutes before the tank overflows. How much time (in minutes) will it take to fill the tank if the three pipes work in this pattern?Solution
Pipes M, N, and S fill the tank in 25, 50, and 100 minutes, respectively. Their rates are 1/25, 1/50, and 1/100 tank per minute. N and S fill in 10 minutes: Part filled = 10 × (1/50 + 1/100) = 10 × 3/100 = Remaining part: 1 – 3/10 = 7/10 Let x be the time taken after N is shut. For the last 15 minutes, only M works: Part filled by M in 15 minutes=15/25 = 3/5 Remaining part before the last 15 minutes: 7/10 – 3/5 = 7/10 – 6/10 = 1/10 Let t be the time when both M and S work: t × (1/25 + 1/100) = 1/10 t × 5/100 = 1/10  ⟹  t = 2 Total time: 10 + 2 + 15 = 27 minutes.
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