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Machine A does 1/6 of the work in 1 hours, B does 1/12 and C does 1/18.

Machine A works for 2 hours and hence does 2*1/6 = 1/3 of the work

Machine B works for 4 hours and hence does 4*1/12 = 1/3 of the work

Machine C has to do 1/3 of the work. At its rate, is will take [1/3]/[1/18] = 18/3 = 6 Hours so to do.

Machine C has already worked for 4 hours when Machine B was switched off. It will work for 6 - 4 = 2 Additional Hours = Answer B.
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Working alone, machines A, B and C can complete a task in 6, 12 and 18 hours respectively. All three machines are started together to complete the task but machine A is switched off after 2 hours and machine B is also switched off after another 2 hours. Machine C completes the remaining task. How much more time does machine C take to complete the task?

A. 30 minutes
B. 2 hours
C. 2 hours 30 minutes
D. 4 hours
E. 4 hours 30 minutes
Let the Total work be 36 Units

Efficiency of A is 6 Units/Hr , Efficiency of B is 3 Units/Hr & Efficiency of C is 2 Units/Hr

Work completed by A , B & C in 2 Hours is 22 Units ; Work left is 14 Units
Work completed by B & C in next 2 Hours is 10 Units ; Work left is 4 Units
Time required to complete 4 Units by C working at 2 Units/Hr is 4/2 = 2Units/Hr, Answer must be (B)
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