At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?
Assuming A completes a units of work in a day and B completes B units of work in a day and the total work = 1 unit
Hence, 12(a+b)=1.........(1)
Also, 9($$\ \frac{\ a}{2}$$+3b)=1 .........(2)
Using both equations, we get, 12(a+b)= 9($$\ \frac{\ a}{2}$$+3b)
=> 4a+4b=$$\ \frac{\ 3a}{2}$$+9b
=> $$\ \frac{\ 5a}{2}$$=5b
=> a=2b
Substituting the value of b in equation (1),
12($$\ \frac{\ 3a}{2}$$)=1
=> a=$$\ \frac{\ 1}{18}$$
Hence, the number of days required = 1/($$\ \frac{\ 1}{18}$$)=18
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