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A and B together can finish a task in 12 days. If A worked half as efficiently as he usually does and B works thrice as efficiently as he usually does, the task gets completed in 9 days. How long would A take to finish the task if he worked independently?
12 days
24 days
27 days
18 days
Let A take ‘a’ days to complete the task and B take ‘b’ days to complete the task. Thus in one day, A will complete (1/a)th of the task. Similarly in one day, B will complete (1/b)th of the task. So in one day, if A and B work together they will complete (1/a + 1/b)th of the task. Given that A and B together take 12 days to complete the task, then in one day A and B together complete (1/12)th of the task. Thus, 1/a + 1/b = 1/12 ……Eqn. 1 If A worked half as efficiently as he usually does, then A will take twice the time as he usually takes, i.e., 2a days. Thus in one day, A completes (1/2a)th of the task. Similarly if B worked thrice as efficiently as he usually does, then B will take one-third the time he usually takes, i.e., b/3 days. Thus in one day, B completed (1/(b/?/3))th or (3/b)th of the task. Thus when both of them work together, they will complete (1/2a + 3/b)th of the task, given that A and B take 9 days to complete the task. Thus, 1/a + 1/b = 1/12 ……Eqn. 1 1/2a + 3/b = 1/9 …… Eqn. 2 1/2a + 3/b = 1/9 …… Eqn. 2 Solving Equations 1 and 2 for ‘a’ we should get the answer, From equation (1) we get 12(a + b) = ab From equation (2), we get 9(b + 6a) = 2ab Substituting ab as 12(a + b) in equation (2) we get 9b + 54a = 2 x 12 x ( a + b) 9b + 54a = 24a + 24b; Or, 30a = 15b, Or, b = 2a Now, 12(a + b) = ab, or 12 x 3a = 2a2 a = 18 days.
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