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Two workers ‘A’ and ‘B’ working together completed a job in 5 days. Had ‘A’ worked twice as efficiently as he actually did and ‘B’ worked one-third as efficiently as he actually did, the work would have completed in 3 days. In how many days could ‘A’ alone complete the job?

Two workers ‘A’ and ‘B’ working together completed a job in 5 days. Had ‘A’ worked twice as efficiently as he actually did and ‘B’ worked one-third as efficiently as he actually did, the work would have completed in 3 days. In how many days could ‘A’ alone complete the job? Correct Answer <span class="math-tex">\(6\frac{1}{4}\:days\)</span>

Let A can alone complete the work in ‘x’ days.

⇒ A’s 1-day work = 1/x

∵ A and B together completed the work in 5 days.

⇒ B’s 1-day work = 1/5 - 1/x

Now, if A worked twice & B worked 1/3rd in a day, time taken = 3 days

⇒ 2(1/x) + (1/3)(1/5 - 1/x) = 1/3

⇒ 2/x + 1/15 - 1/3x = 1/3

⇒ 5/3x = 4/15

⇒ x = 5/3 × 15/4 = 25/4 days

∴ A can alone complete the work in 25/4 days 

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