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A and B together can complete a work in 15 days, B and C together can complete the same work in 36 days while A and C together can complete the same job in \(16\frac{4}{{11}}\) days. Then, A alone can complete 2/5th of the same work in how many days?

A and B together can complete a work in 15 days, B and C together can complete the same work in 36 days while A and C together can complete the same job in \(16\frac{4}{{11}}\) days. Then, A alone can complete 2/5th of the same work in how many days? Correct Answer 8 days

Given:

A + B = 15 days

B + C = 36 days

A + C = 16 (4/11) days = 180/11 days

Concept used:

In this type of time and work question, we assume the total work to be done as the LCM of the number of days taken to each of the persons to complete the work.

⇒ Total work = Efficiency × Number of days

Calculation:

Suppose the total work is (LCM of 15, 36 and 180 / 11)

Fraction LCM formula = LCM of numerator/HCF of denominator

⇒ LCM of (15, 36 and 180)/HCF of (1, 1, 11)

⇒ LCM = 180/1 = 180

180 units

⇒ (A + B)’s 1 day work = 180 / 15 = 12 units

⇒ (B + C)’s 1 day work = 180 / 36 = 5 units

⇒ (C + A)’s 1 day work = 180 / (180 / 11) = 11 units

⇒ One day work of (A + B + C) = (12 + 5 + 11) / 2 = 14 units

⇒ (A + B + C) - (B + C) = A

⇒ Efficiency of A = 14 - 5 = 9

⇒ ⅖ of total work = ⅖ × 180 = 72 units

∴ Time taken by A = 72 / 9 = 8 days

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