154238 wrote:
A can complete a project in 20 days and B can complete the same project in 30 days. If A and B start working on the project together and A quits 10 days before the project is completed, in how many days will the project be completed?
(A) 18 days
(B) 27 days
(C) 26.67 days
(D) 16 days
(E) 12 days
Instead of taking the work as 1 unit, we take it as the LCM of 20 and 30 = 60 units.
Since A can complete the work in 20 days, per day he does 60/20 = 3 units of work
Since B can complete the work in 30 days, per day he does 60/30 = 2 units of work
From here, we can proceed in 2 ways:
Method 1: Let the number of days in which the project is completed be \(n\)
Thus, B worked for the entire \(n\) days
Work done by B = \(2n\) units
A worked for \((n - 10)\) days
Work done by A = \(3(n - 10)\) units
Since total work in 60 units:
\(2n + 3(n - 10) = 60\)
\(n = 18\) days
Answer AMethod 2: Since A quit 10 days before completion, B must have worked alone for those 10 days
Work done by B in those 10 days = 10 * 2 = 20 units
Since total work is 60 units, A and B must have completed the remaining 60 - 20 = 40 units together
Work done by A and B in 1 day = 3 + 2 = 5 units
Thus, time for which A and B worked to compete 40 units = 40/5 = 8 days
Thus, total project duration = 8 + 10 = 18 days
Answer A _________________
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