A, B, and C can complete a project alone in 12 days, 18 days, and 24 days, respectively. If they all work together for 4 days, what fraction of the project will remain incomplete?
\(\frac{5}{18}\)
If a person completes a project alone in n days, the person does \(\frac{1}{n}\) of it in one day; the one-day works of people working together add up.
One-day work of A, B and C together = \(\frac{1}{12} + \frac{1}{18} + \frac{1}{24} = \frac{6 + 4 + 3}{72} = \frac{13}{72}\).
Work done in 4 days = \(4 \times \frac{13}{72} = \frac{52}{72} = \frac{13}{18}\).
So the fraction that remains incomplete = \(1 - \frac{13}{18} = \frac{5}{18}\).
Check: take the project as 72 units (LCM of 12, 18, 24); A, B and C do 6, 4 and 3 units a day, i.e. 13 units a day, so 52 units in 4 days, leaving 20 units, and 20/72 = 5/18.
Hence, \(\frac{5}{18}\).
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