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A can complete a piece of work in 10 days, B in 15 days and C in 20 days. A and C worked together for 2 days and then A was replaced by B. In how many days, altogether, was the work completed?

Solution

✅ Correct Option: 2

One day's work: A =110= \frac1{10}, B =115= \frac1{15}, C =120= \frac1{20}.

A and C in 2 days: 2(110+120)=3102\left(\frac1{10}+\frac1{20}\right) = \frac{3}{10}. Remaining =710= \frac{7}{10}.

B and C together do 115+120=760\frac1{15}+\frac1{20} = \frac{7}{60} per day, needing 7/107/60=6\frac{7/10}{7/60} = 6 days.

Total =2+6=8= 2 + 6 = 8 days.

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