A can complete $$33\frac{1}{3}$$% of a piece of work in 10 days and B can complete one-fourth of the same work in 30 days. A, B and C together can complete the same work in 12 days. In how many days will the work be completed if A works on all days, and is assisted by B and C on alternate days, starting with B assisting A on the first day?

Correct option is B

Given:

A completes $$33\frac{1}{3}\%$$​ of the work in = 10 days

B can complete $$\frac{1}{4}$$​ th of the work in = 30 days

A, B and C can complete the work in = 12days

Formula Used:

Let total work to fill the tank be 1.

Then work done by a pipe in 1 hour $$=\frac{1}{Total\ time\ to\ complete}$$​​

Solution:

A does $$33\frac{1}{3}\%$$​ of the work in 10 days

​$$33\frac{1}{3}\% = \frac{100}{3}\% = \frac{100}{3}\times \frac{1}{100} = \frac{1}{3}$$​​

Time taken by A to complete $$\frac{1}{3}$$​rd of the work in 10 days

Hence time taken to complete the whole work = $$10 \times 3$$​= 30 days

B does the $$\frac{1}{4}$$​ work in = 30days

Time taken by B to complete the whole work =$$30 \times \frac{4}{1}$$​ = 120 days

A’s one day work =$$\frac{1}{30}$$​​

B’s one day work = $$\frac{1}{120}$$​​

A,B and C one day’s work =$$\frac{1}{12}$$​​

C’s one day’s work $$= \frac{1}{12} - \left(\frac{1}{30} +\frac{1}{120}\right)$$​​

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