$A$ alone can do a piece of work in $15$ days and $B$ alone can do the same work in $12$ days. $A$ works alone for the first four days. In how many days will $A$ and $B$ together finishing the remaining work?

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Solution

$A$ alone can do the work in $15$ days $= \frac{1}{15} t h$ work in one day

$B$ alone can do the work in $12$ days $= \frac{1}{12} t h$ work in one day.

$A$ works for $4$ days.
Whole work completion $= 1$
Work done by $A$ in $4$ days $= 4 \times \frac{1}{15} = \frac{4}{15}$

Work remaining $= 1 - \frac{4}{15} = \frac{15 - 4}{15} = \frac{12}{15} = \frac{4}{5}$

$A$ and $B$ together can do $= \frac{1}{15} + \frac{1}{12} = \frac{4 + 5}{60} = \frac{9}{60} = \frac{3}{20}$ in one day

To finish the remaining work they need $= (\frac{4}{5}) \div \frac{3}{20} = \frac{4}{5} \times \frac{20}{3}$

$= \frac{16}{3} = 5 \frac{1}{3}$ days

They need $5 \frac{1}{3}$ days to finish the work.

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A alone can do a piece of work in 15 days and B alone can do the same work in 12 days.A works alone for the first four days. In how many days will A and B together finishing the remaining work?

$A$ alone can do a piece of work in $15$ days and $B$ alone can do the same work in $12$ days. $A$ works alone for the first four days. In how many days will $A$ and $B$ together finishing the remaining work?