A alone can complete a work in 16 days and B | vedclass

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(D) Work done by $A$ in $1$ day $= \frac{1}{16}$.Work done by $B$ in $1$ day $= \frac{1}{12}$.Work done in a cycle of $2$ days (starting with $A$) $=

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$13\frac{3}{4}$

Vedclass · Answer

(D) Work done by $A$ in $1$ day $= \frac{1}{16}$.Work done by $B$ in $1$ day $= \frac{1}{12}$.Work done in a cycle of $2$ days (starting with $A$) $= \frac{1}{16} + \frac{1}{12} = \frac{3+4}{48} = \frac{7}{48}$.In $6$ cycles ($12$ days),work done $= 6 	imes \frac{7}{48} = \frac{42}{48} = \frac{7}{8}$.Remaining work $= 1 - \frac{7}{8} = \frac{1}{8}$.On the $13$th day,$A$ works. Work done by $A$ in $1$ day is $\frac{1}{16}$.Remaining work after $13$ days $= \frac{1}{8} - \frac{1}{16} = \frac{1}{16}$.On the $14$th day,$B$ works. $B$ completes $\frac{1}{12}$ work in $1$ day,so $B$ completes $\frac{1}{16}$ work in $\frac{1/16}{1/12} = \frac{12}{16} = \frac{3}{4}$ days.Total time $= 13 + \frac{3}{4} = 13\frac{3}{4}$ days.

$A$ alone can complete a work in $16$ days and $B$ alone in $12$ days. Starting with $A$,they work on alternate days. The total work will be completed in (in days):

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