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A. $ 9\dfrac{1}{5} $ days

B. $ 9\dfrac{2}{5} $ days

C. $ 9\dfrac{3}{5} $ days

D. 10 days

Answer
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In the question, it is given that A does the job in 16 days and B does it in 12 days.

A person D can do a job in x days means that the person can do $ \dfrac{1}{x} $ portion of the work in 1 day.

Using the above statement, we can find the work done by A and B in 1 day individually.

Work done by A in 1 day when x = 16 is equal to work per day = $ \dfrac{1}{16} $

Work done by B in 1 day when x = 12 is equal to work per day = $ \dfrac{1}{12} $

Let C do the work in y days. Then

Work done by C in 1 day when x = y is equal to work per day = $ \dfrac{1}{y} $

In the question, it is given that A, B, C combined do the work in 4 days.

Work done by A, B, C combined in 1 day when x = 4 is equal to work per day = $ \dfrac{1}{4} $

As the individual work done is additive in nature, we can write that

Work done by A,B,C combined in one day is equal to the sum of individual works done by them in one day. That is

\[\begin{align}

& \dfrac{1}{4}=\dfrac{1}{16}+\dfrac{1}{12}+\dfrac{1}{y} \\

& \dfrac{1}{y}=\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{12} \\

\end{align}\]

By L.C.M and simplifying, we get

\[\begin{align}

& \dfrac{1}{y}=\dfrac{1\times 12-1\times 3-1\times 4}{48}=\dfrac{5}{48} \\

& \dfrac{1}{y}=\dfrac{5}{48} \\

\end{align}\]

By taking reciprocal, we get

$ y=\dfrac{48}{5}=\dfrac{45+3}{5}=9\dfrac{3}{5} $

So C takes $ 9\dfrac{3}{5} $ days to finish the work.

$ \begin{align}

& \dfrac{1}{y}=\dfrac{1}{16}+\dfrac{1}{12}+\dfrac{1}{4}=\dfrac{3+4+12}{48}=\dfrac{17}{48} \\

& y=\dfrac{48}{17} \\

\end{align} $

which is a totally different answer. So, we should know what works are added to finish the whole work. This can be achieved by understanding the question properly.