Question

# A can lay a railway track between two given stations in 16 days and B can do the same job in 12 days. With the help of C, they did the job in 4 days only. Then, C alone can do the job in:A. $9\dfrac{1}{5}$ daysB. $9\dfrac{2}{5}$ daysC. $9\dfrac{3}{5}$ daysD. 10 days

Hint: To do this problem, we should know the concept of work vs days problems. A person A can do a job in x days means that the person can do $\dfrac{1}{x}$ portion of the work in 1 day. By using this analogy, we can get the work done by A, B individually in a day. Similarly we can find the work done by A, B, C combined in a day. By assuming that C takes x days to finish the work which implies that C can do $\dfrac{1}{x}$ the portion of work in 1 day. The works of A, B, C are additive and we can use that to get the value of x.

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.
So, the correct answer is “Option C”.

Note: Students can make a mistake by thinking that the work done by C can be equal to the sum of the works done by A, B and A,B,C combined. That leads to an answer of
\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.