# Moderate Time and Work Solved QuestionAptitude Discussion

 Q. $A$, $B$ and $C$, can complete a piece of work individually in 15, 30 and 40 days respectively. They started the work together and the $A$ and $B$ let 2 days and 4 days before the completion of the work respectively. In how many days was the work completed?
 ✖ A. $10\dfrac{13}{15}$ ✖ B. $12$ ✖ C. $15$ ✔ D. $10\dfrac{2}{15}$

Solution:
Option(D) is correct

Let it takes $x$ days to complete the work

The $A$ worked for $(x−2)$ days, $B$ for $(x−4)$ days and $C$'s for $x$ days.

$\dfrac{x}{40}+\dfrac{x-4}{30}+\dfrac{x-2}{15}=1$

$\Rightarrow x=\dfrac{152}{15}=10\dfrac{2}{15}$

## (1) Comment(s)

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The above equation makes logical sense. However,

Doing this by efficiency method,

A,B,C if they work together, they will complete the work in 8 days. So, B left after 4th day and A left after 6th day.

Computing for the first four days,

Total work completed will be half. (Since all are there.)

A and C work together for two days after which A leaves. Meaning.

Solving the above yields,

60(Work remaining) - 22(Work done in the two days i.e Eff. of A + Eff. of C equals 8 * (2 Number of Days) + 3 * 2)

thus work remaining is 38 which will be done by C.

Time taken will be, 38/3 ~ 12.66

Thus total days to complete the work will be 4 + 2 + 12.66 = 18.66.

Can anyone explain why the difference?