# Time Work Question

If $A$, $B$ and $C$ can do a piece of work in $8$, $10$, $12$ days respectively. They started working together. After working for $1$ day $A$ left the work while after $2$ days $A$ joins & $B$ left the work. After $3$ days $B$ joins and $C$ left the work. There is a holiday after $4$ days and $5th$ day they started working together. In how many days work would be finished?

From the calculations I did I got the answer as $9/37$ but it is not present in the multiple choice questions answer list. So I am a bit confused here if I am missing something which is causing me to get the wrong answer.

Edited :

Here are the Ans to be opted from -

1. $5\frac{9}{37}$
2. $5\frac{8}{37}$
3. $8\frac{111}{120}$
4. $5\frac{8}{120}$
• Your question is somewhat ambiguous. Does "after 3 days" mean "on the start of day 4" or something else? In other words, are those days measured from the start of the project or from the previously mentioned day? Similarly, is the answer the count from the start of the project? (If so, 9/37 is clearly wrong since it is much less than the 5 days mentioned in the question.) – Rory Daulton May 13 '18 at 9:54
• Out of curiosity, can you give a reference for "it is not present in the multiple choice questions answer list"? – Dave L. Renfro May 13 '18 at 18:15
• @RoryDaulton I really don't have any other info from question paper, that is all mentioned. and I believe those days are measured from the start of the project. so they all (A, B, C) worked only for 3 days. – Abhi May 14 '18 at 6:28
• @DaveL.Renfro I have added the available ans. to be opted from. – Abhi May 14 '18 at 6:39
• Sorry, in looking at what I wrote, I can see that I was not clear enough. By "reference", I meant a book citation or a web page citation. Your question is fine without this, although including the answer choices makes it better. However, I was curious about whether this (including the answer choices) appears in any of the official guides or test prep guides for the GRE, GMAT, NMAT, etc. – Dave L. Renfro May 14 '18 at 10:20

For each day A we add $\frac{15}{120}$ , for B we add $\frac{12}{120}$ and for C we add $\frac{10}{120}$

now we make a little table for the first few days ('x' for working day 'o' for not)

Day: 1 - 2 - 3 -4 - 5 - 6 - 7 ...
A....: x - o - o - x - x - x - x ...
B....: x - x - x - o - o - o - x ...
C....: x - x - x - x - x - x - o ...

and we sum up column 1, then 2, and so on, eventually at the $5$th day we reach$\frac{131}{120}$ and the work is done.

The simplest way is to assume a total work of 120 units (lcm of 8,10,12),, then A works @ 15, B works @12 and C @10 units/day

Day $1: A+B+C$

Day $2: B+C$

Day $3: A+C$

Day $4: A+B$

Day $5$ is a holiday. Work done so far $=3(A+B+C) = 3(15+12+10)=111$

Can you now figure out how much more time $A+B+C$ will need to complete the work ?