# Work and time, when work is split into parts

I'm stuck on a particular type of work and time problems.

For example,

1) A,B,C can complete a work separately in 24,36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?

A simpler version of the same type of problem is as follows:

2) A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is?

My attempt at problem 2:

A's 1 day work=1/14 and B's 1 day work= 1/21

Assume that it takes 'd' days to complete the entire work when both A and B are working together. Then,

(1/14 + 1/21)*d= 1

-> d=42/5 days.

But it is stated that 3 days before the completion of the work, A left. Therefore, work done by both in (d-3) days is:

(1/14 + 1/21)*(42/5 - 3)= 9/14

Remaining work= 1- 9/14 = 5/14 which is to be done by B alone. Hence the time taken by B to do (5/14) of the work is:

(5/14)*21 = 7.5 days.

Total time taken to complete the work = (d-3) + 7.5 = 12.9 days.

However, this answer does not concur with the one that is provided.

My Understanding of problem 1:

Problem 1 is an extended version of problem 2. But since i think i'm doing problem 2 wrong, following the same method on problem 1 will also result in a wrong answer.

Where did i go wrong?

-

You asked where you went wrong in solving this problem:

A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is?

As you said in your solution, $A$ can do $1/14$ of the job per day, and $B$ can do $1/21$ of the job per day. On each day that they work together, then, they do $$\frac1{14}+\frac1{21}=\frac5{42}$$ of the job. Up to here you were doing fine; it’s at this point that you went astray. You know that for the last three days of the job $B$ will be working alone. In those $3$ days he’ll do $$3\cdot\frac1{21}=\frac17$$ of the job. That means that the two of them working together must have done $\frac67$ of the job before $A$ left. This would have taken them

$$\frac{6/7}{5/42}=\frac67\cdot\frac{42}5=\frac{36}5\text{ days}\;.$$

Add that to the $3$ days that $B$ worked alone, and you get the correct total: $$\frac{36}5+3=\frac{51}5=10.2\text{ days}\;.$$

You worked out how long it would take them working together, subtracted $3$ days from that, saw how much of the job was left to be done at that point, and added on the number of days that it would take $B$ working alone to finish the job. But as your own figures show, $B$ actually needs $7.5$ days to finish the job at that point, not $3$, so he ends up working alone for $7.5$ days. This means that $A$ actually left $7.5$ days before the end of the job, not $3$ days before. You have to figure out how long it takes them to reach the point at which $B$ can finish in $3$ days.

1) A,B,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?

Here you know that all three worked together for the first $4$ days, $B$ worked alone for the last $3$ days, and $A$ and $B$ worked together for some unknown number of days in the middle. Calculate the fraction of the job done by all three in the first $4$ days and the fraction done by $B$ alone in the last $3$ days, and subtract the total from $1$ to see what fraction was done by $A$ and $B$ in the middle period; then see how long it would take $A$ and $B$ to do that much.

-
 Yes, in short i misinterpreted the question. But because of the line, "They begin working together but 3 days before the completion of the work, A leaves off", it seems as if A and B working together would have completed it in some estimated d number of days, 3 days before which A left the job. Hence obviously B would require >3 days to complete the job. How do i avoid such misinterpretations in these types of problems? again, an excellent answer. Thanks! – user85030 Sep 25 '12 at 12:46 @user85030: You’re welcome! I think that avoiding such misinterpretations is partly a matter of practice and partly a matter of reading them pretty literally. Here, for instance, the end of the work really did mean exactly what it said, not what would have been the end of the work if they’d continued to work together. – Brian M. Scott Sep 25 '12 at 21:52 Can you please have a look at this:- math.stackexchange.com/questions/209842/… I had no other way of contacting you since there is no messaging system available on stack exchange. – user85030 Oct 9 '12 at 19:26

In problem 2 you are misinterpreting the phrase "$A$ left 3 days before the work was done." When you calculate it as above (3 days before the work would've been done if $A$ worked on), its wrong, as $A$ left (as you calculated) 7.5 days before the work was done.

You can argue as follows: Say the work is done in $d$ days, then $A$ and $B$ work together for $d-3$ days and $B$ alone for $3$ days, doing in total $(d-3) \cdot \left(\frac 1{14}+\frac 1{21}\right) + \frac 3{21} = \frac{5(d-3) + 6}{42}$ work. So we must have $5(d-3) = 36$, so $5d = 51$, that is $d = 57/5$. For 1), you can argue along the same lines.

-

Problem $1.)$

Let $n$ be the required number of days.

$A,B,C$'s $1$ day work is $1/24,1/36,1/48$ respectively.

Work done by $C=4/48$

Work done by $B=n/36$

Work done by $A=(n-3)/24$

Sum of all the work is $1$ which gives

$$\frac{1}{12}+\frac{n}{36}+\frac{n-3}{24}=1$$

Solving which you will get your answer.

Problem $2.)$ can be solved using similar approach

-