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?
 A: 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.
Added:

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.
A: 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
A: 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.
A: 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?
ans--
       A,B AND C ONE DAY WORK=(1/24+1/36+1/48)=13/144 
       FOUR DAYS WORK OF A,B AND C IS =[4*(13/144)]=13/36
       AFTER FOUR DAYS REMAINING WORK =[1-(13/36)]=23/36
       IN  LAST 3 DAYS A WORKING ALONE IS =[3*(1/24)]=1/8
       REST OF WORK IS ([(23/36)-(1/8)]=37/72) DONE BY A AND B TOGETHER
       A AND B ONE DAY WORK IS=[(1/24)+(1/36)]=5/72
       TIME TAKEN THEM TO COMPLETE THE WORK=[(37/72)/(5/72)]=37/5
       TOTAL TIME TO COMPLETE THE WORK=[(37/5)+3+4]=72/5
A: A better approach to the problem.
Take the LCM of 14 and 21 which will give you the total amount of work.
LCM (14,21) = 42.
A completes in 14 days.So he does 42/14 in 1 day.Similarly B does 42/21 in 1 day.
A=3,B=2.Since A left 3 days prior to the completion of the work, so his 3 days work is absent.Lets add his 3 days pending work to the total amount of work.
In 1 day he does 3 units of work, so in 3 days he will do 9 units of work.
Total work = 42+9 = 51.
Both will now take 3+2 to complete the total work.
A+B = 51/5 = 10.2.
