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?
Best Answer
You asked where you went wrong in solving this problem:
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:
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.