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.
We shall assume that the mason's initial error was due to a misjudgement and that the workers are always productive at a fixed rate.
Let $x,y,p$ be the initial number of workers, the desired completion duration in days, and the required percentage, respectively.
workers $(w)$ |
days $(d)$ |
jobs $(j)$ |
$x$ |
$\displaystyle\frac54y$ |
$1$ |
$x$ |
$\displaystyle\frac p{100}\left(\frac54y\right)$ |
$\displaystyle\frac p{100}$ |
$\displaystyle\frac43x$ |
$\displaystyle y-\frac p{100}\left(\frac54y\right)$ |
$\displaystyle\frac{100-p}{100}$ |
The joint proportionality among $w,d$ and $j,$ is such that $\displaystyle\frac{w_id_i}{j_i}$ has a fixed value.
Thus, $$\frac{x\left(\frac54y\right)}1=\frac{\left(\frac43x\right)\left(y-\frac p{100}\left(\frac54y\right)\right)}{\frac{100-p}{100}}\\p=20.$$
Best Answer
If the number of expected days (ploughing 120 hectares a day) is $x$ you have $120x = A$ where $A$ is the total area.
Instead the farmer increased the workload after two days to $150$, as you stated.
In the first two days they plowed $240$ hectares, so in this new set up the total area can be seen as $240 + 150n = A$ where $n$ is the total number of days they ploughs $150$ hectares. You can see that $x - 2 = n + 2$, so $n = x-4$.
You now have $240 + 150(x-4) = 120x$ which you can rearrange to give $x = 12$. From here it follows that $A = 1440$, and $n = 8$. (Actual days spent ploughing $=10$)