49 Painters are required to complete a painting job in 12 days . However, due to unforeseen circumstances , there was a delay and only one third of the job was completed in 5 days . Assuming that all Painters worked at the same initial rate , how Many more Painters are needed for the rest of the job to be completed as scheduled ?
I'm not quite sure on how to approach this qn ..
On
So the difficulty to this question is the phrasing of it. You know:
-1/3 of the JOB was done in 5 days with 49 painters.
We let the following variables:
-$p$ be the number of painters
In this case, the rate, $r_1$, of the first batch of workers are:
$$r_1=\frac{1/3 JOB}{\text{5 days} \cdot \text{49 painters}}=\frac{1}{15\cdot49}\frac{JOB}{\text{painter}\cdot\text{day}}$$
Then, we know there is 2/3 of the JOB left to be done in 7 days. So we state that the rate of the first batch times the remaining time and size of the first painter batch, with the rate of the second batch times the remaining time and size of the second painter batch, should equal 2/3:
$$r_1\cdot49\text{painters}\cdot7\text{days}+r_2\cdot p\text{ painters}\cdot7\text{days}=\frac{2}{3}$$
Therefore, since we know the second batch of painters will have the same rate, $r_1=r_2$, we say:
$$(49+p)\text{painters}\cdot7\text{days}\cdot r_1=\frac{2}{3}$$
So we simplify:
$$(49+p)\text{painters}=\frac{2}{21r_1}\text{painters}$$
$$p=\frac{2}{21r_1}-49$$
Insert $r_1$ and simplify and you should obtain the number of painters
Without delay the question hints at this rate: $$ 49 \cdot r \cdot 12 = W = 1 \Rightarrow \\ r = 1/(49\cdot 12) $$ where $W$ is the amount of work to be done.
They are now left with $1-1/3$ of the work and $12-5$ days, this has to do be done with $49+k$ painters, which can do $$ (49 + k) r (12-5) = (1- 1/3) $$ So insert the rate $r$ and solve for the number of extra painters $k$.
Spoiler: