## 1 Answer

If $M_{1}$ person can do $W_{1}$ work in $D_{1}$ days working $T_{1}$ hours in a day and $M_{2}$ person can do $W_{2}$ work in $D_{2}$ days working $T_{2}$ hours in a day then the relationship between them is :

$$ \boxed {\frac{M_{1} \ast D_{1} \ast T_{1}}{W_{1}} = \frac{M_{2} \ast D_{2} \ast T_{2}}{W_{2}}} $$

Let $` W\text{’} \; \text{unit}$ be the capacity of the tank.

Now, $ \frac{(10A + 45B) \ast 30}{W} = \frac{(8A + 18B) \ast 60}{W} $

$ \Rightarrow 10A + 45B = 16A + 36B $

$ \Rightarrow 6A = 9B $

$ \Rightarrow 2A = 3B $

$ \Rightarrow \frac{A}{B} = \frac{3}{2} = k \;(\text{let}) $

$ \Rightarrow \boxed{ A = 3k, \; B = 2k} $

And, $\frac{(7A +27B) \ast \text{Time}}{W} = \frac{(8A+18B) \ast 60}{W} $

$ \Rightarrow [ 7(3k) + 27(2k)] \ast \text{Time} = [ 8(3k) + 18(2k) \ast 60] $

$ \Rightarrow (21k + 54k) \ast \text{Time} = (24k + 36k) \ast 60 $

$ \Rightarrow 75k \ast \text{Time} = 60k \ast 60 $

$ \Rightarrow \boxed{ \text{Time} = 48 \; \text{minutes}} $

Correct Answer $:48$

$\textbf{PS:}$

- Work Done $=$ Time Taken $\ast$ Rate of work
- Total Work Done $=$ Total Time $\ast$ Efficiency