For easy understanding we can assume $24 \; \text{hours}$ clock.
Let $`t\text{’}$ be the time when the tank is emptied.
On Monday $\text{A}$ alone completed filling the tank at $8 \; \text{pm} \; (20 \; \text{in 24 hours clock}).$
So, time taken by $\text{A}$ to fill the tank $ = (20 – t) \; \text{hours}.$
On Tuesday $\text{B}$ alone completed filling the tank at $ 6 \; \text{pm} ( 18 \text{ in 24 hours clock}).$
So, time taken by $\text{B}$ to fill the tank $ = (18 – t) \; \text{hours}.$
On Wednesday $\text{A}$ alone worked till $ 5 \; \text{pm} \; (17 \; \text{in 24 hours clock}),$ and then $\text{B}$ worked alone from $5 \; \text{pm}$ to $ 7 \; \text{pm} \; ( 2 \; \text{hours}) $
So, time taken by $\text{A} = (17 – t) \; \text{hours},$ and time taken by $ \text{B} = 2 \; \text{hours}$ to fill the tank.
Let $\text{A}$ and $\text{B}$ be the rate of works (efficiency) of $\text{A}$ and $\text{B}$ respectively.We can that, the capacity of the tank will be the same each day.
So, $(20-t) \text{A} = (18-t) \text{B} = (17-t) \text{A+2B}\quad \longrightarrow (1) $
Taking first two terms,
$(20-t) \text{A} = (18-t) \text{B} $
$ \Rightarrow 20 \text{A – At} = 18 \text{B – Bt} $
$ \Rightarrow \text{At – Bt} = 20\text{A} – 18\text{B} \quad \longrightarrow (2) $
Taking last two terms.
$(18 – t) \text{B} = (17-t) \text{A+2B} $
$ \Rightarrow \text{18B – B}t = 17 \text{A- A}t + 2 \text{B} $
$ \Rightarrow \text{A}t – \text{B}t = 17 \text{A} – 16 \text{B} $
$ \Rightarrow \text{20A – 18B = 17A – 16B} \quad [ \because \text{From equation (2)}] $
$ \Rightarrow \text{3A = 2B} $
$ \Rightarrow \frac{\text{A}}{\text{B}} = \frac{2}{3} = k \; \text{(let)}$
$ \Rightarrow \boxed {\text{A} = 2k, \; \text{B} = 3k} $
Now, from equation $(1),$ we get
$ (20-t) \text{A} = (18-t) \text{B} $
$ \Rightarrow (20 -t) 2k = (18 -t) 3k $
$ \Rightarrow 40 – 2t = 54 – 3t $
$ \Rightarrow \boxed{t=14 = 2\; \text{pm}} $
Total work $ = 2k \times (20 – t) = 3k (18 – t) $
$ = 2k \times (20 – 14) = 3k \times (18 – 14) $
$ = 12k = 12k \; \text{units} $
On Thursday, when both pumps were used simultaneously, time taken $ = \frac{12k}{5k} = \frac{12}{5} = 2.4 \; \text{hours} $
We know that,
- $ 1 \; \text{hour} \longrightarrow 60 \; \text{minutes} $
- $ 0.4 \; \text{hour} \longrightarrow 60 \times \frac{0.4}{10} = 24 \; \text{minutes} $
So, time taken by $\text{A and B} = 2 \; \text{hours 24 minutes}. $
$\therefore$ The total time taken by both the pumps to fill the tank $ = 14 + 2 \;\text{hours 24 minutes}$
$\quad = 2 \; \text{pm + 2 hours 24 minutes} $
$\quad = \boxed{4 : 24 \; \text{pm}} $
Correct Answer $: \text{C}$