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A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump $\text{A}$ or pump $\text{B}$ or both start working until the tank is full. On Monday, $\text{A}$ alone completed filling the tank at $8$ pm. On Tuesday, $\text{B}$ alone completed filling the tank at $6$ pm. On Wednesday, $\text{A}$ alone worked till $5$ pm, and then B worked alone from $5$ pm to $7$ pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?

1. $4:36$ pm
2. $4:12$ pm
3. $4:24$ pm
4. $4:48$ pm

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 $: \text{C}$

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