CAT 2021 Set-2 | Quantitative Aptitude | Question: 20

Question

Two pipes $\text{A}$ and $\text{B}$ are attached to an empty water tank. Pipe $\text{A}$ fills the tank while pipe $\text{B}$ drains it. If pipe $\text{A}$ is opened at $2 \; \text{pm}$ and pipe $\text{B}$ is opened at $3 \; \text{pm},$ then the tank becomes full at $10 \; \text{pm}.$ Instead, if pipe $\text{A}$ is opened at $2 \; \text{pm}$ and pipe $\text{B}$ is opened at $4 \; \text{pm},$ then the tank becomes full at $6 \; \text{pm}.$ If pipe $\text{B}$ is not opened at all, then the time, in minutes, taken to fill the tank is

• A. $140$
• B. $264$
• C. $144$
• D. $120$

Answer 1

Let’s draw the table for better understanding.

$\begin{array}{llll}  & \;\;\textbf{Pipe A} &\;\; \textbf{Pipe B} \\ \text{Time:} & \underbrace{8\;\text{hours}}_{\text{2 pm to 10 pm}} & \underbrace{7\;\text{hours}}_{\text{3 pm to 10 pm}} \\  & \underbrace{4\;\text{hours}}_{\text{2 pm to 6 pm}} & \underbrace{2\;\text{hours}}_{\text{4 pm to 6 pm}} \\ \text{Efficiency:} & x\;\text{litres/hour} & -y\;\text{litres/hour} \\ \text{Capacity of tank:} & 8x-7y  \qquad = & 4x-2y \end{array}$

Now, $8x – 7y = 4x – 2y$

$\Rightarrow 4x = 5y$

$\Rightarrow \boxed{y = \frac{4x}{5}}$

The capacity of tank $ = 8x – 7y = 4x – 2y$

$ \qquad = 8x – 7 \left(\frac{4x}{5} \right) = 4x – 2 \left(\frac{4x}{5} \right)$

$\qquad = \frac{40x – 28x}{5} = \frac{20x – 8x}{5} $

$\qquad = \frac{12x}{5} \; \text{litres} = \frac{12x}{5} \; \text{litres}$

$\therefore$ The time taken by $\text{A}$ to fill the tank if $\text{B}$ is not opened at all $ = \dfrac{ \frac{12x}{5} \; \text{litres}}{x \; \text{litres/hour}}$

$\qquad  = \frac{12}{5} \; \text{hours}= \frac{12}{5} \times 60= 12 \times 12 = 144 \; \text{minutes}.$

Correct Answer $: \text{C}$

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$\textbf{PS:}$ 

• Total Work Done $=$ Number of Days $\times$ Efficiency
• Efficiency and Time are inversely proportional to each other.