Question

Two pipes A and B together fill a cistern in $\frac{8}{3}$ minutes. If A takes 4 minutes more than B, then time taken by A to fill the cistern is .

• A. 5 minutes
• B. 8 minutes
• C. 4 minutes

Answer 1

The correct option is B 8 minutes

Let pipe B take $x$ minutes to fill the cistern.
Then, the time taken by pipe A to do the same is $x+4$ minutes.

$\therefore$ Portion of the cistern filled by pipe B in one minute $=\frac{1}{x},$ and

portion of the cistern filled by pipe A in one minute $=\frac{1}{x+4}.$

Since the two pipes together fill the cistern in $\frac{8}{3}$ minutes, the portion of the cistern filled in one minute by both pipes together $=\frac{3}{8}.$

$\Rightarrow \frac{1}{x}+\frac{1}{x+4}=\frac{3}{8}$

$\Rightarrow \frac{x+4+x}{x(x+4)}=\frac{3}{8}$

$\Rightarrow 8\times (2x+4)=3x(x+4)$

$\Rightarrow 16x+32=3x^2+12x$

$\Rightarrow 3x^2-4x-32=0$

$\Rightarrow 3x^2-12x+8x-32=0$

$\Rightarrow 3x(x-4)+8(x-4)=0$

$\Rightarrow (3x+8)(x-4)=0$

$\Rightarrow x=\frac{-8}{3}\text{or}4$

Since time cannot be negative, we have the time taken by pipe B to fill the cistern is $x=4$ minutes.

Thus, the time taken by pipe A to fill the cistern is $x+4=4+4=8$ minutes.