Question

Pipe A takes $\frac{3}{4}$ of the times required by pipe B to fill the empty tank individually. When an outlet pipe C is also opened simultaneously with pipe A and pipe B, it takes$\frac{3}{4}$ more time to fill the empty tank than it takes, when only pipe A and pipe B are opened together, If it takes to fill 33 hours when all the three pipes are opened simultaneously, then in what time pipe C can empty the full tank operating alone?

• A. 66 hours
• B. 50 hours
• C. 44 hours
• D. can't be determined

Answer 1

The correct option is C 44 hours

Let pipe A fill the tank in 3x hours then pipe B fill it in 4x hours.
Therefore, in 1 hour they will fill = $\frac{1}{3x}+\frac{1}{4x}=\frac{7}{12x}$
i.e., they will take $\frac{12x}{7}hours$
When an outlet pipe C is also opened then it takes
$=\frac{12x}{7}+\frac{12x}{7}\times \frac{3}{4}=\frac{12x}{7}\left(\frac{7}{4}\right)=3xhours$
Now, in one hour pipe A, B and C working together fill
$=\frac{1}{3x}+\frac{1}{4x}-\frac{1}{C}=\frac{1}{3x}\Rightarrow \frac{1}{12x}-\frac{1}{C}=\frac{1}{3x}\Rightarrow \frac{1}{C}=(\frac{1}{12x}-\frac{1}{3x})\Rightarrow C=4x$
Hence, in 4x hours pipe C can empty the whole tank.
Now, since $3x=33\Rightarrow x=11$
$\therefore 4x=4\times 11=44hours$