Question

One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 min, then the slower alone will be able to fill the tank in:

A) 81 min B) 108 min C) 144 min D) 192 min

Answer 1

$\sf\large\underline{Let:}$

$\tt{\implies Time\:taken\:_{(slower\:pipe)}=x\:min}$

$\tt{\implies Time\:taken\:_{(faster\:pipe)}=x/3\:min}$

$\sf\large\underline{To\: Find:}$

$\tt{\implies Time\:taken\:_{(slower\:pipe\: alone)}=?}$

$\sf\large\underline{Given\: that,}$

• One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36]

$\tt{\implies Time\:taken\:in\:1\:min\:_{(slower\:pipe)}=\frac{1}{x}}$

$\tt{\implies Time\:taken\:in\:1\:min\:_{(faster\:pipe)}=\frac{1}{x/3}}$

$\tt{\implies Time\:taken\:in\:1\:min\:_{(faster\:pipe)}=\frac{3}{x}}$

$\sf\small\underline{Now,\: according\:to\: above\: information:}$

$\tt{\implies Pipe\:_{(faster)}+Pipe\:_{(slower)}=\dfrac{1}{36}}$

$\tt{\implies \dfrac{3}{x}+\dfrac{1}{x}=\dfrac{1}{36}}$

$\tt{\implies \dfrac{3+1}{x}=\dfrac{1}{36}}$

$\tt{\implies \dfrac{4}{x}=\dfrac{1}{36}}$

$\tt{\implies x=36\times\:4}$

$\tt{\implies x=144\:min}$

$\sf\large{Hence,}$

$\bf{\implies Time\:taken\:_{(slower\:pipe\: alone)}=144\:min}$

$\tt{\underbrace{ANSWER=OPTION-(C)}}$