Question

A tank fills completely in $2hours$ if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes $3hours$ more than the larger one to fill the tank. How much times does each tap take to fill the tank completely?

Answer 1

Assume smaller tap take 'A' hours to fill larger tap " 'B' hours "

$(\frac{1}{A}+\frac{1}{B})\times 2=1$

$A=3+B$

$\frac{1}{3+B}+\frac{1}{B}=\frac{1}{2}$

$\frac{B+3+B}{3B+B^2}=\frac{1}{2}$

$4B+6=3B+B^2$

$B^2-B-6=0$

$B^2-3B+2B-6=0$

$B(B-3)2(B-3)=0$

$(B-3)(B+2)=0$

$B=3-2soB=3(+ve)$

Smaller tap $=6hrs.$

larger tap $=3hrs.$