Two pipes A and B can fill a tank in $$3\frac{\text{3}}{\text{7}}$$ hours when opened simultaneously. If B alone can take 2 hours less than…
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Two pipes A and B can fill a tank in $$3\frac{\text{3}}{\text{7}}$$ hours when opened simultaneously. If B alone can take 2 hours less than A alone takes to fill the tank completely. How much time does A alone take to fill the tank?
Solution
Let us assume that A alone takes 'x' hours to fill the tank.
B alone takes 'x - 2' hours to fill the tank.
Part of the tank filled by pipe A in 1 hour = $\frac{1}{x}$
Part of the tank filled by pipe B in 1 hour = $\frac{1}{x-2}$
Together they fill the tank in $3\frac{3}{7}$ hr = $\frac{24}{7}$ hr
$\frac{24}{7}(\frac{1}{x} + \frac{1}{x-2}) = 1$
On simplifying, we get
$24(2x-2) = 7x(x-2)$
$7x^2-62x+48 = 0$
$(x-8)(x-\frac{6}{7})=0$
$\therefore x = 8$ ( $x$ can't take the value of $\frac{6}{7}$ as $x-2$ is negative )
Hence, A alone takes 8 hr to fill the tank completely.