Pipe A can fill a tank in 6 hours and pipe B can fill it in 7 hours. If they are opened in alternate hours and if pipe A is opened first, in how many hours the tank shall be full ?

A can fill tank $=\frac{1}{6}parts/hour$

B can fill tank $=\frac{1}{7}part/hour$

(A+B)’s work for 2 hours $=\frac{1}{6}+\frac{1}{7}=\frac{7+6}{42}=\frac{13}{42}$

(A+B)’s work for 4 hours $=2\times \frac{13}{42}=\frac{26}{42}$

(A+B)’s work for 6 hours $=\frac{39}{42}$

Remaining part $\Rightarrow 1-\frac{39}{42}\Rightarrow \frac{42-39}{42}=\frac{3}{42}=\frac{1}{14}$

Which is done by ‘A’ is $\Rightarrow \frac{1}{14}\times 6=\frac{3}{7}\text{\hspace{0.17em}\hspace{0.17em}minutes}$

$\therefore Totalhours\Rightarrow 6\left(\frac{3}{7}\right)\text{\hspace{0.17em}\hspace{0.17em}hours}$