Three taps A, B and C can fill a tank in 8, 10 and 16 hours respectively. If A is open all the time and B and C are open for 1 hour each alternatively, the tank will be full in :                       

Total work is L.C.M of 8, 10, 16 i.e., 80 units

A can fill the tank $=\frac{80}{8}=10parts/hour$

B can fill the tank $=\frac{80}{10}=8parts/hour$

C can fill the tank $=\frac{80}{16}=5parts/hour$

A & B in 1 hour  $=10+8=18parts/hour$

A & C in 1 hour  $=10+5=15parts/hour$

In ‘2’ hours $\Rightarrow$ tank filled by  $=18+15=33parts$

After 4 hours $\Rightarrow$tank filled by $=31\times 2=66parts$

Remaining = 80 – 66 = 14 parts

$\therefore$ Remaining quantity filled by A + B $\Rightarrow \frac{14}{18}=\frac{7}{9}hours$

$\therefore$ Total time$=4+\frac{7}{9}hours$

$=4\left(\frac{7}{9}\right)hours$