Pipe A can fill a tank in 4 hours and pipe B can fill it in 6 hours. If they are opened on alternate hours and if pipe A is opened first then in how many hours, the tank shall be full?

‍Correct Answer: 423hrs

 

A→4 hours

 

B→6 hours

1/4+1/6

= 3+2/12

 

 

According to the question

 

=For the first hour tap A is opened and B for second hour

 

= Work done by both in 2 hours

 

(3l/h+2l/h)×2=10 units

2 h.                   5 lit

×2.                   ×2

4 h.                   10 lit

Remaining part =12 – 10 = 2 litre

 

Again 5th hour A will be opened

 

Tap A will fill the 2 litre water with its efficiency =2/3 hours

 

Therefore tank will be filled in

 

=(4+2/3) hours=4 2/3hours