Question

Two pipes S and T can fill the tank in 6 hrs and 4 hrs respectively. If they are opened in alternative hours, and if pipe S is opened first in how many hours the tank shall be full? 

1. 4 hrs
2. 5 hr
3. 2 hrs
4. 6 hrs
5. None of these

Answer 1

Correct Answer - Option 2 : 5 hr

Given:

Pipe S can fill the tank alone in 6 hr

Pipe T can fill the tank alone in 4 hr

Concept used:

Let, Pipe A can fill a tank in x hrs

⇒ Pipe A can fill a tank in 1 hr = 1/x

Calculation:

Pipe S fill the tank in 1 hr = 1/6

Pipe T fill the tank in 1 hr = 1/4

Pipe (S + T) fill the tank in 2hrs when opened alternatively = (1/6) + (1/4) = 5/12

Pipe (S + T) fill the tank in 4hrs when opened alternatively = 2 × (5/12) = 10/12 = 5/6

⇒ Remaining part = (1 - 5/6) = 1/6

Now, it is S’s turn 1/6 part is filled by S in 1 hr

So, total time = (4 + 1) hours = 5 hours

∴ The time to fill the tank is 5 hours.