Question

The Pipe M can fill the cistern in 8 hours and pipe N can empty the cistern in 9 hours respectively. These pipes are opened alternately and pipes M is opened first, in how many hours, the cistern fill-in?

1. 130 hours
2. 124 hours
3. 135 hours
4. 127 hours
5. None of these

Answer 1

Correct Answer - Option 4 : 127 hours

Given:

Pipe M can fill the cistern in = 8 hours

Pipe N can empty the cistern in = 9 hours

Calculation:

Let the total capacity of cistern be LCM (8, 9) = 72 litres

Now,

Cistern filled by M in 1 hour = 72/8

⇒ 9 litre/hour

Cistern empty by N in 1 hour = 72/9

⇒ (– 8) litre/hour

Cistern filled by (M + N) in 1 hour = (9 – 8)

⇒ 1 litre/hour

Now,

According to the question,

For first hour Pipe M is opened = 9 litre

For second hour Pipe N is opened = (– 8) litre

For third hour Pipe M is opened = 9 litre

For fourth hour Pipe N is opened = (– 8) litre

Similarly, So…. On

Now,

We conclude that –

2 hours   =      1 litre

× 63                × 63

126 hours    =   63 litre

+1 hours      =   +9 litre

127 hours =      72 litres

∴ The Cistern will be filled in 127 hours.