The efficiency of pipe R is 60% more than the efficiency of pipe S. Pipe Q alone can fill 30% of the empty tank in 6 hours. Pipe R alone can fill an empty tank completely in (y-5) hours. The ratio between the efficiencies of pipe P and R is 5:6 respectively. Pipe P and Q together can fill 25% of the empty tank in 3 hours. Find out the time taken by pipe S alone to fill 55% of the tank and the value of ‘y’.

Solution

Let’s assume the capacity of the tank is 600 units.

Pipe Q alone can fill 30% of the empty tank in 6 hours.

Time taken by Pipe Q alone to fill the tank completely = (6/30)x100

= 20 hours

Efficiency of pipe Q = 600/20 = 30 units/hour    Eq.(i)

Pipe P and Q together can fill 25% of the empty tank in 3 hours.

Time taken by Pipe P and Q together to fill the tank completely = (3/25)x100

= 12 hours

Efficiency of pipe P and Q together = 600/12 = 50 units/hour    Eq.(ii)

Efficiency of pipe P = Eq.(ii)-Eq.(i)

= 50-30

= 20 units/hour

The ratio between the efficiencies of pipe P and R is 5:6 respectively.

Efficiency of pipe R = (20/5)x6

= 24 units/hour

The efficiency of pipe R is 60% more than the efficiency of pipe S.

24 = (100+60)% of the efficiency of pipe S

24 = 160% of the efficiency of pipe S

efficiency of pipe S = 24/1.6

 = 240/16

= 15 units/hour

Pipe R alone can fill an empty tank completely in (y-5) hours.

(Efficiency of pipe R) x (time taken by pipe R to fill an empty tank completely) = total capacity of the tank

24 x (y-5) = 600

4 x (y-5) = 100

(y-5) = 25

y = 25+5

y = 30

Time taken by pipe S alone to fill 55% of the tank = 55% of (total capacity of the tank)/(efficiency of pipe S)

= 55% of (600)/(15)

= 55% of 40

= 22 hours