Question

There is an empty tank to be filled with water using the three pipes 'x', 'y' & 'z'. Pipe 'x' & 'y' alone can fill the tank completely in 30 hours and 45 hours respectively. If the pipes 'x'
& 'y' are opened in alternate hours along with 'z' (i.e. in the first hour pipes 'x' & 'z' fill the tank, in the second hour pipes 'y' & 'z' fill the tank & so on), then the tank is 70% filled in 15 hours. If the supply of through pipe 'z' is 75% of the original supply when it is opened
along with pipe 'x' or pipe 'y', then what is the time taken by pipe 'z' alone to fill the tank completely?

1) 45 hours
2) 40.5 hours
3) 56.25 hours
4) 50 hours​

Answer 1

x and y pipes filled tank in 30 to 45 hours

so x+y=30

y-x=z 45-30=15

so x=15 y=15 z=15

pipe z contains 45 litres

Answer 2

ans - 40.5 hrs

let the capacity of tank be 90 L(lcm of 30 &45)

efficiency of x = 3 L/hr and y= 2 L/hrs.

now 1st hour = x+z , 2nd hour = y+z, so in 15 hours,

8x + 7y +15z = 70% of 90 = 63

putting value of x &y ,

we will get z= 5/3 L/hr which is just 75%.

100% efficiency of z = 20/9 L/hr.

now , total capacity = efficiency * time taken

90 = 20/9 * time taken by Z

time taken by Z = 40.5 hrs