2 days = 48 hours = x

3 days = 72 hours = y

4 days = 96 hours = z

6 hours = a

(x + y)/2= 60

(60+ z)/2=78

(78+6)/2=42

Answer= 42 hours

you've calculated a weighted average ... i think. this is more like finding the factorial of the pipes (wrong), not the sum of the pipes and their "volume/time" flow rate.

by your calculations, it would vary depending on the order calculated:

(a + x) / 2 = (6 + 48) / 2 = 54 / 2 = 27

(27 + y) / 2 = (27 + 72) / 2 = 99 / 2 = 49.5

(49.5 + z) / 2 = (49.5 + 96) / 2 = 145.5 / 2 = 72.75 hours????

why would your answer change depending on plumbing order?

step back from the problem and as mentioned earlier; if all four are ON, then they can't be slower than the fastest tap at 6 hours. then reverse determine a flow-rate ... say the reservoir is 1 litre. you should get calculations like:

relative flowrates:

6

*hpl*, 48

*hpl*, 72

*hpl*, 96

*hpl*(

*hpl*= hours per litre)

invert to get

*lph*from

*hpl*(and find lowest common denominator for easy math)

6

*hpl*==> 1/6

*lph*= 16/96

*lph*= 48/288

*lph*(

*lph*= litres per hour)

2

*days*==> 48

*hpl*==> 1/48

*lph*= 2/96

*lph*= 6/288

*lph*

3

*days*==> 72

*hpl*==> 1/72

*lph*= 4/288

*lph*

4

*days*==> 96

*hpl*==> 1/96

*lph*= 3/288

*lph*

combined flowrates

(48+6+4+3)/288

*lph*= 61/288

*lph*===> 288/61

*hpl*= 4.72 hours per litre