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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:
6hpl, 48hpl, 72hpl, 96hpl (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
