Notorious Tank Problem
The notorious Tank Problem that one finds in every single collection of arithmetical and algebraic problems. No doubt, you all remember those classic scholastically dry problems of this order:
"The tank has two pipes -- one leading to fill the tank to the brim and the other out. the first needs five hours to fill the tank and the second ten hours to drain it dry. How long will it take the tank to fill up when the stop cocks are out of both pipes ?"
This problem has a venerable history, dating right back some twenty centuries to Heron of Alexandria. Here is one from his collection for your free time in the weekend.
Four Fountains are there and a reservoir vast.
In but one day the first doth fill it to the brim.
The second two days and nights must play to do the same.
The third takes thrice the time as did the first.
The fourth comes last with four days and nights.
Now tell me when the reservoir will fill,
when all four play at once.
It is two thousand years now that the tank problem has posed and -- such is the force of habit ! -- has been solved wrongly in all this time.
"The tank has two pipes -- one leading to fill the tank to the brim and the other out. the first needs five hours to fill the tank and the second ten hours to drain it dry. How long will it take the tank to fill up when the stop cocks are out of both pipes ?"
This problem has a venerable history, dating right back some twenty centuries to Heron of Alexandria. Here is one from his collection for your free time in the weekend.
Four Fountains are there and a reservoir vast.
In but one day the first doth fill it to the brim.
The second two days and nights must play to do the same.
The third takes thrice the time as did the first.
The fourth comes last with four days and nights.
Now tell me when the reservoir will fill,
when all four play at once.
It is two thousand years now that the tank problem has posed and -- such is the force of habit ! -- has been solved wrongly in all this time.
Labels: friday special, funphysics, mathemagic





12 Comments:
I never heard of it >.<
as a consequence ill say, for the first part, That it would take 10 hours. Because in the time it takes to drain 50%, anothr 100% has been added.0-50+100=50%.
As for the second part i cant seem to understand it properly. ill look into it.
Well i think i got it. ^^
the first takes 1 day.
the 2nd takes 2 days = twice as long = 1/2 the amount of water gained*.
etc etc.
*in comparison to the first
to cut it short, the ratios for the amount of water gained (in ascending order) is 4:3:2:1.
4+3+2+1=10.
1/10 = water added from 4th pipe to make it full.
So 4/10 = water from 1st pipe.
So by the time the first has added 4/10 of the water the reservoir is full.
the 1st takes 1 day, 4/10s of 1 day is 24/10*4 =
2.4*4 =
9.6 hours =
9 hours and 36 minutes.
I think thats it :D
The speed of fountains is 1, 1/2, 1/3 and 1/4 in units of tank/day. Add them up - 25/12 tanks/day. So it takes 12/25 of a day to fill the tank with all four running.
Doing a bit of arithmetic, this works out to 11 hours 31 minutes 12 seconds.
man we just did this is fluid mech... except i wasnt listening...
i think m^3/s= V/t is the formula...
I think that...
Fountain 1 fills it in 24 hours, therefore the hourly fill rate is 1/24
Fountain 2 fills it in 48 hours, therefore the hourly fill rate is 1/48
Fountain 3 fills it in 72 hours, therefore the hourly fill rate is 1/72
Fountain 4 fills it in 96 hours, therefore the hourly fill rate is 1/96.
Therefore Time(1/24+1/48+1/72+1/96)=1
Time=11.52
= 11 Hours 31 Minutes and 12 Seconds
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My guess for the first tank puzzle is: 5 hours to fill and 10 hours to empty. Both turned on the empty time is half that to fill. So I think a total of 7.5 hours to fill.
it takes ten hours
it will take 10 hours
flow rate= amount/time
x/y= x/5 - x/10
y=10 (by solving)
x=amount and y=time
it takes 9.93 hours or 9 hr 55 min
pipe 1= 1/24 an hour
pipe 2= 1/48 an hour
pipe 3= 1/36 an hour
pipe 4= 1/96 an hour
make common fractions
1 hour you get
12/288
8/288
6/288
3/288
or
29/288 an hour
288/29=9.93 hours
9 hr 55 min 48 sec
the one above is almost right
except 3 days have 72 hours:
it takes 10 hr 39 min 36 sec
pipe 1= 1/24 an hour
pipe 2= 1/48 an hour
pipe 3= 1/72 an hour
pipe 4= 1/96 an hour
make common fractions
1 hour you get
12/288
8/288
4/288
3/288
or
27/288 an hour
288/27=10.66 hours
10 hr 39 min 36 sec
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