Re: Archimedes bath problem
let A = flow rate for tap A in units per hour
(2 hrs)(A + 2A + 3A) + (2.75 hrs)(A + 3A) = 1 tub filled
23A = 1
A = (1 tub filled)/(23 hrs)
using all 3 ...
(A + 2A + 3A)t = 1 tub filled
solve for t
Re: Archimedes bath problem
Hello, Natasha1!
Quote:
Archimedes has a bath which is filled using three taps labelled A, B and C.
The rate of flow of B is twice that of A; the rate of flow of C is three times that of A.
Yesterday, Archimedes filled his bath as follows:
First, he used all three taps for 2 hours. Then he turned off B and finished filling the bath
in an additional 2 hours and 45 mins using taps A and C only.
Today Archimedes is going to fill his bath (currently empty) using all three taps.
How long will this take him?
Let 
= number of hours for A to fill the bath alone.
In one hour, A can fill 
of the bath.
Then B can fill the bath alone in 
hours.
In one hour, B can fill 
of the bath.
And C can fill the bath alone in 
hours.
In one hour, C can fill 
of the bath.
In 
hours, the three taps can fill:  + 2(\tfrac{2}{a}) + 2(\tfrac{3}{a}) \,=\,\tfrac{12}{a})
of the bath.
In 
hours, A and C can fill:  + \tfrac{11}{4}(\tfrac{3}{a}) \,=\,\tfrac{11}{a})
of the bath.
These two quantities can fill the entire bath: . 
. . Hence: . 
A takes 
hours to fill the bath.
. . In one hour, A fills 
of the bath.
. . In 
hours, A fills 
of the bath.
B takes 
hours to fill the bath.
. . In one hour, B fills 
of the bath.
. . In hours, B fills 
of the bath.
C takes 
hours to fill the bath.
. . In one hour, C fills 
of the bath.
. . In hours, C fills 
of the bath.
In hours, the three taps will fill the entire bath.
. . 
. . 
Therefore, the three taps will take 
hours.
